Medical Imaging Signals And Systems Solutions Manual

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Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter 11
Chapter 12
Chapter 13

Medical Imaging

Signals and Systems (2e):

Solutions Manual

Version 1.0 (March 27, 2014)

Jerry L. Prince

Electrical and Computer Engineering

Whiting School of Engineering

Johns Hopkins University

Jonathan M. Links

Environmental Health Sciences

Bloomberg School of Public Health

Johns Hopkins University

PRENTICE HALL

Upper Saddle River, New Jersey 07458

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Contents

Preface 1

Signals and Systems 2

Image Quality 37

Physics of Radiography 60

Projection Radiography 69

Computed Tomography 95

The Physics of Nuclear Medicine 133

Planar Scintigraphy 140

Emission Computed Tomography 161

The Physics of Ultrasound 175

Ultrasound Imaging Systems 187

Physics of Magnetic Resonance 207

Magnetic Resonance Imaging 215

iii

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Preface

These solutions have been prepared over the past 16 years or so during the process of teaching our course on

Medical Imaging Systems at Johns Hopkins University. We would like to thank the following individuals for

their help: William R. Brody, Elliot R. McVeigh, John I. Goutsias, Ergin Atalar, Scott Reeder, Bradley Wyman,

Chris Constandinides, Rong Xue, Christopher Yeung, Maryam Rettmann, Duygu Tosun, M. Faisal Beg, Xiao Han,

Li Pan, Pramodsingh H. Thakur, Vijay Parthasarathy, Tara Johnson, Abd El-Monem El-Sharkawy, Minnan Xu,

Khaled Abd-Elmoniem, Lotta Ellingsen, Jing Wan, Snehashis Roy, Harsh Agarwal, Issel Lim, Xian Fan, Nan Li,

Sahar Soleimanifard, Nathanael Kuo, Min Chen, Jeffrey Pompe, and Daniel Tward. Special thanks to Xiaodong

Tao, who created and solved a large number of problems speciﬁcally for the ﬁrst edition and to Zhen Yang and

Chuyang Ye for the problems they checked, corrected, and solved for this edition. As always, tremendous thanks to

Aaron Carass, who patiently solved countless LaTeX and CVS problems that were encountered in the preparation

of both the textbook and this solutions manual.

Jerry L. Prince

Jonathan M. Links

March 27, 2014

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Signals and Systems

SIGNALS AND THEIR PROPERTIES

Solution 2.1

(a) δs(x, y) = P∞

m=−∞ P∞

n=−∞ δ(x−m, y −n) = P∞

m=−∞ δ(x−m)·P∞

n=−∞ δ(y−n), therefore it is a

separable signal.

(b) δl(x, y)is separable if sin(2θ)=0. In this case, either sin θ= 0 or cos θ= 0,δl(x, y)is a product of a

constant function in one axis and a 1-D delta function in another. But in general, δl(x, y)is not separable.

exp(j2πωt). Therefore, e(x, y)is a separable signal.

(d) s(x, y)is a separable signal when u0v0= 0. For example, if u0= 0,s(x, y) = sin(2πv0y)is the product

of a constant signal in xand a 1-D sinusoidal signal in y. But in general, when both u0and v0are nonzero,

s(x, y)is not separable.

Solution 2.2

(a) Not periodic. δ(x, y)is non-zero only when x=y= 0.

(b) Periodic. By deﬁnition

comb(x, y) = ∞

m=−∞

∞

n=−∞

δ(x−m, y −n).

For arbitrary integers Mand N, we have

comb(x+M, y +N) = ∞

m=−∞

∞

n=−∞

δ(x−m+M, y −n+N)

=∞

p=−∞

∞

q=−∞

δ(x−p, y −q) [let p=m−M, q =n−N]

=comb(x, y).

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So the smallest period is 1 in both xand ydirections.

sin(2πx) cos(4πy) = sin(2π(x+Tx)) cos(4πy).

Solving the above equation, we have 2πTx= 2kπ for arbitrary integer k. So the smallest period for xis

Tx0= 1. Similarly, we ﬁnd that the smallest period for yis Ty0= 1/2.

(d) Periodic. Let f(x+Tx, y) = f(x, y), we have

sin(2π(x+y)) = sin(2π(x+Tx+y)).

So the smallest period for xis Tx0= 1 and the smallest period for yis Ty0= 1.

(e) Not periodic. We can see this by contradiction. Suppose f(x, y) = sin(2π(x2+y2)) is periodic; then there

exists some Txsuch that f(x+Tx, y) = f(x, y), and

sin(2π(x2+y2)) = sin(2π((x+Tx)2+y2))

= sin(2π(x2+y2+ 2xTx+T2

x)) .

In order for the above equation to hold, we must have that 2xTx+T2

x=kfor some integer k. The solution

for Txdepends on x. So f(x, y) = sin(2π(x2+y2)) is not periodic.

(f) Periodic. Let fd(m+M, n) = fd(m, n). Then

sin π

5mcos π

5n= sin π

5(m+M)cos π

5n.

Solving for M, we ﬁnd that M= 10kfor any integer k. The smallest period for both mand nis therefore

10.

(g) Not periodic. Following the same strategy as in (f), we let fd(m+M, n) = fd(m, n), and then

sin 1

5mcos 1

5n= sin 1

5(m+M)cos 1

5n.

The solution for Mis M= 10kπ. Since fd(m, n)is a discrete signal, its period must be an integer if it is

to be periodic. There is no integer kthat solves the equality for M= 10kπ for some M. So, fd(m, n) =

sin 1

5mcos 1

5nis not periodic.

Solution 2.3

(a) We have

E∞(δs) = Z∞

−∞ Z∞

−∞

δ2

s(x, y)dx dy

= lim

X→∞ lim

Y→∞ ZX

−XZY

−Y

∞

m=−∞

∞

n=−∞

δ(x−m, y −n)dx dy

= lim

X→∞ lim

Y→∞(2bXc+ 1)(2bYc+ 1)

=∞,

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4CHAPTER 2: SIGNALS AND SYSTEMS

where bXcis the greatest integer that is smaller than or equal to X. We also have

P∞(δs) = lim

X→∞ lim

Y→∞

4XY ZX

−XZY

−Y

δ2

s(x, y)dx dy

= lim

X→∞ lim

Y→∞

4XY ZX

−XZY

−Y

∞

m=−∞

∞

n=−∞

δ(x−m, y −n)dx dy

= lim

X→∞ lim

Y→∞

(2bXc+ 1)(2bYc+ 1)

4XY

= lim

X→∞ lim

Y→∞ 4bXcbYc

4XY +2bXc+ 2bYc

4XY +1

4XY 

= 1 .

(b) We have

E∞(δl) = Z∞

−∞ Z∞

−∞ |δ(xcos θ+ysin θ−l)|2dx dy

=Z∞

−∞ Z∞

−∞

δ(xcos θ+ysin θ−l)dx dy

1

=









Z∞

−∞

|sin θ|dx, sin θ6= 0

Z∞

−∞

|cos θ|dy, cos θ6= 0

E∞(δl) = ∞.

Equality 1comes from the scaling property of the point impulse. The 1-D version of Eq. (2.8) in the text is

δ(ax) = 1

|a|δ(x). Suppose cos θ6= 0. Then

δ(xcos θ+ysin θ−l) = 1

|cos θ|δx+ysin θ

cos θ−l

cos θ.

Therefore, Z∞

−∞

δ(xcos θ+ysin θ−l)dx =1

|cos θ|.

We also have

P∞(δl) = lim

X→∞ lim

Y→∞

4XY ZX

−XZY

−Y|δ(xcos θ+ysin θ−l)|2dx dy

= lim

X→∞ lim

Y→∞

4XY ZX

−XZY

−Y

δ(xcos θ+ysin θ−l)dx dy .

Without loss of generality, assume θ= 0 and l= 0, so that we have sin θ= 0 and cos θ= 1. Then it follows

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that

P∞(δl) = lim

X→∞ lim

Y→∞

4XY ZX

−XZY

−Y

δ(x)dx dy

= lim

X→∞ lim

Y→∞

4XY ZY

−Y(ZX

−X

δ(x)dx)dy

= lim

X→∞ lim

Y→∞

4XY ZY

−Y

1dx

= lim

X→∞ lim

Y→∞

4XY

= lim

X→∞

= 0 .

E∞(e) = Z∞

−∞ Z∞

−∞ |exp [j2π(u0x+v0y)]|2dx dy

=Z∞

−∞ Z∞

−∞

1dx dy

=∞.

And also

P∞(e) = lim

X→∞ lim

Y→∞

4XY ZX

−XZY

−Y|exp[j2π(u0x+v0y)]|2dx dy

= lim

X→∞ lim

Y→∞

4XY ZX

−XZY

−Y

1dx dy

= 1 .

(d) We have

E∞(s) = Z∞

−∞ Z∞

−∞

sin2[2π(u0x+v0y)] dx dy

2

=Z∞

−∞ Z∞

−∞

1−cos[4π(u0x+v0y)]

2dx dy

=Z∞

−∞ Z∞

−∞

2dx dy −Z∞

−∞ Z∞

−∞

cos[4π(u0x+v0y)]

2dx dy

3

=∞.

Equality 2comes from the trigonometric identity cos(2θ)=1−2 sin2(θ). Equality 3holds because

the ﬁrst integral goes to inﬁnity. The absolute value of the second integral is bounded, although it does not

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6CHAPTER 2: SIGNALS AND SYSTEMS

converge as Xand Ygo to inﬁnity. We also have

P∞(s) = lim

X→∞ lim

Y→∞

4XY ZX

−XZY

−Y

sin2[2π(u0x+v0y)] dx dy

= lim

X→∞ lim

Y→∞

4XY ZY

−Y(ZX

−X

1−cos[4π(u0x+v0y)]

2dx)dy

= lim

X→∞ lim

Y→∞

4XY ZY

−YX+sin[4π(u0X+v0y)] −sin[4π(−u0X+v0y)]

8πu0dy

4

= lim

X→∞ lim

Y→∞

4XY ZY

−YX−sin(4πu0X) cos(4πv0y)

4πu0dy

= lim

X→∞ lim

Y→∞

4XY 2XY −2 sin(4πu0X) sin(4πv0Y)

(4π)2u0v0

In order to get 4, we have used the trigonometric identity sin(α+β) = sin αcos β+ cos αsin β. The rest

of the steps are straightforward.

Since s(x, y)is a periodic signal with periods X0= 1/u0and Y0= 1/v0, we have an alternative way to

compute P∞by considering only one period in each dimension. Accordingly,

P∞(s) = 1

4X0Y0ZX0

−X0ZY0

−Y0

sin2[2π(u0x+v0y)] dx dy

4X0Y02X0Y0−2 sin(4πu0X0) sin(4πv0Y0)

(4π)2u0v0

4X0Y02X0Y0−2 sin(4π) sin(4π)

(4π)2u0v0

SYSTEMS AND THEIR PROPERTIES

Solution 2.4

Suppose two LSI systems S1and S2are connected in cascade. For any two input signals f1(x, y),f2(x, y), and

two constants a1and a2, we have the following:

S2[S1[a1f1(x, y) + a2f2(x, y)]] = S2[a1S1[f1(x, y)] + a2S1[f2(x, y)]]

=a1S2[S1[f1(x, y)]] + a2S2[S1[f2(x, y)]] .

So the cascade of two LSI systems is also linear. Now suppose for a given signal f(x, y)we have S1[f(x, y)] =

g(x, y), and S2[g(x, y)] = h(x, y). By using the shift-invariance of the systems, we can prove that the cascade of

two LSI systems is also shift invariant:

S2[S1[f(x−ξ, y −η)]] = S2[g(x−ξ, y −η)] = h(x−ξ, y −η).

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This proves that two LSI systems in cascade is an LSI system

To prove Eq. (2.46) we carry out the following:

g(x, y) = h2(x, y)∗[h1(x, y)∗f(x, y)]

=h2(x, y)∗Z∞

−∞ Z∞

−∞

h1(ξ, η)f(x−ξ, y −η)dξ dη

=Z∞

−∞ Z∞

−∞

h2(u, v)Z∞

−∞ Z∞

−∞

h1(ξ, η)f(x−u−ξ, y −v−η)dξ dηdu dv

=Z∞

−∞ Z∞

−∞

h2(u, v)h1(ξ, η)f(x−u−ξ, y −v−η)dξ dηdu dv

=Z∞

−∞ Z∞

−∞

h1(ξ, η)Z∞

−∞ Z∞

−∞

h2(u, v)f(x−ξ−u, y −η−v)du dvdξ dη

=h1(x, y)∗[h2(x, y)∗f(x, y)] .

This proves the second equality in (2.46). By letting α=u+ξ, and β=v+η, we have

g(x, y) = Z∞

−∞ Z∞

−∞

h2(u, v)h1(ξ, η)f(x−u−ξ, y −v−η)dξ dηdu dv

=Z∞

−∞ Z∞

−∞ Z∞

−∞ Z∞

−∞

h2(α−ξ, β −η)h1(ξ, η)dξ dηf(x−α, y −β)dα dβ

= [h1(x, y)∗h2(x, y)] ∗f(x, y),

which proves the second equality in (2.46).

To prove (2.47) we start with the deﬁnition of convolution

g(x, y) = Z∞

−∞ Z∞

−∞

h2(ξ, η)h1(x−ξ, y −η)dξ dη

=h1(x, y)∗h2(x, y).

We then make the substitution α=x−ξand β=y−ηand manipulate the result

g(x, y) = Z−∞

+∞Z−∞

+∞

h2(x−α, y −β)h1(α, β)(−dα) (−dβ)

=Z+∞

−∞ Z+∞

−∞

h1(α, β)h2(x−α, y −β)dα dβ

=Z+∞

−∞ Z+∞

−∞

h1(ξ, η)h2(x−ξ, y −η)dξ dη

=h2(x, y)∗h1(x, y),

where the next to last equality follows since αand βare just dummy variables in the integral.

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8CHAPTER 2: SIGNALS AND SYSTEMS

Solution 2.5

1. Suppose the PSF of an LSI system is absolutely integrable.

Z∞

−∞ Z∞

−∞ |h(x, y)|dx dy ≤C < ∞(S2.1)

where Cis a ﬁnite constant. For a bounded input signal f(x, y)

|f(x, y)| ≤ B < ∞,for every (x, y),(S2.2)

for some ﬁnite B, we have

|g(x, y)|=|h(x, y)∗f(x, y)|

=Z∞

−∞ Z∞

−∞

h(x−ξ, y −η)f(ξ, η)dξdη

≤Z∞

−∞ Z∞

−∞ |h(x−ξ, y −η)|·|f(ξ, η)|dξdη

≤BZ∞

−∞ Z∞

−∞ |h(x, y)|dx dy

≤BC < ∞,for every (x, y)(S2.3)

So g(x, y)is also bounded. The system is BIBO stable.

2. We use contradiction to show that if the LSI system is BIBO stable, its PSF must be absolutely integrable.

Suppose the PSF of a BIBO stable LSI system is h(x, y), which is not absolutely integrable, that is,

Z∞

−∞ Z∞

−∞ |h(x, y)|dx dy

is not bounded. Then for a bounded input signal f(x, y)=1, the output is

|g(x, y)|=|h(x, y)∗f(x, y)|=Z∞

−∞ Z∞

−∞ |h(x, y)|dx dy,

which is also not bounded. So the system can not be BIBO stable. This shows that if the LSI system is BIBO stable,

its PSF must be absolutely integrable.

Solution 2.6

(a) If g0(x, y)is the response of the system to input PK

k=1 wkfk(x, y), then

g0(x, y) =

k=1

wkfk(x, −1) +

k=1

wkfk(0, y)

k=1

wk[fk(x, −1) + fk(0, y)]

k=1

wkgk(x, y)

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where gk(x, y)is the response of the system to input fk(x, y). Therefore, the system is linear.

(b) If g0(x, y)is the response of the system to input f(x−x0, y −y0), then

g0(x, y) = f(x−x0,−1−y0) + f(−x0, y −y0);

while

g(x−x0, y −y0) = f(x−x0,−1) + f(0, y −y0).

Since g0(x, y)6=g(x−x0, y −y0), the system is not shift-invariant.

Solution 2.7

(a) If g0(x, y)is the response of the system to input PK

k=1 wkfk(x, y), then

g0(x, y) = K

k=1

wkfk(x, y)! K

k=1

wkfk(x−x0, y −y0)!

i=1

j=1

wiwjfi(x, y)fj(x−x0, y −y0),

while K

k=1

wkgk(x, y) =

k=1

wkfk(x, y)fk(x−x0, y −y0).

Since g0(x, y)6=PK

k=1 gk(x, y), the system is nonlinear.

On the other hand, if g0(x, y)is the response of the system to input f(x−a, y −b), then

g0(x, y) = f(x−a, y −b)f(x−a−x0, y −b−y0)

=g(x−a, y −b)

and the system is thus shift-invariant.

(b) If g0(x, y)is the response of the system to input PK

k=1 wkfk(x, y), then

g0(x, y) = Z∞

−∞

k=1

wkfk(x, η)dη

k=1

wkZ∞

−∞

fk(x, η)dη

k=1

wkgk(x, y),

where gk(x, y)is the response of the system to input fk(x, y). Therefore, the system is linear.

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10 CHAPTER 2: SIGNALS AND SYSTEMS

On the other hand, if g0(x, y)is the response of the system to input f(x−x0, y −y0), then

g0(x, y) = Z∞

−∞

f(x−x0, η −y0)dη

=Z∞

−∞

f(x−x0, η −y0)d(η−y0)

=Z∞

−∞

f(x−x0, η)dη.

Since g(x−x0, y −y0) = R∞

−∞ f(x−x0, η)dη, the system is shift-invariant.

Solution 2.8

From the results in Problem 2.5, we know that an LSI system is BIBO stable if and only if its PSF is absolutely

integrable.

(a) Not stable. The PSF h(x, y)goes to inﬁnite when xand/or ygo to inﬁnity. R∞

−∞ R∞

−∞ |h(x, y)|dx dy =

R∞

−∞ R∞

−∞(x2+y2)dx dy =R∞

−∞ hR∞

−∞ x2dxidy +R∞

−∞ hR∞

−∞ y2dyidx. Since R∞

−∞ x2dx =R∞

−∞ y2dy is

not bounded, then R∞

−∞ R∞

−∞(x2+y2)dx dy is not bounded.

(b) Stable. R∞

−∞ R∞

−∞ |h(x, y)|dx dy =R∞

−∞ R∞

−∞(exp{−(x2+y2)})dx dy =hR∞

−∞ e−x2dxi2=π, which is

bounded. So the system is stable.

−∞ R∞

−∞ x2e−y2dx dy =R∞

−∞ x2hR∞

−∞ e−y2dyidx =R∞

−∞ √πx2dx is

unbounded. So the system is not stable.

Solution 2.9

(a) g(x) = R∞

−∞ f(x−t)f(t)dt.

(b) Given an input as af1(x) + bf2(x), where a, b are some constant, the output is

g0(x)=[af1(x) + bf2(x)] ∗[af1(x) + bf2(x)]

=a2f1(x)∗f1(x)+2abf1(x)∗f2(x) + b2f2(x)∗f2(x)

6=ag1(x) + bg2(x),

where g1(x)and g2(x)are the output corresponding to an input of f1(x)and f2(x)respectively.

Hence, the system is nonlinear.

g1(x) = f1(x)∗f1(x)

=Z∞

−∞

f1(x−t)f1(t)dt

=Z∞

−∞

f(x−t−x0)f1(t−x0)dt.

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Changing variable t0=t−x0in the above integration, we get

g1(x) = Z∞

−∞

f(x−2x0−t0)f1(t0)dt0

=g(x−2x0).

Thus, if the input is shifted by x0, the output is shifted by 2x0. Hence, the system is not shift-invariant.

CONVOLUTION OF SIGNALS

Solution 2.10

(a)

f(x, y)δ(x−1, y −2) = f(1,2)δ(x−1, y −2)

= (1 + 22)δ(x−1, y −2)

= 5δ(x−1, y −2)

(b)

f(x, y)∗δ(x−1, y −2) = Z∞

−∞ Z∞

−∞

f(ξ, η)δ(x−ξ−1, y −η−2) dξ dη

=Z∞

−∞ Z∞

−∞

f(x−1, y −2)δ(x−ξ−1, y −η−2) dξ dη

=f(x−1, y −2) Z∞

−∞ Z∞

−∞

δ(x−ξ−1, y −η−2) dξ dη

=f(x−1, y −2)

= (x−1) + (y−2)2

(c)

Z∞

−∞ Z∞

−∞

δ(x−1, y −2)f(x, 3)dx dy 1

=Z∞

−∞ Z∞

−∞

δ(x−1, y −2)f(1,3)dx dy

=Z∞

−∞ Z∞

−∞

δ(x−1, y −2)(1 + 32)dx dy

= 10 Z∞

−∞ Z∞

−∞

δ(x−1, y −2)dx dy

2

= 10

Equality 1comes from the Eq. (2.7) in the text. Equality 2comes from the fact:

Z∞

−∞ Z∞

−∞

δ(x−1, y −2)dx dy =Z∞

−∞ Z∞

−∞

δ(x, y)dx dy = 1.

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12 CHAPTER 2: SIGNALS AND SYSTEMS

(d)

δ(x−1, y −2) ∗f(x+ 1, y + 2) 3

=Z∞

−∞ Z∞

−∞

δ(x−ξ−1, y −η−2)f(ξ+ 1, η + 2)dξ dη

4

=Z∞

−∞ Z∞

−∞

δ(x−ξ−1, y −η−2)f((x−1) + 1,(y−2) + 2)dξ dη

=Z∞

−∞ Z∞

−∞

δ(x−ξ−1, y −η−2)f(x, y)dξ dη

5

=f(x, y) = x+y2

3comes from the deﬁnition of convolution; 4comes from the Eq. (2.7) in text; 5is the same as 2in part

(c). Alternatively, by using the sifting property of δ(x, y)and deﬁning g(x, y) = f(x+ 1, y + 2), we have

δ(x−1, y −2) ∗g(x, y) = g(x−1, y −2)

=f(x−1+1, y −2 + 2)

=f(x, y)

=x+y2.

Solution 2.11

(a)

f(x, y)∗g(x, y) = Z∞

−∞ Z∞

−∞

f(ξ, η)g(x−ξ, y −η)dξ dη

=Z∞

−∞ Z∞

−∞

f1(ξ)f2(η)g1(x−ξ)g2(y−η)dξ dη

f(x, y)∗g(x, y) = Z∞

−∞

f1(ξ)g1(x−ξ)dξZ∞

−∞

f2(η)g2(y−η)dη.

Hence, their convolution is also separable.

(b)

f(x, y)∗g(x, y) = (f1(x)∗g1(x)) (f2(y)∗g2(y)) .

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Solution 2.12

g(x, y) = f(x, y)∗h(x, y)

=Z∞

−∞ Z∞

−∞

f(x−ξ, y −η)h(ξ, η)dξdη

=Z∞

−∞ Z∞

−∞

(x−ξ+y−η)exp{−(ξ2+η2)}dξdη

= (x+y)Z∞

−∞ Z∞

−∞

e−ξ2−η2dξdη −Z∞

−∞ Z∞

−∞

ξe−ξ2−η2dξdη −Z∞

−∞ Z∞

−∞

ηe−ξ2−η2dξdη

= (x+y)Z∞

−∞

e−ξ2dξ2

−Z∞

−∞

e−η2Z∞

−∞

ξe−ξ2dξdη −Z∞

−∞

e−ξ2Z∞

−∞

ηe−η2dηdξ

=π(x+y)(S2.4)

We get (S2.4) by noticing that since ξis an odd function and e−ξ2is an even function, we must have

Z∞

−∞

ξe−ξ2dξ = 0 .

Also, Z∞

−∞

e−ξ2dξ =√π .

FOURIER TRANSFORMS AND THEIR PROPERTIES

Solution 2.13

(a) See the solution to part (b) below. The Fourier transform is

F2{δs(x, y)}=δs(u, v)

(b)

F2{δs(x, y; ∆x, ∆y)}=Z∞

−∞ Z∞

−∞

δs(x, y; ∆x, ∆y)e−j2π(ux+vy)dx dy

δs(x, y; ∆x, ∆y)is a periodic signal with periods ∆xand ∆yin xand yaxes. Therefore it can be written

as a Fourier series expansion. (Please review Oppenheim, Willsky, and Nawad, Signals and Systems for the

deﬁnition of Fourier series expansion of periodic signals.)

δs(x, y; ∆x, ∆y) = ∞

m=−∞

∞

n=−∞

Cmnej2π(mx

∆x+ny

∆y),

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14 CHAPTER 2: SIGNALS AND SYSTEMS

where

Cmn =1

∆x∆yZ∆x

−∆x

2Z∆y

−∆y

δs(x, y; ∆x, ∆y)e−j2π(mx

∆x+ny

∆y)dx dy

∆x∆yZ∆x

−∆x

2Z∆y

−∆y

∞

m=−∞

∞

n=−∞

δ(x−m∆x, y −n∆y)e−j2π(mx

∆x+ny

∆y)dx dy.

In the integration region −∆x

2<x<∆x

2and −∆y

2< y < ∆y

2there is only one impulse corresponding to

m= 0,n= 0. Therefore, we have

Cmn =1

∆x∆yZ∆x

−∆x

2Z∆y

−∆y

δ(x, y)e−j2π(0·x

∆x+0·y

∆y)dx dy

∆x∆y.

We have:

δs(x, y; ∆x, ∆y) = 1

∆x∆y

∞

m=−∞

∞

n=−∞

ej2π(mx

∆x+ny

∆y).

Therefore,

F2{δs}=Z∞

−∞ Z∞

−∞

δs(x, y; ∆x, ∆y)e−j2π(ux+vy)dx dy

=Z∞

−∞ Z∞

−∞

∆x∆y

∞

m=−∞

∞

n=−∞

ej2π(mx

∆x+ny

∆y)e−j2π(ux+vy)dx dy

=∞

m=−∞

∞

n=−∞

∆x∆yZ∞

−∞ Z∞

−∞

ej2π(mx

∆x+ny

∆y)e−j2π(ux+vy)dx dy

=∞

m=−∞

∞

n=−∞

∆x∆yF2nej2π(mx

∆x+ny

∆y)o

=∞

m=−∞

∞

n=−∞

∆x∆yδu−m

∆x, v −n

∆y

5

=∞

m=−∞

∞

n=−∞

∆x∆y·∆x∆yδ(u∆x−m, v∆y−n)

F2{δs}=δs(u∆x, v∆y)

Equality 5comes from the property δ(ax) = 1

|a|δ(x).

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(c)

F2{s(x, y)}=Z∞

−∞ Z∞

−∞

s(x, y)e−j2π(ux+vy)dx dy

=Z∞

−∞ Z∞

−∞

sin[2π(u0x+v0y)]e−j2π(ux+vy)dx dy

=Z∞

−∞ Z∞

−∞

2jhej2π(u0x+v0y)−e−j2π(u0x+v0y)ie−j2π(ux+vy)dx dy

2jZ∞

−∞ Z∞

−∞

ej2π(u0x+v0y)e−j2π(ux+vy)dx dy

−Z∞

−∞ Z∞

−∞

e−j2π(u0x+v0y)e−j2π(ux+vy)dx dy

2jZ∞

−∞ Z∞

−∞

e−j2π[(u−u0)x+(v−v0)y]dx dy

−Z∞

−∞ Z∞

−∞

e−j2π[(u+u0)x+(v+v0)y]dx dy

F2{s(x, y)}=1

2j[δ(u−u0, v −v0)−δ(u+u0, v +v0)] .

We used Eq. (2.69) twice to get the last equality.

(d)

F2(c)(u, v) = Z∞

−∞ Z∞

−∞

c(x, y)e−j2π(ux+vy)dx dy

=Z∞

−∞ Z∞

−∞

cos[2π(u0x+v0y)]e−j2π(ux+vy)dx dy

=Z∞

−∞ Z∞

−∞

2[ej2π(u0x+v0y)+e−j2π(u0x+v0y)]e−j2π(ux+vy)dx dy

2Z∞

−∞ Z∞

−∞

ej2π(u0x+v0y)e−j2π(ux+vy)dx dy

+Z∞

−∞ Z∞

−∞

e−j2π(u0x+v0y)e−j2π(ux+vy)dx dy

2Z∞

−∞ Z∞

−∞

e−j2π[(u−u0)x+(v−v0)y]dx dy

+Z∞

−∞ Z∞

−∞

e−j2π[(u+u0)x+(v+v0)y]dx dy

F2(c)(u, v) = 1

2[δ(u−u0, v −v0) + δ(u+u0, v +v0)].

We used Eq. (2.69) twice to get the last equality.

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16 CHAPTER 2: SIGNALS AND SYSTEMS

(e)

F2(f)(u, v) = Z∞

−∞ Z∞

−∞

f(x, y)e−j2π(ux+vy)dx dy

=Z∞

−∞ Z∞

−∞

2πσ2e−(x2+y2)/2σ2e−j2π(ux+vy)dx dy

=Z∞

−∞ Z∞

−∞

2πσ2e−(x2+j4πσ2ux)/2σ2e−(y2+j4πσ2vy)/2σ2dx dy

=Z∞

−∞

√2πσ2e−(x2+j4πσ2ux)/2σ2dxZ∞

−∞

√2πσ2e−(y2+j4πσ2vy)/2σ2dy

=Z∞

−∞

√2πσ2e−(x+j2πσ2u)2/2σ2e(j2πσ2u)2/2σ2dx·

Z∞

−∞

√2πσ2e−(y+j2πσ2v)2/2σ2e(j2πσ2v)2/2σ2dy

=e−2π2σ2u2Z∞

−∞

√2πσ2e−(x+j2πσ2u)2/2σ2dx·

e−2π2σ2v2Z∞

−∞

√2πσ2e−(y+j2πσ2v)2/2σ2dy

=e−2π2σ2u2·e−2π2σ2v2

F2(f)(u, v) = e−2π2σ2(u2+v2).

Solution 2.14

The Fourier transform of f(x)is

F(u) = Z∞

−∞

f(x)e−j2πuxdx.

(a) Assuming f(x)is real and f(x) = f(−x),

F∗(u) = Z∞

−∞ f(x)e−j2πux∗dx

=Z∞

−∞

f∗(x)ej2πuxdx

=Z∞

−∞

f∗(−ξ)e−j2πuξ dξ, let ξ=−x

=Z∞

−∞

f(ξ)e−j2πuξdξ, since f(−x) = f(x)and f(x)is real

=F(u).

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(b) Similarly, assuming f(x)is real and f(x) = −f(−x),

F∗(u) = Z∞

−∞

f∗(−ξ)e−j2πuξ dξ

=Z∞

−∞ −f(ξ)e−j2πuξ dξ, since f(−x) = −f(x)

=−F(u).

Solution 2.15

In deriving the symmetric property F∗(u) = F(u), we used the fact that f(x)is real. If f(x)is a complex

signal, we have f∗(−ξ) = f∗(ξ)instead of f∗(−ξ) = f(ξ). Therefore,

F∗(u) = Z∞

−∞ f(x)e−j2πux∗dx

=Z∞

−∞

f∗(−ξ)e−j2πuξ dξ, let ξ=−x

=Z∞

−∞

f∗(ξ)e−j2πuξ dξ,

=F {f∗(x)}

Solution 2.16

(a) Conjugate property: F2(f∗)(u, v) = F∗(−u, −v).

F2(f∗)(u, v) = Z∞

−∞ Z∞

−∞

f∗(x, y)e−j2π(ux+vy)dx dy

=Z∞

−∞ Z∞

−∞

f(x, y)ej2π(ux+vy)dx dy∗

=Z∞

−∞ Z∞

−∞

f(x, y)e−j2π[(−u)x+(−v)y]dx dy∗

= [F(−u, −v)]∗

=F∗(−u, −v).

Conjugate symmetry property: If f(x, y)is real, F(u, v) = F∗(−u, −v). Since f(x, y)is real, f∗(x, y) =

f(x, y). Therefore,

F∗(−u, −v) = F2{f∗(x, y)}=F2{f(x, y)}=F(u, v).

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18 CHAPTER 2: SIGNALS AND SYSTEMS

(b) Scaling property: F2(fab)(u, v) = 1

|ab|F2(f)u

a,v

b.

F2(fab)(u, v) = Z∞

−∞ Z∞

−∞

f(ax, by)e−j2π(ux+vy)dx dy

=Z∞

−∞ Z∞

−∞

f(ax, by)e−j2π[u(ax)/a+v(by)/b]1

ab d(ax)d(by)

|ab|Z∞

−∞ Z∞

−∞

f(p, q)e−j2π[(u/a)p+(v/b)q]dp dq

|ab|F2(f)u

a,v

b.

F2(f∗g)(u, v) = Z∞

−∞ Z∞

−∞ Z∞

−∞ Z∞

−∞

f(ξ, η)g(x−ξ, y −η)dξ dηe−j2π(ux+vy)dx dy.

Interchange the order of integration to yield

F2(f∗g)(u, v) = Z∞

−∞ Z∞

−∞

f(ξ, η)Z∞

−∞ Z∞

−∞

g(x−ξ, y −η)e−j2π(ux+vy)dx dydξ dη

=Z∞

−∞ Z∞

−∞

f(ξ, η)Z∞

−∞ Z∞

−∞

g(x−ξ, y −η)

e−j2π[u(x−ξ)+v(y−η)]e−j2π(uξ+vη)dx dydξ dη

=Z∞

−∞ Z∞

−∞

f(ξ, η)e−j2π(uξ+vη)Z∞

−∞ Z∞

−∞

g(x−ξ, y −η)

e−j2π[u(x−ξ)+v(y−η)]dx dydξ dη

=Z∞

−∞ Z∞

−∞

f(ξ, η)e−j2π(uξ+vη)Z∞

−∞ Z∞

−∞

g(p, q)e−j2π[up+vq]dp dqdξ dη

=Z∞

−∞ Z∞

−∞

f(ξ, η)e−j2π(uξ+vη)F2(g)(u, v)dξ dη

=F2(g)(u, v)·Z∞

−∞ Z∞

−∞

f(ξ, η)e−j2π(uξ+vη)dξ dη

F2(f∗g)(u, v) = F2(g)(u, v)· F2(f)(u, v).

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(d) Product property: F2(fg)(u, v) = F(u, v)∗G(u, v).

F2(fg)(u, v) = Z∞

−∞ Z∞

−∞

f(x, y)g(x, y)e−j2π(ux+vy)dx dy

=Z∞

−∞ Z∞

−∞ Z∞

−∞ Z∞

−∞

G(ξ, η)ej2π(xξ+yη)dξ dηf(x, y)e−j2π(ux+vy)dx dy

=Z∞

−∞ Z∞

−∞

G(ξ, η)Z∞

−∞ Z∞

−∞

f(x, y)ej2π(xξ+yη)e−j2π(ux+vy)dxdydξ dη

=Z∞

−∞ Z∞

−∞

G(ξ, η)Z∞

−∞ Z∞

−∞

f(x, y)e−j2π[(u−ξ)x+(v−η)y]dx dydξ dη

=Z∞

−∞ Z∞

−∞

G(ξ, η)F(u−ξ, v −η)dξ dη

=F(u, v)∗G(u, v).

Solution 2.17

Since both the rect and sinc functions are separable, it is sufﬁcient to show the result for 1-D rect and sinc

functions. A 1-D rect function is

rect(x) = 









1,for |x|<1

0,for |x|>1

F{rect(x)}=Z∞

−∞

rect(x)e−j2πuxdx

=Z1/2

−1/2

e−j2πuxdx

=Z1/2

−1/2

cos(2πux)dx −jZ1/2

−1/2

sin(2πux)dx, ejθ = cos θ+jsin θ

=Z1/2

−1/2

cos(2πux)dx

=sin(πu)

πu

= sinc(u).

Therefore, we have F{sinc(x)}= rect(u). Using Parseval’s Theorem, we have

E∞=Z∞

−∞ Z∞

−∞ krect(x, y)k2dx dy

=Z1/2

−1/2Z1/2

−1/2

dx dy

= 1

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20 CHAPTER 2: SIGNALS AND SYSTEMS

For the sinc function, P∞= 0, because E∞is ﬁnite.

Solution 2.18

Since the signal is separable, we have

F[f(x, y)] = F1D[sin(2πax)]F1D[cos(2πby)] ,

F1D[sin(2πax)] = 1

2j[δ(u−a)−δ(u+a)] ,

F1D[cos(2πby)] = 1

2[δ(v−b) + δ(v+b)] .

So,

F[f(x, y)] = 1

4j[δ(u−a)δ(v−b)−δ(u+a)δ(v−b) + δ(u−a)δ(v+b)−δ(u+a)δ(v+b)] .

Now we need to show that δ(u)δ(v) = δ(u, v)(in a generalized way):

δ(u)δ(v) = 0,for u6= 0,or v6= 0

Therefore,

Z∞

−∞ Z∞

−∞

f(u, v)δ(u)δ(v)du dv =Z∞

−∞ Z∞

−∞

f(u, v)δ(u)duδ(v)dv =Z∞

−∞

f(0, v)δ(v)dv =f(0,0) .

Based on the argument above δ(u)δ(v) = δ(u, v), and

F[f(x, y)] = 1

4j[δ(u−a, v −b)−δ(u+a, v −b) + δ(u−a, v +b)−δ(u+a, v +b)] .

The above solution can also be obtained by using the relationship:

sin(2πax) cos(2πby) = 1

2[sin(2π(ax −by)) + sin(2π(ax +by))] .

Solution 2.19

A function f(x, y)can be expressed in polar coordinates as:

f(x, y) = f(rcos θ, r sin θ) = fp(r, θ).

If it is circularly symmetric, we have fp(r, θ)is constant for ﬁxed r. The Fourier transform of f(x, y)is deﬁned as:

F(u, v) = Z∞

−∞ Z∞

−∞

f(x, y)e−j2π(ux+vy)dx dy

=Z∞

0Z2π

fp(r, θ)e−j2π(ur cos θ+vr sin θ)r dr dθ

=Z∞

fp(r, θ)Z2π

e−j2π(ur cos θ+vr sin θ)dθr dr .

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Letting u=qcos φand v=qsin φ, the above equation becomes:

F(u, v) = Z∞

fp(r, θ)Z2π

e−j2πqr cos(φ−θ)dθr dr .

Since F(u, v)is also circularly symmetric, it can be written as Fq(q, φ)and is constant for ﬁxed q. In particular,

Fq(q, φ) = Fq(q, π/2), and therefore

Fq(q, φ) = Fq(q, π/2) = Z∞

fp(r, θ)Z2π

e−j2πqr sin θdθr dr .

Now we will show that (2.108) holds.

Z2π

e−j2πqr sin θdθ =Z2π

cos(2πqr sin θ)dθ −jZ2π

sin(2πqr sin θ)

1

= 2 Zπ

cos(2πqr sin θ)dθ

= 2πJ0(2πqr).

Equality 1holds because cos(−θ) = cos(θ), and sin(θ) = −sin(θ).

Based on the above derivation, we have proven (2.108).

Solution 2.20

The unit disk is expressed as f(r) = rect(r)and its Hankel transform is

F(q)=2πZ∞

f(r)J0(2πqr)r dr

= 2πZ∞

rect(r)J0(2πqr)r dr

= 2πZ1/2

J0(2πqr)r dr .

Now apply the following change of variables

s= 2πqr ,

r=s

2πq ,

dr =ds

2πq ,

to yield

F(q) = 1

2πq2Zπq

J0(s)sds .

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22 CHAPTER 2: SIGNALS AND SYSTEMS

From mathematical tables, we note that

J0()d =xJ1(x).

Therefore,

F(q) = J1(πq)

=jinc(q).

TRANSFER FUNCTION

Solution 2.21

(a) The impulse response function is shown in Figure S2.1.

−3

−2

−1

−3

−2

−1

Figure S2.1 Impulse response function of the system. See Problem 2.21(a).

(b) The transfer function of the function is the Fourier transform of the impulse response function:

H(u, v) = F{h(x, y)}

=F{e−πx2}F{e−πy2/4},since h(x, y)is separable

= 4e−π(u2+4v2).

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Solution 2.22

(a) The 1D proﬁle of the bar phantom is:

f(x) = 1,k−1

2w≤x≤k+1

0,k+1

2w≤x≤k+3

2w,

where kis an integer. The response of the system to the bar phantom is:

g(x) = f(x)∗l(x) = Z∞

−∞

f(x−ξ)l(ξ)dξ .

At the center of the bar, we have

g(0) = Z∞

−∞

f(0 −ξ)l(ξ)dξ

=Zw/2

−w/2

cos(αξ)dξ

αsin αw

2.

At the point halfway between two adjacent bars, we have

g(w) = Z∞

−∞

f(w−ξ)l(ξ)dξ

=Zw/2

w−π/2α

cos(αξ)dξ +Zw+π/2α

3w/2

cos(αξ)dξ

= 2 Zw/2

w−π/2α

cos(αξ)dξ

αhsin αw

2−sin αw −π

2i.

(b) From the line spread function alone, we cannot tell whether the system is isotropic. The line spread function

is a “projection” of the PSF. During the projection, the information along the ydirection is lost.

l(x) = Z∞

−∞

h(x, y)dy

=h1D(x)Z∞

−∞

h1D(y)dy .

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24 CHAPTER 2: SIGNALS AND SYSTEMS

Therefore h1D(x) = cl(x)where 1/c =R∞

−∞ h1D(y)dy. Hence,

1/c =Z∞

−∞

cl(y)dy ,

1/c2=Zπ/2α

−π/2α

cos(αy)dy ,

1/c2= 2/α .

Therefore,

h(x, y) = 





2cos(αx) cos(αy)|αx| ≤ π/2and |αy| ≤ π/2

0otherwise

The transfer function is

H(u, v) = F2D{h(x, y)}

=Z∞

−∞ Z∞

−∞

h(x, y)ej2πuxdxej2πuydy

=Z∞

−∞ Z∞

−∞

h1D(x)h1D(y)ej2πuxdxej2πuydy

=Z∞

−∞ Z∞

−∞

h1D(x)ej2πuxdxh1D(y)ej2πuydy

=Z∞

−∞

h1D(x)ej2πuxdx Z∞

−∞

h1D(y)ej2πuy dy

=H1D(u)H1D(v),

which is also separable with H(u, v) = H1D(u)H1D(v). We have

H1D =rα

2F1D{l(x)}

=rα

2F1D{cos(αx)}∗F1D nrect αx

πo

=rπ

2hsinc π

α(u−α/2π)+ sinc π

α(u+α/2π)i.

Therefore, the transfer function is

H(u, v) = π

2hsinc π

α(u−α/2π)+ sinc π

α(u+α/2π)i

hsinc π

α(v−α/2π)+ sinc π

α(v+α/2π)i.

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APPLICATIONS, EXTENSIONS AND ADVANCED TOPICS

Solution 2.23

(a) The system is separable because h(x, y) = e−(|x|+|y|)=e−|x|e−|y|.

(b) The system is not isotropic since h(x, y)is not a function of r=px2+y2.

Additional comments: An easy check is to plug in x= 1, y = 1 and x= 0, y =√2into h(x, y). By

noticing that h(1,1) 6=h(0,√2), we can conclude that h(x, y)is not rotationally invariant, and hence not

isotropic.

Isotropy is rotational symmetry around the origin, not just symmetry about a few axes, for example the x-

and y-axes. h(x, y) = e−(|x|+|y|)is symmetric about a few lines, but it is not rotationally invariant.

When we studied the properties of Fourier transform, we learned that if a signal is isotropic then its Fourier

transform has a certain symmetry. Note that the symmetry of the Fourier transform is only a necessary, but

not sufﬁcient, condition for the signal to be isotropic.

g(x, y) = h(x, y)∗f(x, y)

=Z∞

−∞ Z∞

−∞

h(ξ, η)f(x−ξ, y −η)dξ dη

=Z∞

−∞ Z∞

−∞

e−(|ξ|+|η|)δ(x−ξ)dξ dη

=Z∞

−∞

e−(|x|+|η|)dη

=e−|x|Z∞

−∞

e−|η|dη

=e−|x|Z0

−∞

eηdη +Z∞

e−ηdη

= 2e−|x|.

(d) The response is

g(x, y) = h(x, y)∗f(x, y)

=Z∞

−∞ Z∞

−∞

h(ξ, η)f(x−ξ, y −η)dξ dη

=Z∞

−∞ Z∞

−∞

e−(|ξ|+|η|)δ(x−ξ−y+η)dξ dη

=Z∞

−∞

e−|η|Z∞

−∞

e−|ξ|δ(x−ξ−y+η)dξdη

=Z∞

−∞

e−|η|e−|x−y+η|dη .

1. Now assume x−y < 0, then x−y+η < η. The range of integration in the above can be divided into

three parts (see Fig. S2.2):

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26 CHAPTER 2: SIGNALS AND SYSTEMS

-( )x-y

h<0

+h<0x-y

h>0

+h<0x-y

h>0

+h>0x-y

Figure S2.2 For x−y < 0the integration interval (−∞,∞)can be partitioned into three segments. See Prob-

lem 2.23(d).

I. η∈(−∞,0). In this interval, x−y+η < η < 0.|η|=−η,|x−y+η|=−(x−y+η);

II. η∈[0,−(x−y)). In this interval, x−y+η < 0≤η.|η|=η,|x−y+η|=−(x−y+η);

III. η∈[−(x−y),∞). In this interval, 0≤x−y+η < η.|η|=η,|x−y+η|=x−y+η.

Based on the above analysis, we have:

g(x, y) = Z∞

−∞

e−|η|e−|x−y+η|dη

=Z0

−∞

e−(|η|+|x−y+η|)dη +Z−(x−y)

e−(|η|+|x−y+η|)+Z∞

−(x−y)

e−(|η|+|x−y+η|)

=Z0

−∞

ex−y+2ηdη +Z−(x−y)

ex−ydη +Z∞

−(x−y)

e−(x−y+2η)dη

2ex−y−(x−y)ex−y+1

2ex−y

= [1 −(x−y)]ex−y.

2. For x−y≥0,η < x −y+η. The range of integration in the above can be divided into three parts (see

Fig. S2.3):

-( )x-y 0h

h<0

+h<0x-y

h<0

+h>0x-y

h>0

+h>0x-y

Figure S2.3 For x−y > 0the integration interval (−∞,∞)can be partitioned into three segments. See Prob-

lem 2.23(d).

I. η∈(−∞,−(x−y)). In this interval, η < x −y+η < 0.|η|=−η,|x−y+η|=−(x−y+η);

II. η∈[−(x−y),0). In this interval, η < 0≤x−y+η.|η|=−η,|x−y+η|=x−y+η;

III. η∈[0,∞). In this interval, 0≤η < x −y+η.|η|=η,|x−y+η|=x−y+η.

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Based on the above analysis, we have:

g(x, y) = Z∞

−∞

e−|η|e−|x−y+η|dη

=Z−(x−y)

−∞

e−(|η|+|x−y+η|)dη +Z0

−(x−y)

e−(|η|+|x−y+η|)+Z∞

e−(|η|+|x−y+η|)

=Z−(x−y)

−∞

ex−y+2ηdη +Z0

−(x−y)

e−(x−y)dη +Z∞

e−(x−y+2η)dη

2e−(x−y)+ (x−y)e−(x−y)+1

2e−(x−y)

= [1 + (x−y)]e−(x−y).

Based on the above two steps, we have:

g(x, y) = (1 + |x−y|)e−|x−y|.

Solution 2.24

(a) Yes, it is shift invariant because its impulse response depends on x−ξ.

(b) By linearity, the output is

g(x) = e−(x+1)2

2+e−(x)2

2+e−(x−1)2

Solution 2.25

(a) The impulse response of the ﬁlter is the inverse Fourier transform of H(u), which can be written as

H(u)=1−rect u

2U0.

Using the linearity of the Fourier transform and the Fourier transform pairs

F {δ(t)}= 1 ,

F {sinc(t)}= rect(u),

we have

h(t) = F−1{H(u)}

=δ(t)−2U0sinc(2U0t).

(b) The system response to f(t) = cis 0, since f(t)contains only a zero frequency component while h(t)passes

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28 CHAPTER 2: SIGNALS AND SYSTEMS

only high frequency components. Formal proof:

f(t)∗h(t) = f(t)∗[δ(t)−2U0sinc(2U0t)]

=f(t)−2U0f(t)∗sinc(2U0t)

=c−cZ∞

−∞

2U0sinc(2U0t)dt

=c−cZ∞

−∞

sinc(τ)dτ

= 0 .

The system response to f(t) = 1, t ≥0

0, t < 0is

f(t)∗h(t) = f(t)∗[δ(t)−2U0sinc(2U0t)]

=f(t)−2U0f(t)∗sinc(2U0t)

=f(t)−Z∞

−∞

f(x)2U0sinc(2U0(t−x))dx

=f(t)−Z∞

2U0sinc(2U0(t−x))dx

=f(t) + Z−∞

2U0sinc(2U0(y))dy

=f(t)−Zt

−∞

2U0sinc(2U0(y))dy

=









1−Z0

−∞

2U0sinc(2U0(y))dy +Z0

2U0sinc(2U0(y))dy t < 0

1−Z0

−∞

2U0sinc(2U0(y))dy −Zt

2U0sinc(2U0(y))dy t > 0

=









−1

2+Z0

2U0sinc(2U0(y))dy t < 0

1−1

2−Zt

2U0sinc(2U0(y))dy t > 0

=









−1

2+Z0

2U0sinc(2U0(y))dy t < 0

2−Zt

2U0sinc(2U0(y))dy t > 0

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Solution 2.26

(a) The rect function is deﬁned as

rect(t) = 1,|t| ≤ 1/2

0,otherwise .

So we have

rect t

T=1,|t| ≤ T/2

0,otherwise

and

rect t+ 0.75T

0.5T=1,|t+ 0.75T| ≤ T/4

0,otherwise .

Therefore,

h(t) = 









−1/T, −T < t < −T/2

1/T, −T/2< t < T/2

−1/T, T/2< t < T

0,otherwise

The impulse response is plotted in Fig. S2.4.

Figure S2.4 The impulse response h(t). See Problem 2.26(a).

The absolute integral of h(t)is R∞

−∞ |h(t)|2dt = 2/T . So The system is stable when T > 0. The system is

not causal, since h(t)6= 0 for −T < t < 0.

(b) The response of the system to a constant signal f(t) = cis

g(t) = f(t)∗h(t) = Z∞

−∞

f(t−τ)h(τ)dτ =cZ∞

−∞

h(τ)dτ = 0 .

g(t) = f(t)∗h(t) = Z∞

−∞

f(t−τ)h(τ)dτ =Zt

−∞

h(τ)dτ

g(t) = 









0, t < −T

−t/T −1,−T < t < −T/2

t/T, −T/2< t < T/2

−t/T + 1, T/2< t < T

0, t > T

The response of the system to the unit step signal is plotted in Figure S2.5.

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30 CHAPTER 2: SIGNALS AND SYSTEMS

Figure S2.5 The response of the system to the unit step signal. See Problem 2.26(c).

(d) The Fourier transform of a rect function is a sinc function (see Problem 2.17). By using the properties of the

Fourier transform (scaling, shifting, and linearity), we have

H(u) = F {h(t)}

=−0.5e−j2πu(−0.75T)sinc(0.5uT ) + sinc(uT )−0.5e−j2πu(0.75T)sinc(0.5uT )

= sinc(uT )−cos(1.5πuT ) sinc(0.5uT ).

(e) The magnitude spectrum of h(t)is plotted in Figure S2.6.

−20 −15 −10 −5 0 5 10 15 2

0.2

0.4

0.6

0.8

1.2

1.4

|H(u)|

T = 0.25

T = 0.1

T = 0.05

Figure S2.6 The magnitude spectrum of h(t). See Problem 2.26(e).

(f) From the calculation in part (d) and the plot in part (c), it can be seen that |H(0)|= 0. So the output of the

system does not have a DC component. The system is not a low pass ﬁlter. The system is not a high-pass

ﬁlter since it also ﬁlters out high frequency components. As T→0, the pass band of the system moves to

higher frequencies, and the system tends toward a high-pass ﬁlter.

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Solution 2.27

(a) The inverse Fourier transform of ˆ

H(%)is

h(r) = F−1{ˆ

H(%)}

=Z∞

−∞

H(%)ej2πr%d%

=Z%0

−%0|%|ej2πr%d%

=Z%0

%ej2πr%d% −Z0

−%0

%ej2πr%d%

=Z%0

%ej2πr%d% +Z−%0

%ej2πr%d%

=Z%0

%ej2πr%d% +Z%0

%e−j2πr%d%

=Z%0

%ej2πr% +e−j2πr%d%

= 2 Z%0

%cos(2πr%)d%

= 2 "%sin(2πr%)

2πr 

%=0 −Z%0

sin(2πr%)

2πr d%#

= 2 "%sin(2πr%0)

2πr +cos(2πr%)

4π2r2

%=0#

2π2r2[cos(2πr%0)+2πr%0sin(2πr%0)−1] .

(b) The response of the ﬁlter is g(r) = f(r)∗ˆ

h(r), hence G(%) = F(%)ˆ

H(%). i) A constant function f(r) = c

has the Fourier transform

F(%) = cδ(%).

The transfer function of a ramp ﬁlter has a value zero at %= 0. So the system response has the Fourier

transform

G(%)=0.

Therefore, the responses of a ramp ﬁlter to a constant function is g(r)=0. ii) The Fourier transform of a

sinusoid function f(r) = sin(ωr)is

F(%) = 1

2jhδ(%−ω

2π)−δ(%+ω

2π)i.

Hence,

G(%) = 





4πj hδ%−ω

2π−δ%+ω

2πi %0=ω

0otherwise

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32 CHAPTER 2: SIGNALS AND SYSTEMS

Therefore, the response of a ramp ﬁlter to a sinusoid function is

g(r) = 





2πsin(ωr)%0=ω

0otherwise

Solution 2.28

Suppose the Fourier transform of f(x, y)is F(u, v). Using the scaling properties, we have that the Fourier

transform of f(ax, by)is 1

|ab|Fu

a,v

b. The output of the system is

g(x, y) = F1

|ab|Fu

a,v

b

=Z∞

−∞ Z∞

−∞

|ab|Fu

a,v

be−j2π(ux+vy)du dv

|ab|Z∞

−∞ Z∞

−∞

F(ξ, η)ej2π(aξ(−x)+bη(−y))|ab|dξ dη .

Given the inverse Fourier transform

f(x, y) = Z∞

−∞ Z∞

−∞

F(u, v)ej2π(ux+vy)du dv

we have Z∞

−∞ Z∞

−∞

F(ξ, η)ej2π(aξ(−x)+bη(−y))|ab|dξdη =|ab|f(−ax, −by).

Therefore, g(x, y) = f(−ax, −by)is a scaled and inverted replica of the input.

Solution 2.29

The Fourier transform of the signal f(x, y)and the noise η(x, y)are:

F(u, v) = F {f(x, y)}

=|ab|F {sinc(ax, by)}

=|ab|1

|ab|rect u

a,v

b

= rect u

a,v

b

=1,|x|<|a|/2and |y|<|b|/2

0,otherwise ,

E(u, v) = F {η(x, y)}

2[δ(u−A, v −B) + δ(u+A, v +B)] .

Using the linearity of Fourier transform, the Fourier transform of the measurements g(x, y)is

G(u, v) = rect u

a,v

b+1

2[δ(u−A, v −B) + δ(u+A, v +B)] ,

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which is plotted in Figure S2.7. In order for an ideal low pass ﬁlter to recover f(x, y), the cutoff frequencies of the

Figure S2.7 The Fourier transform of g(x, y). See Problem 2.29.

ﬁlter must satisfy

|a|/2< U < A and |b|/2< V < B .

The Fourier transform of h(x, y)is rect u

2U,v

2V; therefore, the impulse response is

h(x, y) = F−1nrect u

2U,v

2Vo= 4UV sinc(2Ux) sinc(2V y).

For given aand b, we need A > |a|/2and B > |b|/2. Otherwise we cannot ﬁnd an ideal low pass ﬁlter to exactly

recover f(x, y).

Solution 2.30

(a) The continuous Fourier transform of a rect function is a sinc function. Using the scaling property of the

Fourier transform, we have:

G(u) = F1D{g(x)}= 2 sinc(2u).

A sinc function, sinc(x), is shown in Figure 2.4(b).

(b) If the sampling period is ∆x1= 1/2, we have

g1(m) = g(m/2) = 1,−2≤m≤2

0,otherwise .

Its DTFT is

G1(ω) = FDTFT{g1(m)}

=ej2ω+ejω + 1ej0ω+e−jω + 2e−j2ω

= 1 + 2 cos(ω) + 2 cos(2ω).

The DTFT of g1(m)is shown in Figure S2.8.

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34 CHAPTER 2: SIGNALS AND SYSTEMS

Figure S2.8 The DTFT g1(m). See Problem 2.30(b).

Figure S2.9 The DTFT g2(m). See Problem 2.30(c).

g2(m) = g(m) = 1,−1≤m≤1

0,otherwise .

Its DTFT is

G2(ω) = FDTFT{g2(m)}

=ejω + 1ej0ω+e−jω

= 1 + 2 cos(ω).

The DTFT of g2(m)is shown in Figure S2.9.

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(d) The discrete version of signal g(x)can be written as

g1(m) = g(x−m∆x1), m =−∞,··· ,−1,0,1,··· ,+∞.

The DTFT of g1(m)is

G1(ω) = FDTFT{g1(m)}

+∞

m=−∞

g1(m)e−jωm

+∞

m=−∞

g(m∆x1)e−jωm

=Z∞

−∞

g(x)δs(x; ∆x1)e−jω x

∆x1dx .

In the above, δs(x; ∆x1)is the sampling function with the space between impulses equal to ∆x1. Because of

the sampling function, we are able to convert the summation into integration. The last equation in the above

is the continuous Fourier transform of the product of g(x)and δs(x; ∆x1)evaluated as u=ω/(2π∆x1).

Using the product property of the continuous Fourier transform, we have:

G1(ω) = F{g(x)} ∗ F{δs(x; ∆x1)}|u=ω/(2π∆x1)

=G(u)∗comb(u∆x1)|u=ω/(2π∆x1).

The convolution of G(u)and comb(u∆x1)is to replicate G(u)to u=k/∆x1. Since u=ω/(2π∆x1),

G1(ω)is periodic with period Ω=2π.

(e) The proof is similar to that for the continuous Fourier transform:

FDTFT{x(m)∗y(m)}=FDTFT {x(m)∗y(m)}

=FDTFT (∞

n=−∞

x(m−n)y(n))

=∞

m=−∞

e−jωm ∞

n=−∞

x(m−n)y(n)

=∞

n=−∞ "∞

m=−∞

e−jωmx(m−n)#y(n)

=∞

n=−∞

e−jωn "∞

k=−∞

e−jωkx(k)#y(n)

(let k=m−n)

=∞

n=−∞

e−jωnFDTFT{x(m)}y(n)

=FDTFT{x(m)}FDTFT{y(m)}.

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36 CHAPTER 2: SIGNALS AND SYSTEMS

(f) First we evaluate the convolution of g1(m)with g2(m):

g1(m)∗g2(m) = 









3,−1≤m≤1

2, m =±2

1, m =±3

0,otherwise

Then by direct computation, we have

FDTFT{g1(m)∗g2(m)}= 3 + 3 ×2 cos(ω)+2×2 cos(2ω) + 2 cos(3ω)

= 3 + 6 cos(ω) + 4 cos(2ω) + 2 cos(3ω).

On the other hand, we have

FDTFT{g1(m)}= 1 + 2 cos(ω) + 2 cos(2ω)

and

FDTFT{g2(m)}= 1 + 2 cos(ω).

So, the product of the DTFT’s of g1(m)and g2(m)is

FDTFT{g1(m)}FDTFT{g2(m)}= [1 + 2 cos(ω)][1 + 2 cos(ω) + 2 cos(2ω)]

= 1 + 4 cos(ω) + 2 cos(2ω)

+4 cos2(ω) + 4 cos(ω) cos(2ω)

= 1 + 4 cos(ω) + 2 cos(2ω)

+41 + cos(2ω)

2+ 4cos(ω) + cos(3ω)

= 3 + 6 cos(ω) + 4 cos(2ω) + 2 cos(3ω).

Therefore,

FDTFT{g1(m)∗g2(m)}=FDTFT{g1(m)}FDTFT{g2(m)}.

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Image Quality

CONTRAST

Solution 3.1

g(x, y) = Z∞

−∞ Z∞

−∞

h(ξ, η)f(x−ξ, y −η)dξdη

=AH(0,0) + B

2jZ∞

−∞ Z∞

−∞

h(ξ, η)ej2πu0(x−ξ)dξdη

−B

2jZ∞

−∞ Z∞

−∞

h(ξ, η)e−j2πu0(x−ξ)dξdη

=AH(0,0) + B

2jej2πu0xZ∞

−∞ Z∞

−∞

h(ξ, η)e−j2πu0ξdξdη

−B

2je−j2πu0xZ∞

−∞ Z∞

−∞

h(ξ, η)ej2πu0ξdξdη

=AH(0,0) + B

2jej2πu0xH(u0,0) −e−j2πu0xH(−u0,0)

=AH(0,0) + B|H(u0,0)|sin(2πu0x).

Solution 3.2

(a) The PSF of the medical imaging system is isotropic, so we have:

MTF(u) = |H(u, 0)|

=|F{h}(u, 0)|

From Table 2.1, we have F{h}(u, v) = Z∞

−∞ Z∞

−∞

2πe−(x2+y2)/2e−j2π(ux+vy)dx dy =e−2π2(u2+v2). The

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38 CHAPTER 3: IMAGE QUALITY

MTF associated with the system is:

MTF(u) = e−2π2u2.

(b) See Figure S3.1.

Figure S3.1 The modulation transfer function of the system. See Problem 3.2(b).

a value MTF(0.5) = e−2π20.52= 0.0072. So the percentage change in modulation caused by this system is

100 ×(1 −0.0072)% = 99.28%.

Solution 3.3

(a) Given h1(x)we ﬁrst ﬁnd the Fourier Transform H1(u)as follows:

H1(u) = Z∞

−∞

e−x2/5e−j2πuxdx

=Z∞

−∞

e−(x2+j10πux)/5dx

=Z∞

−∞

e−(x2+j10πux−25π2u2)/5e−5π2u2dx

=e−5π2u2Z∞

−∞

e−(x+j5πu)2/5dx

=√5πe−5π2u2.

Hence, the MTF is given as:

MTF1(u) = |H1(u)|

H1(0)

=e−5π2u2.

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(b) The Fourier transform of the second system h2(x)can be computed by analogous methods, and is found to

H2(u) = √10πe−10π2u2.

Since the two systems are in serial cascade, the overall system transfer function is

H(u) = H1(u)H2(u)

=√5π√10πe−5π2u2e−10π2u2

=√50πe−15π2u2.

Hence, the MTF is MTF(u) = e−15π2u2.

Solution 3.4

Let h(x, y)denote the PSF of the nonisotropic medical imaging system, and we assume h(x, y)is normalized to

1; i.e., Z∞

−∞ Z∞

−∞

h(ξ, η)dξdη = 1 .

Given the input

f(x, y) = A+Bsin(2π(ux +vy))

=A+B

2hej2π(ux+vy)−e−j2π(ux+vy)i,

the output g(x, y)of the system is given by

g(x, y) = Z∞

−∞ Z∞

−∞

h(ξ, η)f(x−ξ, y −η)dξdη

=A+B

2jZ∞

−∞ Z∞

−∞

h(ξ, η)ej2π[u(x−ξ)+v(y−η)] dξdη

−B

2jZ∞

−∞ Z∞

−∞

h(ξ, η)e−j2π[u(x−ξ)+v(y−η)] dξdη

=A+B

2jej2π(ux+vy)Z∞

−∞ Z∞

−∞

h(ξ, η)e−j2π(uξ+vη)dξdη

−B

2je−j2π(ux+vy)Z∞

−∞ Z∞

−∞

h(ξ, η)ej2π(uξ+vη)dξdη

g(x, y) = A+B

2jhej2π(ux+vy)H(u, v)−e−j2π(ux+vy)H(−u, −v)i.

Assuming that h(x, y)is a real function, we have H(u, v) = H∗(−u, −v) = |H(u, v)|exp(jφ), where φdenotes

the phase angle of H(u, v). Hence,

g(x, y) = A+B

2j|H(u, v)|hej[2π(ux+vy)+φ]−e−j[2π(ux+vy)+φ]i

=A+B|H(u, v)|sin(2π(ux +vy) + φ).

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40 CHAPTER 3: IMAGE QUALITY

The output g(x, y)is again sinusoidal with gmax =A+B|H(u, v)|, and gmin =A−B|H(u, v)|. Therefore, the

modulation of g(x, y)is

mg=B

A|H(u, v)|=mf|H(u, v)|.

Thus, the MTF of the system is given by

MTF(u, v) = mg

=|H(u, v)|.

Solution 3.5

(a) By multiplying the image with a constant α, the intensities of the background and the target become fb=αIo,

and ft=αIt. The local contrast of the processed image is:

C0=ft−fb

=αIt−αIo

αIo

=It−Io

=C ,

where Cis the local contrast of the original image.

(b) By subtracting a constant Isfrom the image, the intensities of the background and the target become fb=

Io−Is, and ft=It−Is. The local contrast of the processed image is:

C00 =(It−Is)−(Io−Is)

Io−Is

=It−Io

Io−Is

=CIo

Io−Is

> C .

So, by subtracting a constant 0< Is< Iofrom the image will improve the local contrast, while scaling the intensity

will not change the local contrast.

RESOLUTION

Solution 3.6

The proﬁle of the impulse response as a function of the polar angle θcan be expressed as:

h(r, θ) = e−π(r2cos2θ+(r2sin2θ)/4) .(S3.1)

For a ﬁxed θ, the maximal value occurs at r= 0 with h(0, θ)=1. Solving for rin

h(r, θ) = e−π(r2cos2θ+(r2sin2θ)/4) = 1/2

yields

r1/2=sln 2

πcos2θ+ sin2θ/4.

The FWHM is therefore

FWHM = 2 ∗r1/2= 2sln 2

πcos2θ+ sin2θ/4.

which is plotted as a function of θin Figure S3.2.

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Figure S3.2 FWHM as a function of θfor an anisotropic system. See Problem 3.6.

Solution 3.7

(a) We have

l(x1/2)

l(0) =1

2= cos(αx1/2)⇒αx1/2=π

3⇒x1/2=π/3

α=π

6cm .

FWHM is twice x1/2, so

FWHM = 2x1/2=π

3cm .

(b) The resolution of the system is the inverse of the FWHM:

FWHM =3

πcm−1.

Solution 3.8

(a) We have h0(x) = e−x2/2and ha(x) = e−ax2/2. We need to ﬁnd aso that xaat half maximum of ha(x)is

half of x0at half maximum of h0(x), i.e. xa=x0/2.

e−x2

0/2= 1/2

e−ax2

a/2= 1/2)⇒e−a(x0/2)2/2=e−ax2

0/8= 1/2⇒ax2

0/8 = x2

0/2⇒a= 4 .

The impulse response for the new system is:

hnew(x) = e−2x2.

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42 CHAPTER 3: IMAGE QUALITY

(b) Yes. The resolution improves. The system with smaller FWHM can distinguish objects that are closer

together.

broader pass band.

Solution 3.9

(a) The line spread function (LSF) l(x)is deﬁned as the output of the system to a line impulse function f(x, y) =

δ(x):

l(x) = Z∞

−∞

h(x, η)dη

=Z∞

−∞

2πe−(x2+η2)/2dη

√2πe−x2/2Z∞

−∞

√2πe−η2/2dη

√2πe−x2/2.

(b) FWHM is the full width at half maximum. The maximum value of LSF occurs at x= 0, i.e., l(0) = 1

√2π.

Solve l(xh) = l(0)/2(we can ignore the constant 1

√2πand solve e−x2

h/2= 1/2) for xhto get xh=

1.1774 mm. So FWHM = 2xh= 2.3548 mm.

Solution 3.10

(a) We have

max

xh1(x) = h1(0) = 1 .

Solving

h1(x01) = e−x2

01/2=1

yields

x01=√2 ln 2 .

Thus, the FWHM of subsystem h1(x)is

FWHM1= 2x01= 2√2 ln 2 ≈2.35 .

Similarly,

max

xh2(x) = h2(0) = 1 .

Solving

h2(x02) = e−x2

02/200 =1

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yields

x02= 10√2 ln 2 .

Thus, the FWHM of subsystem h2(x)is

FWHM2= 2x02= 20√2 ln 2 ≈23.55 .

(b) The PSF of the overall system is given by

h(x) = h1(x)∗h2(x)

=Z∞

−∞

e−ξ2/2e−(x−ξ)2/200 dξ

=Z∞

−∞

e−(x2−2ξx+ξ2+100ξ2)/200 dξ

=Z∞

−∞

e−101(ξ−x/101)2/200e−x2/202 dξ

=Ce−x2/202 ,

where C=Z∞

−∞

e−101ξ2/200dξ is a constant. Clearly,

max

xh(x) = h(0) = C .

Thus, from h(x0) = e−x2

0/202 =1

2, we get the FWHM of the overall system as

FWHMtotal = 2√202 ln 2 ≈23.67 ≈FWHM2.

Alternatively, since the subsystems have PSFs that are in exponential form, one can directly compute the

FWHM of the overall system as

FWHMtotal =qFWHM2

1+FWHM2

=r2√2 ln 22+20√2 ln 22=√808 ln 2

≈23.67 .

Solution 3.11

A bar phantom, with bars parallel to y-axis, can be modeled as

b(x, y) = X

rect x−2kw

w,

where wis the width and the separation of bars. Since b(x, y)is constant in yfor any ﬁxed x, it sufﬁces to consider

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44 CHAPTER 3: IMAGE QUALITY

the proﬁle of the system and the phantom for y= 0:

b1D(x) = X

rect x−2kw

w,

h1D(x) = rect x

∆.

The output of the system is the convolution of b(x, y)and h(x, y):

g(x, y) = b(x, y)∗h(x, y)

=ZZ b(x−ξ, y −η)h(ξ, η)dξdη

=Zrect η

∆Zb1D(x−ξ) rect ξ

∆dξdη

=Zrect η

∆dη Zb1D(x−ξ) rect ξ

∆dξ

= ∆ Zb1D(x−ξ)h1D(ξ)dξ

= ∆b1D(x)∗h1D(x).

(a) If w= ∆, the separation of the bars is just wide enough to contain f1D(x). The minimal value of g1D(x) =

b1D(x)∗h1D(x)is 0, which occurs at x= (2k+ 1)∆ when h1D(ξ)completely overlaps with the separations

of b1D(x−ξ). The maximal value of g1D(x)is ∆, which occurs at x= 2k∆when h1D(ξ)completely

overlaps with the bars of b1D(x−ξ). The values between extreme values change linearly from 0 to ∆. This

situation is shown in Figure S3.3. Based on the above analysis, we have

Figure S3.3 w= ∆. See Problem 3.11(a).

g1D(x) = ∆−(x−2k∆),2k∆≤x < (2k+ 1)∆

x−(2k−1)∆,(2k−1)∆ ≤x < 2k∆.

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So,

g(x, y) = b(x, y)∗h(x, y)

=∆2−∆(x−2k∆),2k∆≤x < (2k+ 1)∆

∆x−(2k−1)∆2,(2k−1)∆ ≤x < 2k∆.

(b) If w= 0.5∆, no matter what xis, h1D(x−ξ)always overlaps with one complete bar (or parts of two adjacent

bars that add up to 1 complete bar) of b1D(ξ)(see Figure S3.4). So, the output of the system is

Figure S3.4 w= 0.5∆. See Problem 3.11(b).

g(x, y) = b(x, y)∗h(x, y)=0.5∆2.

h1D(x−ξ)at most overlaps with one bar of b1D(ξ), and it at least overlaps with part of a bar of width ∆−w.

So the maximal value of g(x, y)is ∆wand the minimal value is ∆(∆ −w). The contrast of the output image

of the bar phantom is therefore

C(w) = 2w−∆

∆.

RANDOM VARIABLES AND NOISE

Solution 3.12

Evaluate the expectation of Mas follows:

E{M}=EN−µN

σN

=E{N−µN}

σN

(because σNis a constant)

=E{N} − µN

σN

(because µNis a constant)

=µN−µN

σN

= 0 .

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46 CHAPTER 3: IMAGE QUALITY

Evaluate the variance of Mas follows:

σ2

M1

=E{M2} − (E{M})2

=E{M2}(because E{M}= 0 from above)

=E(N−µN)2

σ2

N

=E{(N−µN)2}

σ2

=σ2

σ2

= 1 .

In order to get equality 1, we used the following property of variance:

σ2

M=E{(M−µM)2}

=E{M2−2µMM+µ2

=E{M2} − 2µME{M}+µ2

=E{M2} − µ2

Solution 3.13

Let X=

i=1

Xi. The mean of Xis

µ=E[X] = E"N

i=1

Xi#=

i=1

E[Xi] =

i=1

µi,

where we used the linearity of the expectation operator E. The variance of Xis

σ2=Eh(X−µ)2i

=E

 N

i=1

Xi−

i=1

µi!2

=E

 N

i=1

(Xi−µi)!2



i=1

E[(Xi−µi)2] +

i=1

j=1,j6=i

E[(Xi−µi)(Xj−µj)]

Since Xi, i = 1,··· , N are independent, E[(Xi−µi)(Xj−µj)] = 0 if j6=i. Therefore,

σ2=

i=1

E[(Xi−µi)2] =

i=1

σ2

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Solution 3.14

If Xi, i = 1,··· , N are not independent, then

µ=E[X] =

i=1

µi

still holds, since in deriving this equality, we used only the linearity of the expectation operator. The equality

for the variance, however, does not hold because when Xi, i = 1,··· , N are not independent the statement

E[(Xi−µi)(Xj−µj)] = 0 is not necessarily true.

Solution 3.15

The PDF of the uniform random variable is given by

pX(x) = 









(b−a),for a≤x<b

0,otherwise

Thus,

µX=Z∞

−∞

xpX(x)dx

=Zb

b−adx =b2−a2

2(b−a)

=a+b

and

σ2

X=Z∞

−∞

(x−µX)2pX(x)dx

=Zb

(x−a+b

2)21

b−adx

=Zb

(x−a+b

2)21

b−ad(x−a+b

b−aZb−(a+b)/2

a−(a+b)/2

t2dt

3(b−a)"b−a

23

−a−b

23#

=(b−a)2

12 .

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48 CHAPTER 3: IMAGE QUALITY

Solution 3.16

For the system with PSF h1(x, y), the output power SNR is given by (3.63)

SNRp1=Z∞

−∞ Z∞

−∞ |h1(x, y)∗f(x, y)|2dx dy

Z∞

−∞ Z∞

−∞

NPS(u, v)du dv

By applying Parseval’s theorem, we have

SNRp1=Z∞

−∞ Z∞

−∞ |h1(x, y)∗f(x, y)|2dx dy

Z∞

−∞ Z∞

−∞

NPS(u, v)du dv

=Z∞

−∞ Z∞

−∞ |H1(u, v)F(u, v)|2du dv

Z∞

−∞ Z∞

−∞

NPS(u, v)du dv

=Z∞

−∞ Z∞

−∞ |H1(u, v)|2|F(u, v)|2du dv

Z∞

−∞ Z∞

−∞

NPS(u, v)du dv

=Z∞

−∞ Z∞

−∞

MTF2

1(u, v)|F(u, v)|2du dv

Z∞

−∞ Z∞

−∞

NPS(u, v)du dv

Similarly, we have the output power SNR for the second system

SNRp2=Z∞

−∞ Z∞

−∞

MTF2

2(u, v)|F(u, v)|2du dv

Z∞

−∞ Z∞

−∞

NPS(u, v)du dv

Since MTF1(u, v)≤MTF2(u, v), we have MTF2

1(u, v)|F(u, v)|2≤MTF2

2(u, v)|F(u, v)|2. Therefore

Z∞

−∞ Z∞

−∞

MTF2

1(u, v)|F(u, v)|2du dv ≤Z∞

−∞ Z∞

−∞

MTF2

2(u, v)|F(u, v)|2du dv .

So SNRp1≤SNRp2, the output power SNR of the second system, the one with larger MTF, is higher. Therefore,

the second system is better in terms of image quality.

Solution 3.17

(a) The noise in the output g(x, y)is n0(x, y)

n0(x, y) = h(x, y)∗n(x, y).

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Its mean is

E{n0(x, y)}=E{h(x, y)∗n(x, y)}

=EZ∞

−∞ Z∞

−∞

h(ξ, η)n(x−ξ, y −η)dξdη

=Z∞

−∞ Z∞

−∞

h(ξ, η)E{n(x−ξ, y −η)}dξdη

= 0 .

Its variance is

E{n0(x, y)n0(x, y)}=E[h(x, y)∗n(x, y)]2

=EZ∞

−∞ Z∞

−∞

h(ξ, η)n(x−ξ, y −η)dξ dη

Z∞

−∞ Z∞

−∞

h(p, q)n(x−p, y −q)dp dq

=EZ∞

−∞ Z∞

−∞

h(ξ, η)n(x−ξ, y −η)

h(p, q)n(x−p, y −q)dp dq dξ dη

=Z∞

−∞ Z∞

−∞

h(ξ, η)h(p, q)

E{n(x−ξ, y −η)n(x−p, y −q)}dp dq dξ dη

=Z∞

−∞ Z∞

−∞

h(ξ, η)h(p, q)

σ2

nδ(p−ξ, q −η)dp dq dξ dη

=σ2

nZ∞

−∞ Z∞

−∞

h2(ξ, η)dξ dη

=σ2

nH0,

where, H0=R∞

−∞ R∞

−∞ h2(ξ, η)dξ dη.

(b) The power SNR for the input image is

SNRin =R∞

−∞ R∞

−∞ f2(x, y)dx dy

σ2

The power SNR for the output image is

SNRout =R∞

−∞ R∞

−∞[h(x, y)∗f(x, y)]2dx dy

H0σ2

in order for the SNR to be improved by the system.

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50 CHAPTER 3: IMAGE QUALITY

SAMPLING THEORY

Solution 3.18

(a) We have

fs(t) = f(t)δs(t; ∆T)

=∞

m=−∞

f(t)δ(t−m∆T)

=∞

m=−∞

f(m∆T)δ(t−m∆T).

Since

f(t) = 









sin 2πt

T,0≤t≤T

0,otherwise

and ∆T= 0.25T, then

fs(t) = δ(t−0.25T)−δ(t−0.75T).

Also

fd(m) = f(m∆T) = 





1, m = 1

−1, m = 3

0,otherwise

(b) The signal fh(t)is referred to as a zero-order hold. By deﬁnition,

fh(t) = 





1,0.25T≤t < 0.5T

−1,0.75T≤t<T

0,otherwise

= rect t−0.375T

0.5T−rect t−0.825T

0.5T.

Using the properties of the Fourier transform, we have

Fh(f) = F(fh(t))

=Z∞

−∞

fh(t)e−j2πf tdt

= 0.25Tsinc(0.25T f)e−j2π(0.375T f )−0.25Tsinc(0.25T f)e−j2π(0.825T f)

= 0.25Tsinc(0.25T f)he−j2π(0.375T f )−e−j2π(0.825T f)i.

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fs(t)=0,

fd(m)=0,

fh(t)=0,

Fh(f)=0.

Solution 3.19

Since the Nyquist sampling periods for 1-D band-limited signals f(x)and g(x)are ∆fand ∆g, the highest

frequency of f(x)and g(x)are 1

2∆fand 1

2∆g. In order to ﬁnd the Nyquist sampling periods, we need to ﬁnd the

highest frequency for each of the signals.

(a) A shift in location does not change the frequency components of a signal, so the magnitude spectrum of

f(x−x0)is the same as that of f(x). The Nyquist sampling period of f(x−x0)is ∆f.

(b) The Fourier transform of f(x) + g(x)is F[f(x) + g(x)] = F[f(x)] + F[g(x)]. The highest frequency of

f(x) + g(x)is max( 1

2∆f,1

2∆g), so the Nyquist sampling period of f(x) + g(x)is min(∆f,∆g).

2∆f, The Nyquist

sampling period of f(x)∗f(x)is ∆f.

(d) The Fourier transform of f(x)g(x)is F[f(x)] ∗ F[g(x)], The highest frequency of f(x)g(x)is 1

2∆f+1

2∆g,

and the Nyquist sampling period is ∆f∆g

∆f+∆g.

(e) If f(x)≥0,kf(x)k=f(x), the Nyquist sampling period of kf(x)kis ∆f. But in general, the operation

of taking absolute value will reverse part of the original signal f(x), and therefore introduce high frequency

component. In general case, kf(x)kis no longer bandlimited, even though f(x)is.

Solution 3.20

The sampling frequencies are 1

∆x= 1.5and 1

∆y= 1.5. From the sampling theorem, in order to avoid aliasing,

the cutoff frequencies of the low-pass ﬁltered signal f∗hmust satisfy:

U≤1

2∆x= 0.75,and V≤1

2∆y= 0.75 .

Thus, the ideal low-pass ﬁlter h(x, y)that gives the maximum possible frequency content must have a frequency

response as

H(u, v) = 1, if |u| ≤ 0.75 and |v| ≤ 0.75

0,otherwise .

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52 CHAPTER 3: IMAGE QUALITY

H(u, v)is one inside a square region and zero outside. The PSF of the required anti-aliasing low-pass ﬁlter can be

computed as:

h(x, y) = F−1

2(H(u, v)) = Z∞

−∞ Z∞

−∞

H(u, v)ej2π(ux+vy)dudv

=Z0.75

−0.75 Z0.75

−0.75

ej2π(ux+vy)dudv

=Z0.75

−0.75

ej2πux du·Z0.75

−0.75

ej2πvy dv

=exp[j2π(0.75)x]−exp[j2π(−0.75)x]

j2πx ·exp[j2π(0.75)y]−exp[j2π(−0.75)y]

j2πy

=sin(1.5πx)

πx ·sin(1.5πy)

πy .

From Table 2.1, we know that

F2(f)(u, v) = e−π(u2+v2).

Thus, the total spectrum energy of f(x, y)is

Etotal =Z∞

−∞ Z∞

−∞ |F2(f)(u, v)|2du dv

=Z∞

−∞ Z∞

−∞

e−2π(u2+v2)du dv

=2πσ21

2πσ2Z∞

−∞ Z∞

−∞

e−(u2+v2)/2σ2du dv with σ2=1

4π

= 2π·1

4π

= 0.5.

The spectrum that is kept by the low pass ﬁlter has energy of

Epreserve =Z0.75

−0.75 Z0.75

−0.75

e−2π(u2+v2)dudv

=Z0.75

−0.75

e−2πu2du ·Z0.75

−0.75

e−2πv2dv

= 1

√2πZ0.75√2π

−0.75√2π

e−t2dt!2

=1

√2π

√π

22erf 0.75√2π2

2herf 0.75√2πi2

≈1

2[0.992]2

≈0.492 ,

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where erf(·) is the error function. Thus, the percentage of the spectrum energy that is preserved is

Epreserve

Etotal

=0.492

0.5= 98.4% .

Since the spectrum of f(x, y), which is F2(f)(u, v) = e−π(u2+v2), is non-zero for all (u, v)∈(−∞,∞)×

(−∞,∞), it is impossible to sample f(x, y)free of aliasing without using an anti-aliasing ﬁlter.

Solution 3.21

(a) Impulse response: h(x, y) = rect( x

w)rect( y

w).

MTF: H(u, v) = w2sinc(wu)sinc(wv)and H(0,0) = w2.

Thus MTF(u, v) = H(u,v)

H(0,0) = sinc(wu)sinc(wv).

Horizontal FWHM = w.

(b) Since H(u, v) = w2sinc(wu)sinc(wv)and H(0,0) = w2, considering the main lobe of the sinc function

and from Nyquist sampling theory, us≥2/w,vs≥2/w. Then ∆X≤w/2and ∆Y≤w/2. This means

that aliasing occurs in the sampling scheme using detectors of dimensions w×wand separation of w. Since

∆X=wand ∆Y=wthen us= 1/w, vs= 1/w then the object must be limited to −1/2wand 1/2w. i.e

the object must have width Wx= 1/w and Wy= 1/w so that no aliasing occurs.

of four of the small detectors is done so that the new detector size is 2w×2w. That means that separation

between detectors of ∆X≤wand ∆Y≤wwill guarantee no aliasing (detectors must overlap). This can be

achieved as in Figure S3.5 by using the fact that we can overlap the resultant detector by sequentially using

the small detectors for overlap.

Figure S3.5 Overlapping of the detectors. See Problem 3.21.

(d) The impulse response function now is h(x, y) = rect( x

2w)rect( y

2w)and the (horizontal) FWHM = 2w.

(e) Sequential grouping can be done after the image is acquired by the summation of 4 pixel values using the

same scheme described in (c) to get an aliasing free image.

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54 CHAPTER 3: IMAGE QUALITY

Solution 3.22

(a) System 1 has a PSF that is a rectangle of width 0.5. Its FWHM is therefore 0.5.

System 2’s FWHM can be found by

2=e−πx2

=⇒ −log 2 = −πx2

=⇒x2=log 2

=⇒x=±rlog 2

π.

Therefore the FWHM is 2qlog 2

π∼0.9394.

System 1 has the better resolution.

(b) The MTF is deﬁned by the absolute value of the transfer function divided by its value at zero frequency. The

transfer function for system 1 is therefore given by

rect(x)↔sinc(u),

rect(2x)↔1

2sinc u

2,

h1(x) = rect(2x)↔H1(u) = 1

2sinc u

2.

And the transfer function for system 2 is given by

h2(x) = e−πx2↔e−πu2=H2(u)

The MTFs are then

MTF1(u) =

2sinc u

2

= sinc u

2,

MTF2(u) = e−πu2.

real/even transfer function, the only effect is to rescale the amplitude of the sinusoid by the transfer function

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at frequency 1

4. Therefore

g1(x) = H1(2) cos(4πx)

2sinc(2/2)

2sinc(1)

= 0 ,

and

g2(x) = H2(2) cos(4πx)

=e−π22cos(4πx)

∼3.4873 ×10−6cos(4πx).

Therefore you should use system 2 to image this signal because system 1 will not respond to it at all.

(d) We must sample at a rate greater than twice the highest frequency in our signal. The highest frequency (the

only frequency) is 4. Therefore we must sample at a rate greater than 4. This corresponds to a period less

than 1/4.

ARTIFACTS, DISTORTION, AND ACCURACY

Solution 3.23

Both noise and artifacts degrade the image, making correct detection and delineation of anatomical features

difﬁcult. The main technical difference between the two is that artifacts are reproducible scan after scan, whereas

noise will come out differently with each scan. On the other hand, noise is well-modeled using probability and

random variables, so that the broad characteristic of noise— for example, mean and variance—will be the same.

Artifacts are deterministic, and can originate from a variety of sources that, in principle, can be modeled and

removed. For example, some artifacts appear because of instrumentation failure or calibration problems. Artifacts

can also appear because the image reconstruction method fails to adequately model the true physics of the imaging

modality. Finally, artifacts might arise due to inadequacies in data collection—aliasing, for example.

Solution 3.24

Suppose the center of the ball has coordinates (x, 0,0). When the source is inside the ball or on the ball surface,

i.e. x≤r, the shadow of the ball will cover the entire detector plane. In this case, the radius of the image is ∞.

When x>r, by simple geometry, we have the radius of the image on the detector plane, R:

R=dtan θ=r

√x2−r2⇒R=dr

√x2−r2,

where θis the angle between the xaxis and the tangent plane of the ball through the source. The size distortion is

measured by the ratio R/r =d/√x2−r2. When dis ﬁxed, we can increase xto reduce the ratio R/r. And the

largest xwe can get is d−r, in which case R/r =d/√d2−2dr.

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56 CHAPTER 3: IMAGE QUALITY

Solution 3.25

(a) If we take measurements on rectangular grids in the image plane, the locations of sample points are (k∆x, lδy),

where k, and lare integers, and ∆x, δy are spacing in xand ydirections. The corresponding coordinates of

the samples in the physical domain can be obtained by solving the equations above, yielding

ξ(k, l) = k∆x

1+(l∆y)2/50 ,

η(k, l) = l∆y .

The (k, l)-th sample on the image plane needs to be placed at (ξ(k, l), η(k, l)) on the physical domain to

correct the geometric distortion.

(b) If we take measurements on rectangular grids in the physical domain, the locations of sample points in the

physical domain are (k∆ξ, lδη). On the image plane, we need to sample points at

x(k, l) = k∆ξ+1

50k∆ξ(l∆η)2,

y(k, l) = l∆η .

Solution 3.26

(a) The pdf of the test value for normal and diseased subjects are

pN(t) = 1

p2σ2

e−(x−µ0)2/2σ2

pD(t) = 1

p2σ2

e−(x−µ1)2/2σ2

(b) When the threshold is set to be t0= (µ0+µ1)/2, the sensitivity and the speciﬁcity are

sensitivity = a

a+c

=R∞

t0pD(t)dt

Rt0

−∞ pD(t)dt +R∞

t0pD(t)dt

2+erf µ1−t0

σ1

2+erf µ1−µ0

2σ1,

speciﬁcity = d

b+d

2+erf µ1−µ0

2σ0.

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sensitivity(t) = 









2+erf µ1−t

σ1, t ≤µ1

2−erf t−µ1

σ1, t > µ1

(d) The diagnostic accuracy is

DA = a+d

a+b+c+d=











2h1−erf µ0−t

σ0+erf µ1−t

σ1i, t < µ0

2h1 + erf t−µ0

σ0+erf µ1−t

σ1i, µ0≤t≤µ1

2h1 + erf t−µ0

σ0−erf t−µ1

σ1i, t > µ1

APPLICATIONS, EXTENSIONS AND ADVANCED TOPICS

Solution 3.27

We consider image quality to be characterized by contrast and resolution. Since resolution is typically character-

ized using the FWHM, from (3.22) we know that the FWHM of the overall system can be determined approximately

from the FWHMs of the individual subsystems according to

FWHMtotal =qFWHM2

1+FWHM2

2+···+FWHM2

It follows that

FWHMtotal ≥FWHMi, for all 1≤i≤K .

Thus the resolution of the overall system is worse than each of the individual subsystems.

Both contrast and resolution can be characterized using the MTF. The MTF of the overall system is given by

MTF(u, v) = MTF1(u, v)MTF2(u, v)···MTFK(u, v),

in terms of the individual subsystem MTFs MTFi(u, v),i= 1,2, ..., K . For most medical imaging systems,

MTF(u, v)≤1for all (u, v). Assuming this is true for all the subsystems, that is,

MTFi(u, v)≤1,for i= 1,2, . . . , K ,

then it follows that

MTF(u, v)≤MTFi(u, v),for i= 1,2, . . . , K .

Therefore, from this standpoint as well, the contrast and resolution of the overall system is inferior to each individual

subsystem.

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58 CHAPTER 3: IMAGE QUALITY

Solution 3.28

(a) Let λ= (x−x0)/(x1−x0), and µ= (y−y0)/(y1−y0). Linear interpolation gives

f(E) = (1 −λ)f(A) + λf (B),

f(F) = (1 −λ)f(C) + λf (D).

Then f(P)can be obtained by linear interpolation of f(E)and f(F):

f(P) = (1 −µ)f(F) + µf (E)

=µ(1 −λ)f(A) + µλf(B) + (1 −µ)(1 −λ)f(C) + (1 −µ)λf(D).

(b) Similarly,

f(G) = (1 −µ)f(C) + µf (A),

f(H) = (1 −µ)f(D) + µf (B),

and

f(P) = (1 −λ)f(G) + λf (H)

= (1 −λ)(1 −µ)f(C) + (1 −λ)µf(A) + λ(1 −µ)f(D) + λµf(B).

Comparing the coefﬁcients for f(A),f(B),f(C), and f(D), we can see that the results from (a) and (b) are

the same.

four corners x0= 3, y0= 3,x0= 3, y1= 4,x1= 4, y0= 3, and x1= 4, y1= 4. By our deﬁnition in

Problem 3.25, λ= 0.735, µ = 0.5. The value for ξ= 3, η = 3.5is

f0(ξ= 3, η = 3.5) = 0.1325(f(3,4) + f(3,3)) + 0.3675(f(4,4) + f(4,3)),

where f(m, n)are the measurements on the image plane.

Solution 3.29

(a) For a given threshold value twith µ0≤t≤µ1, the sensitivity and the speciﬁcity are given as:

sensitivity = 1

2+erf µ1−t

σ1,

speciﬁcity = 1

2+erf t−µo

σ1.

The ROC curve is shown in Figure S3.6.

(b) The perfect diagnostic test should have both sensitivity and speciﬁcity equal to 1. In this case, the ROC curve

is a point (0,1) on the coordinate system of Figure S3.6.

the diseased patients will be diagnosed correctly, while 1.79% of the normal patients will be wrongfully

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Figure S3.6 The ROC curve. See Problem 3.29.

diagnosed as diseased. Using the relationship between sensitivity and threshold, the corresponding threshold

value is topt = 5.24, which is different from (µ0+µ1)/2because the two groups of subjects have different

variances.

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Physics of Radiography

PHYSICS OF ATOMS

Solution 4.1

(a) From tables (internet or physics or chemistry textbooks),

mass of carbon-12 = 1.99264663 ×10−26 kg .

From the information given in the problem statement, we calculate

mass of (6p + 6n + 6e) = 2.0090759569 ×10−26 kg .

The mass defect is therefore

mass defect of carbon atom = 2.0090759569 ×10−26 kg −1.99264663 ×10−26 kg

= 1.6429326956 ×10−28 kg .

To ﬁnd this in atomic mass units

mass defect of carbon atom = 1.6429326956 ×10−28 kg ×6.0221415 ×1026 u/kg

= 0.098939732 u.

(b) We have

E=mc2

= 1.6429326956 ×10−28 kg ×(2.99792458 ×108m/s)2

= 1.47659426 ×10−11 J.

Since 1eV = 1.60217653 ×10−19 J, we also have

E= 9.21617711 ×107eV .

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Solution 4.2

(a) The mass-equivalent energy of an electron at rest is

E=m0c2.

Since m0= 9.10938 ×10−31 kg is the mass of an electron at rest and c= 2.99792 ×108m/s, we have

E= 9.10938 ×10−31 ×(2.99792 ×108)2kg −m2/s2

= 8.18697 ×10−14 J

= 5.11 ×105eV = 511 keV.

(b) Ignoring relativity the kinetic energy of an electron at speed v=1

10 cis

Ek=1

2m0v2=1

200m0c2= 2.558 keV .

So when the effect of relativity is ignored, the potential needed to accelerate an electron to a speed equal to

1/10 the speed of light is 2.558 kV. This is not accurate since at 1/10 the speed of light, the effect of relativity

cannot be ignored.

KE =mc2−m0c2= 120 keV ,

where mis the relativistic mass of the electron after acceleration, which is given by

m=m0

p1−(v/c)2.

Therefore, we can carry out the following steps to ﬁnd v:

120 keV =mc2−m0c2

=m0

p1−(v/c)2c2−m0c2

=m0c2 1

p1−(v/c)2−1!

120

511 =1

p1−(v/c)2−1

v= 0.5867c.

Thus, at 120 keV the speed of the electrons hitting the anode is over 1/2 that of the speed of light.

Solution 4.3

From Eq. (4.3), we have:

KE =E−E0

=mc2−m0c2,

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62 CHAPTER 4: PHYSICS OF RADIOGRAPHY

where m,m0are the mass and the rest mass of the particle, respectively. They are related by the following equation

(Eq. (4.1) in the text):

m=m0

p1−v2/c2.

When vc,v2/c2is close to 0. By using Taylor’s expansion of function f(x) = 1

√1−xin the neighborhood of

x= 0, we have the following approximation for m:

m≈m01 + 1

c2.

Using this approximation yields

KE =mc2−m0c2

≈m01 + 1

c2c2−m0c2

≈1

2m0v2.

Notice that when vc, the mass mis approximately equal to the rest mass m0, so we have:

KE ≈1

2mv2,

which is the usual expression for kinetic energy of a mass in motion.

IONIZING RADIATION

Solution 4.4

Characteristic radiation is produced by electrons that drop to lower energy states (more inner orbits) after they

have been excited to higher energy states (more outer orbits). The differential in energy lost by the electron is given

off as an x-ray—characteristic radiation. Because electrons exist in discrete energy states that are speciﬁc to a given

atom, characteristic radiation can only be emitted at a collection of discrete energy levels within the EM spectrum.

Therefore, the intensity spectrum for characteristic radiation comprises a discrete spectrum—that is, spectral lines.

On the other hand, Bremsstrahlung radiation is caused by interaction of an energetic electron with a nucleus of an

atom. Speciﬁcally, the nucleus, having a positive charge, will tend to attract the electron, having a negative charge,

causing the electron to slow down and be deﬂected from its original path. The electron loses energy as a result,

which is radiated away as an x-ray with energy equal to that lost by the electron. An electron can lose all its energy,

by collision into the atomic nucleus, or any smaller amount, by smaller deﬂection. Therefore, unlike characteristic

radiation, the energy spectrum of bremsstrahlung radiation is continuous. Since lower energy losses are more likely,

and direct collision with a nucleus is very unlikely, the bremsstrahlung spectrum is zero at the incident energy of

the electrons and grows larger with decreasing energy.

Solution 4.5

(a) Ionization is the ejection of an electron from an atom. In order to eject an electron, the incident radiation must

have sufﬁcient energy to overcome the binding energy of the electron. The smallest binding energy among

atoms having smaller atomic numbers is that of the sole electron in the hydrogen atom. Its binding energy is

13.6 eV. Therefore, a radiation having energy above 13.6 eV is capable of ionizing the hydrogen atom, which

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makes it ionizing radiation. If the radiation has energy smaller than 13.6 eV it is not capable of ionizing the

hydrogen atom or any other atom (with smaller atomic number), and is therefore considered non-ionizing.

(There are larger atoms having electrons with binding energy less than 13.6 eV, but these are rare in nature

and even rarer in the human body.)

(b) Ionization is the ejection of an electron from an atom, while excitation is the process of raising the energy of

an electron within the electron cloud, without causing ejection. Excitation rearranges the electrons within the

shells, but this is only a temporary effect, since the electrons will seek a lower energy conﬁguration, and in

the process generate characteristic radiation.

Solution 4.6

(a) The frequencies and the wavelengths of EM waves are related by the formula:

λ=c

ν,

where c= 3.0×108meters/sec is the speed of light. For λ= 4 nanometers, we have

ν=c

=3.0×108m/s

4×10−9m

= 7.5×1016 Hz .

Similarly, for λ= 400 nanometers we have ν= 7.5×1014 Hz. So the frequency range for ultraviolet light

is 7.5×1014 Hz ∼7.5×1016 Hz.

(b) The energy of a photon is given by

E=hν ,

where h= 6.626 ×10−34 Joule-sec is Planck’s constant. So for ultraviolet light with frequency ν= 7.5×

1014 Hz, the energy is E=hν = 6.626 ×10−34 ×7.5×1014 = 4.97 ×10−19 Joule. Since 1eV =

1.6×10−19 Joule, we have that E= 4.97 ×10−19 Joule = 3.1eV. Similarly, for ultraviolet light with

frequency ν= 7.5×1016 Hz, the energy is E= 310 eV. So the photon energy range for ultraviolet light is

3.1–310 eV.

to calculate that when the frequency of the ultraviolet light is ν= 3.284 ×1015 Hz, the photon energy

is E=hν = 13.6eV. So ultraviolet light is ionizing radiation when its frequency is greater or equal to

ν0= 3.284 ×1015 Hz. Ultraviolet light with a frequency lower than that is not ionizing radiation. Or

equivalently, when the wavelength is larger than λ0=c

ν0= 91.35 nanometers, ultraviolet light is not

ionizing radiation.

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64 CHAPTER 4: PHYSICS OF RADIOGRAPHY

Solution 4.7

(a) Electron density is:

EC =NAZ

where

NA= 6.022 ×1023 ,

Z= 1 (for hydrogen) ,

Wm= 1 gram/mole (for hydrogen) ,

where the last fact follows from the fact that the atomic weight of a hydrogen atom is approximately 1 u.

Therefore,

ED =6.022 ×1023 ×1

1gram/mole(of electrons) ≈6×1023 electrons/g = 6 ×1026 electrons/kg .

(b) Except for the hydrogen atom, all other low atomic number materials have nearly equal numbers of neutrons

as protons. Therefore, since the weight of these other atoms is doubled, while the number of electrons remains

tied to the number of protons, the electron density is approximately halved from that of hydrogen.

atomic number and from the differing hydrogen content in various materials.

ATTENUATION OF EM RADIATION

Solution 4.8

(a) Let I0denote the incident intensity, and Ixthe exiting intensity. Denote the thickness of the shielding material.

From the problem speciﬁcation, we know that

= 1 −99.5% = 0.005.

Since Ixand I0are related by

Ix=I0e−µx ,

we have

e−µx =Ix

= 0.005 .

Then, we solve for xas follows

x=−1

µln0.005 = ln200

µ≈5.3

µ,

which is the required thickness of the shielding material.

(b) The desired range of an ionizing beam in tissue would be centimeters to tens of centimeters, which is about

the distance that the beam would have to travel through the body. If the range is larger, then the incident beam

would travel through the body with almost no attenuation, and no contrast would be obtained. If the range is

too short, all beam energy would be absorbed by the body, and no image can be formed.

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Solution 4.9

Let df =dN

N=−µdx. By solving df

dN =1

N, we have f= ln N+c1, and by solving df

dx =−µ, we have

f=−µx +c2, where c1and c2are two arbitrary constants. Therefore, f= ln N+c1=−µx +c2. This leads to

ln N=−µx +c, where cis an arbitrary constant. So we have N=N0e−µx, with a constant N0.

Solution 4.10

Suppose that the x-ray photons hit the phantom on a unit area is N0. On the screen where it is not blocked by

the bars, the photons detected on a unit area is also N0. The thickness of the bars is 0.4 cm, which is 4 times the

HVL, so the x-ray photons passing through the bars is (1/2)4= 1/16 of those entering the bars. So the screen that

is blocked by the bars detects N0/16 photons on a unit area. The contrast of the image on the screen is

C=N0−N0/16

N0+N0/16 =15

17 .

Solution 4.11

From Eq. (4.8), the energy of the scattered photon is given by

hν0=hν

1 + hν

m0c2(1 −cos θ)

where m0c2= 511 keV. Thus, the larger θis, the smaller the energy of the scattered photon. Using the facts

1 Angstrom = 10−10meter, h= 6.626 ×10−34joule-sec, and 1 joule = 6.241 ×1015 keV, we can compute the

energy of the source x-ray photon as

hν =hc/λ

=6.626 ×10−34joule-sec ×3×108m/s ×6.241 ×1015 keV/joule

8.9×10−2×10−10m

≈139.4keV .

The energy of a photon that has been scattered by 25◦is

hν0=hν

1 + hν

m0c2(1 −cos 25◦)

=139.4keV

1 + 139.4keV

511 keV (1 −cos 25◦)

≈135.9keV .

Thus, to eliminate all photons that have scattered more than 25 degrees, the system should only accept photon

energy between 135.9keV and 139.4keV.

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66 CHAPTER 4: PHYSICS OF RADIOGRAPHY

Solution 4.12

From Eq. (4.8), the energy of the scattered photon is given by:

hν0=hν

1 + hν

m0c2(1 −cos θ)

where m0c2= 511 keV.

(a) We use the provided numbers and solve for θas follows:

99 = 100

1 + (1 −cos θ)100/511

=⇒θ= cos−11−511

100 ×99

= 18.49◦.

(b) We use the provided numbers to solve for Eas follows:

E=100

1 + (1 −cos(25◦))100/511

= 98.20 keV .

The range is therefore 98.20–100 keV.

Solution 4.13

(a) We use the provided linear attenuation coefﬁcient and solve for HVL as follows:

=e−µHVL =1

=⇒µHVL = ln 2 ,

=⇒HVL = ln 2

µ= 0.3 cm .

Therefore, the HVL of the NaI crystal at 140 keV is 0.3 cm.

(b) Plugging in the provided numbers yields

E0=E

1 + (1 −cos θ)E/(511 keV)

=140 keV

1 + (1 −0) ×140/511

=140

1+0.274

= 109.89 keV .

Therefore, the energy of the scattered photon is 109.89 keV.

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Therefore, the scattered photons are more likely to be absorbed than the incident photons because the scattered

photon has lower energy than the incident photon.

Solution 4.14

(a) We are not told the linear attenuation coefﬁcient of the shielding material. But we know that it blocks 90% of

the incident radiation. So

N=N0e−µ∆x

=1−0.9

1=e−1.5 cmµ.

Therefore,

µ=−ln 0.1

1.5 cm .

The deﬁnition of HVL is 1

2=e−HVLµ,

HVLµ=−ln 0.5.

Plugging in our expression for µderived above yields

HVL

1.5 cm =ln 0.5

ln 0.1,

which can be solved for HVL as follows

HVL = ln 0.5

ln 0.1×1.5 cm = 0.45 cm .

(b) From Eq. (4.8), the energy of the scattered photon is given by:

hν0=hν

1 + hν

m0c2(1 −cos θ)

where m0c2= 511 keV. Therefore,

E0=102.2

1 + (1 −0) ×102.2/511

=102.2

1+0.2

= 85.17 keV .

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68 CHAPTER 4: PHYSICS OF RADIOGRAPHY

Solution 4.15

(a) Use of Beer’s law yields

I=I0e−µd =I0e−0.3×1= 0.7408I0

(b) If 1/2 of the incident x-rays are blocked then I=I0/2. Then using I=I0e−µd and solving for dyields

d=ln 0.5

−µ= 2.31 cm .

detector because of Compton scattering. Those that were directed at the detector and scattered away will do so

in both geometries, so they have no impact on the relative number of detected photons in the two geometries.

Therefore more photons will be detected, in general.

RADIATION DOSIMETRY

Solution 4.16

From Example 4.7, we know that in order to keep the dose equivalent to be under 10 mrems, the lung should

have an exposure less than 10.8 mR. Since the exposure follows an inverse square law for point sources, the smallest

distance the patient should be away from the source should be

r10

10.8×10−3×1 = 30.5cm .

Solution 4.17

The effective dose is given by (4.38):

Deffective =X

organs

Hjwj= 0.002Hbone + 0.002Hmuscle .

From Sections 4.6.1-4.6.5, we have

Deffective = 0.002DboneQ+ 0.002DmuscleQ

= 0.002fboneXQ + 0.002fmuscleXQ

= 0.002 ×0.87(µ/ρ)bone + (µ/ρ)muscle

(µ/ρ)air

XQ .

For x-ray at 20 keV, Q≈1(see http://physics.nist.gov/PhysRefData/XrayMassCoef/tab4.html) and

(µ/ρ)bone = 0.4cm2/g,(µ/ρ)muscle = 0.82cm2/g,(µ/ρ)air = 0.78cm2/g.

So,

Deffective = 0.00174 (µ/ρ)bone + (µ/ρ)muscle

(µ/ρ)air

X= 2.06 mrems .

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Projection Radiography

INSTRUMENTATION

Solution 5.1

The system is shift variant in the zdirection because of the divergence of the x-ray. The system is shift invariant

in xand ydirections if the object is inﬁnitesimally thin in the zdirection. Otherwise the system is shift variant in

general.

The intensity of x-rays incident on the detector at (x, y)is given by

I(x, y) = ZEmax

S0(E0)E0exp (−Zr(x,y)

µ(s;E0, x, y)ds)dE0,

where S0(E)is the spectrum of the incident x-rays. When two objects with linear attenuation coefﬁcients µ1(s;E0, x, y)

and µ2(s;E0, x, y)are presented, the intensity of x-rays on the detector is

Isum(x, y) = ZEmax

S0(E0)E0exp (−Zr(x,y)

(µ1(s;E0, x, y) + µ2(s;E0, x, y)) ds)dE0.

In general, Isum(x, y)6=I1(x, y) + I2(x, y), where Ii(x, y)is the intensity of x-rays on the detector when only the

i-th object is presented. So in general the system is not linear.

When monoenergetic x-rays are used, we can remove the outer integral and have

Isum(x, y) = S0(E0)E0exp (−Zr(x,y)

(µ1(s;E0, x, y) + µ2(s;E0, x, y)) ds).

Once again, this is not a linear system.

Solution 5.2

(a) The highest energy is determined by the peak x-ray tube voltage. For example, if the peak voltage is pkV,

then the peak x-ray energy will be pkeV. The energy spectrum is determined by several factors. First, it will

be zero above pkeV. Second, it will be the sum of characteristic x-ray spectrum and a bremsstrahlung x-ray

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70 CHAPTER 5: PROJECTION RADIOGRAPHY

spectrum. The characteristic x-ray spectrum depends on the atoms in the anode of the x-ray tube, and their

relative proportions. The bremsstrahlung x-ray spectrum has a typical shape, linearly increasing from zero at

the peak energy as energy decreases.

(b) Low energy photons are undesirable because they are usually completely absorbed by the body. Therefore,

they contribute to dose but not image quality. Measures that can be taken to reduce the number of low energy

photons entering the body include: restriction (which works on all photons regardless of energy) and ﬁltering.

Filtering occurs as x-rays pass through objects between the anode and the body, including the glass tube and

surrounding oil and objects placed between the tube and the patient, typically containing plastics and metals.

If copper is used, then aluminum usually follows because copper produces characteristic x-rays at 8 keV,

which would otherwise form a new low energy x-ray source.

materials. It is caused by the selective attenuation of low-energy x-rays in a polyenergetic x-ray beam. This

occurs because most materials have larger attenuation coefﬁcients at lower x-ray energies.

Solution 5.3

The mass attenuation coefﬁcient of aluminum at 80 kVp is µ/ρ = 0.02015 m2/kg. The density of aluminum is

ρ= 2,699 kg/m3. Therefore,

µ(Al) = 0.02015 m2/kg ×2,699 kg/m3

= 54.38 m−1.

For the new material at 80 kVp: µ/ρ = 0.08 m2/kg, ρ= 5,000 kg/m3. So,

µ(new) = 0.08 m2/kg ×5,000 kg/m3

= 400 m−1.

Since attenuation is determined by the exponential factor e−µx, the x-ray attenuation is equal if the exponents are

equal. Hence, the following relation must be satisﬁed:

µ(Al)x(Al) =µ(new)x(new) .

The equivalent thickness of the new material to 2.5 mm of aluminum at 80 kVp is therefore given by

x(new) =54.38 m−1×2.5mm

400 m−1

= 0.34 mm .

From Example 5.1, we know that the copper thickness equivalent to 2.5 mm of aluminum at 80 kVp is x(Cu) =

0.2mm. For the ﬁlter of same cross section area, the copper ﬁlter weighs 0.2×10−3A×8,960 kg/m3= 1.792Akg

and the ﬁlter made of the new material weighs 0.34 ×10−3A×5,000 kg/m3= 1.7Akg. So the ﬁlter made of the

new material is lighter.

Solution 5.4

(a) Iodine and barium are used as contrast agents for two reasons. First, they are bio-compatible—that is, they

are both nontoxic and can be directed to a useful target in the body. Second, they exhibit K-edges in the

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diagnostic x-ray range. Because of their K-edges, they are highly attenuating in the x-ray energy range

immediately above the K-edge, far more attenuating than both tissues and bone. This means that they will

provide exquisite contrast between the agent and the body.

(b) Figure S5.1(a) demonstrates the beneﬁts of an airgap in scatter reduction. Scattering path 1shows a photon

that, when scattered, would hit the standard detector but miss the detector in both cases of a small airgap and

large airgap. Scattering path 2shows a photon that, when scattered, would hit both the standard detector

and positions with a small airgap, but would miss the detector positioned with a large airgap. This example

shows that larger airgaps reject scatter better.

Figure S5.1 See Problem 5.4(b).

The problem with airgaps is demonstrated in Figure S5.1(b). In this ﬁgure, and extended object is projected

onto the three detector positions, demonstrating edge blurring as the result of depth dependent magniﬁcation.

Clearly, the blurring is smallest in the standard detector position and largest for the largest airgap. This shows

that, when using a large airgap for scatter rejection, the objects will appear with more geometric distortion

and/or edge blurring.

Solution 5.5

(a) Compton scattering is random phenomenon. Hence it will cause a random fog throughout the projection

radiograph. It contributes to the loss in SNR and contrast in the resulting image.

(b) The H & D curve has a toe, shoulder and a linear region. When the x-ray exposures are in the toe or shoulder

regions, the optical density of ﬁlm remains constant; thereby reducing the contrast of the resulting image. So,

it is better for the x-ray exposures to be in the linear portion of the H & D curve.

tributing to the dose. So, it is necessary to ﬁlter out the low energy photons coming out of the x-ray source.

(d) If wand hare the width and height of the lead strips in the grid, then the maximum scatter angle θis given

by θ= tan−1(w/h) = tan−1(1/8) = 0.1244 radians.

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72 CHAPTER 5: PROJECTION RADIOGRAPHY

IMAGE FORMATION

Solution 5.6

Let I0be the intensity of the incident beam. Let Icbe the intensity of the x-ray beam falling at the center of the

imaging screen, while Ixbe the intensity at a point on the screen where then intensity falls off to 95% of that at the

center, thus giving a 5% variation in the image intensity. Thus, we have Ix= 0.95I0. If the linear attenuation of

the slab is µand its thickness is L, then

Ic=I0e−µL ,

Ix=I0cos3θe−µL/ cos θ.

Assuming θis small, then µL/ cos θ≈µL and

= cos3θ ,

cos3θ= 0.95 ,

cos θ= 0.983 ,

θ= 10.56◦.

The maximal size is 2dtan θ= 2 ×2×0.19 = 0.746 m.

Solution 5.7

(a) Assume the source-to-object distance is z, and source-to-detector distance is d, then the magniﬁcation of the

object is simply

M=d

(b) One can reduce the magniﬁcation and distortion effects by either moving the object closer to the detector

panel or moving the x-ray source further away from the object and the detector.

Solution 5.8

(a) The weighting aims to compensate for the cos3θdependency and the path length factor. As we know, the

image intensity is given by

Id(xd, yd) = I0cos3θe−µL

cos θ,

where cos θ= 1/p1 + r2

d/d2. We want to transform this relationship back to I0

d=I0exp(−µL)with a

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weighting function that is independent of µand L. The derivation can be done as follows:

I0cos3θ=e−µL/ cos θ,

Id

I0cos3θcos θ

=e−µL ,

I0Id

I0cos3θcos θ

=I0e−µL ,

Id·I0

IdId

I0cos3θcos θ

=I0e−µL .

Thus, the weighting function should be chosen as

w(cos θ) = I0

IdId

I0cos3θcos θ

where cos θ=1

p1 + r2

d/d2. Now substitute the expression for Idinto above expression and after some

simpliﬁcation we ﬁnd

w(cos θ) = 1

cos3θe−cos θ−1

cos θµL .

This correction will hold as long as µ(x, y, z) = µ(z). That is, we assume that we are imaging an object in

which µonly varies in the zdirection.

(b) Assume that the image of the object-of-interest lies in the center of the detector, i.e, it has small rdwhile the

background region has large rd. Assume also that initially It< Ib. Then the image contrast will be improved

after the correction. Under the same assumptions, the SNR is also improved because the image contrast is

improved.

Solution 5.9

(a) Let us consider a 2-D cross section of the system through the y-zplane as shown in Figure S5.2.

The image on the screen will have 3 regions. In the center of the image, between points −aand a, the

appearance of the image is governed by the inverse square law, obliquity, and path length variations. So we

have

Id(x, y) = I0cos3θe−µL/ cos θ.

The value of acan be obtained as follows:

d=w/2

z0+L/2,

a=wd

2z0+L.

When the x-rays are passing through the edges of the object, however, there is a loss of object path length,

and corresponding reduction in attenuation. This corresponds to the region between aand b, and between −b

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74 CHAPTER 5: PROJECTION RADIOGRAPHY

Figure S5.2 The cross section of the prism through y-zplane. See Problem 5.9(a).

and −aon the screen and the intensity is given as

Id(x, y) = I0cos3θe−µa(z0−z0+L/2)/cos θ.

The value of band z0are obtained as follows:

d=w/2

z0−L/2,

b=wd

2z0−L,

z0=wd

2x.

Beyond, band −b, the rays miss the prism completely. In this case,

Id(x, y) = I0cos3θ .

In summary, we have

Id(x, y) =











I0cos3θexp(−µaL/ cos θ)

if 0≤x≤wd

2z0+Land 0≤y≤wd

2z0+L

I0cos3θexp h−µawd

2 max(|x|,|y|)−z0+L/2/cos θi

if wd

2z0+L≤x≤wd

2z0−Lor wd

2z0+L≤y≤wd

2z0−L

I0cos3θ

if x > wd

2z0−Land y > wd

2z0−L

.(S5.1)

(b) The plot along y= 0, is shown in Figure S5.3.

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Figure S5.3 The intensity of the image in the detector plane along y= 0. See Problem 5.9(b).

From (5.32), we have

D= Γ log10(X/X0).

But Xis exposure, not intensity. In a given material, however, the ratio of exposures is equal to the ratio of

intensities. So, we also have

D= Γ log10(Id/I0),

where it is understood that this applies only in the linear range of the H&D curve. Accordingly, I0must

be the intensity at which I0∆tyields the “fog” level on the ﬁlm, where ∆tis the duration of the exposure.

Therefore, the developed ﬁlm will have optical density

D(x, y) = Γ log10(Id(x, y)/I0).

Solution 5.10

Most of the derivation is included in the text preceding the equation. Here we provide a review, and ﬁll in the

missing details. Assume the intensity surrounding a given point on the source is IS. Then the inverse square law

predicts the following intensity at the center of the detector

I0=IS

4πd2.

Moving away from the center of the detector a distance rleads to an additional cos2θloss of intensity; but while

moving away, the unit area increases as well, leading to an additional cos θloss of intensity. Put together, we have

Id=IScos3θ

4πd2.

We now incorporate an inﬁnitesimally thin object tz(x, y), at range z(measured from the source) such that if z=d

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76 CHAPTER 5: PROJECTION RADIOGRAPHY

it provides a further multiplicative attenuation of the source intensity, as follows

Id=IScos3θ

4πd2td(x, y).

If the object is moved away from the detector, it will cast a wider shadow on the detector. This fact is captured

mathematically using the magniﬁcation M=M(z) = d/z, and by scaling the xand yaxes as follows

Id=IScos3θ

4πd2tz(x/M, y/M ).

The last phenomenon that must be included is due to the extended source. Suppose that the source location is not a

point but instead is a small area having a source intensity distribution given by s(x, y). If this source were viewed

through a small hole (which blocks all other transmission) on the z-axis (where x=y= 0) at range z, it would

make an inverted and scaled image of the source intensity, as follows

Id(x, y) = s(x/m, y/m)

4πd2m2,

where m=m(z) = 1−M(z). This represents a response to the impulse transmittivity δz(x, y). If the transmittivity

were not unity, then the response would be attenuated by the transmittivity as follows

Id(x, y) = s(x/m, y/m)

4πd2m2tz(0,0) ,

Now suppose the impulse transmittivity (hole) were moved to position (ξ/M, η/M). Assume that θis small so

that both the source shape distortion due to obliquity and the difference in source magniﬁcation as compared to that

at the detector origin can be ignored. Then the detected image is simply an inverted and scaled source intensity,

shifted to a new position

Id(x, y) = cos3θ s((x−ξ)/m, (y−η)/m)

4πd2m2tz(ξ/M, η/M),

This image represents an approximate impulse response to an impulse in transmittivity at (ξ/M, η/M )within the

thin object at range z. The whole response is the superposition of these individual responses:

Id(x, y) = Z∞

−∞ Z∞

−∞

cos3θ s((x−ξ)/m, (y−η)/m)

m2tz(ξ/M, η/M)dξ dη .

=cos3θ

4πd2m2s(x/m, y/m)∗tz(x/M, y/M ),

which is the desired expression.

IMAGE QUALITY

Solution 5.11

If mphotons are incident on a detector and each one has a probability pof getting detected, independently from

other photons, then the probability that nout of those mphotons are detected has a binomial distribution and is

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given as:

P{nout of mphotons are detected}=m

npn(1 −p)m−n.

Now, the PMF of the number of photons detected D(t)can be computed as follows:

P{D(t) = n}=∞

m=n

P{mphotons are ﬁred by the x-ray tube} · P{nout of mphotons are detected}

=∞

m=n

e−µt(µt)m

m!m

npn(1 −p)m−n

=∞

m=n

e−µt(µt)m

n!(m−n)!pn(1 −p)m−n

=∞

m=n

e−µt(µt)m

n!(m−n)!pn(1 −p)m−n.

Substituting k=m−n, we get

P{D(t) = n}=∞

m=n

e−µt(µt)n+k

n!k!pn(1 −p)k

=e−µt(µt)n

n!pn∞

k=0

(µt)k

k!(1 −p)k

=e−µt(µtp)n

∞

k=0

[µt(1 −p)]k

=e−µt(µtp)n

n!eµt(1−p).

Here we have used the identity et=∞

k=0

k!. By simple rearrangement, we get the PMF of D(t)as

P{D(t) = n}=e−µpt(µpt)n

n!.

Thus, D(t)also follows a Poisson distribution.

Solution 5.12

(a) Since the object is located at z= 3d/4, the magniﬁcation of the object is

m=−d−z

=−d−3d/4

3d/4

=−1/3.

The PSF of an extended source when the object is magniﬁed by mis given by h(x/m). Let the PSF for any

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78 CHAPTER 5: PROJECTION RADIOGRAPHY

arbitrary magniﬁcation mbe h1(x) = e−ax2/m2. Since h1(x) = e−x2/5is the PSF of the extended source,

when m=−1/3we have

ax2/m2=x2/5,

=⇒a=m2/5

= (1/3)2/5

= 1/45 .

Hence at any arbitrary range z, the PSF of the extended source is given by

h1(x) = e−x2

45m2=e−x2z2

45(d−z)2.

(b) Since the PSF is h1(x) = e−x2/45m2the Fourier transform is

H1(u) = e−45m2/π2u2√45m2.

Hence, the transfer function of the overall blurring is

H(u) = √450me−(45m2+10)π2u2,

and the MTF is given by

MTF(u) = e−(45m2+10)π2u2.

h(x) = mp450(45m2+ 10)πe−x2

45m2+10 .

At x=FWHM/2we have e−x2

45m2+10 = 1/2, and therefore

FWHM = 2(45m2+ 10) ln 2 .

Solution 5.13

Scatter fraction, denoted by SF, is deﬁned as

SF =Is

Is+Ib

where Ibdenotes the background intensity and Isdenotes the intensity contributed by scattering. The new image

contrast C0(with scattering) is related to the original scatter-free contrast Cby

C0=It+Is−(Ib+Is)

Ib+Is

=C(1 −SF).

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Thus, when SF = 0.35,

0.35 =C(1 −0.35) = 0.08 ×0.65 = 0.052 ,

and when SF = 0.8,

0.8=C(1 −0.8) = 0.08 ×0.2=0.016 .

Using the relationship SNR =C√ηN, we can compute the SNR in both cases as follows (assuming η= 1)

SNR0.35 =C0.35pηN = 0.052p1×1,000 = 1.64; and

SNR0.8=C0.8pηN = 0.016p1,000 = 0.51 .

If the detector absorption efﬁciency ηis halved, the SNRs become

SNR0

0.35 = 0.052p0.5×1,000 = 1.16 ;

SNR0

0.35 = 0.016p0.5×1,000 = 0.36 .

This problem shows that SNR can be altered in two ways:

•Increase the scatter fraction, which causes an increase in the noise level;

•Decrease the absorption efﬁciency, which causes a decrease in the signal amplitude.

Solution 5.14

Suppose an x-ray burst with an average of ¯

Nphotons is incident upon a detector having quantum efﬁciency QE.

Then the average number of photons stopped by the detector is QE ¯

N. The intrinsic SNR’s of the stopped photons

is pQE ¯

N. By physical law, the signal itself—whatever measured and derived quantity that might be—must have

an SNR lower than the intrinsic SNR. Therefore, we ﬁnd that the maximum DQE is

DQEmax = pQE ¯

√¯

N!2

=QE ,

which was to be proven.

Solution 5.15

By deﬁnition,

DQE =ˆ

λ,

where ˆ

λis the noise-equivalent quanta,

λ=d

SNR2

="λd

pσ2

N#2

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80 CHAPTER 5: PROJECTION RADIOGRAPHY

where λdis the number of photons detected at the detector every second and σ2

Nis the variance of the noise. For

this problem, λd= 10,000 and λ= 10,000. Now substitute these numbers into the above equations to get

DQE =λ2

σ2

Nλ=10,000

σ2

So, the variance of the noise as a function of DQE is:

σ2

N=10,000

DQE .

The function is plotted in Figure S5.4.

Figure S5.4 The variance of the detector’s output (σ2

N) as a function of DQE. See Problem 5.15.

If the variance of the output noise is 2,000, we require DQE =10,000

2,000 = 5. This answer may look incorrect at

the ﬁrst glance, since in the text we say that “Clearly, 0≤DQE ≤1.” But, let’s think about what it means that the

variance of the output noise is 2,000 in the setting of the problem. Given that we have detected 10,000 photons per

second at the detector, the amplitude signal to noise ratio is, by deﬁnition,

SNRa=10,000

√2,000 = 223.6.

But the output signal to noise ratio of an ideal detector is only SNRa=√10,000 = 100 <d

SNRa, which means

this detector does better than the ideal detector. This is impossible in reality, where 0≤DQE ≤1.

Solution 5.16

Assume that each point on the detector sees a Poisson random variable with parameter a. Since this input is

white noise, it has a ﬂat noise power spectrum, and is given by the variance of the random variable, a. Thus, the

frequency dependent (power) SNR of the input is

SNRp(in) = a .

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From Table 2.1, we can determine the frequency response of the nonideal detector to be

H(u, v) = e−2π2(u2+v2).

The frequency response of the detector will selectively ﬁlter out frequencies in the power spectrum according to the

square of the transfer function. Therefore, the frequency dependent (power) SNR of the output is

SNRp(out) = a|H(u, v)|2.

Putting this together, and using (5.40), we ﬁnd that the DQE is given by

DQE(u, v) = (SNRout)2

(SNRin)2

=SNRp(out)

SNRp(in)

=a|H(u, v)|2

=e−4π2(u2+v2).

Solution 5.17

Figure S5.5 shows the orientation of the plastic hollow cylinder with respect to the source and detector. The

distortion is due to depth dependent magniﬁcation. The circular cross section closest to the source gets magniﬁed

more than the other cross section.

Figure S5.5 Oﬀ axis cylinder with depth dependent magniﬁcation artifact. See Problem 5.17.

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82 CHAPTER 5: PROJECTION RADIOGRAPHY

Solution 5.18

From basic trigonometry,

tan φ=s2+dtan α2

d(S5.2)

=s1+dtan α1

d(S5.3)

=s+d1tan α1

(S5.4)

=s+d2tan α2

.(S5.5)

So, we see that s2+dtan α2=s1+dtan α1; therefore,

= 1 + dtan α1−tan α2

(S5.6)

= 1 + tan α1−tan α2

tan φ−tan α1(using (S5.3)) (S5.7)

= 1 + d1

tan α1−tan α2

s(using (S5.4)) .(S5.8)

Equating (S5.4) and (S5.5), we get tan α1−tan α2=s(d1−d2

d1d2). Substituting this back into (S5.8) yields

m=s1

=d2

Therefore, when d1 = 40 cm and d2 = 80 cm,m= 2.

Solution 5.19

(a) For |y| ≤ r, we have

I(d, y) = Ipe−µA·2(r−√r2−y2)−µB·2√r2−y2.

For r≤ |y| ≤ a, we have

I(d, y) = Ipe−µA·2r.

(b) From the deﬁnition of local contrast, with appropriate substitution we have

C=It−Ib

=Ipe−µB·2r−Ipe−µA·2r

Ipe−µA·2r

=e−(µB−µA)·2r−1.

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µB< µA.

(d) The boundary of B (a circle) can be represented as

(x−(z+r))2+y2=r2.

The line connecting the source (0,0) and the point (d, r

√z2+2rz d)can be represented as

y=r

√z2+ 2rz x .

The intersection (x0, y0)of the line and the circle satisﬁes

((x0−(z+r))2+y2

0=r2

y0=r

√z2+2rz x0

Thus,

(x0−(z+r))2+ ( r

√z2+ 2rz x0)2=r2,

=⇒(z+r)2

z2+ 2rz x2

0−2(z+r)x0+z2+ 2rz = 0 .

Since the quadratic discriminant is

∆ = 4(z+r)2−4(z2+ 2rz)(z+r)2

z2+ 2rz ,

= 0

the line is tangent to the circle. The geometry, where cosθ=√z2+2rz

r+z, is shown in Figure S5.6. The intensity

at (d, r

√z2+2rz d)is

I(d, r

√z2+ 2rz d) = I0cos3θe−µA·2r/cosθ

=I0 √z2+ 2rz

r+z!3

e−µA·2rr+z

√z2+2rz .

(e) The magniﬁcation is M(z) = d

z+r. Thus, the point is (z+r, z+r

dy0).

APPLICATIONS

Solution 5.20

(a) You would change the peak tube voltage, or kVp. To generate the ﬁrst ﬁlm, you would use a tube voltage

= 30 kVp, and to generate the second ﬁlm, you would use a tube voltage = 100 kVp.

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84 CHAPTER 5: PROJECTION RADIOGRAPHY

Figure S5.6 See Problem 5.19.

(b) If you did not change anything else, the second ﬁlm, taken at 100 kVp, would be more exposed. That is

because the body is more transparent at higher x-ray energies, so more x-rays would get through to expose

the ﬁlm. The high energy x-rays, when stopped by the intensifying screen, will generate more light output as

well adding to the exposure. It is true that the intensifying screen is also more transparent at higher energies,

but the x-ray spectrum at 100 kVp also contains lower energy x-rays that would contribute to the overall

exposure.

as the x-ray energy increases. Therefore, Compton scattering will be more of “a problem” at 100 keV versus

30 keV, yielding lower contrast images.

(d) This depends on what kind of ﬁltration is used. Ordinarily, when using a 100 kVp source, ﬁltration would

remove lower energy x-rays. In this case, the higher energy source would be more penetrating and the dose

for the 100 kVp source would be lower. However, if the complete 100 kVp spectrum is allowed to be incident

on the patient, then the 100 kVp source would generate more dose to the patient.

(e) The subtracted optical density is

D(x, y) = D(x, y;Eh)−D(x, y;El)

= Γ log10(Xh/X0)−Γ log10(Xl/X0)

= Γ log10 Xh

Xl.

Since Xh> Xl(in general), D(x, y)will be a nonnegative image revealing the relative additional “trans-

parency” of tissues at the higher x-ray energies. (Note: If the two energies were used on “opposite sides” of

the k-edge of a contrast agent, then the difference D(x, y;El)−D(x, y;Eh)would be used instead, since the

attenuation at the higher energy would be larger.)

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Solution 5.21

(a) We have b= 20 cm and

t(E) = e−µd = 10−(E−150)2

5,000 .

So,

µd =(E−150)2

5,000 ln 10 =⇒µ(E) = (E−150)2

5,000dln 10 .

(b) Intrinsic contrast is C=µt−µb

µt+µb. The object is shown in Figure S5.7. The two linear attenuation coefﬁcients

Figure S5.7 See Problem 5.21(b).

are

µt= 0.15 cm−1,

µb=(75 −150)2

5,000dln 10 = 0.13 cm−1.

So, the intrinsic contrast is

C=0.15 −0.13

0.15 + 0.13 = 0.071 .

Ithrough =I0e−µb×15 cme−µt×5 cm

=I0e−0.13×15 cme−0.15×5 cm

= 0.067I0,

and

Imiss =I0e−µb×20 cm

= 0.074I0.

The contrast is therefore given by

C=0.067 −0.074

0.067 + 0.74 =−0.05 .

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86 CHAPTER 5: PROJECTION RADIOGRAPHY

Solution 5.22

(a) The energy spectrum is shown in Figure S5.8. The spectrum is just the number of photons, viewed as a

continuous plot. 104and 105represent integrals (area or mass) of spectrum, which has units photons-keV.

Figure S5.8 See Problem 5.22.

(b) Consider Figure S5.9. The total number of x-ray photons per cm that hit the detector as a function of position

Figure S5.9 See Problem 5.22.

xis

A1:N1

d= 0.5×104e−0.2×2= 3,351 ,

d= 0.5×105e−0.4×2= 22,466 .

The total is = 25,817. For the other parts we have

A2:N1

h= 0.5×104e−0.3×1= 3,704 ,

h= 0.5×105e−0.1×1= 45,242 .

and

A3:N1

d= 3,704e−0.5×1= 2,247 ,

d= 45,242e−0.4×1= 30,327 .

Therefore, the total is = 32,574.

that A is the target and B is the background is I=αN. Therefore

It−Ib

=Nt−Nb

Nt−Nb

=−6,757

58,391 =−0.12 = C .

(d) The optical density, given by D= Γ log X

X0, as a function of position xassuming x-ray ﬁlm is shown in

Figure S5.11.

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Figure S5.10 See Problem 5.22.

Figure S5.11 See Problem 5.22.

Solution 5.23

(a) The object magniﬁcation is

M=d/z = 60/40 = 1.5.

(b) The source magniﬁcation is

m=−d−z

z= 1 −M=−0.5.

Id(xd, yd) = K[S0e−(xd/m)2δ(yd/m)] ∗hδxd

M−w

2+δxd

M+w

2i

=KS0nhe−(xd/m)2δ(yd/m)i∗δxd

M−w

2+he−(xd/m)2δ(yd/m)i∗δxd

M+w

2o.

To evaluate this, we ﬁrst compute the convolution

he−(xd/m)2δ(yd/m)i∗δxd

M−w

2

as follows:

he−(xd/m)2δ(yd/m)i∗δxd

M−w

2=Z∞

−∞ Z∞

−∞

e−xd−u

m2

δyd−v

mδu

M−w

2du dv

=M|m|Z∞

−∞

e−xd−u0M

m2

δu0−w

2du0Z∞

−∞

δyd

m−v0dv0

=M|m|e−xd−wM

m2

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88 CHAPTER 5: PROJECTION RADIOGRAPHY

Similarly,

he−(xd/m)2δ(yd/m)i∗δxd

M−w

2=M|m|e−xd+wM

m2

Hence,

Id(xd, yd) = KM |m|S0e−xd−wM

m2

+KM |m|S0e−xd+wM

m2

= 0.75KS0he−4(xd−0.75w)2+e−4(xd+0.75w)2i.

(d) As shown in (c), the image of the line phantom is the sum of two Gaussian-shaped functions that are centered

at wM/2and −wM/2respectively, and both have a variance of m2/2. With the same reasoning as in

determining FWHM, we know that in order for the two Gaussians to be distinguishable, the following must

be true

KM |m|S0e−0−wM

m2

≤1

2max

KM |m|S0e−xd−wM

m2

2KM |m|S0.

Hence, wM

2m2

≥ln 2

and

w≥2|m|√ln 2

Thus, the minimum value that wcan take is 2|m|√ln 2/M ≈0.555 cm.

Here is an alternative solution. Notice that the image of the line phantom Id(xd, yd)is simply its image

under an ideal point source (t0(xd, yd) = t(xd/M, xd/M)) blurred (convoluted) by the magniﬁed source

distribution (s0(xd, yd) = s(xd/m, yd/m)) with a suitable scaling (K). The image of the phantom under a

point source would still be two parallel lines with the same orientation, but the spacing is magniﬁed to wM .

The magniﬁed source still has the same form, and can be computed as

s0(xd, yd) = s(xd/m, yd/m) = S0e−(xd/m)2δ(yd/m) = |m|S0e−x2

d/m2δ(yd).

By the deﬁnition of FWHM, we know that in order for the images of the two lines to be distinguishable, the

spacing wM must be greater than the FWHM of s0(xd, yd), which is computed to be 2|m|√ln 2. Hence the

minimum value of wis 2|m|√ln 2/M ≈0.555 cm.

Solution 5.24

(a) Since the number of photons are uniformly shed upon the side of the whole tissue, it is clear that 1/4(=

0.5/2.0) of the total incident photons will go through the blood vessel, and the remaining 3/4Niphotons

only pass through the soft tissue. Hence the total number of photons can be computed as follows

Nt=Ni

4e−µvesselLvessel−µtissue(Ltissue−Lvessel)+3Ni

4Nie−µtissueLtissue

=Ni

4e−0.5µvessel e−1.5µtissue +3Ni

4e−2µtissue .

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At 15 keV,

Nt=4×106

4e−0.5×3.0e−1.5×4.0+3×4×106

4e−2×4.0

≈1,006 + 553

= 1,559.

At 40 keV,

Nt=4×106

4e−0.5×0.2e−1.5×0.4+3×4×106

4e−2×0.4

≈4.96 ×105+ 1.35 ×106

= 1.84 ×106.

We see from this analysis that, at the lower energy level 15 keV, more photons are absorbed because the linear

attenuation coefﬁcients of the tissue and the blood vessel are both higher at 15 keV than at 40 keV.

(b) Since the incident photons are uniformly shed upon the tissue, the photon density p, that is, the number of

photons per unit area, is a constant and can be computed as p=Ni/A, where Ais the area of the side of

the tissue. Notice that the value of pdoes not affect the local contrast computation, which can be shown as

follows. The background intensity Nbis simply

Nb=pe−µtissueLtissue =pe−2.0µtissue .

The object intensity Nois given by

No=pe−µvesselLvessel−µtissue(Ltissue−Lvessel)

=pe−0.5µvessel−1.5µtissue .

Hence, the local contrast can be computed as

C=No−Nb

=pe−0.5µvessel−1.5µtissue −pe−2.0µtissue

pe−2.0µtissue

=e−0.5(µvessel−µtissue)−1.

At 15 keV, the local contrast is

C15 =e−0.5×(3.0−4.0) −1≈0.649 .

At 40 keV, the local contrast is

C40 =e−0.5×(0.2−0.4) −1≈0.105 .

Hence, the local contrast is higher (better) at 15 keV.

Note: If you confused local contrast and contrast, the answer you would get differs. At 15 keV, the contrast

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90 CHAPTER 5: PROJECTION RADIOGRAPHY

of the blood vessel is

C15 =No−Nb

No+Nb

=pe−0.5µvessel e−1.5µtissue −pe−2µtissue

pe−0.5µvessel e−1.5µtissue −pe−2µtissue

=e−0.5µvessel −e−0.5µtissue

e−0.5µvessel −e−0.5µtissue

=1−e−0.5(µtissue−µvessel)

1 + e−0.5(µtissue−µvessel)

=1−e−0.5×(4.0−3.0)

1 + e−0.5×(4.0−3.0)

≈0.2449.

Similarly, at 40 keV the local contrast of the blood vessel is

C40 =1−e−0.5×(0.4−0.2)

1 + e−0.5×(0.4−0.2) ≈0.05 .

In this case, the contrast is still higher at 15 keV.

tenuation coefﬁcients of the soft tissue and the blood vessel. As can be seen from the table, at 15 keV this

difference does not change much after the contrast agent is injected into the blood vessel. Hence, it would be

expected that the local contrast (in its absolute value) does not change much. (The new contrast is actually

0.393 in absolute value.)

At 40 keV, the linear attenuation coefﬁcient of the contrast agent is hugely different from that of the soft

tissue and the original blood vessel. Thus, it can be expected that the local contrast of the blood vessel will

be largely changed (improved) after the contrast agent is injected in. (The new contrast is actually 0.999 in

absolute value.)

(d) The explanation is that the contrast agent material has K-shell electrons whose binding energy is slightly

lower than 40 keV but higher than 15 keV. When x-ray photons with an energy of 40 keV enter the material,

photoelectric interaction will cause electrons from the K-shell to be ejected and the x-ray photons will be

completely absorbed. This effect, called K-edge absorption, signiﬁcantly increases the attenuation coefﬁcient

of the contrast agent.

Solution 5.25

(a) The intensity of x-ray beam is given by

I(E) = NE

A∆t.

When the x-ray source is an ideal point source, we need to take the magniﬁcation into account. The distance

between the source and the object is 2z0, the distance between the source and the detector is 3z0, so the

object magniﬁcation is M=d

z= 1.5. The intensity proﬁle on the detector along the x-axis is shown in

Figure S5.12.

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Figure S5.12 Intensity proﬁle along xaxis for ideal point x-ray source. See Problem 5.25.

(b) The dark bars on the phantom are treated as target. So, the contrast is

C=It−Ib

=1/4−1

1=−3

H(u, v) = F{h(x, y)}

=F{sinc(αx)}F{sinc(βy)}

αrect u

α·1

βrect v

β.

The system is an ideal low pass ﬁlter. The highest frequencies of the output signal in xand ydirections are

U0=α

2and V0=β

2, respectively. According to the sampling theorem, the maximum sampling periods for

the output signals of the system are (∆x)max =1

2U0=1

α, and (∆y)max =1

2V0=1

β.

(d) The imaging equation of a projection radiography system is

Id=I0ZEmax

S0(E)Eexp −Zx

µ(s, E)dsdE.

Since the x-ray photons are monochromatic, and the target in the phantom is homogeneous, the above equa-

tion can be simpliﬁed to

=e−µ(160 keV)x=1

4.(S5.9)

Then

−µ(160 keV) = ln 640( keV)

160( keV)cm−1= ln(4) cm−1.

Solving for xin Equation (S5.9) yields

x=ln(4)

ln(4) = 1 cm .

Solution 5.26

(a) SNR decreases because the differences between the attenuation of different body tissues decreases as the

energy increases.

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92 CHAPTER 5: PROJECTION RADIOGRAPHY

(b) The dose is reduced due to the compensatory change in the exposure time.

•I

•N

•I

(d) The airgap is most effective in reducing the scatter fraction in case of small ﬁeld size.

(e) The image noise ultimately limits the contrast sensitivity of an x-ray imaging system.

(f) The average photon energy is mainly determined by the material in the beam path.

Solution 5.27

(a) The best contrast agent to be used in this case is iodine because it has the k-shell energies within the energy of

the source. This increases probability of photoelectric effect and hence higher linear attenuation coefﬁcient

that barium. Therefore the use of iodine will provide better contrast in the image.

(b) (i) Contrast before contrast agent is applied: Let I0be the intensity at the middle of the detector when

nothing is put between the source and the detector. The intensity at the center of the tumor is

It=I0e−(µtumor)(2R)e−(µtissue )w

=I0e−(0.75×0.2)−(1×1) = (0.3166)I0.

The intensity at the edge of the tumor is

Ib=I0cos3θ1e−(µtissue)w

cos θ1.

From Figure P5.8 we see that tan θ1=R/(D−Dtd −w−R). Therefore θ1= 2.5714 ×10−4radians and

cos θ1= 1. This implies that we can neglect the cos3θ1effect in this problem. Accordingly,

Ib=I0e−(µtissue)w

=I0e(−1×1) = (0.3679)I0,

and

Contrast before contrast agent is applied =I0(0.3166 −0.3679)

I0(0.3679) =−0.1394 .

(ii) Contrast After contrast agent is applied: The intensity at the center of the tumor is

It=I0e−(µtumor with contrast agent )(2R)e−(µtissue )w

=I0e−(10×0.2)−(1×1) = (0.0498)I0.

The intensity at the edge of the tumor remains the same since the photons do not pass through the tumor.

Accordingly,

Ib= (0.3679)I0,

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and

Contrast before contrast agent is applied =I0(0.0498 −0.3679)

I0(0.3679) =−0.8646 .

in the tumor. But, for the photon to have the lowest possible energy while hitting the detector, it should be

scattered the most—that is, it should have the largest scattering angle.

The dashed line in Figure S5.13 shows the path that yields a Compton scattered photon, which will hit the

detector at the lowest possible energy. Among all the Compton scattered photon trajectories (with single

scattering events) the dashed line path will have the largest scatter angle.

Figure S5.13 θis the maximum Compton scattering angle. The dotted line shows the trajectory of the photon

having the lowest possible energy reaching the detector. See Problem 5.27.

(d) In this part, we have to ﬁnd the energy of the Compton scattered photon which follows the dashed line

trajectory in part (c). To ﬁnd the energy, we should ﬁnd the scatter angle θ(see Figure S5.13). The angle θ

can be written as sum of θaand θb. From the geometry of the setup,

tan θa=(l/2) + (h/2)

Dtd

tan θb=(h/2)

D−Dtd

Given the above relationships, and using the given values of l,h,D, and Dtd, we ﬁnd that θa= 1.3790 radians

and θb= 0.0038 radians. Therefore, θ= 0.0038+1.3790 = 1.3828 radians is the maximum scattering angle.

This implies that

Minimum energy of Compton photon =E

1 + (1 −cos(θ)) E

moc2

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94 CHAPTER 5: PROJECTION RADIOGRAPHY

Substituting, E= 35 keV (energy of incoming photon), θ= 1.3828 radians, and moc2= 511 keV yields

minimum energy = 33.1536 keV .

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Computed Tomography

INSTRUMENTATION

Solution 6.1

(a) (hW−hA) = a(hW

m−hA

m), hence

a= (hW−hA)/(hW

m−hA

m),

b=hW−hW

m(hW−hA)/(hW

m−hA

=hA−hA

m(hW−hA)/(hW

m−hA

m).

(b) hW= 0;hA= -1000.

Solution 6.2

(a) The x-ray source detector apparatus rotates at a speed of 4πradians/s, so it takes 0.5 s to rotate a full circle

(2π). During this period of time, the patient table moves 2cm/s×0.5s= 1 cm. So the pitch of the helix is

1 cm.

(b) It takes 0.5 s for the imaging devices to rotate a full circle of 2π, and it takes 1 ms to measure a projection.

So the system can measure at most 500 projections over a 2πangle.

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96 CHAPTER 6: COMPUTED TOMOGRAPHY

RADON TRANSFORM

Solution 6.3

Proof: An operator Ris linear if R(af1+bf2) = aR(f1) + bR(f2). Let

Rf=Z∞

−∞ Z∞

−∞

f(x, y)δ(xcos θ+ysin θ−`)dx dy.

Then we have

R(af1+bf2) = ZZ[af1(x, y) + bf2(x, y)]δ(xcos θ+ysin θ−`)dx dy

=aZZ f1(x, y)δ(xcos θ+ysin θ−`)dx dy

+bZZ f2(x, y)δ(xcos θ+ysin θ−`)dx dy

=aRf1+bRf2.

which was to be proved.

Solution 6.4

Let u=x−x0,v=y−y0, then du =dx,dv =dy. We get

ZZ f(x−x0, y −y0)δ(xcos θ+ysin θ−`)dx dy

=ZZ f(u, v)δ[(u+x0) cos θ+ (v+y0) sin θ−`]du dv

=ZZ f(u, v)δ[ucos θ+vsin θ−(`−x0cos θ−y0sin θ)] du dv

=g(`−x0cos θ−y0sin θ, θ),

where g(`, θ)is the Radon transform of f(x, y).

Solution 6.5

Since f(x, y)is rotationally symmetric, g(`, θ) = g(`, 0). Hence,

g(`, θ) = ZL(`,θ)

f(x, y)ds

=Z∞

−∞ Z∞

−∞

e−x2−y2δ(xcos(θ)−`)dxdy

=Z∞

−∞

e−`2−y2dy

=e−l2Z∞

−∞

e−y2dy .

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Since 1

√2πσ2Z∞

−∞

e−x2/σ2dx = 1 ,

then Z∞

−∞

e−y2dy =√π .

Thus,

g(`, θ) = √πe−`2.

Solution 6.6

Z∞

−∞

g(`, θ)d` =Z∞

−∞

h`(`)hθ(θ)d`

=hθ(θ)Z∞

−∞

h`(`)d` .

On the other hand

Z∞

−∞

g(`, θ)d` =Z∞

−∞ ZZ f(x, y)δ(xcos θ+ysin θ−`)dx dy d`

=ZZ f(x, y)Z∞

−∞

δ(xcos θ+ysin θ−`)d` dx dy

=ZZ f(x, y)dx dy .

Thus,

hθ(θ)Z∞

−∞

h`(`)d` =ZZ f(x, y)dx dy .

Since the right-hand side does not depend on θ, the left-hand side cannot depend on θeither. Hence, hθ(θ)must be

a constant.

Solution 6.7

(a) The Fourier transform of f(x, y)is

F(u, v) = 0.5(δ(u−f0, v) + δ(u+f0, v)) .

In order to use the projection slice theorem, we must change to polar coordinates. We use the following steps

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98 CHAPTER 6: COMPUTED TOMOGRAPHY

for a shifted impulse function:

δ(u−f0, v) = δ(u−f0)δ(v)

=δ(%cos θ−f0)δ(%sin θ)

=δ(%[1 −θ2

2! +···]−f0)δ(%[θ−θ3

3! +···])

=δ(%−f0)δ(%θ)

=δ(%−f0)1

|%|δ(θ).

In this derivation we have used a Taylor series approximation for cosine and sine and the scaling property of

the impulse function. This derivation can be repeated for the impulse shifted in the opposite direction; then

we apply the projection slice theorem, yielding

G(%, θ) = F(%cos θ, % sin θ) = 0.5

|%|[δ(%−f0) + δ(%+f0)]δ(θ).

Now we write the expression for ﬁltered backprojection, plug in this Radon transform, and simplify as fol-

lows:

f(x, y) = Zπ

0Z∞

−∞ |%|G(%, θ)ej2π%`d%`=xcos θ+ysin θ

dθ

=Zπ

0Z∞

−∞ |%|0.5

|%|[δ(%−f0) + δ(%+f0)]δ(θ)ej2π%`d%`=xcos θ+ysin θ

dθ

=Zπ

0Z∞

−∞

0.5[δ(%−f0) + δ(%+f0)]δ(θ)ej2π%`d%`=xcos θ+ysin θ

dθ

=Zπ

0Z∞

−∞

0.5[δ(%−f0) + δ(%+f0)]ej2π%`d%`=xcos θ+ysin θ

δ(θ)dθ

=Zπ

[cos(2πf0`)]`=xcos θ+ysin θδ(θ)dθ

=Zπ

cos(2πf0[xcos θ+ysin θ])δ(θ)dθ

= cos(2πf0x).

The last step follows from the sifting property of the impulse function. This proves that ﬁltered backprojection

produces the correct result.

(b) i) Using the result from part (a) and the linearity of the Radon transform, it follows that the Radon transform

of f(x, y) = cos 2πax + cos 2πby is

G(%, θ) = 0.5

|%|[δ(%−a) + δ(%+a)]δ(θ) + 0.5

|%|[δ(%−b) + δ(%+b)]δ(θ).

From the linearity of the inverse Radon transform (i.e., ﬁltered backprojection), we can follow the same steps

carried out in part (a) to prove that ﬁltered backprojection will yield f(x, y) = cos 2πax + cos 2πby.

ii) The function f(x, y) = cos 2π(ax +by)is a rotated version of cos(2πf0x), which was solved in part (a).

Although the math can be carried out in analogous fashion, it is easier to simply use the form found in (a)

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and carefully apply it here. The frequency of this sinusoid (distance from the origin is f0=√a2+b2and its

rotation from the x-axis is θ0= tan−1(b/a). Therefore, its Radon transform is given by

G(%, θ) = 0.5

|%|[δ(%−f0) + δ(%+f0)]δ(θ−θ0).

In proving that ﬁltered backprojection gives the right answer, it is only necessary to evaluate the last step

differently.

f(x, y) = Zπ

[cos(2πf0`)]`=xcos θ+ysin θδ(θ−θ0)dθ

=Zπ

cos(2πf0[xcos θ+ysin θ])δ(θ−θ0)dθ

= cos(2πf0[xcos θ0+yθ0)

= cos(2π[xa +yb]) .

The last step follows from the facts that a=f0cos θ0and b=f0θ0from the geometry.

Solution 6.8

(a) We write

µ(x, y) = µ0rect(x) rect(y)

=µ0if −1/2≤x≤1/2and −1/2≤y≤1/2;

0otherwise .

(b) F2D{µ}(u, v) = µ0sinc(u)sinc(v).

function

g(`, θ) = Z∞

−∞Z∞

−∞

µ(x, y)δ(xcos θ+ysin θ−`)dx dy

=Z1/2

−1/2Z1/2

−1/2

µ0δ(xcos θ+ysin θ−`)dx dy .

(d) Writing the projection slice theorem yields

G(%, θ) = F2D{µ}(%cos θ, % sin θ) = µ0sinc(%cos θ)sinc(%sin θ).

(e) We want to ﬁnd g(`, θ) = F−1

1D {G(%, θ)}.By symmetry, we have

G(%, θ +π

2) = G(%, θ),

G(%, θ +π) = G(%, θ),

G(%, π

2−θ) = G(%, θ),

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100 CHAPTER 6: COMPUTED TOMOGRAPHY

since sinc(−x) = sinc(x). Hence, we only need to compute g(`, θ)for 0≤θ≤π/4.

First, if θ= 0,G(%, θ) = µ0sinc(%), and hence

g(`, 0◦) = F−1

1D {µ0sinc(%)}=µ0rect(`).

If 0< θ ≤π/4,

g(`, θ) = F−1

1D {µ0sinc(%cos θ)sinc%sin θ}

=µ0

|sin θcos θ|rect( `

cos θ)∗rect( `

sin θ).

The convolution of two rect functions, rect(x/a)∗rect(x/b)for 0< b ≤a, can be easily computed to be:

rect(x/a)∗rect(x/b) = 









x+1

2(a+b)−1

2(a+b)≤x≤ −1

2(a−b)

b−1

2(a−b)≤x≤1

2(a−b)

−x+1

2(a+b)1

2(a−b)≤x≤1

2(a+b)

0otherwise

Since cos θ > sin θfor θ∈(0, π/4], hence

g(`, θ) = µ0

|sin θcos θ|×









`+1

2(cos θ+ sin θ)−1

2(cos θ+ sin θ)≤`≤ −1

2(cos θ−sin θ)

sin θ−1

2(cos θ−sin θ)≤`≤1

2(cos θ−sin θ)

−`+1

2(cos θ+ sin θ)1

2(cos θ−sin θ)≤`≤1

2(cos θ+ sin θ)

0otherwise

for 0< θ ≤π/4.

(f) θ= 30◦,sin θ= 1/2,cos θ=√3/2. From (e), we get

g(`, 30◦) = 4µ0

√3×









`+1+√3

4−1+√3

4≤`≤ −√3−1

2−√3−1

4≤`≤√3−1

−`+1+√3

√3−1

4≤`≤1+√3

0otherwise

for 0< θ ≤π/4. Therefore,

b30◦(x, y) = g(xcos 30◦+ysin 30◦,30◦) = g(√3x+y

2,30◦)

=4µ0

√3×









√3x+y

2+1+√3

4−1+√3

4≤√3x+y

2≤ −√3−1

2−√3−1

4≤√3x+y

2≤√3−1

−√3x+y

2+1+√3

√3−1

4≤√3x+y

2≤1+√3

0otherwise

The sketch is straightforward; note that g(`, 30◦)is trapezoid shaped, not a triangle.

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101

Solution 6.9

We start with the convolution integral

g(x, y) = Zξ

Zη

f(ξ, η)h(x−ξ, y −η)dξ dη .

Now, we perform the following steps:

R{g}=ZyZxZηZξ

f(ξ, η)h(x−ξ, y −η)dξdηδ(xcos θ+ysin θ−`)dxdy

=ZηZξ

f(ξ, η)ZyZx

h(x−ξ, y −η)δ(xcos θ+ysin θ−`)dxdy dξdη

=ZηZξ

f(ξ, η)Zy0Zx0

h(x0, y0)δ(x0cos θ+y0sin θ−[`−ξcos θ−ηsin θ]) dx0dy0dξdη

=ZηZξ

f(ξ, η)R{h}(`−ξcos θ−ηsin θ, θ)dξdη

=ZηZξ

f(ξ, η)Z`0R{h}(`−`0, θ)δ(ξcos θ+ηsin θ−`0)d`0dξdη

=Z`0R{h}(`−`0, θ)ZηZξ

f(ξ, η)δ(ξcos θ+ηsin θ−`0)dξdη d`0

=Z`0R{h}(`−`0, θ)R{f}(`0, θ)d`0

=R{h} ∗ R{f}

which was to be proved.

CT RECONSTRUCTION

Solution 6.10

(a) We carry out the following steps:

gs(`, θ +π/2) = ZZ s(x, y)δ(xcos(θ+π/2) + ysin(θ+π/2) −`)dxdy

=ZZ s(x, y)δ(−xsin θ+ycos θ−`)dxdy

=ZZ s(−v, u)δ(ucos θ+vsin θ−`)dudv (u=y, v =−x)

=ZZ s(u, v)δ(ucos θ+vsin θ−`)dudv

=gs(`, θ).

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102 CHAPTER 6: COMPUTED TOMOGRAPHY

(b) We carry out the following steps:

gs(`, −θ) = ZZ s(x, y)δ(xcos(−θ) + ysin(−θ)−`)dxdy

=ZZ s(x, y)δ(xcos θ−ysin θ−`)dxdy

=ZZ s(u, −v)δ(ucos θ+vsin θ−`)dudv (u=x, v =−y)

=ZZ s(u, v)δ(ucos θ+vsin θ−`)dudv

=gs(`, θ).

˜gs(`, θ) = 









gs(`, θ) 0 ≤θ < π

gs`, π

2−θπ

4≤θ < π

which covers 0≤θ < π/2. Then

gs(`, θ) = 









˜gs(`, θ) 0 ≤θ < π

˜gs`, θ −π

2π

2≤θ≤π

covers 0≤θ < π.

(d) See Figure S6.1.

-1 1 t

à2

gs(t; 0)

gs(t; ù=4)

gs(t; ù=8)

Sketch only

Figure S6.1 Sketch of projections at diﬀerent angles. See Problem 6.10(d).

(e) See Figure S6.2. From simple rotations, we have `1=−cos θ−sin θ,`2=−cos θ+sin θ,`3= cos θ−sin θ,

and `4= cos θ+ sin θ. By similar triangles, we have:

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103

Figure S6.2 See Problem 6.10(e).

`1≤`≤`2:`−`1

`2−`1

=g(`, θ)

2

cos θ

⇒g(`, θ) = 2

cos θ`+ cos θ+ sin θ

2 sin θ

=`+ cos θ+ sin θ

cos θsin θ

`2≤`≤`3:g(`, θ) = 2

cos θ

`3≤`≤`4:`4−`

`4−`3

=g(`, θ)

2

cos θ

⇒g(`, θ) = cos θ+ sin θ−`

cos θsin θ.

and g(`, θ) = 0 elsewhere.

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104 CHAPTER 6: COMPUTED TOMOGRAPHY

Solution 6.11

(a) Carry out the following steps:

mp(θ) = Zg(`, θ)d`

=ZZZ f(x, y)δ(xcos θ+ysin θ−`)dxdydt

=ZZ f(x, y)Zδ(xcos θ+ysin θ−`)d`dxdy

=ZZ f(x, y)dxdy

=m .

(b) Carry out the following steps:

cp(θ) = 1

mZ`g(`, θ)d`

mZ`ZZ f(x, y)δ(xcos θ+ysin θ−`)dxdyd`

mZZ f(x, y)Z`δ(xcos θ+ysin θ−`)d`dxdy

mZZ f(x, y) [xcos θ+ysin θ]dxdy

= cos θ1

mZZ f(x, y)x dxdy

+ sin θ1

mZZ f(x, y)y dxdy

=cxcos θ+cysin θ .

4=m= 1 and cx= 0 since f(x, y)is symmetric about the y-axis.

cy=ZZ yf(x, y)dxdy

=Z0

−1Z1+x

y dxdy +Z1

0Z1−x

y dxdy

=Z0

−1

(1 + x)2

2dx +Z1

(1 −x)2

2dx

Hence,

cpπ

4=cysin θ=1

√2

2≈0.2357.

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105

Solution 6.12

(a) The object is shown in Figure S6.3.

Figure S6.3 See Problem 6.12(a).

(b) The number of photons as a function of `for θ= 0◦and θ= 90◦are shown in Figures S6.4 and S6.5.

Figure S6.4 θ= 0◦. See Problem 6.12(b).

Figure S6.5 θ= 90◦. See Problem 6.12(b).

Figure S6.6 θ= 0◦. See Problem 6.12(c).

(d) The backprojection image at θ= 0◦is shown in Figure S6.8. In the shaded area, the value is zero, on x= 0,

the value is ln 2, and on x= 2, the value is ln 4.

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106 CHAPTER 6: COMPUTED TOMOGRAPHY

Figure S6.7 θ= 90◦. (See Problem 6.12(c).)

Figure S6.8 See Problem 6.12(d).

Solution 6.13

(a) We have

g(`, 60◦) = 





√3µa

2+`,−a

2≤`≤0

√3µa

2−`,0≤`≤a

0,otherwise

The projection g(`, 60◦)is shown in Figure S6.9.

Figure S6.9 See Problem 6.13(a).

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107

(b) We have

b60◦0,a

4=g0 cos 60◦+a

4sin 60◦,60◦

=g √3a

8,60◦!

=√3µ a

2−√3a

=√3µa

8(4 −√3) .

f(t) = K, −p

2≤t≤p

0,otherwise .

The convolution f(t)∗f(t)is given by:

f(t)∗f(t) = 





K2(p+t),−p≤t≤0

K2(p−t),0≤t≤p

0,otherwise

Comparing with g(`, 60◦), we see that it is a convolution of function f(`)with itself, where K2=√3µ, and

p=a/2:

g(`, 60◦) = q√3µrect 2`

a∗q√3µrect 2`

a.

By the projection slice theorem, we get F(%cos θ, % sin θ) = G(%, θ) = F{g(`, θ)}. Since g(`, 60◦)is

expressed as a convolution of a rect function with itself, we have:

F{g(`, 60◦)}=F{q√3µrect 2`

a}=√3µa2

4sinc2a%

2.

Therefore,

F(%cos 60◦, % sin 60◦) = √3µa2

4sinc2a%

2.

Solution 6.14

(a) Deﬁne ∧(x) = rect(x)∗rect(x). By the convolution theorem, its Fourier transform is

F{(∧(x)}(%) = sinc2(%).

We see that W(%) = ∧(%/%0). So by the duality of the Fourier transform and the scaling theorem, we have

F−1{W(%/%0)}(`) = %0sinc2(%0`).

Multiplying |%|by W(%) = ∧(%/%0)corresponds to convolving c(`) = F−1{|%|}(`)by F−1{W(%/%0)}(`),

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108 CHAPTER 6: COMPUTED TOMOGRAPHY

which yields

˜c(`) = c(`)∗%0sinc2(%0`).

(b) We have

lim

%0→∞ %0sinc2%0`=δ(`),

Therefore,

lim

%0→∞ ˜c(`) = c(`),

which means that the exact solution is produced.

Solution 6.15

First expand the integral:

Z2π

0Z∞

%Gθ(%)e+j2π%ω·xd% dθ

=Zπ

0Z∞

%Gθ(%)e+j2π%ω·xd% dθ

| {z }

+Z2π

πZ∞

%Gθ(%)e+j2π%ω·xd% dθ

| {z }

Now make the substitution φ=θ−πin I2:

I2=Zπ

0Z∞

%Gφ+π(%)e+j2π%[xcos(φ+π)+ysin(φ+π)] d% dφ .

From the geometry, gφ+π(`) = gθ(−`), which implies Gφ+π(%) = Gθ(−%). Therefore,

I2=Zπ

0Z∞

%Gθ(−%)e+j2π%[−xcos(φ)−ysin(φ)] d% dφ .

Now let q=−%and θ=φto yield:

I2=−Zπ

0Z∞

0−qGθ(q)e+j2πqω·xdq dθ .

Now let %=qand switch around the limits with the minus sign:

I2=Zπ

0Z0

−∞ −%Gθ(%)e+j2π%ω·xd% dθ .

Now, I1and I2can be added together to yield the desired result.

Solution 6.16

(a) We have F{δ(x, y)}= 1, so G(%, θ)=1. Therefore, g(`, θ) = F−1{G(%, θ)}=δ(`).

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109

(b) We write

fδ

b=Zπ

δ(xcos θ+ysin θ)dθ

=Zπ

δ(rcos φcos θ+rsin φsin θ)dθ

=Zπ

δ(rcos(θ−φ)) dθ

=Zπ/2

−π/2

δ(rsin(θ−φ)) dθ .

Since sin θ≈θfor small θ,sin(θ−φ)≈θ−φfor θ≈φ. Hence

fδ

b≈Zπ/2

−π/2

δ(r(θ−φ))dθ, for θ≈φ .

Since δ(at) = δ(t)

|a|(this is the scaling theorem for the impulse function), we have

fδ

b=Zπ/2

−π/2

|r|δ(θ−φ)dθ, for θ≈φ

|r|if φ∈−π

2,π

2.

If φ6∈ −π

2,π

2, it still works by noticing that

Zπ/2

−π/2

g(xcos θ+ysin θ, θ)dθ =Z3π/2

π/2

g(xcos(θ−π) + ysin(θ−π), θ −π)dθ

=Z3π/2

π/2

g(−xcos θ−ysin θ, θ −π)dθ

=Z3π/2

π/2

g(xcos θ+ysin θ, θ)dθ ,

since g(`, θ −π) = g(−`, θ).

|r|=1

px2+y2.

By the Fourier shift theorem, we know that F{δ(x−x0, y −y0)}=e−j2π(ux0+vy0). Hence,

G(%, θ) = e−j2π%(cos θx0+sin θy0)

and

g(`, θ) = F−1{G(%, θ)}=δ(`−x0cos θ−y0sin θ).

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110 CHAPTER 6: COMPUTED TOMOGRAPHY

We have

fδ

b=Zπ

δ(xcos θ+ysin θ−x0cos θ−y0sin θ)dθ

=Zπ

δ((x−x0) cos θ+ (y−y0) sin θ)dθ .

Let r=p(x−x0)2+ (y−y0)2and φ= tan−1((y−y0)/(x−x0))—i.e., the radius and angle are measured

from (x0, y0). Then

fδ

b=Zπ

δ(rcos(θ−φ))dθ =1

|r|.

Therefore,

fδ

b=1

p(x−x0)2+ (y−y0)2.

(d) Deﬁne Ras the Radon transform operator and Bas the backprojection operator. Both are linear operators

and the composition BR was shown in part (c) to be shift-invariant. Therefore convolution still holds. The

impulse response was found in part (b) to be 1/px2+y2. Therefore,

fb=f∗1

px2+y2.

(e) In principle, all one needs to do is ﬁnd the Fourier transform of 1/px2+y2and apply its inverse to fb.

Deﬁne H(u, v) = F1

√x2+y2, then Fb(u, v) = F(u, v)H(u, v). Hence, F(u, v) = Fb(u, v)/H(u, v)

provided that H(u, v)6= 0.

The problem is that the Fourier Transform of 1/r is 1/g where g=√u2+v2. Therefore, the inverse ﬁlter is

q=√u2+v2, which has inﬁnite gain at inﬁnite frequencies. In other words, it is the worst type of high-pass

ﬁlter.

Solution 6.17

(a) bθ(x, y) = g(xcos θ+ysin θ, θ).

(b) We have

`=xcos θ+ysin θ= 1 cos 30◦+ 2 sin 30◦

= 0.866 + 1 = 1.866 .

Therefore,

b30◦(1,2) = g(1.866,30◦)≈0.155 .

(d) Since 210◦= 30◦+ 180◦, this is the “opposite” projection, and therefore

g(`, 210◦) = g(−`, 30◦) = 0.155 .

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111

Figure S6.10 See Problem 6.17(e).

(e) See Figure S6.10. The image always has the same value along the lines with a slope of tan 120◦=−√3/3.

(f) No, because to determine b30◦(1,2), we need `= 1.866 as shown in (b), which is not an integer. An

approximate value might be to choose `= 2, which yields 0.135.

(g) We have that `= 2 ×0.866 + 1 ×0.5 = 2.232, which, again, is not an integer. Thus, the exact value still

cannot be determined. As an approximation, we might choose `= 2, and the approximate value is again

0.135.

Solution 6.18

The ramp ﬁlter is deﬁned as

c(`) = Z∞

−∞ |%|e+j2π%` d% .

Letting

`=D0sin γ ,

yields

c(D0sin γ) = Z∞

∞|%|e+j2π%D0sin γd% .

Now let

%0=%D0sin γ

γ=%a ,

which yields

c(D0sin γ) = Z∞

∞|%0

a|e+j2π%0γ1

|a|d%0.

Rearranging terms yields

c(D0sin γ) = 1

a2Z∞

∞|%0|e+j2π%0γd%0,

which yields the correct result after substituting the following

a=D0sin γ

γ.

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112 CHAPTER 6: COMPUTED TOMOGRAPHY

IMAGE QUALITY

Solution 6.19

Beam width Whas the effect of convolving projection with rect `

W. Ignoring sampling (at ﬁrst) and pretending

that we don’t know about the distortion, CBP yields

f(x, y) = Zπ

0(gθ(`)∗rect `

W∗c(`)`=xcos θ+ysin θ

dθ

f(x, y) = Zπ

0Z∞

−∞ (gθ(`)∗rect `

W∗c(`)δ(xcos θ+ysin θ−`)d` dθ.

We let gθ(`) = δ(`) = 2-D Radon transform of δ(x, y)to ﬁnd the impulse response. But δ(`)∗rect `

W∗c(`) =

rect `

W∗c(`). Therefore, the impulse response in the inverse 2-D Radon transform of gθ(`) = rect `

W, or the

function h(x, y)whose 2-D Radon transform is rect `

W. The function has support on the disk with diameter W

centered at the origin, but is not constant within, as shown in Figure S6.11.

Figure S6.11 See Problem 6.19.

The easiest way to determine h(x, y)is via the projection-slice theorem. Since

Frect `

W=|W|sinc(W %)

and since all projections are the same, we conclude that

H(%) = |W|sinc(W %) = |W|sin(πW %)

πW % .

H(%)is the radial part of F(h(x, y)). The inverse transform of H(%)also has circular symmetry and is given by

the inverse Hankel transform:

h(r)=2πZ∞

H(%)J0(2π%r)% d% .

where J0is the Bessel function of order 0. From the Hankel transform table we ﬁnd

rect r

2a

√a2−r2←→ sin(2πa%)

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113

Hence

h(r) = 1

πW rect r

W

sW

22

−r2

which gives

h(x, y) = 1

πW rect px2+y2

sW

22

−(x2+y2)

Finally, we conclude that ˜

f(x, y) = f(x, y)∗h(x, y).

Additional comments: h(x, y)is a low-pass ﬁlter since its Fourier transform decays as a sinc in %. Hence, ˜

f(x, y)

is a blurred version of f(x, y)as expected. However, h(x, y)has ﬁnite support, so that the blurring is strictly local-

in fact contributions occur only from over the disk of radius W

2. But h(x, y)approaches asymptotically at r=W

which means that the contribution to blurring at exactly the radius W

2can be very strong and one might expect to

see circular artifacts of radius W

2near bright point objects. In a real system CBP is done for sampled data. The

convolution gθ(`)∗sinc(`/W )is a continuous convolution, however, followed by discrete sampling. Therefore we

might write

f(x, y) = π

j=1

i=1 hgθ(s)∗sinc s

wis=iT C(xcos θj+ysin θj−iT ).

But there is not more we can say analytically about ˆ

f(x, y)versus f(x, y).

Solution 6.20

Assume a rectangular windowed ramp ﬁlter is used. The SNR can be computed using Eq. (6.74). By assumption,

M= 100,C= 0.05,¯µ= 0.15 cm−1. Since the detectors are touching each other, k= 1. Since the cylinder has a

diameter of 20 cm, and the detector dimension is 2.0mm ×2.0mm, the number of measurements per projection is

20 cm

2 mm = 100 .

Hence,

0.1 R/projection ×1

100 projection/measurement = 0.001 R/measurement .

The worst-case intersection length of a beam with the water is 20 cm. Therefore, the worst-case ¯

Nis

N= 2.5×1010 photons

cm2R×0.04 cm2×0.001 R

measuremente−0.15×10

≈50 ×103photons/measurement.

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114 CHAPTER 6: COMPUTED TOMOGRAPHY

Thus,

SNR ≈0.4kC ¯µp¯

NMw

= 0.4×1×0.05 ×0.15 cm−1p50 ×103×100 ×0.2

≈1.3.

Also, SNR = 20 log10 1.3≈2.5 dB, since the SNR is not a power ratio as deﬁned.

Solution 6.21

(a) Since SNR-in-dB = 20 log10 SNR, from assumption we get

SNR = 1020 dB/20 = 10 .

Since SNR =C¯µ

σµ= 10, we have

σµ=0.005

10 ×0.15 cm−1= 1.5×10−5cm−1.

Thus,

σ2

µ= 5.625 ×10−9cm−2=2π2

M¯

Since %0= 1/d, and T=d, we have

σ2

µ=2π2

M¯

Furthermore, since d= 100 cm/D and M=D, then

σ2

µ=2π2D

3(100)2¯

Thus, the photons-per-projection is

Pp=¯

ND =2π2

3(100)2

σ2

=2π2

3(100)2

3002

5.625 ×10−9

= 1.053 ×1010(minimum) .

(b) Photons-per-scan is

Ps=DPp=2π2

3(100)2

5.625 ×10−9.

Since

2.5×1010 photons

cm2R×0.125 m2×10000 cm2

m2×2R = 6.25 ×1013 (maximum) ,

then

D3=(6.25 ×1013)(5.625 ×10−9)(3)(100)2

2π2≈5.34311 ×108.

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115

Then, D≈811.455, and hence Dmax = 811.

Solution 6.22

(a) SNR is given by

SNR = C¯µ

π%−3/2

0r3

2(¯

N/T )M .

Since T=d,d=L/D, and ¯

N=¯

Nf/D, we have

T=¯

d=¯

L/D =¯

Nf/D

L/D =¯

Let

K=C¯µ

πr3

2r¯

then SNR = K%−3/2

0√M. But M= 1.5D, then SNR = √1.5K%−3/2

0D1/2. Since %0= min{d−1, %max}=

min{D/L, %max}, there can be two cases:

When D≤L%max,

SNR = √1.5KD

L−3/2

D1/2=√1.5KL3/2D−1,

or when D≥L%max,

SNR = √1.5K%−3/2

max D1/2.

(b) SNR increases away from L%max in either direction. Thus, either D= 1 or D=Jgives the maximum SNR,

but not D≈L%max. At D= 1, SNR=√1.5KL3/2; at D=J, SNR=√1.5K[J/(2L)]−3/2J1/2. (Note that

L%max =LJ/(2L) = J/2, which lies between 1 and J.) So,

R=√1.5KL3/2

√1.5K(2L)3/2J−1=J

23/2=J

2.8.

Since Jis an image pixel size, R1, hence SNR is biggest at D= 1. SNR may be maximum but resolution

is poor. SNR can be improved at large D’s by lowering %max.

Solution 6.23

(a) Every projection looks the same, like that shown in Figure S6.12(a). Accordingly, the sinogram looks like

that shown in Figure S6.12(b).

(b) The observed sinogram can be modeled as

y(`, θ) = g(`, θ)(1 −rect(`/h))

=g(`, θ)−g(`, θ) rect(`/h),

where his some small distance, equal to the width of a detector. The inverse Radon transform is a linear

operator, so the reconstruction will be

f(x, y) = f(x, y)− R−1{g(`, θ) rect(`/h)}.

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116 CHAPTER 6: COMPUTED TOMOGRAPHY

Figure S6.12 See Problem 6.23.

Assume the width of the detector his small, so that we can approximate the above equation by

f(x, y) = f(x, y)− R−1{g(0, θ) rect(`/h)}.

Now, we need to ﬁnd frect(x, y) = R−1{g(0, θ) rect(`/h)}. All the projections are the same rect function.

The 1-D Fourier transform of any projection is a sinc function, independent of θ:

Grect(%) = F1D{g(0, θ) rect(`/h)}

=hg(0, θ) sinc(h%).

By using the projection slice theorem, the 2-D Fourier transform of frect(x, y)is circularly symmetric. In

this case, frect(x, y)and Grect(%)is related by Hankel transform (Section 2.7). With some abuse of notation,

we have

frect(r) = H−1{Grect(%)}

=H{hg(0, θ) sinc(h%)}

=g(0, θ)

2 rect(r/h)

πp1−4r2/h2.

So the reconstructed image looks like that shown in Figure S6.13. The disk in the center has a diameter h,

the intensity is −frect(r).

Figure S6.13 See Problem 6.23(b).

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117

z(`, θ) = g(`, θ)(1 −rect `−`0

h

=g(`, θ)−g(`, θ) rect `−`0

h.

Again, assume his small, we have

z(`, θ) = g(`, θ)−g0rect `−`0

h,

where g0=g(`0, θ). Let f0

rect =R−1g0rect `−`0

h. So the 1-D Fourier transform of a projection is:

rect(%) = F1D g0rect `−`0

h

=hg0sinc(h%)e−j2π%`0.

The inverse Hankel transform of G0

rect(%)is:

rect(r) = H−1{G0

rect(%)}

= 2πZ∞

hg0sinc(h%)e−j2π%`0J0(2π%r)% d%.

The function f0

rect(r)is a complex function, which means that the sinogram g0rect `−`0

his not a valid

radon transform of a real image. The explicit expression of f0

rect(r)is hard to obtain. Numerical simulation

shows that the reconstructed image will have parts of a circle with radius `0around the image center.

APPLICATIONS AND ADVANCED TOPICS

Solution 6.24

(a) Plugging the form of f(x, y)into the observation equation yields

gi=ZLi

j=1

fjφj(x, y)ds =

j=1

fjZLi

φj(x, y)ds, i = 1,...,m.

Therefore,

Hij =ZLi

φj(x, y)ds .

So, 









=





H11 H12 ··· H1n

H21 H22

....

Hm1Hmn

















(b) Consider each case:

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118 CHAPTER 6: COMPUTED TOMOGRAPHY

(1) H−1exists but v6= 0: Since y=Hf +v,Hf =y−vand H−1Hf =H−1(y−v)or

f=H−1y−H−1v .

This provides a reconstruction formula, but will give a noisy solution.

(2) v= 0 but m<n: We have y=Hf, but there are fewer measurements than unknowns. Therefore,

y=H(f+˜

f)for any ˜

fin the nullspace of H. Hence, there is no unique solution to the inverse

problem.

(3) v= 0 but m>n: We have y=Hf, but there are more measurements than unknowns. If the system is

truly noise-free, then some of thee extra measurements will be redundant. In this case, if H−1exists,

there will be a unique solution.

f= (HTH)−1HTy ,

which is a standard result of least squares minimization from linear algebra.

(d) The image vector has the dimensions 2562×1. The output vector has dimensions (360×512)×1. Therefore,

we will be required to invert a matrix of dimensions 2562×2562, which is too large to solve directly.

Solution 6.25

(a) There are four important points `1,`2,`3, and `4, and three ranges (see Fig. S6.14).

Figure S6.14 See Problem 6.25(a).

•`1≤`≤`2: From similar triangles:

gθ(`) = (`−`1)/(`2−`1)

cos θ

•`2≤`≤`3:

gθ(`) = 1

cos θ

which is independent of `.

•`3≤`≤`4:

gθ(`)(`4−`)/(`4−`3)

cos θ

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119

Rotation by θgives the values:

`1=−1

2cos θ−1

2sin θ ,

`2=−1

2cos θ+1

2sin θ ,

`3=1

2cos θ−1

2sin θ ,

`4=1

2cos θ−1

2sin θ .

(b) A projection is shown in Figure S6.15.

Figure S6.15 See Problem 6.25(b).

−∞

gθ(`)d` =1=Z∞

−∞Z∞

−∞

f(x, y)dx dy

(d) We have g0(`) = g90◦(`), as shown in Figure S6.16(a). Therefore,

Figure S6.16 See Problem 6.25(d) and (e).

fb(x, y) = Zπ

gθ(xcos θ+ysin θ)dθ

fb(x, y) = π

i=1

gθi(xcos θi+ysin θi)

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120 CHAPTER 6: COMPUTED TOMOGRAPHY

Here V= 2; therefore,

fb(x, y) =











π−1/2≤x, y ≤1/2

π/2−1/2≤y≤1/2, x > 1/2

π/2−1/2≤y≤1/2, x < −1/2

π/2−1/2≤x≤1/2, y > 1/2

π/2−1/2≤x≤1/2, y < −1/2

0otherwise

(e) No, it is generally not possible. From the projection slice theorem, we know that, given a ﬁnite number of

slices, there is always some “blank” space between the central slices passing radially through the origin. See

Figure S6.16(c).

Solution 6.26

For the given two energy photons, we have

µ1(100 keV) = 1.0e−1≈0.3679 cm−1,

µ1(140 keV) = 1.0e−1.4≈0.2466 cm−1,

µ2(100 keV) = 2.0e−1≈0.7358 cm−1,

µ2(140 keV) = 2.0e−1.4≈0.4932 cm−1.

(a) The incident intensity of the x-ray burst is

I0= 106×100 keV + 0.5×106×140 keV = 1.7×108photon keV .

(b) For −30 cm ≤x≤ −10 cm, the photons are only attenuated by µ1,

Id(x) = 106×100e−60×0.3679 + 0.5×106×140e−60×0.2466 ≈26.285 photon keV .

For −10 cm ≤x≤10 cm, the photons are attenuated by 40 cm of µ1and 20 cm of µ2,

Id(x) = 106×100e−40×0.3679−20×0.7358 + 0.5×106×140e−40×0.2466−20×0.4932 ≈0.1894 photon keV .

For 10 cm ≤x≤10 cm, the photons are also only attenuated by µ1, as previously,

Id(x)≈26.285 photon keV .

C=Ido −Idb

Idb

=0.1894 −26.285

26.285

≈ −0.9928 .

The projection g(x, 0) = −lnId(x)

I0.Hence,

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121

For −30 cm ≤x≤ −10 cm,

g(x, 0) = −ln( 26.285

1.7×108)≈15.682 .

For −10 cm ≤x≤10 cm,

g(x, 0) = −ln( 0.1894

1.7×108)≈20.615 .

For 10 cm ≤x≤10 cm,

g(x, 0) ≈15.682 .

The local contrast computed by g(x, 0) is

C=go−gb

=20.615 −15.682

15.682

≈0.3146 .

(d) The detector is of ﬁnite length so its response is no longer a delta function; instead, the response is a rect

function. The measured projection I0

d(x)is equal to the previous projection convolved with the detector

response:

d(x) = Id(x)∗rect(x).

Since the detector width is still quite small relative to the object size, the local contrast remains the same in

most parts, but is reduced near x=−10 and x= 10, where the original step transition is blurred to a ramp.

Solution 6.27

(a) θ0= 0,θ1=π/4,θ2= 2π/4 = π/2, and θ3= 3π/4. See Figure S6.17.

Figure S6.17 See Problem 6.27(a).

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122 CHAPTER 6: COMPUTED TOMOGRAPHY

Figure S6.18 See Problem 6.27(b).

(b) See Figure S6.18.

projection slice theorem and by the linearity of the Radon transform. Hence, the fact that G2=G1means

that F2(%cos θj, % sin θj) = 0 for θj,j= 0, . . . , M −1. Then

F2(u, v) = F2{cos 2πfxxcos 2πfyy}

=F1{cos 2πfxx}F1{cos 2πfyy}

4[δ(u−fx) + δ(u+fx)][δ(v−fy) + δ(v+fy)]

4[δ(u−fx, v −fy) + δ(u−fx, v +fy) + δ(u+fx, v −fy) + δ(u+fx, v +fy)] .

But (fx, fy)is a unit vector pointing in the θ= 3π/16 direction. A picture of this 2-D Fourier transform

is shown in Figure S6.19. Since F2(u, v)does not intersect the lines over which we sample the Fourier

Figure S6.19 See Problem 6.27(c).

transform of f1, the value of D4f2will be zero. Hence, G2=G1.

(d) 3π/16 = πi/16 for i= 3 and 3 does not go into 16 without a remainder. Hence M= 16.

(e) No, at least in theory. Sampling the plane with a ﬁnite number of lines will always permit us to place delta

functions in the appropriate spots to deﬁne ghost functions. In practice, functions of inﬁnite extent are not

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123

available. Therefore, F2will always have some spread and a line will hit it. Because of this, “nearly” ghost

functions will have to be high frequency functions in order to “ﬁt between the sampling lines.”

(f) Yes. f2will still be a ghost function. Low-pass ﬁltering a projection does not change the geometry over

which Fis sampled. It would just ﬁlter along the sampled lines.

Solution 6.28

(a) We have

g(`, 0◦) = 









µ1×20 = 2 −30 cm ≤`≤ −10 cm

µ2×20 = 4 −10 cm ≤`≤10 cm

µ3×20 = 6 10 cm ≤`≤30 cm

0otherwise

which is shown in Figure S6.20.

Figure S6.20 See Problem 6.28(a).

(b) We have

g(`, 90◦) = µ1×20 + µ2×20 + µ3×20 = 12,−10 cm ≤`≤10 cm ,

which is shown in Figure S6.21.

Figure S6.21 See Problem 6.28(b).

g(`, 45◦) = 





5`+ 4√2−20√2cm ≤`≤10√2cm

−3

5`+ 12√2 10√2cm ≤`≤20√2cm

0otherwise

which is shown in Figure S6.22.

(d) We have

b45◦(x, y) = g(xcos 45◦+ysin 45◦,45◦)

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124 CHAPTER 6: COMPUTED TOMOGRAPHY

Figure S6.22 See Problem 6.28(c).

b45◦(1,1) = g(1 cos 45◦+ 1 sin 45◦,45◦)

=g(√2,45◦)

5√2+4√2

=21

5√2

≈5.94 ,

which is shown in Figure S6.23.

Figure S6.23 See Problem 6.28(d).

(e) The ﬁeld of view should cover the object in any angle. Thus the smallest possible circular FOV will have the

diameter equal to the diagonal of the object:

d=p202+ 602=√4000 = 63.2cm

r=d

2= 31.6cm .

The geometry is shown in Figure S6.24. We can solve for xas follows

m=p(1.5−r)2−r2=p(1.5−0.316)2−0.3162= 1.14 m

1.5=r

=⇒x=r×1.5

m= 0.415 m

The length of detector array should be 2x= 0.83 m.

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125

Figure S6.24 See Problem 6.28(e).

(f) From the “rule of thumb” of CT, M=D=J= 256. Therefore,

M=63.2cm

256 = 0.25 cm

So, the pixel size is 0.25 cm ×0.25 cm.

Solution 6.29

(a) Since this is a ﬁrst generation CT scanner, collimation is technically not required. However, it is best to

collimate the source to a pencil beam in order that (1) radiation dose to regions not affecting the measurements

is reduced, (2) single Compton scattering events cannot be detected (and thereby contribute to measurement

errors).

(b) A circle with diameter 56.57 cm will contain the square.

will be acquired. The CT “rule of thumb” says that M=D=J. Therefore, since M= 720, there should be

D= 720 line integrals per projection. The reconstructed image should cover the FOV, which was determined

in Part (a) to be a circle with diameter d= 56.57 cm. By the CT “rule of thumb,” J= 720. Therefore, the

pixel size is square with side dimension equal to 565.7mm/720 = 0.78 mm.

(d) The most fundamental expression for SNR given the present scenario is

SNR =C¯µ

π%−3/2

0r3

2(¯

N/T )M .

By problem assumption, we will not violate the “rule of thumb.” Therefore, Tand Mwill remain unchanged.

This still leaves some ﬂexibility in selection of ¯

Nand %0. We could, for example, keep %0unchanged and

quadruple the number of incident x-rays. This would quadruple the number of x-rays ¯

Nincident on the

detector array, ¯

N0= 4 ¯

N .

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126 CHAPTER 6: COMPUTED TOMOGRAPHY

This solution would increase the dose to the patient. We could, on the other hand, keep the number of incident

x-rays constant and use a cutoff frequency equal to

0=1

√4%0= 0.63%0.

This solution would reduce the resolution of the resultant scan.

(e) The sketch of g(`, 45◦)is shown in Figure S6.25(a).

Figure S6.25 See Problem 6.29.

(f) The sketch of b45◦(x, y)is shown in Figure S6.25(b). Assume that (10,10) is in units of cm. Then this point

projects to

`= 10 cos 45◦+ 10 sin 45◦

= 20√2

= 14.14 cm .

This is halfway between the origin and the corner at `= 28.285 cm. Therefore, the projection value will be

1/2 of that at the origin, i.e., b45◦(10,10) = 2.825.

Solution 6.30

(a) We have M=D=Jand %0= 1/d. Therefore, k= 1. In order to resolve two point source separated by

1 mm, the pixel size can not be bigger than 0.707 mm. See Figure S6.26. We also have 60 cm

0.707 mm/pixel =

848.6pixels, so the minimum number of pixels is 849. See Figure S6.27.

(b) The detector length is 925 ×0.8mm = 740 mm, and 180 = d+ 30 ⇒d= 150 cm.

tan θ=37

180 ⇒θ= 11.6156◦

sin θ=x

150 ⇒x= 30.2cm.

Since x= 30.2cm, the circular FOV with radius 30 cm ﬁts.

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127

Figure S6.26 See Problem 6.30(a).

Figure S6.27 See Problem 6.30(b).

SNR = 0.4kC ¯µLD−3/2q¯

Nfm

k= 1 since %0= 1/d

C=0.25 −0.2

0.2= 0.25

¯µ= 0.2cm−1

L= 74 cm

D= 925

Nf= 1.5×1011

m= 925 .

Plugging these numbers in yields

SNR = 619.677

SNR(dB) = 20 log10 619.677 = 55.84 dB .

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128 CHAPTER 6: COMPUTED TOMOGRAPHY

Solution 6.31

(a) The radius of the circular FOV is 25 cm. Therefore, a right triangle can be formed for which the hypotenuse

is 70 cm and the side opposite 1/2 of the fan angle is 25 cm. Therefore,

θ1/2= sin−125

70 = 20.925◦.

The fan angle is therefore 2×20.925 = 41.85◦.

(b) The distance between source and detector is Q= 105 cm and the total circumference of a circle with radius

Qis C= 2πQ = 2π×105 cm = 659.7cm. Therefore, the arclength of the detector array is

L=41.85◦

360◦659.7cm = 76.69 cm .

Since there are 703 detectors over this range, we have that the spacing between detectors is

∆d=766.9mm

703 = 1.09 mm .

increment is therefore ∆θ= 360◦/1000 = 0.36◦. Since each line will pass through the origin, the value of

the lateral position of each of these lines is `= 0. Therefore, the following line integrals are acquired:

g(0,0.36◦m), m = 0,...,999 .

The acquired data are shown in Figure S6.28(a). One half of the lines are repeated.

Figure S6.28 See Problem 6.31.

(d) From the work we did in Part (a), we see that the starting angle is −20.925◦and the lateral displacement is

−25 cm. The angles increment exactly as in Part (c) and the lateral displacement never changes. Therefore,

the following line integrals are acquired

g(−25,−20.925◦+ 0.36◦m), m = 0,...,999 .

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129

These are at the limits of the lateral displacement in the sinogram. But, rather than repeating, these acquire

both the positive and negative max displacements, as shown in Figure S6.28(b).

(e) Cycle through each of the detectors in order from left to right. The sinogram will be ﬁlled in vertical columns

as shown in Figure S6.28(b). When scanning the central detector, turn off the tube for the second half of the

rotation to avoid redundancy.

(f) This scanner has a fanbeam source collimation, but only acquires information from one detector at a time.

Therefore, the patient is getting irradiated repeatedly in a slice, and the vast majority of that radiation is not

getting used to image the slice. This is like having 703 CT scans, just to obtain one image.

Solution 6.32

(a) Use the following steps:

g(`, θ) = Rδ(x, y)

=Z∞

−∞ Z∞

−∞

δ(x, y)δ(xcos θ+ysin θ−`)dxdy

=δ(xcos θ+ysin θ−`)|x=0, y=0

=δ(−`) = δ(`)δ(`)is an even function .

The sinogram is shown in Figure S6.29.

Figure S6.29 The Radon transform of δ(x, y). See Problem 6.32.

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130 CHAPTER 6: COMPUTED TOMOGRAPHY

(b) The shift theorem is proven as follows:

Rf(x−x0, y −y0) = Z∞

−∞ Z∞

−∞

f(x−x0, y −y0)δ(xcos θ+ysin θ−`)dxdy

Let ξ=x−x0, η =y−y0

(x=ξ+x0, y =η+y0, dx =dξ, dy =dη).

Rf(x−x0, y −y0) = Z∞

−∞ Z∞

−∞

f(ξ, η)δ((ξ+x0) cos θ+ (η+y0) sin θ−`)dξdη

=Z∞

−∞ Z∞

−∞

f(ξ, η)δ(ξcos θ+ηsin θ−(`−x0cos θ−y0sin θ))dξdη

=g(`−x0cos θ−y0sin θ, θ).

Rδ(x−1, y) = δ(`−cos θ).

The trajectory is plotted in Figure S6.30.

Figure S6.30 The Radon transform of δ(x−1, y). See Problem 6.32.

(d) The acquired sinogram is shown in Fig. S6.31.

Figure S6.31 Acquired sinogram. See Problem 6.32.

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131

(e)

Z∞

−∞

`g(`, θ)d` =Z∞

−∞

`Z∞

−∞ Z∞

−∞

f(x, y)δ(xcos θ+ysin θ−`)dxdyd`

=Z∞

−∞ Z∞

−∞

f(x, y)Z∞

−∞

`δ(xcos θ+ysin θ−`)d`dxdy

=Z∞

−∞ Z∞

−∞

f(x, y)(xcos θ+ysin θ)dxdy

=Z∞

−∞ Z∞

−∞

f(x, y)xcos θdxdy +Z∞

−∞ Z∞

−∞

f(x, y)ysin θdxdy

= cos θZ∞

−∞ Z∞

−∞

f(x, y)xdxdy + sin θZ∞

−∞ Z∞

−∞

f(x, y)ydxdy

=qxcos θ+qysin θ .

(f) The sinogram acquired can be expressed as

g(`, θ) = δ(`) 0 ≤θ≤π/2

δ(`−cos θ)π/2< θ ≤π

Let us calculate the ﬁrst moment of each projection:

Z∞

−∞

`g(`, θ)d` =(R∞

−∞ `δ(`)d` 0≤θ≤π/2

R∞

−∞ `δ(`−cos θ)d` π/2< θ ≤π

=0 0 ≤θ≤π/2

cos θ π/2< θ ≤π

From the results in part (e) we have

qxcos θ+qysin θ= 0 for 0≤θ≤π/2⇒qx=qy= 0

qxcos θ+qysin θ= cos θfor π/2< θ ≤π⇒qx= 1, qy= 0

Since qxand qyare quantities calculated from f(x, y), they do not depend on θ. The above two results

contradict. So the acquired sinogram cannot be the Radon transform of any object.

Solution 6.33

(a) The energy spectrum of the x-ray beam after it passes through the material is shown in Figure S6.32

(b) The two measurements should ideally be identical because the basic measurement of CT is the line integral

of the linear attenuation coefﬁcient. If the x-ray was perfectly monochromatic, then both the measurements

will be exactly the same, since the line integrals in θ= 90◦projection and θ= 270◦projection are the same.

But in practice, we have polychromatic x-ray source and because of beam hardening, the effective energy of

the x-ray beam and hence the linear attenuation coefﬁcient is different for different projection. So the two

measurements are different in practice.

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132 CHAPTER 6: COMPUTED TOMOGRAPHY

Figure S6.32 See Problem 6.33(a).

(d) gL

t1=line integral of µat lower photon energy,

t1=line integral of µat higher photon energy.

At lower energy, µis larger and therefore the line integral is larger. At higher energy, µis smaller and

therefore the line integral is smaller.

(e)

µ(Al,75 keV)=0.7cm−1µ(H2O,75 keV) = 0.1866 cm−1.

g75

t1= 0.7×2+0.1866 ×8=2.8928,

g75

t2= 0.7×8+0.1866 ×2=5.9732.

(f) The set of linear equations we get is

2.8928 = aL(3.16) + aH(2.2) ,

5.9732 = aL(6.79) + aH(4.3) .

Solving the equations, we have

aL= 0.52, aH= 0.568 .

(g)

µ(object,75 keV) = Zπ

00.52gL(`, θ)+0.568gH(`, θ)∗˜c(`)dθ .

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The Physics of Nuclear Medicine

FUNDAMENTALS OF ATOMS

Solution 7.1

The mass of an electron meis 0.000548 u. So 1u = 1

0.000548 me. The equivalent energy of an electron is 511 keV.

So the equivalent energy of 1 u is 1

0.000548 ×511 keV = 931 MeV.

Solution 7.2

The mass defect of a deuteron is 1.007276+1.008665−2.01355 = 0.002391u. Its binding energy is 0.002391 u×

931 MeV/u= 2.228 MeV.

RADIOACTIVE DECAY AND ITS STATISTICS

Solution 7.3

The PMF of a Poisson distribution with parameter ais given by

Pr[N=k] = ake−a

k!.

Its mean is given by

µN=∞

k=0

kPr[N=k]

=∞

k=0

kake−a

k!=∞

k=1

ake−a

(k−1)!

=a∞

k=1

a(k−1)e−a

(k−1)! =a∞

k=0

ake−a

=a .

133

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134 CHAPTER 7: THE PHYSICS OF NUCLEAR MEDICINE

The variance is

σ2

N=∞

k=0

k2Pr[N=k]−a2because σ2=E[X2]−(E[X])2.

Evaluate the summation as follows:

∞

k=0

k2Pr[N=k] = a∞

k=1

ka(k−1)e−a

(k−1)!

=a"1 + ∞

k=2

(k−1)a(k−1)e−a

(k−1)! #

=a[1 + a]

=a+a2.

So the variance of a Poisson random variable with parameter ais

σ2

N=a .

Solution 7.4

(a) Using (7.8), the decay constant λis found as

λ=0.693

T1/2

=0.693

13 ×3600 sec ≈1.4808 ×10−5sec−1.

The radioactivity Ais then

A=λN = 1.4808 ×10−5×109= 1.4808 ×104dps.

(b) Since Nt=N0e−λt, then

N24 h= 109×exp(−1.4808 ×10−5×24 ×3600) ≈2.78 ×108atoms.

large mean value, the Poisson distribution can be well approximated by a Gaussian distribution with the same

mean and variance.

Thus,

PN=108=1

√2π×2.78 ×108exp −(108−2.78 ×108)2

2×2.78 ×108≈0.

Solution 7.5

At t= 0, the number of technetium-99m atoms is 1×1012. Since the half-life of technetium-99m is 6 hours

(Table 7.1), the decay constant is

λ=0.693

t1/2

=0.693

6×3600 sec = 3.21 ×10−5sec−1.

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135

The radioactivity at t= 0 is

A0= 1 ×1012 ×3.2×10−5sec−1= 3.2×107Bq = 0.86mCi .

The intensity measured is:

I0= 8.91 ×109keV

sec ·m2.

One hour later, the radioactivity becomes

A1=A0e−λt =A0e−0.693/6= 0.89A0.

So the intensity measured at t= 1 hour is

I1= 7.94 ×109keV

sec m2.

Solution 7.6

(a) A0=1 Ci= 3.7×1010Bq and At=A0e−λt = 1Bq. So

e−λt =1

3.7×1010 = 2.7×10−11 ,

which is solved as

−λt = ln 2.7×10−11=−24.334

=⇒t=24.334

Since T1/2=0.693

λ=τ, we have λ=0.693

τ, and t= 35.114τ. It takes t= 35.114τfor a radioactive sample

with activity 1 Ci to decay to activity 1 Bq if the half-life is τ.

(b) The radioactive tracers used in nuclear medicine should have a half-life on the order of minutes to hours,

about the time it takes to perform study. If longer, activity remains in patient. If shorter, activity disappears

before scan is completed.

Solution 7.7

(a) The radioactive source decays according to

At=A0e−λt .

The intensity at range rfrom this source is

It=AtE

4πr2,

where the time-dependency is made explicit using a subscript tand Eis the gamma-ray energy. A point

(x, y)on the detector is at a distance

r=pR2+x2+y2

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136 CHAPTER 7: THE PHYSICS OF NUCLEAR MEDICINE

from the source. Therefore, the intensity on the detector face is

It(x, y) = AtE

4π(R2+x2+y2),|x|,|y|< D/2.

(b) The average intensity is

Iav

t=1

D2ZD/2

−D/2ZD/2

−D/2

It(x, y)dx dy

D2ZD/2

−D/2ZD/2

−D/2

AtE

4π(R2+x2+y2)dx dy

≈AtE

4πR2,

where the last approximation holds if RD.

Solution 7.8

(a) DF is deﬁned as DF =e−λt. And decay constant λis given by

A1/2

2=e−λT1/2.

Taking the natural logarithm of the above equation yields −λT1/2=−ln 2 = −0.693, and λ=0.693

T1/2. So

the decay factor is

DF =e−0.693t/T1/2.

(b) From above, we have τ=1

λ=T1/2

0.693 = 1.443T1/2.

Solution 7.9

(a) The half-life of 99mTc is 6 hours. It is 8 hours from 8 a.m. to 4 p.m. Therefore, using the relation between

the decay constant and the half-life,

λ=0.693

T1/2

we can write

A4p.m.=A8a.m.e−λt = 2e−0.693×8/6≈0.7939 mCi/ml .

(b) To get 1.5 mCi radioactivity, we need a volume of

V=1.5mCi

0.7939 mCi/ml ≈1.89 ml .

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137

Solution 7.10

(a) First we ﬁnd the decay constant as follows

Nt=N0e−λt

9.9212 ×106= 108e−λ864000

λ=−ln((9.9212 ×106)/108)/864000

λ= 2.6742 ×10−6sec−1.

Using the relationship between the half-life and the decay constant, we ﬁnd

t1/2= ln(2)/λ

= 259198 ≈259200 sec (or 3 days)

(b) For ∆t << t1/2, the average number of disintegrations = Poisson rate ×∆t=N0λ∆t= 2.6742 disintegra-

tions.

Prob(∆N > 2) = 1 −Prob(∆N= 2) −Prob(∆N= 1) −Prob(∆N= 0)

= 1 −(a)2e−a

2! −(a)1e−a

1! −(a)0e−a

= 1 −(a2

2+a+ 1)e−a

= 1 −(2.67422

2+ 2.6742 + 1)e−2.6742

= 0.50003 .

Solution 7.11

Determine the decay constant of 21

11Ms as follows:

1/2 = eλt1/2,

t1/2= 2 hours ,

λ=ln(1/2)

−2= 0.347 hr−1.

Determine the amount of 21

11Ms left at 5 pm as follows:

∆t= 4 hours ,

N=N0eλt

= 8 g ×e−0.347×4

= 2 g .

Subtract to determine the amount that has decayed:

8 g −2 g = 6 g .

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138 CHAPTER 7: THE PHYSICS OF NUCLEAR MEDICINE

Solution 7.12

(a) First determine the decay constant:

At=A0e−λt ,

1 mCi/ml = 3 mCi/ml e−λ3600 ,

λ=−ln 1

3×1

3600 = 3.05 ×10−4s−1.

Then ﬁnd the half-life:

t1/2=ln 2

λ=ln 2

3.05 ×10−4s−1= 2271.3 s = 0.63 h = 37.86 min .

(b) Compute the radioactivity:

At= 3 mCi/ml e−λ×4×3600

= 3 mCi/ml e−3.05×10−4s−1×4×3600s

= 0.6371 mCi/ml .

V=1.5 mCi

0.6371 mCi/ml = 2.3544 ml .

RADIOTRACERS

Solution 7.13

(a) Explanation for each:

(i) Eγ= 30 Kev, t1/2= 7 hours: This is a bad choice for medical imaging because the energy of the gamma

rays are low and the body will absorb most of the emitted gamma rays.

(ii) Eγ= 150 Kev, t1/2= 5 hours: This is a good choice for medical imaging purposes because its half-

life is long enough to enable imaging and short enough to weaken strongly before the patient leaves the

hospital. The gamma ray energy is high so that it is somewhat transparent in the body but still detectable by

conventional detectors.

(iii) Eγ= 200 Kev, t1/2= 10 days: The energy would be a pretty good choice for this one. The half-life

would be good for biological processes that take a week or so for the radiotracer to reach its destination. It is

too long, however, for most processes.

(b) Activity follows the radioactive decay law. If activity reduces to 1/4 after 5 hours then 5 hours is twice the

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139

half life. Accordingly,

t1/2= 2.5 hours ,

λ=0.693

t1/2

= 7.7×10−5s−1,

N0=A0

λ=4.4×1010

7.7×10−5= 5.19 ×1014 .

Solution 7.14

A radiotracer is chosen ﬁrst for its properties of biodistribution, and then its physical imaging properties. The

two radiotracers are not equivalent if they distribute in the body in different ways and most likely they cannot be

interchanged.

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Planar Scintigraphy

INSTRUMENTATION

Solution 8.1

(a) For the diagrams of an Anger gamma camera, see Figures 8.1, 8.2, 8.3, and 8.4. An Anger gamma camera

consists of a multi-hole lead collimator, a sodium iodide scintillation crystal, an array of PMTs on the crystal,

a positioning logic network, a pulse height analyzer, a gating circuit, and a computer. The functions of each

of these parts are:

The collimator provides an interface between the patient and the scintillation crystal, by allowing only those

photons traveling in an appropriate direction (i.e., those that can pass through the holes without being ab-

sorbed in the lead) to interact with the crystal;

The scintillation crystal emits light photons after deposition of energy in the crystal of ionizing radiation;

The photomultiplier tubes do two things: converting light signals into electrical signals and amplifying these

signals;

The positioning logic network determines both where the event occurred on the face of the crystal and the

combined output of all the tubes, which represents the light output of the crystal (which in turn represents

the energy deposited by the gamma photon). These output signals are denoted as Xand Yfor the estimated

two-dimensional position of the event and Zfor the total light output. The amplitude of a given tube’s output

is directly proportional to the amount of light (number of scintillation photons) its photocathode receives.

The tubes closest to the scintillation event will have the largest output pulses, while those farther away will

have smaller output pulses. By analyzing the spatial distribution of pulse heights, the location of a single

scintillation event (X, Y )can be determined quite accurately.

The pulse height analyzer is used to distinguish photons been Compton scattered from those are not by an-

alyzing the energy deposited in the crystal via the Z-pulse whose height is proportional to the total energy

deposited in the crystal. The pulse height analyzer is used to set an acceptable window around the photopeak

in the spectrum of the Z-pulse.

The gating circuit is used to compensate for the imperfect photopeak localization and further reduce the

scattered photons being accepted as a valid event.

The computer is used to record the location of each event and form images.

(b) When we select radionuclides in nuclear medicine, the following issue must be considered:

140

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141

•The radionuclides must be “clean” gamma ray emitters, which means that they do not emit alpha or beta

particles.

•The radionuclides must emit gamma rays with appropriate energy. The energy cannot be too low because

low-energy gamma rays are more likely to be absorbed by the body; therefore, increase patient dose

without contributing to the images. Also, the energy cannot be too high since high-energy gamma rays

are less likely to be detected.

•The radionuclides should have a half-life on the order of minutes to hours.

•The radionuclides should be useful and safe to trace in the body.

•The radionuclides should emit gamma rays as monochromatic as possible.

Solution 8.2

(a) Use Beer’s law for calculating the path length win the septa to allow less than 60% incident photons to

pass through. If µis the linear attenuation coefﬁcient for lead at 140 keV, then e−µw ≤0.60. This gives

w≥0.51/µ. From geometry, the collimator septa thickness his related to path length wfor gamma-rays

incident at 45◦by h=w/√2=0.36/µ.

(b) Here, using the above two equations to ﬁnd dand , the values of lare found by the roots of a quadratic

equation to be l= 8 mm, 30 mm. Choose l= 30 mm, then d= 0.22 mm.

Sensitivity decreases too, as less photons reach the detector. Also collimators with large lmay be heavy.

(d) Increasing the thickness of the scintillator will increase the sensitivity and compensate to some extent its

decrease due to long holes. The disadvantage of increasing crystal thickness is that the intrinsic resolution of

the crystal degrades.

Solution 8.3

(a) Note that 20% pulse-height window is 10% on either side.

150 keV ×0.1 = 15 keV ,

150 keV −15 keV = 135 keV .

Since

hν0=hν

1 + hν

m0c2(1 −cos θ),

we have

135 keV =140 keV

1 + 140 keV

511 keV (1 −cos θ).

Solving for θ, we get θ= 30.14◦.

(b) For a window centered at the photopeak, the maximum acceptable scattering angle for a 140 keV photon is

53.54◦, as shown in Example 8.2. Do a similar computation, we can see that photons with energy hν =

364 keV can be scattered by an angle θ= 32.43◦and still be accepted by a 20% window centered at the

photopeak.

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142 CHAPTER 8: PLANAR SCINTIGRAPHY

(b), we can see that an offset window centered at a higher energy can further reduce scatter.

Solution 8.4

(a) The intensity at radius rfrom the source is

I=AE

4πr2,

where Eis the gamma ray photon energy.

(b) Hole size does not matter since “per unit area” is already ﬁgured into the intensity. Therefore, the intensity is

the same as in part (a).

I=AE

4π(2r)2=AE

16πr2.

This is 1/4 the intensity of part (a).

Solution 8.5

The septal thickness of a collimator depends on the minimum required path length for adequate attenuation. That

is, the septa must be thick enough that photons traveling through them have a high probability of being absorbed.

From a geometric point of view, we can deﬁne a minimum septal thickness from a minimum path length as

h=2dw

l−w.(S8.1)

If septal penetration is to be less than 5%, the transmission factor from Beer’s Law [(4.24)]for the minimum path

length is:

e−µw ≤0.05 .(S8.2)

We note that e−3≈0.05, so this implies µw ≥3. We can thus substitute this deﬁnition into (S8.1) for septal

thickness:

h≥6d

µl −3.(S8.3)

The µfor lead at 140 keV is 21.43 cm−1. For comparison, at 511 keV, it is 1.746 cm−1.

Solution 8.6

(a) Energy of 30◦Compton scattered 140 keV photon.

E0=140 Kev

1 + (1 −cos(30◦))140 Kev/511 Kev = 135.04 Kev (S8.4)

Acceptance window = 2140−135.04

140 100 = 7.08%.

(b) See Figure S8.1.

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143

Figure S8.1 Plot of response. See Problem 8.6(b).

should not arrive while the response from the ﬁrst is at 80%. Time for the response to fall to 80% is t=

20/140 ×140 ×0.2=4ns.

The acceptance window is 20% this means that second photon should not arrive while the response from the

ﬁrst photon is above 10%. Otherwise the net response will be over 110% and the acceptance window will

reject the second photon. Time for the response to fall to 10% is t= 20/140 ×140 ×0.9 = 18 ns.

This means that arrival of two photons should be 18 ns apart so that both the photons are accepted as separate

event.

(d)

Probability of at least one disintegration = 1 −Probability of no disintegration

= 1 −e−λN0∆t

= 1 −e−A0∆t.

Plug in the known numbers as follows

0.5=1−e−0.25×A×3.7×1010dps×18×10−10 s,

and solve the equation to get A= 0.0416 Ci = 41.6mCi.

(e) The height of the Z-pulse is 80 + 30 + 20 + 5 = 135. Find the center of mass as follows:

xlocation = (−1.5×80 + 1.5×30 + (−1.5) ×20 + 1.5×5)/135 = −0.722 cm ,

ylocation = (1.5×80 + 1.5×30 + (−1.5) ×20 + (−1.5) ×5)/135 = 0.9442 cm .

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144 CHAPTER 8: PLANAR SCINTIGRAPHY

IMAGE FORMATION

Solution 8.7

Refer to Figure S8.2.

Figure S8.2 Converging and diverging collimators. See Problem 8.7.

(a) Let the converging collimator have a focal point located at A, which is at a distance Don the left side of

the detector, as shown in the Figure S8.2(a). Let the coordinate system be such that the origin is located

at the focal point A. Consider a point Bon the detector at a distance d, as shown and at an angle θ. The

coordinates of this point Bare (dcos θ, d sin θ, D). The photons reaching this point will travel along a line

passing through the focal point Aand the point on the detector B. Hence, these photons will experience

an attenuation obtained by integrating the linear attenuation coefﬁcient µ(x, y, z)along this line. Consider

a point Pon this line, at a distance z0from the origin, as shown in the ﬁgure. The coordinates of Pare

(rcos θ, r sin θ, z0) = (z0dcos θ/D, z0dsin θ/D, z0). Hence the intensity at B, due to an event occurring at

a location z, is given as:

Id=AE

4πl2exp (−ZD

µ(z0dcos θ/D, z0dsin θ/D, z0)dz0),

where, l=p(zd cos θ/D −dcos θ)2+ (zd sin θ/D −dsin θ)2+ (z−D)2. This is the intensity due to a

single event occurring at a depth z, along the line. Integrating the activity over all possible events along the

line, we get

I(dcos θ, d sin θ) = ZD

−∞

A(zd cos θ/D, zd sin θ/D, z)E

4πl2

exp (−ZD

µ(z0dcos θ/D, z0dsin θ/D, z0)dz0)dz .

(b) Let the diverging collimator have a focal point located at A, which is at a distance Don the right side of

the detector as shown in Figure S8.2(b). Let the coordinate system be such that the origin is located at

the focal point A. Consider a point Bon the detector at a distance d, as shown and at an angle θ. The

coordinates of this point Bare (dcos θ, d sin θ, D). The photons reaching this point will travel along a line

colinear with the focal point Aand the point on the detector B. Hence, these photons will experience an

attenuation obtained by integrating the linear attenuation coefﬁcient µ(x, y, z)along this line. Consider a

point Pon this line, at a distance z0from the origin, as shown in the ﬁgure. The coordinates of Pare

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145

(rcos θ, r sin θ, z0) = (z0dcos θ/D, z0dsin θ/D, z0). Hence the intensity at B, due to an event occurring at

a location z, is given as:

Id=AE

4πl2exp (−Z−D

µ(z0dcos θ/D, z0dsin θ/D, z0)dz0).

This is the intensity due to a single event occurring at a depth z, along the line. Integrating the activity over

all possible events along the line, we get

I(dcos θ, d sin θ) = Z−D

−∞

A(zd cos θ/D, zd sin θ/D, z)E

4πl2

exp (−Z−D

µ(z0dcos θ/D, z0dsin θ/D, z0)dz0)dz .

Solution 8.8

(a) By simple computation, we have the outputs of the PMTs are:

a1= 21.10 a2= 21.10 a3= 12.13

a4= 21.10 a5= 21.10 a6= 12.13

a7= 12.13 a8= 12.13 a9= 8.13 .

(b) The Z-pulse is

i=1

ai= 141.05 .

The estimated position is

X=1

i=1

aixi=−0.16 cm ,

Y=1

i=1

aiyi= 0.16 cm .

event position estimation uses a linear model, while (P8.1) is nonlinear.

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146 CHAPTER 8: PLANAR SCINTIGRAPHY

Solution 8.9

(a) From Figure P8.3(a), when pulse height is 180, the energy deposited is 160 keV. Therefore, we have

hν0=hν

1 + hν

mc2

0(1 −cos θ)

=160

1 + 160

511 (1 −cos 50◦)

= 143.9keV .

Suppose the acceptance window is centered at the photo peak at 160 keV, the upper bound of the energy

window is 160 + (160 −143.9) = 176.1keV. So the acceptance window can be set to be 143.9−176.1keV.

(b) The Z-pulse = 40 + 5 + 15 + 15 + 20 + 45 + 30 = 170. The corresponding energy deposited is 150 keV. It

will be accepted by the acceptance window.

tube 1 2 3 4 5 6 7

(x, y)in mm (0,0) (−2,0) (−1,√3) (1,√3) (2,0) (1,−√3) (−1,−√3)

pulse height 40 5 15 15 20 45 30

and

X=1

k=1

akxk= 0.26 mm, Y =1

k=1

akyk=−0.46 mm

(d) If (X, Y )is set equal to the location of the PMT that has the largest amplitude, this will give a less accurate

estimation of the location of an event. The resolution of the resulting image will be on the order of the size

of tubes.

Solution 8.10

(a) See Figure S8.3.

(b) Note that a 10 percent pulse height window is 5 percent on either side. So the lowest energy that can be

accepted is 140 keV ×(1 −0.05) = 133 KeV.

hv0=hv

1 + hv

m0c2(1 −cos θ)) ,

133 KeV = 140 keV

1 + 140 keV

511 keV (1 −cos θ)) .

And we get θ= 36.11◦.

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147

Figure S8.3 Pulse height spectrum. See Problem 8.10(a).

(d) The position of the event (X, Y ) = (0.91,−1.14) is

X=1

ZPxkak=0×5+2×15+4×25−2×10+0×20+2×45−4×5−2×10+0×40

175 =160

175 = 0.91 ,

Y=1

ZPykak=4×5+2×15+0×25+2×10+0×20−2×45+0×5−2×10−4×40

175 =−200

175 =−1.14 .

(e) Causes of event localization error include edge effects, badly calibrated PMTs, and gamma rays passing

through septa (scattering).

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148 CHAPTER 8: PLANAR SCINTIGRAPHY

IMAGE QUALITY

Solution 8.11

(a) The overall system response is given by the convolution of the system responses of all its subsystems. In

this case, it is the convolution of three rect functions. By assumption, fI(x) = arect(x/rI),fC(x) =

brect(x/rC), and fP(x) = crect(x/rP), where a, b, c denote the individual amplitude (actually, they can be

assumed to be one when computing FWHM). Let’s assume rI≤rC≤rPand rI≤rP−rC.

fCP (x) = fC(x)∗fP(x) = Z∞

−∞

bc rect(x−t

) rect( t

)dt

=bc ZrP/2

−rP/2

rect(x−t

)dt

=









bc(x+rC+rP

2)if −rP+rC

2≤x≤ −rP−rC

bcrCif −rP−rC

2≤x≤rP−rC

bc(rC+rP

2−x)if rP−rC

2≤x≤rP+rC

0otherwise.

Then, ftotal(x) = fI(x)∗fC(x)∗fP(x) = fI(x)∗fCP (x)can be similarly computed to be: ftotal(x) =











abc

2x+rI+rC+rP

22if −rP+rC+rI

2≤x≤ −rP+rC−rI

abcrIx+rC+rP

2if −rP+rC−rI

2≤x≤ −rP−rC+rI

abc −x2

2−rP−rC−rI

2x+(rP−rC−rI)(rI−3rC−rP)

8+rC(rP−rC+rI)

2

if −rP−rC+rI

2≤x≤ −rP−rC−rI

abcrIrCif −rP−rC−rI

2≤x≤rP−rC−rI

abc −x2

2+rP−rC−rI

2x+(rP−rC−rI)(rI−3rC−rP)

8+rC(rP−rC+rI)

2

if rP−rC−rI

2≤x≤rP−rC+rI

abcrI−x+rC+rP

2if rP−rC+rI

2≤x≤rP+rC−rI

abc

2x+rI+rC+rP

22if rP+rC−rI

2≤x≤rP+rC+rI

0otherwise.

The maximum value of ftotal(x)is abcrIrC. Thus, we need to solve for x0such that ftotal(x0) = abcrIrC

which can be easily computed to be x0=±rP

2. Hence the FWHM of the overall system is equal to rP,

which is the largest width of the three sub-systems.

(b) In the case of Gaussian cascade we know that

FWHMtotal =qFWHM2

I+FWHM2

C+FWHM2

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149

and for each subsystem the FWHM is equal to 2√2 ln 2σ·. Hence,

FWHMtotal = 2√2 ln 2qσ2

I+σ2

C+σ2

Solution 8.12

(a) The half-life of technetium-99m is 6h= 360 min. Therefore the radioactive decay formula is

A=A0e−t/360 .

Our images will be acquired over 2h= 120 min. There are 6 images per hour for 2 hours, which makes 12

images total.The counts in the last image are the integration of the activity over the interval 110 < t < 120:

N12 =Z120

110

A0e−t/360

=A0(−360)e−t/360

120

110

=−360A0(e−120/360 −e−110/360)

= 7.265759A0.

This is then solved for A0

A0=2,000,000

7.265759 = 275,263 counts/min .

(b) The total count in image nis

Nn=Z10n

10(n−1)

A0e−t/τ dt

=A0(−τ)e−t/τ 

10n

10(n−1)

=A0τ(e−10(n−1)/τ −e−10n/τ ).

The count per pixels is

n=Nn

J2.

and the SNR per pixel is

SNRp=qNp

=√Nn

=pA0τ(e−10(n−1)/τ −e−10n/τ )

=√275,263 ×360

Jpe−10(n−1)/τ −e−10n/τ

= 77pe−10(n−1)/τ −e−10n/τ .

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150 CHAPTER 8: PLANAR SCINTIGRAPHY

This is calculated for n= 1,...,12, yielding 12.74, 12.57, 12.40, 12.22, 12.06, 11.89, 11.72, 11.56, 11.40,

11.25, 11.09, 10.94.

SNRl=Cp¯

Nb.

In decibels this is

SNRl(dB) = 20 log10 Cp¯

Nb= 5 dB .

Therefore,

log10 Cp¯

Nb= 5/20 ,

Cp¯

Nb= 1.778 ,

p¯

Nb= 17.78 ,

Nb= 316 .

This implies that there must be approximately 5 M counts in the last image. From the result in part (a), we

can deduce that there are approximately 6.8 M counts in the ﬁrst image.

Solution 8.13

(a) The counting rate is at most 128K dps (disintegration per second). Each frame last for 75 ms during each

heart beat, and there are 64 ×64 = 4,096 pixels on each frame. So, during one heart beat, each pixel can get

at most 128,000 dps ×0.075 s

4,096 = 2.34 disintegration

pixel ·heartbeat .

In order to get the required counts, we need

N=1,000

2.34 = 427 heartbeat .

(b) The heart rate is 50 bpm, so the study will take

T=427

50 = 8.54 minutes .

SNR =p1,000 = 31.62 .

(d) If we want to double the SNR, we need to have 4,000 counts per pixel for each frame. Therefore, the study

will be 4 times as long as the one described in parts (a) and (b). The time it takes is

T2= 4 ×T= 34.16 minutes .

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151

Solution 8.14

Consider a source positioned at distance rfrom the collimator (as in Figure S8.4). Because of the collimator’s

geometry, this source will only be “seen” by the scintillation crystal over a certain (horizontal) extent. We will take

1/2 of this range to be the collimator resolution, Rc. This result would be exactly the FWHM if the response to the

point source were a triangle function—a bold assumption, and one that is necessary for this geometric derivation.

By similar triangles, we have

Figure S8.4 See Problem 8.14.

l=Rc

l+b+r.

Rearranging yields the desired result.

Solution 8.15

(a) The radioactivities of A and B are

t=NA

10 ln 2

1/2!exp{−tln 2/tA

1/2}

=N0

10 ln 2

3exp{−tln 2/3},

t=NB

10 ln 2

1/2!exp{−tln 2/tB

1/2}

=N0

20 ln 2

6exp{−tln 2/6}.

The projection is

φ(x, t) = AA

trect x+ 5

10 +AB

trect x−5

10 eµRlR.

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152 CHAPTER 8: PLANAR SCINTIGRAPHY

At time t= 0, we have

φ(x, 0) = N0

10 ln 2

3rect x+ 5

10 +N0

20 ln 2

6rect x−5

10 e−1

= 8.5×10−3N0rect x+ 5

10 + 2.12 ×10−3N0rect x−5

10 .

At time t= 3 hours, we have

φ(x, 3) = N0

10 ln 2

3e−ln 2rect x+ 5

10 +N0

20 ln 2

6e−ln 2rect x−5

10 e−1

= 4.25 ×10−3N0rect x+ 5

10 + 1.5×10−3N0rect x−5

10 .

(b) tmax = 0.

(c)

=Kd2

le(d+h)2

=0.25 ×4

35 ×2.22

= 1.69 ×10−4,

RC=d

l(l+b+|z|) = 2

35(35 + b+ 110) = 2

35(b+ 145) .

(d) φd= [4.25 ×10−3−1.50 ×10−3]×1.69 ×10−4= 4.6×10−7.

(e) Width(P) = 10 + Rc.

Solution 8.16

(a) We have

RC=d

l(l+b+|z|) = 3

100(10 + 2.5 + 50) = 1.875 cm .

(b) We have

RC= 18.75 mm = 2σc√2 ln 2 ⇒σc=18.75

2√2 ln 2 = 7.96 .

Assuming that the PSF is Gaussian, then

hC=e−x2

2σ2

Similarly,

RI= 0.2 mm ⇒σI=0.2

2√2 ln 2 = 0.0849 ,

and

hI=e−x2

2σ2

Then the overall PSF is

h=hC∗hI.

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153

Notice that

F{e−x2

2σ2

c}=√2πσce−π(2πσ2

cu2),

F{e−x2

2σ2

I}=√2πσIe−π(2πσ2

Iu2).

F{h}= 2πσcσIe−π{2π(σ2

c+σ2

I)u2},

and from the inverse Fourier transformation we get

h=s2π

σ2

c+σ2

σcσIe−x2

2(σ2

c+σ2

I).

to the right bottom of the adjacent hole, the angle is denoted by θ. From the geometry, tan θ=l

h+2d= 8.33

and w=h/ cos θ=h/ h+2d

√l2+(h+2d)2= 50.36 mm.

Photon

Adjacent hole

Primary hole

Figure S8.5 See Problem 8.16(c).

(d) If the septal penetration is to be less than 5%, the transmission factor for the minimum path length is e−µw ≤

0.05 ⇒µw ≥3.0. For ﬁxed land d,µh√l2+(h+2d)2

h+2d≥3.0. Simplify the expression by the fact that

lh+ 2d, then h≥6d

µl−3.

(e) The septal penetration degrades the collimator resolution because it blurs the image.

(f) No. Because the attenuation in the septa lead doesn’t change the energy of the photon, so the energy window

wouldn’t help.

(g) Compton scattering.

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154 CHAPTER 8: PLANAR SCINTIGRAPHY

Solution 8.17

(a) Rate of photons emitted = 0.54 ×10−3×3.7×1010 = 2 ×107photons/second.

Assuming uniform emission, rate of photons hitting the detector is given by:

2 arctan(0.5

0.8)

2π×2×107= 0.355 ×107photons/second .

(b) Detector efﬁciency (DE) is deﬁned as

DE = I0−I

where I=I0e−µb.b= 2 cm and µ= 0.64cm−1gives:

DE = 1 −e−0.64×2= 0.7220 .

Thus, detector efﬁciency is 72.2%.

RC=d

l(|y|+b) = 5mm

80 mm (800 mm + 20 mm) = 51.25 mm .

RCis labeled on Figure S8.6.

Figure S8.6 See Problem 8.17(c).

(d) We have

RC= 51.25 mm = 2σc√2 ln 2 ,

RI= 1 mm = 2σI√2 ln 2 .

Therefore, σc=51.25

2√2 ln 2 = 21.65 and σI=1

2√2 ln 2 = 0.42. Assuming that the PSF is Gaussian, we have

hC=e−x2

2σc2,

hI=e−x2

2σI2.

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155

The overall PSF is

h=hC∗hI.

Using properties of the Fourier transform, F{e−x2

2σc2}=√2πσce−2πσc2u2and F{e−x2

2σI2}=√2πσIe−2πσI2u2,

and knowing F{h=hc∗hI}=F{hc}F{hI}, we have

F{h}= 2πσcσIe−2π(σI2+σI2)u2.

From the inverse Fourier transform,

h=r2π

σI2+σI2σcσIe−x2

2(σI2+σI2).

(e) Using similar triangles, d/l =|x|/|y|thus the maximum distance from the center of the detector will be

50 mm. We need to ﬁnd number of holes which will get the ray so d/2 + nh +md = 50 mm. Using

d= 5 mm and h= 2.5mm, detector will get through six holes each side which will be 13 holes in total

(including the middle one).

(f) First guess can be |y|= 80 cm where collimator is as long as the distance of source to detector, but the

following ﬁgure shows the geometry for shortest length. Using similar triangles:

Figure S8.7 See Problem 8.17(f).

d/2 + h+d

|y|=d

Solving the equation yields l0= 40 cm. For this l0, we ﬁnd that Rc= 10.25.

(g) The sensitivity of a collimator is (Kd2

le(d+h)2)2. For simplicity, le=l. Thus ∝1/l2. If sensitivity is at

l= 8 cm then, new sensitivity new at l0= 40 cm will be:

new =(l

)2=e

25 .

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156 CHAPTER 8: PLANAR SCINTIGRAPHY

APPLICATIONS

Solution 8.18

(a) 1. The photon goes through the hole;

2. It enters the scintillation crystal;

3. It has a photoelectric event producing an ejected electron;

4. Collapsing electrons (in this atom and many others) cause light photons to be emitted;

5. The light bounces around in the crystal and exits out the back face;

6. The light enters a PMT;

7. Its energy causes electrons to be emitted at the cathode and enhanced by dynode cascades;

8. The current at the anode is recoded as a small pulse;

9. The total height of all pulses, summed up over all tubes is the Z-pulse;

10. Weighted combinations of the pulse heights give the X−,X+,Y−, and Y+signals;

11. X=X+−X−

2, and Y=Y+−Y−

(b) See Figures S8.8 and S8.9.

Figure S8.8 See Problem 8.18(b).

Figure S8.9 See Problem 8.18(b).

for same reason as X-signal.

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157

(d) The sensitivity is given by

sensitivity =kd2

l(d+h)2

If we double the hole diameter and keep the sensitivity unchanged, we have

kd2

l(d+h)2

=k(2d)2

l2(2d+h)2

⇒l2=4l(d+h)

2d+h⇒l2≈4l ,

where dis the original diameter and dh.

Solution 8.19

(a) A straightforward calculation yields

T=2,000,000 photons

64 ×64 pixels ×4photons/pixel second = 122.07 seconds .

(b) The Z-pulse of two photons is shown in Figure S8.10. The output of the pulse height analyzer is shown in

Figure S8.11.

Figure S8.10 Z-pulse arising from two photons. See Problem 8.19(b).

Figure S8.11 Output of the pulse height analyzer. See Problem 8.19(b).

analyzer at the arrival time of the second photon will be larger than 1.2A. Therefore, the second pulse will be

rejected and will not result in an acceptable event.

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158 CHAPTER 8: PLANAR SCINTIGRAPHY

In order for the second photon to be detectable, the peak voltage at the time when the second photon arrives

cannot exceed 1.2A, the voltage resulted from the ﬁrst photon must be less than 0.2A. Since the voltage

drops linearly from the peak value Ato 0 in 250 µs, it takes 200 µs for the voltage to drop to a value of 0.2A.

The time separation required in order for the second photon to be detected as a separate event is 200 µs.

(d) From part (c), we know that two successive photons must be separated by at least 200 µs in order to be

detected as two events. So the maximal rate of arrival of photons is 1

200 µs= 5,000 photons/second. This

arrival rate is for the entire image (recall how Z-pulse is generated.) For each pixel, the arrival rate is at most

5,000

64×64 = 1.22 photons/second·pixel. This rate is smaller than 4 photons/second·pixel we used in part (a). So

it is not possible to complete the experiment in the time we compute in part (a).

An alternative is as follows: we have a maximum rate of arrival of 5,000 photons/second. In 122.07 seconds,

we can have at most 5,000 ×122.07 ≈610,000 photons, which is less than the required number of photons

to complete the experiment. So not possible.

(e) When an incident photon has undergone Compton scatter, it loses some energy. Under this condition, the

photon might be rejected because its Z-pulse height is too small.

(f) On average each pixel is hit by N=2,000,000 photons

64×64 = 488.28 photons. So the intrinsic SNR in a single

pixel is SNR =√N= 22.1.

Solution 8.20

(a) Rate of photons emitted from O = 0.27 ×10−3×3.7×1010 = 107photons/second. Assuming that the

photons ﬂy uniformly in the x-yplane, the rate of photons hitting the detector is

2 tan−1(0.5

0.5)

2π×107= 0.25 ×107photons/second .

(b) A straightforward calculation yields

detector efﬁciency =fraction of photons blocked by the detector

=I0−I

=I0−I0e−µb

= 1 −e−0.644×2.5

= 80.01% .

Hence, the time to register is 2×105counts =2×105

0.2002×107= 0.1second. Neglecting the time required to

rotate the camera, 10 orientations can be captured in 1 second.

(d) The collimator resolution is

Rc=d

l(l+b+|z|) = 0.005

0.12 (0.12 + 0.025 + |0.5−0.12|)=0.0219 m = 21.9 mm .

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159

(e) Let xnbe the span on the detector within which photons can reach in the nth hole. For the central collimator

hole, that is, n= 0, we have x0=d, since the photons can fall upon the entire detector length d. So the rate

of photons hitting the central hole is

r0=2 tan−1(0.005/2

0.5)

2π×107= 1.6×104photons/second .

For n > 0, by the geometry we have the following relation

d−xn

l=(n−1)(d+h) + h+ 0.5d+ (d−xn)

0.5,

d−xn

l= 2[(n−0.5)d+nh +d−xn],

d−xn= 2l[(n+ 0.5)d+nh]−2lxn,

xn(2l−1) = 2l[(n+ 0.5)d+nh]−d ,

xn=2l[(n+ 0.5)d+nh]−d

(2l−1) .

When n= 1,x1= 2.63 mm. So the rate of photons hitting the hole n= 1 is

r1=tan−1(d/2+h+d

0.5)−tan−1(d/2+h+d−x1

0.5)

2π×107= 8.37 ×103photons/second .

It is the same answer for n=−1.

(f) Evaluating xnat n= 2, we get x2=−0.5mm. So the collimator shadow completely covers the hole, that is,

no photon is able to hit the detector from this hole. Therefore, the photons can enter only in the central three

holes. Since the photons from a point source at the origin are spreading out into the central 3 collimator holes,

the resolution, as deﬁned by FWHM value, is ˆ

Rc=3d+2h

2= 12.5 mm.ˆ

Rcis smaller than Rc= 21.9 mm

computed in part (d), since Rcis computed with ideal geometry, neglecting the effects of septa.

Solution 8.21

(a) It is a straightforward calculation:

Rc=d

l(l+b+z)

100(100 + 25 + 0.5×103)

= 18.75 mm .

(b) The intrinsic PSF is a Gaussian function with σcomputed as follows

RI= 0.2mm = 2σ√2 ln 2 ⇒σ=0.2

2√2 ln 2 = 0.0849 .

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160 CHAPTER 8: PLANAR SCINTIGRAPHY

Roverall =qR2

c+R2

I= 18.75 .

(d) The total count is 120 ×106.

(e) The local contrast is

C=¯

Nt−¯

=8−3

3= 1.667 ,

where ¯

Ntand ¯

Nbare mean target and background counts. The SNR is

SNRlocal =Cp¯

Nb= 2.89 .

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Emission Computed Tomography

SPECT

Solution 9.1

(a) For |`|<2, we have

gSPECT(`, 0◦) = Z√4−`2

−√4−`2

fexp{−(Z√4−`2

µ2dy0+Z5

√4−`2

µ1dy0)}dy

=Z√4−`2

−√4−`2

fexp{−(µ2(p4−`2−y) + µ1(5 −p4−`2))}dy

=fexp{−µ2(p4−`2)}exp{−µ1(5 −p4−`2)}Z√4−`2

−√4−`2

exp{µ2y}dy

=fexp{−µ2(p4−`2)}exp{−µ1(5 −p4−`2)}(exp{µ2p4−`2} − exp{−µ2p4−`2})/µ2

µ2

exp{−µ1(5 −p4−`2)}(1 −exp{−µ2p4−`2}).

Thus,

gSPECT(`, 0◦) = (f

µ2exp{−µ1(5 −√4−`2)}(1 −exp{−µ2√4−`2})|`|<2

0otherwise

Similarly,

gSPECT(`, 180◦) = (f

µ2exp{−µ3(5 −√4−`2)}(1 −exp{−µ2√4−`2})|`|<2

0otherwise

161

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162 CHAPTER 9: EMISSION COMPUTED TOMOGRAPHY

(b) For |`|<2,gPET(`, 0◦) = gPET(−`, 180◦)and

gPET(`, 0◦) = Z√4−`2

−√4−`2

fexp{−(Z√4−`2

−√4−`2

µ2dy0+Z5

√4−`2

µ1dy0+Z−√4−`2

−5

µ3dy0)}dy

= 2fp4−`2exp{−µ3(5 −p4−`2)}exp{−µ1(5 −p4−`2)}exp{−2µ2p4−`2)}.

Thus,

gPET(`, 0◦) = (2f√4−`2exp{−µ3(5 −√4−`2)}exp{−µ1(5 −√4−`2)}exp{−2µ2√4−`2)} |`|<2

0otherwise

gPET(`, 180◦) = (2f√4−`2exp{−µ3(5 −√4−`2)}exp{−µ1(5 −√4−`2)}exp{−2µ2√4−`2)} |`|<2

0otherwise

gSPECT(0,0◦)=0.2799 mCi/cm2,

gSPECT(0,180◦) = 0.3022 mCi/cm2.

(d) Substituting numerical values in the formulas found in (b) yields

gPET(0,0◦) = gPET(0,180◦) = 0.0901 mCi/cm2.

Solution 9.2

(a) For θ= 180◦, we see that will be a g(`, 180◦)rect function with a magnitude determined from the SPECT

imaging equation. The photon energy is 150KeV. For |`|<1,

g(`, 180◦) = Z3

0.2e−Ry

0µsdsdy+Z2

0.4e−Ry

0µsdsdy

=Z3

0.2e−Ry

20.2ds−R2

00.4dsdy+Z2

0.4e−Ry

00.4dsdy

= 0.2×e−0.8×e0.4×1

0.2(e−0.4−e−0.6)+0.4×1−e−0.8

0.4

= 0.0814 + 0.5507

= 0.6321 .

So g(`, 180◦)=0.6321 ×rect( `

2).

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163

Similarly, g(`, 90◦)will be two rect functions, and magnitudes are to determined from SPECT equations.

g1(`, 90◦) = Z1

−1

0.2e−R1

xµsdsdx

=Z1

−1

0.2e−0.2(1−x)dx

= 0.2×e−0.2Z1

−1

e0.2xdx

= 0.3297 .

g2(`, 90◦) = Z1

−1

0.4e−R1

xµsdsdx

=Z1

−1

0.4e−0.4(1−x)dx

= 0.4×e−0.4Z1

−1

e0.4xdx

= 0.5507 .

So g(`, 90◦)=0.3297 ×rect(`−1

2)+0.5507 ×rect(`−2.5).

(b) For PET, the energy of photons is 511 keV. The magnitude of projection is given by

g(`, 0◦) = Z3

f(y)e−R3

0µdsdy

=Z3

f1(y)e−R3

0µdsdy+Z3

f2(y)e−R3

0µdsdy

= (0.2×1+0.4×2) + e−R3

0µdsdy

=e−1×0.1−2×0.3

= 0.4966 .

So g(`, 0◦)=0.4966 ×rect( `

2).

gSPECT(`, θ) = ZR

−∞

f(x(s), y(s))

4π(s−R)2exp{−ZR

µ(x(t), y(t); E)dt}ds ,

gPET(`, θ) = KZR

−R

f(x(s), y(s))ds×exp{−ZR

−R

µ(x(s), y(s); E)ds}.

It is seen that the attenuation term containing µis separable from activity term f(x, y)in PET while it is not

separable in SPECT. So prior to PET imaging, a CT scan of the patient is done and µ(x, y)is found by CT

reconstruction. Then PET imaging is done to obtain g(`, θ). Thus, the line integral of activity is obtained by

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164 CHAPTER 9: EMISSION COMPUTED TOMOGRAPHY

dividing the PET projection by attenuation coefﬁcients and Radon transform of Ais given by

G(`, θ) = ZR

−R

f(x(s), y(s))ds=gPET(`, θ)

exp{−RR

−Rµ(x(s), y(s); E)ds},

⇒f(x, y) = R−1{G(`, θ)}.

where Rdenotes Radon transform. µbeing not separable from SPECT equation, this technique is not appli-

cable for attenuation correction of SPECT.

(d) If real collimators are used, the efﬁciency will be <100%. So less number of photons will be detected. So

the height of the rect functions will be decreased.

Solution 9.3

(a) We have

3N0=N0·e−1λP,

3N0=N0·e−1λQ.

Therefore,

λP= ln3 hour−1,

λQ= ln1.5hour−1,

and

2,P =0.693

ln3 = 0.631 hour ,

2,Q =0.693

ln1.5= 1.709 hour .

(b) We have

AP=1

3N0·λP= 0.366N0= 0.366 ×1015/3,600 dps = 2.75 Ci ,

AQ=2

3N0·λQ= 0.270N0= 0.270 ×1015/3,600 dps = 2.03 Ci .

go=AQ·2 + AP·4=0.270 ×2N0+ 0.366 ×4N0= 2.004N0,

gb=AP·6 = 0.366 ×6N0= 2.196N0.

Therefore, the local contrast is

C=go−gb

=2.004N0−2.196N0

2.196N0

=−0.0874 .

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165

(d) Suppose the 180◦projection in the object and the background considering linear attenuation are g0

oand g0

respectively. Then

o=Z6

AP·(e−(y−3)·1·e−1·1·e−2·2)dy+Z3

AQ·(e−(y−1)·2·e−1·1)dy+Z1

AP·(e−y·1)dy

=Z6

AP·e−y−2dy+Z3

AQ·e−2y+1dy+Z1

AP·e−ydy

=AP·e−2·(e−3−e−6) + AQ·e·1

2·(e−2−e−6) + AP·(e0−e−1)

=AP·(e−5−e−8+ 1 −e−1)+0.5AQ·(e−1−e−5)

= 0.366N0·0.639 + 0.5·0.270N0·0.361

= 0.283N0,

b=Z6

AP·e−y·1dy

= 0.366N0·(1 −e−6)

= 0.365N0.

Therefore, the local contrast is

C=g0

o−g0

=0.283N0−0.365N0

0.365N0

=−0.225 .

(e) The absolute value of the local contrast would be bigger in 180◦than that in 0◦. First of all g0

bare the same

on both projections. Second, AQ< APand µcircle > µsquare. Therefore, inside the object, the radioactivity

gets more attenuated in 180◦than that in 0◦. Thus g0

oin 180◦is smaller than that in 0◦, meaning g0

ois farther

away from g0

bon 180◦than that in 0◦. Since we are considering the absolute value of the local contrast, |C|

would be bigger in 180◦than that in 0◦.

Solution 9.4

(a) First, we write

f(x, y) = (0.5 mCi/cm3|x| ≤ 1, y ≤ −x,

0otherwise.

And

µ(x, y) = 









0.1 cm−1|x| ≤ 1,|y| ≤ 1, y +x≤0,

0.2 cm−1|x| ≤ 1,|y| ≤ 1, y +x > 0,

0otherwise.

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166 CHAPTER 9: EMISSION COMPUTED TOMOGRAPHY

For |l| ≤ 1, the projection can be computed as

gSPECT(l, 0◦) = Z−l

−1

fexp{−Z−l

µ1dy0−Z1

−l

µ2dy0}dy

=Z−l

−1

fexp{−(−l−y)µ1−(1 + l)µ2}dy

=e−(1+l)µ2+lµ1Z−l

−1

feµ1ydy

µ1

e−(1+l)µ2+lµ1(e−lµ1−e−µ1)

= 5e−0.1(2+l)(e−0.1l)−e−0.1))

= 5(e−0.2(1+l)−e−0.3−0.1l)).

The ﬁnal answer is

gSPECT(l, 0◦) = 5(e−0.2(1+l)−e−0.3−0.1l))rect( l

2).

Similarly, for |l| ≤ 1,gSPECT(l, 180◦)can be found as:

gSPECT(l, 180◦) = Zl

−1

fexp{−Zy

−1

µ1dy0}dy

=Zl

−1

fe−(y+1)µ1dy

=fe−µ1Zl

−1

e−µ1ydy

µ1

e−µ1(eµ1−e−lµ1)

= 5(1 −e−(l+1)µ1)

= 5(1 −e−0.1(l+1)).

The ﬁnal answer is gSPECT(l, 180◦) = 5(1 −e−0.1(l+1))rect( l

2).

(b) For |l| ≤ 1,

gPET(l, 0◦) = Z−l

−1

fexp{−Z−l

−1

µ1dy0−Z1

−l

µ2dy0}dy

=Z−l

−1

fe−(−l+1)µ1−(1+l)µ2dy

=fe−((1−l)µ1+(1+l)µ2)Z−l

−1

=f(1 −l)e−((1−l)µ1+(1+l)µ2)

= 0.5(1 −l)e−0.1(3+l).

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167

Thus, gPET(l, 0◦) = 0.5(1 −l)e−0.1(3+l)rect( l

2). By the principle of PET imaging gPET(l, 180◦) =

gPET(−l, 0◦),

gP ET (l, 180◦)=0.5(1 + l)e−0.1(3−l)rect( l

applying a CT scan prior to PET scan to get the activity term. Then the standard CT reconstruction method

can be applied to correctly reconstruct the radioactivity distribution.

Solution 9.5

(a) Since Nij is a Poisson random variable, the variance of Nij is also ¯

Nij . The mean and the variance of gij are

k¯

Nij and k2¯

Nij , respectively. Therefore,

mean[ ˆ

f(x, y)] = kπT

j=1

N/2

i=−N/2

Nij ˜c(xcosθj+ysinjθ−iT ),

var[ ˆ

f(x, y)] = k2π2T2

j=1

N/2

i=−N/2

Nij [˜c(xcosθj+ysinjθ−iT )]2.

(b) Carry out the following steps:

j=1

N/2

i=−N/2

[˜c(xcosθj+ysinjθ−iT )]2

≈Zπ

0Z∞

−∞

[˜c(xcosθ+ysinθ−`)]2d`dθ

=πZ∞

−∞

[˜c(`)]2d`

=πZ∞

−∞ |C(%)|2d%

=πZ%0

−%0

%2d%

=2π%3

var[ ˆ

f(x, y)] = k2π2T2¯

j=1

N/2

i=−N/2

[˜c(xcosθj+ysinjθ−iT )]2

=2π2k2%3

0¯

3M.

(d) SNR = mean[ ˆ

f(x,y)]

√var[ ˆ

f(x,y)] ∝¯

√¯

N=√¯

N. So the ratio is √2.

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168 CHAPTER 9: EMISSION COMPUTED TOMOGRAPHY

PET

Solution 9.6

(a) Let the coordinate system be such that the upper left corner of the matrix is the origin. Consider a left-handed

Cartesian coordinate system, so that the yaxis is positive below the origin. Now the coordinates of the center

of PMT(i, j)in inches are (2(j−1) + 1,2(i−1) + 1). Similarly, the coordinates of the center of subcrystal

C(k, l)are (0.5(l−1) + 1,0.5(k−1) + 0.25). Hence, the distance between the centers of PMT(i, j)and

subcrystal (k, l)is:

r=p[2(j−1) + 1 −0.5(l−1) −0.25]2+ [2(i−1) + 1 −0.5(k−1) −0.25]2,

=p(2j−0.5l−0.75)2+ (2i−0.5k−0.75)2.

Hence, The PMT response is

PMT(i, j) = e−√(2j−0.5l−0.75)2+(2i−0.5k−0.75)2.

(b) Given that k= 4 and l= 5, the above equation simpliﬁes to:

PMT(i, j) = e−√(2j−3.25)2+(2i−2.75)2.

Hence,

PMT(1,1) = e−√2.125,

PMT(1,2) = e−√1.125,

PMT(2,1) = e−√3.125,

PMT(2,2) = e−√2.125.

PMT(1,1) = e−√(1.25−0.5l)2+(1.25−0.5k)2,

PMT(1,2) = e−√(3.25−0.5l)2+(1.25−0.5k)2,

PMT(2,1) = e−√(1.25−0.5l)2+(3.25−0.5k)2,

PMT(2,2) = e−√(3.25−0.5l)2+(3.25−0.5k)2.

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169

Rearranging the equations:

(1.25 −0.5l)2+ (1.25 −0.5k)2=log 1

PMT(1,1)2

(3.25 −0.5l)2+ (1.25 −0.5k)2=log 1

PMT(1,2)2

(1.25 −0.5l)2+ (3.25 −0.5k)2=log 1

PMT(2,1)2

(3.25 −0.5l)2+ (3.25 −0.5k)2=log 1

PMT(2,2)2

Subtracting the equations, we get:

(3.25 −0.5k)2−(1.25 −0.5k)2=log 1

PMT(2,1)2

−log 1

PMT(1,1)2

9−2k=log 1

PMT(2,1)2

−log 1

PMT(1,1)2

k= 4.5−1

2(log 1

PMT(2,1)2

−log 1

PMT(1,1)2).

Similarly, an estimate of kcan be obtained from PMT(1,2) and PMT(2,2) as:

k= 4.5−1

2(log 1

PMT(2,2)2

−log 1

PMT(1,2)2).

Averaging the above two estimates of k, we get:

k= 4.5−1

4(log 1

PMT(2,1)2

−log 1

PMT(1,1)2

+log 1

PMT(2,2)2

−log 1

PMT(1,2)2).

Similarly, an estimate of lcan be obtained as:

l= 4.5−1

4(log 1

PMT(1,2)2

−log 1

PMT(1,1)2

+log 1

PMT(2,2)2

−log 1

PMT(2,1)2).

(d) The worst-case scenario occurs when the event occurs very close to the boundary between two PMTs, for

example, if an event occurs very close to the boundary between C(2,4) and C(2,5) but occurs in crystal

C(2,4), then under noiseless condition, PMT(1,1) will be slightly greater than PMT(1,2). However if a small

additive noise cause the signal PMT(1,2) >PMT(1,1), then the event will be attributed to C(2,5).

Solution 9.7

(a) Since the detectors are designed to stop 75% of the photons, we have 0.75 = e−µd where dis the detector

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170 CHAPTER 9: EMISSION COMPUTED TOMOGRAPHY

Figure S9.1 See Problem 9.8(c).

thickness. Hence we have

for NaI(Tl) :d= ln 0.75/(−µ)

= (ln 0.75)/(−0.343)

= 0.8387 cm ,

for BGO :d= ln 0.75/(−µ)

= (ln 0.75)/(−0.964)

= 0.2984 cm .

(b) The gamma rays photon burst is a random phenomena and can be modeled as a Poisson process. Let the

average gamma ray photons arriving at the detector be λ. Let kbe the fraction of these gamma ray photons

converted into light photons by NaI(Tl). Since, BGO is 13% efﬁcient, the fraction of gamma rays converted

into light photons by BGO is 0.13k. Hence the intrinsic SNRs are:

SNRNaI(tl) =√λk ,

SNRBGO =√0.13λk .

The ratio of intrinsic SNR’s is p1/0.13.

Solution 9.8

(a) No collimators. In a PET scanner, one must be able to detect coincidences at diverse angles.

(b) You would have to add a coincidence detector.

ure S9.1. The line between the two detections would be calculated to show where the decay took place. The

Z-pulse will be used in the calculation of the Xand Ypulses. It could also be used for energy discrimination,

as in a conventional PET scanner.

(d) From the geometry in Figure S9.2,

α= tan−10.15 m

0.75 m = 11.31◦.

(e) Consider reconstruction on a “central plane.” There are many gamma rays that are “lost” between the two

detectors in this geometry. In a conventional PET scanner, the plane is completely surrounded by detectors.

Therefore, on this basis alone, we can expect that the total number of coincidence detections in a given time

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171

Figure S9.2 See Problem 9.8(d).

frame will be smaller given the same dose. Therefore, for the same quality, we would have to increase the

dose. In addition to this argument, there is the fact that PET detectors are more efﬁcient at stopping 511 keV

photons. Therefore, the Anger-based camera will also be less efﬁcient and also require a higher dose.

Solution 9.9

(a) Removing the constants in Eq. 9.6 but keeping the attenuation gives

gSPECT(`, θ) = Z∞

−∞

f(x(s), y(s)) exp{−Z∞

µ(x(s0), y(s0)ds0}ds .

In this problem,

f(x, y) = 0.3mCi/cm3if 0≤x≤1,−3≤y≤3,

0otherwise .

And,

µ(x, y) = 









0.2cm−1if −3≤x≤0,−3≤y≤3,

0.3cm−1if 0≤x≤1,−3≤y≤3,

0.1cm−1if 1≤x≤3,−3≤y≤3,

0otherwise .

When θ= 90◦,x(s) = `cos θ−ssin θ=−s, and y(s) = `sin θ+scos θ=`. Hence,

gSPECT(`, 90◦) = Z1

0.3 exp{−Z1

0.3dx0−Z3

0.1dx0}dx

=e−0.1×2Z1

0.3e−0.3(1−x)dx

=e−0.5(e0.3−e0)

≈0.2122 mCi/cm2,

for −3cm ≤`≤3cm. gSPECT(`, 90◦)=0if ` < −3cm, or ` > 3cm. Similarly,

gSPECT(`, 270◦) = e−0.2×3Z1

f(x, `)e−0.3xdx

=e−0.2×3Z1

0.3e−0.3xdx

=e−0.6(−e−0.3+e0)

≈0.1422 mCi/cm2,

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172 CHAPTER 9: EMISSION COMPUTED TOMOGRAPHY

for −3cm ≤`≤3cm. gSPECT(`, 270◦)=0if ` < −3cm, or ` > 3cm. Note that gSPECT(`, 90◦)6=

gSPECT(`, 270◦).

(b) We have

gPET(`, 90◦) = e−0.2×3e−0.1×2Z1

f(x, `)e−0.3×1dx

=e−0.2×3−0.1×2−0.3×1Z1

0.3dx

= 0.3e−1.1

≈0.0999 mCi/cm2,

for −3cm ≤`≤3cm. gPET(`, 90◦)=0if ` < −3cm, or ` > 3cm. By the principle of PET imaging,

gPET(`, 270◦) = gPET(`, 90◦).

imaging line. Thus, to compensate for the attenuation effect of the object, one can image the object using a

separate source to acquire the line integrals of µ(x, y). Then, the data can be corrected using

PET(`, θ) = gPET(`, θ)

exp{−R∞

−∞ µ(`cos θ−ssin θ, ` sin θ−scos θ)ds}.

The standard CT reconstruction method can then be applied on gc

PET(`, θ)to correctly reconstruct the ra-

dioactivity distribution. This compensation approach is not applicable for the SPECT scan.

Solution 9.10

(a) The perimeter of the circle is

πD = 1.5π≈4.712 m.

The approximate detector width is thus

4.712 m/1,000 = 4.712 mm .

Shallow detectors are less efﬁcient to stop the gamma photons, but incoming gamma photons from all direc-

tions can be equally detected. Deep detectors are more efﬁcient, but they are more direction selective.

(b) Coincidence detection in PET is used to determine the direction of travel of the two back-to-back gamma

photons, and hence to decide on which line the radioactivity occurs. Coincidence is assumed if two events

occur within 2–12 ns in typical PET scanners. Since the radioactivity is not always occur at the center of the

PET scanner, the traveling times of the two back-to-back gamma photons are not the same.

(i) If the time interval is too small, off-the-center radioactivities will not be detected.

(ii) If the time interval is too large, scattered photons will still be counted. Also, two or more distinct positron

decays might be mixed together, and the line of coincidence can no longer be correctly determined.

interval is

T=D/500 = 1,500 mm/500 = 3 mm .

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173

By the “rule of thumb”, the number of pixels on one side of the image should be approximately equal to the

number of samples for each projection angle. Since the number of samples is 1,000/2 = 500 for the PET

scanner, the PET image should have at least 5002= 250,000 pixels. A “wobbling” motion of the PET gantry

can reduce the effective spacing of the detectors and thus increase the resolution of the system.

(d) “Parallax errors” cause degradation of resolution farther away from the center. This is because events can be

detected from oblique angles within the detector body (instead of end-on) which creates uncertainty about

the actual line on which the event occurred.

Solution 9.11

(a) Suppose s0represents the center of the circle.

exp{−ZR

µ(x(s0), y(s0); E)ds0}=N+

exp{−Zs0

−R

µ(x(s0), y(s0); E)ds0}=N−

The number of coincidence events Ncarising from positron annihilations at the center of the circle that will

be detected is

Nc(s0) = N0exp{−ZR

µ(x(s0), y(s0); E)ds0}exp{−Zs0

−R

µ(x(s0), y(s0); E)ds0}

=N0

N+

N−

=N+N−

(b) Carry out the following steps

g(`1,0◦) = 2

3g(0,0◦),

f(`1, y)e−6µsquare =Z6

f(0, y)e−4µsquare−2µcircle ,

−6µsquare =−4µsquare −2µcircle + ln2

µsquare −µcircle =1

2ln3

g(`2,0◦)

g(0,0◦=R6

0f(`2, y)e−5µsquare−µcircle

0f(0, y)e−4µsquare−2µcircle

=e−µsquare+µcircle

=e−1

2ln 3

=r2

= 0.8165 .

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174 CHAPTER 9: EMISSION COMPUTED TOMOGRAPHY

(d) The local contrast is

C=ft−fb

=g(0,0◦)−g(`1,0◦)

g(`1,0◦)

=g(0,0◦)−2

3g(0,0◦)

3g(0,0◦))

=1/3

2/3

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The Physics of Ultrasound

THE WAVE EQUATION

Solution 10.1

By taking the derivatives of w1(z, t)with respect to zand t, we have

∂2w1

∂z2=ξ00(z−ct),∂2w1

∂t2=c2ξ00(z−ct).

It is obvious that ∂2w1

∂z2=1

∂2w1

∂t2,

which is Equation (10.6). For w2(z, t) = ξ(z−ct) + ξ(z+ct), we have:

∂2w2

∂z2=ξ00(z−ct) + ξ00(z+ct),∂2w2

∂t2=c2ξ00(z−ct) + c2ξ00(z+ct).

So, w2(z, t) = ξ(z−ct)+ξ(z+ct)is also a solution to the wave equation (10.6). w2(z, t) = ξ(z−ct)+ξ(z+ct)is

the general solution to (10.6). It has two components, a forward-traveling wave ξ(z−ct)and a backward-traveling

wave ξ(z+ct).

Solution 10.2

A sinusoidal plane wave is given as

p(z, t) = cos[k(z−ct)] .

The wavelength is the spacing between crest. Suppose z1, and z2are positions of two adjacent crests for a given

time t, we have:

k(z1−ct)=2nπ, k(z2−ct) = 2(n+ 1)π ,

where nis an arbitrary integer. Then it is obvious

λ=z2−z1= 2π/k .

175

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176 CHAPTER 10: THE PHYSICS OF ULTRASOUND

Solution 10.3

(a) The acoustic pulse is

φ(t) = (1 −e−t/τ1)e−t/τ2.

It reaches a peak when its derivative goes to zero; that is,

dφ(t)

dt =e−t/τ2(− − (t/τ1)e−t/τ1) + (1 −e−t/τ1)(−t/τ2)e−t/τ2= 0 .

This leads to

(1 −e−t/τ1)(1/τ2) = (1/τ1)e−t/τ1,

1−e−t/τ1= (τ2/τ1)e−t/τ1,

1=(τ2/τ1+ 1)e−t/τ1.

Solving for tyields the time delay

td=−τ1ln 1

τ2/τ1+ 1 .

Plugging in τ2=τ1= 5 µs yields td= 3.5µs. Therefore, the peak pressure will return to the transducer at

3.5 + 64.9 = 68.4µs.

(b) The “generic” backward traveling wave is

φb(z, t) = (1 −e−(t+z/c)/τ1)e−(t+z/c)/τ2.

At time t= 64.9µs, this wave will be (begin) at position z= 0.1m heading in the −zdirection. Incorporat-

ing both the temporal and spatial shift yields

φb(z, t) = (1 −e−(t−64.9µs+(z−0.1 m)/c)/τ1)e−(t−64.9µs+(z−0.1mz)/c)/τ2.

2×64.9µs = 129.8µs.

Solution 10.4

The 3-D wave equation is

∇2=1

∂2p

∂t2where ∇2=∂2p

∂x2

+∂2p

∂x2

+∂2p

∂x2

We have that

r2=x2

1+x2

2+x2

2rdr

dxi

= 2xi⇒dr

dxi

=xi

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177

The new pressure function is p=p(r, t). We have

∂p

∂xi

=∂p

∂r

∂xi

=∂p

∂r

and

∂2p

∂x2

=∂

∂xi∂p

∂r

r

=∂

∂xi∂p

∂r

rxi+1

∂p

∂r

=∂

∂r 1

∂p

∂r ∂r

∂xi

xi+1

∂p

∂r

=x2

∂

∂r 1

∂p

∂r +1

∂p

∂r .

Therefore, using the fact that r2=x2

1+x2

2+x2

3, we have

∇2p=3

∂p

∂r +r∂

∂r 1

∂p

∂r .

Now ∂

∂r (rp) = r∂p

∂r +p

and ∂2

∂r2(rp) = r∂2p

∂r2+∂p

∂r +∂p

∂r =r∂2p

∂r2+ 2∂p

∂r .

So,

r∂

∂r 1

∂p

∂r =r−1

∂p

∂r +1

∂2p

∂r2=−1

∂p

∂r +∂2p

∂r2.

Therefore,

∇2p=3

∂p

∂r −1

∂p

∂r +∂2p

∂r2=2

∂p

∂r +∂2p

∂r2=1

∂2

∂r2(rp).

This is the spherical wave equation.

Solution 10.5

Taking the derivatives of w(r, t) = ξ(r−ct)/r with respect to t, we have:

∂w

∂t =−cξ0(r−ct)

∂2w

∂t2=c2ξ00(r−ct)

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178 CHAPTER 10: THE PHYSICS OF ULTRASOUND

Taking the derivatives of rw(r, t)with respect to r, we have:

∂(rw)

∂r =ξ0(r−ct),

∂2(rw)

∂r2=ξ00(r−ct).

It can be seen that 1

∂2

∂r2(rw) = 1

∂2w

∂t2.

Solution 10.6

Substitute p(r, t)into Equation 10.13, we have:

∂2

∂r2(rp) = 1

c2f00(t−c−1r) + 1

c2g00(t+c−1r),

and ∂2p

∂t2.=1

rf00(t−c−1r) + 1

rg00(t+c−1r).

So 1

∂2

∂r2(rp) = 1

rc2(f00(t−c−1r) + 1

rg00(t+c−1r)) = 1

∂2p

∂t2,

and p(r, t) = 1

rf(t−c−1r) + 1

rg(t+c−1r)is a solution to Equation 10.13.

WAVE PROPAGATION

Solution 10.7

I(x, t) = v(x, t)p(x, t)and I(0, t) = Re{V ejωt}Re{P ejωt}. Let V=Vmejφ and P=Pmejθ. Then

I(0, t) = Vmcos(ωt +φ)Pmcos(ωt +θ)

=VmPm

2[cos(φ−θ) + cos(2ωt +φ+θ)]

Therefore, since high frequency oscillations disappear,

Iav =VmPm

2cos(φ−θ).

But V P ∗=VmPmej(φ−θ). Therefore,

2Re{V P ∗}=VmPm

2cos(φ−θ) = Iav .

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179

Solution 10.8

(a) Equations (10.25) and (10.26) are repeated here for convenience:

cos θt

pt+cos θr

pr=cos θi

pi,

pt−pr=pi.

From these two equations, we can form a matrix equation as:

"cos θt

cos θr

1−1#pt

pr="cos θi

1#pi.

Then we have

pt

pr="cos θt

cos θr

1−1#−1"cos θi

1#pi

="−1−cos θr

−1cos θt

z2#

−cos θt

z2−cos θr

z1"cos θi

1#pi

=





z2cos θi+z2cos θr

z1cos θt+z2cos θr

z2cos θi−z1cos θt

z1cos θt+z2cos θr





pi.

Since θr=θi, we have

pt

pr=





2z2cos θi

z1cos θt+z2cos θi

z2cos θi−z1cos θt

z1cos θt+z2cos θi





pi.

The pressure reﬂectivity Rand pressure transmittivity Tare given by

R=pr

=z2cos θi−z1cos θt

z1cos θt+z2cos θi

T=pt

=2z2cos θi

z1cos θt+z2cos θi

(b) We write

RI=Ir

=p2

r/z1

i/z1

=p2

=R2=z2cos θi−z1cos θt

z1cos θt+z2cos θi2

which is Equation (10.29). Also,

TI=It

=p2

t/z2

i/z1

=z1p2

z2p2

=z1

T2=4z1z2cos2θi

(z2cos θi+z1cos θt)2,

which is Equation (10.30).

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180 CHAPTER 10: THE PHYSICS OF ULTRASOUND

Solution 10.9

(a) We have

P1=1

f(t−c−1r1),

P2=1

f(t−c−1r2),

where

r1=px2+y2+ (z+d)2,

r2=px2+y2+ (z−d)2.

From the geometry, we see that for large r,r1≈r+dcos θand r2≈r−dcos θ, where θis the angle off

the z-axis to the line connecting the origin with (x, y, z), and ris the length of that line. But cos θ=z/r.

Therefore,

P=P1+P2≈1

ft−c−1r+dz

r−1

ft−c−1r−dz

r.

But

f(t) = Re{˜n(t)ej2πf0t},

and r1≈r2(for amplitude). Hence, the complete waveform is

Pc≈1

r˜n(t−c−1r)ej2πf0(t−c−1r)(e−jk dz

r−e+jk dz

r),

where we have used the steady-state approximation for ˜n.

But ejΦ−e−jΦ= 2jsin Φ, so

e−jk dz

r−e+jk dz

r=−2jsin kdz

As dgets very small, sin(kdz/r)≈kdz/r, therefore, since −j=e−jπ/2,

Pc≈2kde−jπ/2z

r2˜n(t−c−1r)ej2πf0(t−c−1r).

One can see that the new pressure is given by

P=z

r2fn(t−c−1r),

where

fn(t−c−1r) = Re{2kde−jπ/2˜n(t−c−1r)ej2πf0(t−c−1r)}.

(b) See Figure S10.1.

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181

Figure S10.1 Field pattern. See Problem 10.9(b).

Solution 10.10

(a) The acoustic pressure of an outward propagating spherical wave is expressed as:

p(r, t) = A0

rφ0(t−c−1r),

where r=px2+y2+z2. When there is attenuation, the expression becomes:

p(r, t) = e−µarA0

rφ0(t−c−1r).

(b) Let

d=px2+y2+z2

be the distance between the wave source and the scatter. Assume the reﬂection coefﬁcient of the scatter is R.

The scatter acts like a new point source. So the scattered wave can be expressed as

ps(x0, y0, z0, t) = Re−µar0

r0A0e−µadφ0(t−c−1r−c−1d),

where r0=p(x0−x)2+ (y0−y)2+ (z0−z)2is the distance between the scatter and any point (x0, y0, z0)

in space.

p(x0, y0, z0, t) = e−µarA0

rφ0(t−c−1r),

where r=p(x0−x0)2+ (y0−y0)2+ (z0−z0)2. The scattered wave by a scatter at (x, y, z)is

ps(x0, y0, z0, t) = Re−µar0

r0A0e−µadφ0(t−c−1r−c−1d),

where r0=p(x0−x)2+ (y0−y)2+ (z0−z)2and d=p(x0−x)2+ (y0−y)2+ (z0−z)2is the dis-

tance between the wave source and the scatter.

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182 CHAPTER 10: THE PHYSICS OF ULTRASOUND

Solution 10.11

(a) The acoustic intensity is related to the acoustic pressure by

I=p2/Z .

So the acoustic intensity for the incident wave, the reﬂected wave, and the transmitted wave are:

Ii=p2

i/Z1,

Ir=p2

r/Z1,

It=p2

t/Z2.

The intensity reﬂectivity and intensity transmittivity are:

RI=Ir

=p2

rZ1

iZ1

=Z2cos θi−Z1cos θt

Z2cos θi+Z1cos θt2

TI=It

=p2

tZ2

iZ1

=4Z1Z2cos2θi

(Z2cos θi+Z1cos θt)2.

(b) Carry out the following steps:

T−R=2Z2cos θi−(Z2cos θi−Z1cos θt)

Z2cos θi+Z1cos θt

=Z2cos θi+Z1cos θt

Z2cos θi+Z1cos θt

= 1 .

pt−pr=pi.

Therefore, T−R= 1. But the relationship between acoustic pressure and acoustic intensity is a nonlinear

relationship and Z26=Z1. So TI6= 1 + RIin general. From the above derivation, we have:

RI=R2, TI=T2Z1/Z2,and T=R+ 1 .

So the relationship between TIand RIis

TI=Z1

Z2pRI+ 12.

DOPPLER EFFECT

Solution 10.12

The frequency fRof the sound received by the moving receiver can be derived by considering the time it takes for

the receiver to observe two successive crests. If the source is producing a sinusoid with frequency f0, the distance

separating two adjacent crests is λ=c/f0. Since the receiver is moving towards the source, the time it takes for

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183

the receiver to observe two successive crests is T=λ/(c+v) = c/(c+v)f0, which is the observed period. So the

frequency observed is fR= (c+v)f0/c.

When the receiver moves away from the source with speed v, the frequency observed is fR= (c−v)f0/c.

Solution 10.13

(a) The observed frequency will be fR= (c+v)f0/c and fR= (c−v)f0/c for receivers moving towards and

away from the source, respectively.

(b) When the receivers moves towards the source, the observed frequency will be fR= 2f0. When the receivers

moves away from the source, the observed frequency is 0. In this situation, the receiver will sit on a point

with constant phase, therefore will not observe the wave.

receivers moves away from the source, the observed frequency is fR= (c−v)f0/c < 0. In this situation,

the receiver will observe a wave from the opposite direction with frequency fR=|c−v|f0/c.

ULTRASOUND FIELD PATTERN

Solution 10.14

By using the properties of Fourier transform, we have:

F{˜n(t)}=ejφNe(ω),

Re{˜n(t)e−jω0t}=1

2˜n(t)e−jω0t+ ˜n∗(t)ejω0t,

where * denotes complex conjugate. The Fourier transform of n(t)is:

F{n(t)}=1

2ejφNe(ω+ω0) + e−jφN∗

e(−ω+ω0).

Solution 10.15

(a) We have

α5 MHz = 8.7µa= 8.7×0.04 cm−1·MHz−1×5MHz = 1.74 dB ·cm−1,

α12MHz = 8.7µa= 8.7×0.04 cm−1·MHz−1×12 MHz = 4.176 dB ·cm−1.

(b) It is considered as far ﬁeld when range is greater than D2/λ. The speed of sound is 1,560 m/s. At 5 MHz,

the wavelength is λ5 MHz = 0.312 mm. At 12 MHz, the wavelength is λ12 MHz = 0.13 mm.

For the 5 MHz transducer, when range is greater than 2cm×2cm

0.312 mm = 128.2 cm, it is considered as far ﬁeld. For

the 12 MHz transducer, the range is 0.4 cm×0.4 cm

0.13 mm = 12.3 cm.

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184 CHAPTER 10: THE PHYSICS OF ULTRASOUND

Solution 10.16

(a) The time it takes to travel from (x0, y0,0) to dand back is

t=2

cqx2

0+y2

0+d2.

Therefore, the time delay is

τ=2

cpx2

m+y2

m+d2−2

cqx2

0+y2

0+d2.

In this way, the scattering from dwill be integrated over the (ﬂat) transducer face at the same time upon

reception. Using the binomial approximation,

τ=2

c"dr1 + x2

m+y2

d2−dr1 + x2

0+y2

d2#

≈2

cd[1 + x2

m+y2

d2]−d[1 + x2

0+y2

d2]

≈2

dc r2

m−(x2

0+y2

0).

(b) The narrowband assumption is

n(t) = Re{˜n(t)e−j2πf0t}.

When shifted in time, and using the steady-state approximation:

n(t−τ) = Re{˜n(t−τ)e−j2πf0(t−τ)}

≈Re{˜n(t)e−j2πf0(t−τ)}.

Therefore, after some simpliﬁcation

n(t−τ−c−1r0−c−1r0

≈Re{˜n(t−2c−1z)ejk

dr2

me−jk

d(x2

0+y2

0)ejk(r0−z)ejk(r0

0−z)}.

ejk

dr2

mis just a ﬁxed phase, which can be “thrown” into ˜n(t−2c−1z). Now split up terms:

e−jk

2d(x2

0+y2

0)ejk(r0−z)

| {z }

transmitpattern

e−jk

2d(x2

0+y2

0)ejk(r0

0−z)

| {z }

receivepattern

which leads to the ﬁeld pattern

q(x, y, z)≈ZZ s(x0, y0)

ze−jk

2d(x2

0+y2

0)ejk

2d[(x−x0)2+(y−y0)2]dx0dy0,

where the the paraxial approximation was also used. Expanding the terms as follows

(x−x0)2=x2−2xx0+x2

0and (y−y0)2=y2−2yy0+y2

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185

and setting z=d, yields

q(x, y, d) = ZZ s(x0, y0)

zejk

2d(x2+y2)e−jk

d(xx0+yy0)dx0dy0

zejk

2d(x2+y2)Sx

λd,y

λd,

which is the desired result.

increase our resolution at the focal point over that of a ﬂat transducer. The spread, after the focal point,

however, will increase. Therefore, we need to choose the focal distance carefully.

Solution 10.17

(a) We have

q0(x, y, z) = 1

zejk(x2+y2)/(2z)S(x

λz ,y

λz ).

And

s(x, y) = rect( x

w) rect( y

h),

S(u, v) = F{s(x, y)}=wh sinc(wu) sinc(hv).

Thus,

q0(x, y, z) = wh

zejk(x2+y2)/(2z)sinc(wx

λz ) sinc(hy

λz ).

(b) We have

sinc(v) = sin πv

πv ,

and the ﬁrst zero is when v= 1. Hence, wx

λz = 1,x=λz/w =s, and thus

z0=sw

λ.

sw/λ ≥h2/λ. Hence,

s≥h2

If one says D=√h2+w2, then s≥(w2+h2)/w.

(d) The ﬁeld pattern simply shifts in xfor each of the ﬁve elements and adds (since everything is linear).

q(x, y, z) =

n=−2

q0(x−ns, y, z)

n=−2

zejk

2z((x−ns)2+y2)sinc w(x−ns)

λz sinc hy

λz .

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186 CHAPTER 10: THE PHYSICS OF ULTRASOUND

(e) Because the central pattern goes to zero at ±s, it also goes to zero at ±ns for n6= 0. Hence beam-width in

the x-direction is 6s. In the ydirection, sinc(hy/λz)gives the pattern. The ﬁrst zero is at

λz = 1 ,

which gives y=λz/h. The beam-width is twice that, and hence the beam-width in y-direction is

2λz0

h=2λ

λ=2sw

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Ultrasound Imaging Systems

ULTRASOUND IMAGE FORMATION AND IMAGING MODES

Solution 11.1

(a) From Table 10.1, we know that Z1= 1.52 ×10−6kgm−2s−1and Z2= 1.35 ×10−6kgm−2s−1. Assume z0

is far away. Therefore the only return is from point (0,0, z0)on the interface. Therefore, θr=θi=θt= 0;

cos θr= cos θi= cos θt= 1.

Pr=Z2−Z1

Z2+Z1

Pi≈ −0.06Pi.

We can neglect the minus sign because of envelope detection. When transducer axis is in direction θ, the

strength of Piat (0,0, z0)depends on the ﬁeld pattern. The transducer face in the x-zplane is s(x) =

rect( x

L). Therefore,

S(u) = Lsinc(Lu),

and

Sz0sin θ

λz0cos θ=Stan θ

λ=Lsinc Ltan θ

λ

describes off-axis ﬁeld pattern. Pulse-echo squares this; hence, the strength of the return is

Pr(θ) = 0.06A0S(tan θ

λ)2

= 0.06A0L2sinc2(Ltan θ/λ).

(b) The signal Pris above threshold when

20 log10 0.06A0sinc2(Ltan θ/λ)

A0≥ −80dB ,

187

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188 CHAPTER 11: ULTRASOUND IMAGING SYSTEMS

Figure S11.1 B-mode image. See Problem 11.1(b).

or sinc2(Ltan θ/λ)≥0.001667. Letting u=Ltan θ

λwe get

Sidelobe u= sinc2(u) =

Main lobe 0 1

1st 1.5 0.045

2nd 2.5 0.0162

3 3.5 0.00827

4 4.5 0.005004

5 5.5 0.0033

6 6.5 0.0023

7 7.5 0.0018

8 8.5 0.0014

9 9.5 0.0011

10 10.5 0.0009

Thus, the system will “see” 7 sidelobes on each side. Since

u=Ltan θ

λ=Lz0tan θ

λz0

=2Lx0

λz0

the sidelobe separation is x0=λz0

2L. See Figure S11.1 for a sketch of the B-mode image.

Solution 11.2

(a) The echo is received at

t=2x

c=2×10 cm

1,500 m/second ≈1.33 ×10−4second .

The round trip distance is 20 cm. Since α= 1 dB/cm, there is a loss of 20 dB:

−20 dB = 20 log10

Thus, Az= 0.1A0= 1.225 N/cm2. See Figure S11.2 for a sketch of the A-mode signal.

(b) See Figure S11.3 for a sketch of the M-mode signal.

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189

Figure S11.2 A-mode signal. See Problem 11.2(a).

Figure S11.3 M-mode image. See Problem 11.2(b).

returning signal.

Figure S11.4 B-mode image. See Problem 11.2(c).

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190 CHAPTER 11: ULTRASOUND IMAGING SYSTEMS

Figure S11.5 Peak-height plot. See Problem 11.2(c).

Solution 11.3

(a) See Figure S11.6.

Figure S11.6 z(t). See Problem 11.3(a).

(b) The pulse will be at range z=ct at time t. The valve will be at range z= 16 + 0.5e−t/τ at time t. At time

t=t0, they coincide. So we have:

ct = 16 + 0.5e−t/τ ,

154,000t= 16 + 0.5e−t/0.01 .

Ignore the motion effect, we have t0≈16

154,000 = 0.104 ms. See Figure S11.7.

Figure S11.7 A-mode signal. See Problem 11.3(b).

(d) Each scan line takes about 0.208 ms and ten scan lines take about 2.08 ms. Since the time constant is 10 ms,

there will be ﬁve images in this time. So the time is adequate to make a B-mode image. (However, we won’t

be able to see this real time.)

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191

Figure S11.8 M-mode signal. See Problem 11.3(c).

Solution 11.4

Ignore the ﬁeld pattern in xand ydirections and assume the scatterers are ideal point scatterers, R(x, y, z) =

δ(z−z1) + δ(z−z2). In Fraunhofer ﬁeld, the approximation of the received ﬁeld is given by (11.20):

R(x, y, z) = R(x, y, z)ej2kz ∗∗∗ hSx

λz ,y

λz i2nez

c/2.

Ignoring Sx

λz ,y

λz , we have

R(x, y, z) = R(x, y, z)ej2kz ∗∗∗ nez

c/2

=δ(z−z1)ej2kz1+δ(z−z2)ej2kz2∗∗∗ rect z

λ+1

2

=rect z−z1

λ+1

2ej2kz1+ rect z−z2

λ+1

2ej2kz2.

(a) Since z2−z1=λ/2, we have

R(x, y, z) = rect z−z1

λ+1

2ej2kz1+ rect z−z2

λ+1

2ej2k(z1+λ/2)

=ej2kz1rect z−z1

λ+1

2+ rect z−z2

λ+1

2ejkλ

= rect z−z1

λ+1

2+ rect z−z2

λ+1

2.

(b) If z2−z1=λ/8, we have

R(x, y, z) = rect z−z1

λ+1

2ej2kz1+ rect z−z2

λ+1

2ej2k(z1+λ/8)

=rect z−z1

λ+1

2+ rect z−z2

λ+1

2ejkλ/4

=rect z−z1

λ+1

2+jrect z−z2

λ+1

2.

The estimated reﬂectivities are shown in Figure S11.9.

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192 CHAPTER 11: ULTRASOUND IMAGING SYSTEMS

Figure S11.9 Estimated reﬂectivities. See Problem 11.4.

ULTRASOUND TRANSDUCER ARRAY

Solution 11.5

(a) From the geometry in Figure 10.5, the widest angle (from the axis) at which an ultrasound transducer will

generate sound is deﬁned by

tan θ=d/2

d2/λ .

Using the relation λ=c/f and the fact that θ= 30◦yields

tan θ= 0.5773502 = 1,540 m/s

f0.2mm .

f=1,540 m/s

0.5773502 ×0.0002 m

= 13.336 MHz .

Any frequency higher than this will be more directive and incapable of generating a wave at θ= 30◦.

(b) The zeroth transducer ﬁres at time 0. The ﬁrst transducer ﬁres at

t1=dsin θ

=0.0002 m×sin 30◦

1,540 m/s

= 64.9×10−9second .

This is the same time delay between each pair of transducers. Since there are 100 delays needed to ﬁre 101

transducers the total time is

time to ﬁre array = 100 ×64.9×10−9second = 6.49 ×10−6second .

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193

Solution 11.6

(a) We have

ti=id sin θ

=0.6mm

1,540 ×103mm/s isin θ

= (0.39 µs)isin θ .

(b) The time it takes for one pulse to go to range Rand back is the pulse repetition interval:

TR=0.40 m

1,540 m/s = 260 µs.

The total angle of the sector is 90◦, and given ∆θ= 1◦, we require 90 pulses in order to cover the ﬁeld.

Therefore, the total time it takes to acquire a frame is

TF= 90 ×TR= 23.4×10−3second .

The frame rate is therefore 42.8 frames/second, which will be ﬂicker-free.

Solution 11.7

A diagram of the described transducer is given in Figure S11.10(a).

Figure S11.10 See Problem 11.7.

(a) Ranges > D2/λ. Evaluate as follows

D= 14 ×1.5mm = 21 mm (same as width of transducer) ,

λ=c

f=1,500 m/s

3.0×106s−1= 500 ×10−6m= 0.5mm ,

λ=212mm2

0.5mm = 882 mm .

Therefore,

Ranges >0.882 m.

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194 CHAPTER 11: ULTRASOUND IMAGING SYSTEMS

(b) As shown in Figure S11.10(b), the total travel is 14 cm = 0.14 m. Therefore,

∆t=0.14 m

1,500m/s = 93.3µs.

number of lines of acquisition) is 100 −14 + 1 = 87. One A-mode burst takes this long:

∆t=40 cm

1,500 m/s =0.4m

1,500 m/s = 266.7µs.

Therefore, the total time to acquire one image is

87 ×266.7µs= 23.2ms/frame ,

and the frame rate is

Frame Rate =1

23.2ms = 43.1frames/s .

(d) One can increase the frequency of the transducer or acquire fewer lines—e.g., just acquire 40 lines from the

center of the transducer.

Solution 11.8

(a) The transmit pulse is shown in Figure S11.11(a). An echo from a silicone-skin interface assuming normal

incidence is shown in Figure S11.11(b). The amplitude of this echo is

Amplitude =Zskin −Zsilicone

Zskin +Zsilicone

=1.5×106−1.4×106

1.5×106+ 1.4×106= 0.0345 .

The time of echo is:

Time of Echo =2×2×10−3

1,500 = 2.666 ×10−6second = 2.6µs.

Figure S11.11(c) shows the superposition of these two signal envelopes on the same graph. The A-mode

signal is not simply the sum of the two A-mode signals since the underlying signals are sinusoidal. In

general, there will be constructive or destructive interference. In the region of overlap, the actual signal is:

cos(2πft)+0.0345 cos(2πf (t−t1)) ,

where f= 2.0×106Hz and t1= 2.666 ×10−6second. Using the complex representation of sinusoids

(phasors), it can be shown that the amplitude of this signal is 1.011. The corresponding A-mode signal is

shown in Figure S11.11(d).

(b) C(gel) = C(skin)=1,550 m/s for no refraction at the gel-skin interface.

T1=2Z(gel)

Z(gel) + Z(silicone),

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195

Figure S11.11 See Problem 11.8.

where

Z(gel) = 106√1.5×1.4 = 1.4491 ×106.

This wave gets reﬂected at the gel-skin interface, where the pressure reﬂectivity is

R2=Z(skin)−Z(gel)

Z(skin) + Z(gel).

The above reﬂected wave gets transmitted back through the gel-silicone interface where the pressure trans-

mittivity is

T3=2Z(silicone)

Z(gel) + Z(silicone).

The amplitude of the echo from the gel-skin interface (and incident on the transducer) is

1×T1×R2×T3= 0.0172 .

The time of this echo is

t2=2s

c(silicone)+2g

c(gel)= 6.537 µs.

This echo is superposed on the previous composite signal in Figure S11.11(e). The A-mode signal magnitude

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196 CHAPTER 11: ULTRASOUND IMAGING SYSTEMS

for two new overlapping intervals needs to be worked out. The ﬁrst interval has three overlapping signals:

cos(2πft)+0.0345 cos(2πf (t−t1)) + 0.0172 cos(2πf(t−t2)) ,

where t2is given above. The magnitude of this wave is 1.02683. In the second overlapping region, the wave

is given by

0.0345 cos(2πf(t−t1)) + 0.0172 cos(2πf (t−t2)) ,

and the resulting magnitude is 0.04817.

(d) The amplitude of the echo from the skin is 0.0172, while the initial amplitude is 1. Therefore,

L= 20 log10 0.0172

=−35.289 dB .

The sign of Lis negative because it is an attenuation or loss. Normally, we say that the system is sensitive to

L= 35.289 dB loss.

(e) c=fλ. So,

λ=c(skin)

2MHz =1,550 m/s

2×106s−1= 7.75 ×10−4m.

Then,

λ=(3 ×10−3m)2

7.75 ×10−4m= 0.0116 m= 1.16 cm .

Therefore, point F is in the far ﬁeld because 5 cm >1.16 cm.

(f) There are three relevant distances, dL,dC, and dR, corresponding to the distances from point F to the left,

center, and right transducers, respectively. These are:

dL=p(5 ×10−2+ 8 ×10−3)2+ (10 ×10−2)2,

dC=p(5 ×10−2)2+ (10 ×10−2)2,

dR=p(5 ×10−2−8×10−3)2+ (10 ×10−2)2.

Now deﬁne:

τL=dC−dL

c(skin)=0.1118 −0.1156

1,550 =−2.435 µs,

τC= 0 ,

τR=dC−dR

c(skin)=0.1118 −0.1085

1,550 = 2.129 µs.

Since we can’t have negative times, add −τLto all values, yielding

τL= 0 ,

τC= 2.45 µs,

τR= 4.579 µs.

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197

ULTRASOUND IMAGING SYSTEM DESIGN AND IMAGE QUALITY

Solution 11.9

(a) Assuming the reﬂection coefﬁcient of the line object is R0, the mathematical expression for the scatter is

R(x, y, z) = R0δ(x, z −5) .

(b) The wavelength in the media at f= 2.5MHz is

λ=c/f = 0.062 cm .

The range at which the beam changes from geometric/Fresnel to far ﬁeld is

z0=D2/λ = 16.2cm .

Since the transducer face is a square, the transition between geometric ﬁeld and the Fresnel ﬁeld occurs at

D2/2λ= 8.1cm. The scatter is located at a range of z= 5 cm. So the geometric assumption applies here.

The estimated reﬂectivity is:

R(x, y, z) = KR0δ(x, z −5)ej2kz ∗∗∗˜s(x, y)nez

c/2(S11.1)

=KR0ej2k5δ(x, z −5) ∗∗∗rect(x, y)nez

c/2(S11.2)

=KR0δ(x, z −5) ∗∗∗rect(x, y)nez

c/2(S11.3)

=KR0rect(x)nez−5

c/2,(S11.4)

where ˜s(x, y) = s(−x, −y) = rect(x, y)is the transducer face indicator function and nez

c/2is the envelop

of the narrowband pulse.

parallel to the y-axis, we need to have them separated by at least 1 cm, which is the width of the transducer.

(d) At range z= 20 cm, the scatter is in the far ﬁeld, we need to use Fraunhofer assumption. The estimated

reﬂectivity is:

R(x, y, z) = R(x, y, z)ej2kz ∗hSx

λz ,y

λz i2nez

c/2(S11.5)

=R0δ(x, z −20)ej2kz ∗hSx

λz ,y

λz i2nez

c/2(S11.6)

=R0δ(x, z −20) ∗hSx

λz ,y

λz i2nez

c/2,(S11.7)

where S(u, v)is the Fourier transform of the face shape indicator function s(x, y) = rect(x, y):

S(u, v) = sinc(u) sinc(v).

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198 CHAPTER 11: ULTRASOUND IMAGING SYSTEMS

We have:

R(x, y, z) = R0sinc2x

λ(z−20)nez−20

c/2Z∞

−∞

sinc2y

λ(z−20)dy .

From the above, for a ﬁxed z, the term that determines the minimal separation of two parallel line scatters at

same range is

sinc2x

λ(z−20).

In order to resolve two line scatters, they must be separated by at least a distance 2d, such that

sinc2d

λ(z−20)<0.5.

The resolution depends on the depth.

Solution 11.10

Issues: Depth of penetration, Azimuth resolution, Speckle. First,

dp=L

2af =100dB

2×1×f=50

f(MHz).

If f= 5 MHz, dp= 10 cm. Hence f= 5 MHz is ruled out. Let dGdenote the depth for geometric imaging. Then

dG=D2/λ.

Let BW denote the beamwidth at 20 cm. In particular, BWGdenotes the beamwidth from the geometric

approximation, BGSbeamwidth when using a square transducer, and BGCis the beamwidth if using a circular

transducer. Then,

BWG=Dif z≤D2/λ

λz

Dif z > D2/λ.

BWS=Dif z≤D2/2λ

λz

Dif z > D2/2λ.

BWC=Dif z≤D2/4λ

λz

Dif z > D2/4λ.

Put these results into a chart:

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199

f= 1 MHz f= 2 MHz

D= 1 cm dp= 50 cm dp= 25 cm

λ= 0.15 cm λ= 0.075 cm

dG= 6.67 cm dG= 13.33 cm

BWG= 3.0cm BWG= 1.5cm

BWS= 3.0cm BWS= 1.5cm

BWC= 3.0cm BWC= 1.5cm

D= 2 cm dp= 50 cm dp= 25 cm

λ= 0.15 cm λ= 0.075 cm

dG= 26.67 cm dG= 53.33 cm

BWG= 2.0cm BWG= 2.0cm

BWS= 1.5cm BWS= 2.0cm

BWC= 1.5cm BWC= 0.75 cm

There can be several choices:

•To get the best resolution for geometric imaging, one should choose D= 1 cm, f= 2 MHz;

•To get the best resolution if using circular transducer, one should choose D= 2 cm, f= 2 MHz;

•To get the best resolution if using square transducer, one should choose D= 1 cm, f= 2 MHz.

Solution 11.11

(a) Given an L×Ltransducer, we have:

s(x, y) = rect x

Lrect y

L.

For, the ﬁrst point scatter, we have,

R(x, y, z) = δ(x)δ(y)δ(z−z0) .

Hence, in the region, where geometric assumptions hold, we have

R0(x, y, z) = KR(x, y, z)ej2kz ∗∗∗s(x, y)nez

c/2

=KR hδ(x)∗rect x

Lihδ(y)∗rect y

Liej2kz δ(z−z0)∗sinc 2πz

c∆T

=KRej2kz0rect x

Lrect y

Lsinc 2π(z−z0)

c∆T.

Similarly, for the second scatter we have:

R0(x, y, z) = KRej2kz0rect x

Lrect y

Lsinc 2π(z−z0−∆Z)

c∆T.

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200 CHAPTER 11: ULTRASOUND IMAGING SYSTEMS

(b) This part involves computing FWHM of a sinc function, which cannot be solved analytically. It involves

transcendental equations, which require numerical solutions. But, note that the FWHM of the sinc can be

approximated by the distance from the origin to the ﬁrst zero the sinc.

S(u, v) = L2sinc(Lu) sinc(Ly).

Using the far ﬁeld approximation, we have:

R0(x, y, z) = Rδ(x)δ(y)δ(z−z0)ej2kz ∗∗∗hSx

λz ,y

λz i2sinc 2πz

c∆T

=Rδ(z−z0)ej2kz ∗hSx

λz ,y

λz i2sinc 2πz

c∆T

=RL4ej2kz0sinc2Lx

λz sinc2Ly

λz sinc 2π(z−z0)

c∆T.

(d) This part involves computing FWHM of a sinc function.

Solution 11.12

(a) From the absorption coefﬁcients in part (a) of Problem 10.15, we have

L5MHz = 40 cm ×1.74 dB ·cm−1= 69.6dB ,

L12MHz = 40 cm ×4.176 dB ·cm−1= 167.04 dB .

(b) From the results of part (b) in Problem 10.15, range z= 20 cm is in the geometric region for the 5 MHz

transducer, while it is in the far ﬁeld for the 12 MHz transducer. So for the 5 MHz transducer, the beamwidth

at 20-cm range equals the dimension of the transducer face, which is w= 2 cm. For the 12 MHz transducer,

the beam width is w=λz

D=0.013 cm×20 cm

0.4 cm = 0.65 cm.

2×20 cm

156,000 cm/s= 256µs. Therefore the maximal repetition rate is 1

256µs= 3,906 Hz.

(d) F=1

TRN=1

256 µs×128 = 30 frames/second.

(e) This will cause geometric distortion since the estimated ranges are wrong due to the poor uniform speed

assumption.

Solution 11.13

(a) The depth of penetration in oil is

dp=L

2α=65dB

2×0.95dB/cm ≈34.2cm .

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201

(b) The wavelength of sound in oil is

λ=coil

f=1,500m/s

106s−1= 1.5×10−3m=0.15 cm .

Therefore,

λ=12

0.15 ≈6.67 cm .

Since z0= 20 cm >6.67 cm, the interface lies in the far-ﬁeld. The beamwidth is approximately

w(z0) = λz0

D=0.15 ×20

1= 3 cm .

Zoil =ρoilcoil = 950 ×1,500 kg m−2s−1= 1.425 ×106kg m−2s−1,

and

Zfat =ρfatcfat = 920 ×1,450 kg m−2s−1= 1.334 ×106kg m−2s−1,

respectively. Thus, the pressure reﬂectivity at the interface is

R=1.334 −1.425

1.334 + 1.425 ≈ −0.033 .

We can ignore the negative sign since it only indicates a phase change. The amplitude attenuation coefﬁcient

inside oil is

µa=α

8.686 dB =0.95dB/cm

8.686 dB ≈0.1094 cm−1.

Thus, the amplitude of the returned pulse is

pr=Rpie−2z0µa= 0.033 ×20 ×e−2×20×0.1094 ≈0.0083(N/cm2).

The amplitude gain is

20log10

= 20log10

0.0083

20 ≈ −67.639(dB) .

Since the amplitude loss, which is 67.6 dB, is beyond the sensitivity of the system, the returning echo is

undetectable by the system. This does not conﬂict with (a), since the depth-of-penetration is computed by

assuming a perfect reﬂection from a target.

(d) The interface position is oscillating between 15 cm and 25 cm, with a period of f−1

0= 0.01s = 10 ms. The

time needed for the sound to travel 25 cm is

25 cm

1,500m/s≈0.167 ms ,

which is negligible compared to the slow motion of the interface.

cos(2π×100Hz ×0.167 ×10−3s) ≈0.9945 ≈1.

Hence, the pulse at t= 0 will hit the interface at z1= 20 −5 = 15cm. The amplitude loss for a return at

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202 CHAPTER 11: ULTRASOUND IMAGING SYSTEMS

15 cm will be

0.95 ×2 + 20 log10 R≈ −58.1(dB) .

Hence, there will be signal shown on the A-mode scan, with a time of return about

tr=2×15 cm

1,500m/s= 0.2ms .

(e) The transducer ﬁres with the same period as the oscillation of the interface. Hence, each pulse hits the

interface at the same depth. The M-mode signal is a horizontal line at approximately 15 cm parallel to the

time axis.

Solution 11.14

(a) The speed of sound in the media is

c=Z

ρ=1.35 ×106kg/m2·s

920 kg/m3= 1,500 m/s .

At frequency of f= 2.5MHz, the wavelength is:

λ=c/f = 0.06 cm .

The far ﬁeld begins at range of

z=D2/λ = 0.5cm ×0.5cm/0.06 cm = 4.17 cm .

(b) Assume the transmitter/preampliﬁer can handle at most an 80 dB loss (this is typical in ultrasound systems),

then we have

2αdp=L .

So, α=L/2dp= 80 dB/2×20 cm = 2 dBcm−1.

see that the lateral resolution of the transducer is related to

sinc2x

λz = sinc2x

0.6cm2.

The FWHM is

FWHM = 0.88 ×0.6=0.53 cm .

(d) With the depth of penetration of 20 cm, the time it takes for the ultrasound wave to make a round trip is:

TR= 2 ×20cm/1,500m/s = 2.67 ×10−4s.

There are N=12 cm/1 mm = 120 scans in one frame. So the maximum frame rate is

F= 1/TRN= 31frames/second .

(e) Because the interface is perpendicular to the transducer axis, the incident wave, the reﬂected wave, and the

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203

transmitted wave all travel in the direction of transducer axis. So θi=θr=θt=π/2. The pressure

reﬂectivity is:

R=Z2−Z1

Z2+Z1

=1.7−1.35

1.7+1.35 = 0.11 .

The amplitude of the reﬂected acoustic pressure is

A= 0.11 ×A010−20×1.5

20 = 0.0035A0.

Solution 11.15

(a) All quantities are in the appropriate units to apply the following equation dp=L

2af =80

2·2= 20 cm.

(b) The pulse repitition rate is given by fr=1

TR=c

2dp=c

2·20 =148,000

40 = 3,700 per second. Remember to

convert from m/s into cm/s. So the frame rate is given by F=fr/256 = 14.4frames per second.

128 cm. The delays are given by

τi=r0−ri

=√52+ 102−p(id −5)2+ 102

148,000 .

For i= 64 we get

τ64 =11.18 −p(64/128 −5)2+ 102

148,000

=11.18 −√20.25 + 100

148,000

=11.18 −10.96

148,000

= 0.00015 s.

(d) We can now use the pulse repetition rate, 3,700 frames per second. The heart has frequency 500 cycles per

min =8.3333 Hz. In M-mode we can sample a signal up to 3,700/2=1,850 per second without aliasing.

So there is no aliasing here. However, there is aliasing in the B-mode image, because we cannot sample a

signal with frequency higher than 14.4/2=7.2per second without introducing aliasing.

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204 CHAPTER 11: ULTRASOUND IMAGING SYSTEMS

Solution 11.16

(a) Carry out the following math:

dp=L

2af ,

dp,3 MHz =90 dB

(2 dB

cm MHZ )(3 MHz) = 15 cm

dp,6 MHz =90 dB

(2 dB

cm MHZ )(6MHz)= 7.5 cm .

(b) Focusing at a depth of z= 5 cm and θ= 20◦gives us xf= 5 tan(20◦) = 1.82 and zf= 5.

If we set the transducer distance to be 2d, then we ﬁnd the ﬁring time for each ias follows:

t(6)

i=qx2

f+z2

f−q(i(2d)−xf)2+z2

=p(1.82 cm)2+ (5 cm)2−p(i(2(0.04 cm)) −1.82 cm)2+ (5 cm)2

154,000 cm/s

t(6)

1=.174 µs

tmin =t(6)

−2=−.369 µs

τ(6)

i=t(6)

i−tmin

τ(6)

i=t(6)

i+.369 µs

τ(6)

1=.543 µs

if iis negative. Also, we are focusing at a depth of z= 10 cm. So xf= 10 tan(20◦)=3.64 and zf= 10.

t(3)

i=qx2

f+z2

f−q((2i−sign(i))(d)−xf)2+z2

=p(3.64 cm)2+ (10 cm)2−p((2i−sign(i))(0.04 cm) −3.64 cm)2+ (10 cm)2

154,000 cm/s

t(3)

2=.262 µs

τ(3)

i=t(3)

i−tmin

τ(3)

i=t(3)

i+.369 µs

τ(3)

2=.631 µs

(d) Since the two transducer types are identical outside of the frequency. The main value we will be comparing

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205

is when the amplitude of the traveling wave is larger for the 3MHz than for the 6Mhz. This amplitude is

described by the decay function (Equation 10.32), Az=A0e−µaz.

We know that the attenuation coefﬁcient, µa, is directly related to frequency in our case by µa=α/8.7 =

af/8.7. Hence:

µ3 MHz

a= 3 dB/cm = 3

8.7= 0.345 np/cm

µ6 MHz

a=6

8.7dB/cm = 0.69 np/cm

so the switchover range is:

αzswitch = 30dB = 3 dB/cm ×2zswitch = 5 cm

(e) Knowing zswitch we can solve for the ratio:

A3 MHz

0e−µ3 MHz

azswitch =A6 MHz

0e−µ6 MHz

azswitch

A3 MHz

0e−0.345 (np/cm)5cm =e0.69 (np/cm)5cm

A3 MHz

A6 MHz

=e−0.69 (np/cm)5cm−(−0.345 (np/cm)5cm)

= 0.178 .

HARMONIC IMAGING

Solution 11.17

(a) A1=A0e−µad=A0e−f0/8.7d.

(b) Use linearity: G(t) = A12

πP∞

n=1(−1)n1

2i(δ(f−nf0)−δ(f+nf0)).

π=A0e−f0/8.7d1

π. Then A3=A0e−f0/8.7d1

πe−2f0/8.7d=A0e−3f0/8.7d1

π.

(d) We observe that −20 log A3

A0=L. So

L=−20 log e−3f0/8.7d1

=−20 ln e−3f0/8.7d1

π/ln(10)

= 20 ·3f0/8.7d/ ln(10) + 20 ln π/ ln(10) ,

80 = 3f0d+ 9.94 ,

d= 23.35/f0

= 23.35/2.5

= 9.34 cm .

(e) We can use the simple formula dp=L

2α=L

2·8.7µa=L

2f1=80

10 = 8 cm. This is a 16% increase in depth of

penetration.

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206 CHAPTER 11: ULTRASOUND IMAGING SYSTEMS

(f) In the Fourier domain we can write our ﬁlter as a difference of two rect functions.

H(f) = rect(f/12MHz)−rect(f/8MHz)

=rect(f/12 ×106)−rect(f/8×106).

The ﬁlter in the time domain is then

h(t) = 12 ×106sinc(12 ×106t)−8×106sinc(8 ×106t).

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Physics of Magnetic Resonance

MAGNETIZATION

Solution 12.1

The magnetic ﬁeld Bat z= 0 and z= 1 cm is

B(0) = 1 Tesla and B(1) = 1.5Tesla .

The Larmor frequencies at these positions are

f(0) = 42.58 MHz and f(1) = 63.87 MHz .

The next time when the magnetization vectors on two planes have same phase is when |2πf (0)t−2πf (1)t|= 2π.

Solve for t, we have

t= 0.047 µs.

Solution 12.2

The static magnetic ﬁled is oriented in z-direction, B(t) = B0ˆz. By substituting the equations (12.12) into

(12.7), we have on the left hand side:

dMx(t)

dt = 2πν0M0sin αsin (−2πν0t+φ),

dMy(t)

dt =−2πν0M0sin αcos (−2πν0t+φ),

dMz(t)

dt = 0 .

On the right hand side:

γM(t)×B(t) = γM(t)×B0ˆz=γB0M0sin αsin (−2πν0t+φ)ˆx−γB0M0sin αcos (−2πν0t+φ)ˆy ,

207

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208 CHAPTER 12: PHYSICS OF MAGNETIC RESONANCE

where ˆx,ˆy, and ˆzare the unit vectors in x,y, and zdirections. Since ν0=γ–B0, we have 2πν0=γB0. It is easy to

see that equations (12.12) are solutions to (12.7).

Solution 12.3

The transverse magnetization is

Mxy(t) =

i=1

Aie−t/T2ie−j(2πνit−φi).

The Fourier transform of this time domain signal yields the NMR spectrum of the sample. This multispectral

character of the signal must be included as a constraint when designing appropriate imaging protocols when fat is

present; this includes most tissues other than brain.

RF EXCITATION AND RELAXATION

Solution 12.4

(a) The tip angle is given by

α=γZt

1(τ)dτ

=γZt

10 1−|τ−T|

Tdτ .

In the range 0≤t≤T, we have

α=γZt

10 1−T−τ

Tdτ

=γZt

10 τ

Tdτ

=γt2

20T.

In the range T≤t≤2T, we have

α=γZT

10 τ

Tdτ +γZt

10 1−τ−T

Tdτ

=γT

20 +γ

10 Zt

T2−τ

Tdτ

=γt

5−γt2

20T−γT

10 .

(b) At the end of the pulse, that is, at t= 2T, the angle is

α=γ2T

5−γT

10 =γT

10 .

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209

To make B1(t)aπ/2pulse, we have

π/2 = γT

10 ⇒T=5π

γ.

Solution 12.5

Both equations are ﬁrst-order, ordinary differential equations and are therefore governed by a simple exponential

growth or decay. The initial value of Mz(t)is Mz(0) and the ﬁnal value Mz(∞)satisﬁes

0 = −Mz(∞)

+M0

so Mz(∞) = M0. The solution is therefore given by

Mz(t)=(Mz(0) −M0)e−t/T1+M0

=M0(1 −e−t/T1) + Mz(0)e−t/T1.

The initial value of Mxy (t)is Mxy(0) and the ﬁnal value is 0. Therefore,

Mxy(t) = Mxy (0)e−t/T1.

Solution 12.6

Use the equation

Mz(t) = M0(1 −e−t/T1) + M0cos αe−t/T1.

The last term of the above equation denotes the initial magnetization after an αpulse. Let

z(t) = M0(1 −e−t/T1) + MSS

zcos αe−t/T1,(S12.1)

where MSS

zis the steady state magnetization deﬁned by

z(TR) = MSS

z.(S12.2)

From (S12.1) and (S12.2), we have

MSS

z=M0

1−e−TR/T1

1−cos αe−TR/T1.

Solution 12.7

(a) It is given in the problem statement that the transverse magnetization is gone before the next imaging pulse

occurs; therefore,

Mxy(0−)=0.

The longitudinal magnetization recovery follows [see Equation (12.40)]

Mz(t) = M0(1 −e−t/T1) + Mz(0+)e−t/T1.

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210 CHAPTER 12: PHYSICS OF MAGNETIC RESONANCE

So, in the steady state, just before the next imaging pulse (at time t=TR) the z-magnetization is

Mz(0−) = M0(1 −e−TR/T1) + Mz(0+)e−TR/T1.

But Mz(0−)and Mz(0+)are also related by the ﬂip angle αas follows

Mz(0+) = Mz(0−) cos α .

Substitution yields

Mz(0−) = M0(1 −e−TR/T1) + Mz(0−) cos αe−TR/T1.

which is solved as follows

Mz(0−) = M0(1 −e−TR/T1)

1−cos αe−TR/T1.

(b) Mz(0−)is the effective longitudinal magnetization just prior to an imaging pulse. With reference to (12.39),

the transverse magnetization after the imaging pulse (ignoring the arbitrary phase and assuming demodulation

at the Larmor frequency) is

Mxy(TE) = Mz(0−) sin αe−TE/T2

=M0(1 −e−TR/T1)

1−cos αe−TR/T1sin αe−TE/T2.

A=Mxy(TE),

M=M0,

R=e−TR/T1,

E=e−TE/T1.

Then

A=EM(1 −R) sin α

1−Rcos α.

To maximize this with respect to α, take the derivative of A

dα =EM(1 −R)

1−Rcos αcos α+ (−1)EM(1 −R) sin α(1 −Rcos α)−2Rsin α ,

set it to zero, and solve for α, as follows

EM(1 −R)

1−Rcos αcos α−Rsin2α

1−Rcos α= 0

cos α=Rsin2α

1−Rcos α

cos α−Rcos2α=Rsin2α

cos α=R(cos2α+ sin2α)

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211

Therefore,

α= cos−1e−TR/T1,

which is known as the Ernst angle.

Solution 12.8

(a) Assume the sample is in equilibrium with magnetizations of Mx(0−) = My(0−) = 0 and Mz(0−) = M0.

After the π-pulse applied at t= 0, the magnetizations are

Mx(0+)=0,

My(0+)=0,

Mz(0+) = −M0.

After a delay of τ, the magnetizations become

Mx(τ−)=0,

My(τ−)=0,

Mz(τ−) = M01−2e−τ/T1.

In the time interval (0, τ), since there is no precession about the B0ﬁeld, there is no FID signal. After the

π/2-pulse, the bulk magnetization is tilted into x-yplane. Assume the π/2-pulse is applied along the y-axis,

we have:

Mx(τ+) = M01−2e−τ/T1,

My(τ+)=0,

Mz(τ+)=0.

The bulk magnetization then precess around the B0ﬁeld to generate FID signal:

Mxy(t) = M01−2e−τ /T1e−(t−τ)/T2e−2πν0(t−τ), t > τ .

(b) From the above derivation, we can see that the strength of the FID signal depends on the delay τ. If we can

take a measure right after the π/2-pulse, the signal strength is M01−2e−τ/T1. Therefore we can use two

different values for the delay τ=τ1and τ=τ2. The signal strength measured right after the π/2-pulse are:

M(1)

xy (τ1) = M01−2e−τ1/T1,

M(2)

xy (τ2) = M01−2e−τ2/T1.

Solve the above equations, we can determine T1.

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212 CHAPTER 12: PHYSICS OF MAGNETIC RESONANCE

BLOCH EQUATIONS AND SPIN ECHOES

Solution 12.9

The Bloch equation is given as

dt =γM ×B−R[M−M0].(S12.3)

M×Band R[M−M0]apply in any frame. Assume that the rotating frame rotates with angular frequency Ω

with respect to the lab frame and that M0is the magnetization vector in the rotating frame (B0is Bin the rotating

frame). From classical mechanics, we have

dt =dM0

dt + Ω ×dM0

dt .(S12.4)

Using Equations S12.3 and S12.4, we get

dM0

dt =γM0×B0−R[M0−M0]−Ω×dM0

dt .

Solution 12.10

(a) Different isochromats in a sample precess in different frequencies around the B0ﬁled. The difference in

precession frequency causes the progressive defocusing of the isochromats after their magnetization vectors

are rotated to the x-yplane (see Figure 12.8). After the π-pulse applied at time t=τ, the magnetization

vectors are ﬂipped in the transverse plane and the faster precessing vectors are lagging behind the slower

ones. The time it takes to rephase (for the faster precessing vectors to catch the slower ones) equals τ. A

phase coherence will be recreated at time t= 2τto generate an echo. Therefore, the π-pulse should be

applied at t=TE/2in order to generate an echo at t=TE.

(b) Suppose the sample is in equilibrium with magnetization Mz(0−) = M0. The π/2-pulse is applied at t= 0.

This pulse rotates the bulk magnetization vector into the x-yplane:

Mxy(0+) = M0.

The series of π-pulses will ﬂip the magnetization in the transverse plane and form echoes at t= (k+1/2)TE,

k= 1,2,···. The magnitude of the transverse magnetization decays with constant T2. So

Mxy(kTE) = M0e−kTE/T2.

Solution 12.11

15◦×2π

360◦= 2π×42.6×106Hz/T×A×10−5s⇒A= 9.78 ×10−5T.

Solution 12.12

(a) Just before applying the πpulse, the phase angle is given by

φ(r, τ−) = −γ(B0+ ∆B(r))τ .

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213

Just after applying the πpulse, the phase will jump by an angle π, hence it is given as:

φ(r, τ+) = π−γ(B0+ ∆B(r))τ .

(b) At the end of time TE= 2τ, the phase will be π.

CONTRAST MECHANISM

Solution 12.13

In proton density weighted images, the image intensity should be proportional to the number of hydrogen nuclei

in the sample. We start with the sample in equilibrium, apply an excitation RF pulse, and image quickly, before the

signal has a chance to decay from T2effects. Thus, a PD-weighted contrast can be obtained by using a long TR,

which allows the tissues to be in equilibrium, and either no echo or a short TE. The preferred tip angle is π/2, in

order to get the maximum signal. Large TEcannot be used because it will introduce large T2decay.

Solution 12.14

(a) The magnitude of the transverse magnetization is given as:

|Mxy(t)|=|MSS sin αejφe−t/T2|

=M0

1−e−TR/T1

1−cos αe−TR/T1sin αe−t/T2.

Now, α=π/2, also the signal is measured just after excitation, so t= 0. Hence,

|Mxy|=M01−e−TR/T1.

The local contrast between GM and CSF can be written as:

C=|Mxy|CSF − |Mxy |GM

|Mxy|CSF

=e−TR/T CSF

1−e−TR/T GM

1−e−TR/T CSF

This function is a monotonically decreasing function of TR. Thus, lower the TR, better the contrast. However,

we cannot have an arbitrarily low TR. It has to be approximately in the range of T1, to allow a reasonable

decay. A good contrast can be obtained by using TR=TGM

1= 760 ms.

(b) The contrasts can be obtained by plugging in values in the above equation.

Solution 12.15

(a) This is not a T2-weighted contrast. In order to obtain a T2-weighted contrast we need to include spin echoes

in the pulse sequence.

(b) We would need to use spin echoes as in Figure 12.9.

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214 CHAPTER 12: PHYSICS OF MAGNETIC RESONANCE

τ. The sampling should be done at twice the pulse period TE, since at that time, the dephasing due to T∗

2is

completely removed and the signal truly represents T2.

TR. In order to bring the tissue back to equilibrium, in between the pulses, TRshould be as long as possible.

In practice, however, 6,000 ms is an unusually long repetition time and is impractical.

TE. The echo time should be approximately equal to the T2values of the tissue being imaged.

α. The tip angle should be π/2to obtain the maximum signal.

Flip angle: During the echo, the transverse vector should be shifted by a phase angle of π.

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Magnetic Resonance Imaging

MR IMAGING INSTRUMENTATION

Solution 13.1

The magnetic ﬁeld is still oriented in the zdirection. But the strength of the ﬁeld is not uniform. At points

with same x-coordinates, the magnetic ﬁelds have the same strength. At points with different x-coordinates, the

magnetic ﬁeld have different strength. The difference depends on the difference in x-coordinates and the magnitude

of the gradient.

Solution 13.2

The functions of RF coils are twofold. (1) During radio frequency excitation, a relatively large amount of current

is generated in the coil using an RF ampliﬁer (with a power requirement of approximately 2 kW for human imaging).

Ideally, this coil then produces a relatively uniform B1ﬁeld throughout the entire imaging volume in order that the

same tip angle is generated in each isochromat in the volume. (2) On reception, an RF coil must pick up very

low-amplitude magnetic ﬁelds, which produce very small currents in the coil. Transmission and receiver RF coils

can be the same coil but are different when high SNR or fast imaging is required.

ENCODING SPATIAL POSITION AND MR IMAGING EQUATION

Solution 13.3

(a) We have

∆ν=γ–Gz∆z= 4.258 kHz ×1G/mm ×10 mm = 42.58 kHz .

(b) We have

B1(t) = A∆νsinc(∆νt)ej2π¯νt ,

where ¯ν= ¯zγ–Gz+γ–B0= 212.9kHz +γ–B0.Therefore, we have:

B1(t) = A×42.58 sinc 42.58tej2π212.9tej2πγ–B0t.

In rotating frame,

B1(t) = A×42.58 sinc 42.58tej2π212.9t,

215

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216 CHAPTER 13: MAGNETIC RESONANCE IMAGING

where Adepends on the tip angle.

Solution 13.4

(a) Start with the Fourier transform pair

F{e−πt2}=e−πu2.

Then since

A0exp{−t2/σ2}=A0exp (πt

√πσ 2),

we can use the Fourier scaling theorem to get

FA0exp{−t2/σ2}=A0√πσ exp −π2σ2u2.

The FWHM is found as follows:

1/2 = exp −π2σ2u2,

ln(1/2) = −π2σ2u2,

u2=1

π2σ2ln 2 ,

u=r1

π2σ2ln 2 = 265 Hz ,

FWHM = 2 ×265 Hz = 530 Hz .

This deﬁnes the frequency interval that is excited ∆ν= 530 Hz, and using Equation (13.12) gives

∆z=530 Hz

42.58 ×106Hz/T10−4Tcm = 1.2mm .

(b) The new gradient strength is G0

z= 0.5Gz. So, the new slice thickness is

∆z=∆ν

γ–G0

=∆ν

γ–0.5Gz

= 2 ×∆z .

Halving the gradient strength doubles the slice thickness. Now suppose that σ0= 0.5σ. Starting from the

original (without using G0

z), we have from previous work that

u0=r1

π2σ02ln 2

σ0

0.5σ0r1

π2ln 2

= 2u .

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217

Therefore, the new frequency range ∆ν0is double what it was before, which doubles the slice thickness. If

used in combination, the slice thickness would be four times thicker.

(d) Doubling the slice thickness improves the SNR by a factor of two. The overall imaging time is slightly

smaller (although this will not affect SNR since the actual ADC time is unaffected). Image resolution will be

degraded in the through-plane direction.

Solution 13.5

The RF signal is given by

s(t) = A∆νsinc(∆νt)ej2π¯νt.

For isochromats whose Larmor frequency is ν, the excitation signal in the rotating coordinate system is

1(t) = s(t)e−j2πνt .

The duration of the above signal is from −∞ to ∞. So the tip angle is:

α(ν) = γZ∞

−∞

1(t)dt

=γZ∞

−∞

A∆νsinc(∆νt)ej2π¯νte−j2πνtdt

=γA∆νZ∞

−∞

sinc(∆νt)e−j2π(ν−¯ν)tdt

=γA Z∞

−∞

sinc(τ)e−j2πν−¯ν

∆ντdτ, let τ= ∆νt

=γA rect ν−¯ν

∆ν.

The slice location zand the Larmor frequency νare related by the following equation:

z=ν−γ–B0

γ–Gz

Therefore,

¯z=¯ν−γ–B0

γ–Gz

∆z=∆ν−γ–B0

γ–Gz

which lead us to

rect ν−¯ν

∆ν= rect z−¯z

∆z.

So, the tip angle is

α(z) = γA rect z−¯z

∆z.

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218 CHAPTER 13: MAGNETIC RESONANCE IMAGING

Solution 13.6

Slice selection uses a narrowband RF excitation during a constant gradient (taken to be in the zdirection without

loss of generality). Spins on the high-frequency side of the slice accumulate transverse phase faster than spins on

the low-frequency side. When the slice selection pulse ends, all spins rotate at the Larmor frequency, but are out

of phase across the slice, and may add destructively during imaging. From Section 13.2.2, we know that the ideal

slice selection excitation signal is (Equation 13.14)

s(t) = A∆νsinc(∆νt)ej2π¯νt .

In addition to required truncation, which we ignore in this derivation, this pulse must be shifted to a positive time,

τp/2, where τpis the duration of the slice selection gradient. This shift in time implies a phase shift of its frequency

content, that is,

S(ν) = Arect ν−¯ν

∆νej2π(ν−¯ν)τp/2.

This frequency spectrum forms the following spatial excitation

S(z) = Arect z−¯z

∆zej2π–

γGz(z−¯z)τp/2.

Application of a zgradient with strength −Gzfor a duration τp/2will exactly cancel this phase accumulation.

Solution 13.7

The reconstruction process will presume the following frequencies

u=γ–Gxt ,

v=γ–Ay.

A Fourier transform function (a discrete matrix, in practice) will be formed as follows

F(u, v) = s0u

γ–Gx

γ–,

=Ae−u/γ–GxT2Z∞

−∞Z∞

−∞

M(x, y; 0+)e−j2πxue−j2πyv dx dy .

Thus, F(u, v)is a product of the Fourier transform of M(x, y; 0+)with another Fourier function. This implies that

the inverse transform of F(u, v)will be M(x, y; 0+)convolved with a spatial kernel that depends only on x. At the

very least, the magnetization will be blurred to some extent in the xdirection because of this term.

To ﬁnd a mathematical description of this impulse response function requires an understanding of the pulse

sequence used. If the readout gradient is large, then the readout time Tswill be short, and the effect of the T2decay

will be minimized. In this case, the Fourier resolution of the scan adequately describes the effect of this term. If the

readout is long, however, then T2could have a signiﬁcant effect.

Given the pulse sequence shown in Figure 13.10, we can assume that frequencies in the range 0≤u≤γ–GxTs

are acquired for all v. The complete Fourier transform is reassembled by conjugate symmetry. In this case, the

function actually multiplying the Fourier transform of M(x, y; 0+)is given by

H(u, v) = exp {−|u|/γ–GxT2},

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219

which is separable as follows:

H1(u) = exp {−|u|/γ–GxT2},

H2(v)=1.

From tables or direct calculation, we have the Fourier transform pair

Fne−|x|o=2

1 + (2πu)2.

Therefore, we have

h1(x) = 2γ–GxT2

1 + (2πγ–GxT2x)2,

h2(y) = δ(y).

Ignoring all constant amplitude terms, the reconstructed image is

f(x, y) = M(x, y; 0+)∗1

1 + (2πγ–GxT2x)2δ(y),

h1(x)is a “Gaussian-like” function that gets wider as the product GxT2gets smaller. Thus, this demonstrates that

slower readouts or faster T2’s will cause blurring in the readout direction because of transverse relaxation.

It is worth noting that if the pulse sequence scans across the v-axis, acquiring both positive and negative u

frequencies, then conjugate symmetry will not hold because of this term. In this case, the reconstruction will be

both blurred, and will be a complex-valued image.

Solution 13.8

(a) The phase that is accumulated during a time-varying xgradient pulse is

φ(t) = γZt

Gx(τ)x(τ)dτ .

Therefore, the phase accumulation for the given gradient waveform and xtrajectory is

φ(T) = γZT

G(τ)(x0+vτ)dτ

=γZT

G(τ)x0dτ +γZT

G(τ)vτ dτ

= 0 + γZT/2

(−G)vτ dτ +γZT

T/2

(G)vτ dτ

=−Gγ vτ2

2

T/2

+Gγ vτ2

2

T/2

=Gγ vT 2

(b) Intuitively, we see that the ﬁnal phase should be zero. This is because (1) the pulse has zero area, so the term

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220 CHAPTER 13: MAGNETIC RESONANCE IMAGING

related to x0will cancel out and (2) starting at t=T, the pulse is symmetric in comparison to the ﬁrst half,

starting at t= 0, so the accumulated phase due to velocity in the second half should be the negative of that

of the ﬁrst half. We now go through the calculations:

φ(2T)

γ=Z2T

G(τ)(x0+vτ)dτ

=ZT/2

(−G)(x0+vτ)dτ +Z3T /2

T/2

G(x0+vτ)dτ +Z2T

3T/2

(−G)(x0+vτ)dτ

=ZT/2

(−G)x0dτ +ZT/2

(−G)vτdτ +Z3T /2

T/2

Gx0dτ +Z3T/2

T/2

Gvτdτ

+Z2T

3T/2

(−G)x0dτ +Z2T

3T/2

(−G)vτdτ

=−Gx0

2+−Gvτ2

2

T/2

+Gx0T+Gvτ2

2

3T/2

T/2

+ (−G)x0

2+−Gvτ2

2

3T/2

=−Gv(T/2)2

2+Gv(3T/2)2

2−Gv(T/2)2

2+−Gv(2T)2

2−−Gv(3T/2)2

=−GvT 2

8+9GvT 2

8−GvT 2

8+−4GvT 2

2−9(−G)vT 2

= (−1+9−1−16 + 9)GvT 2

= 0 .

in part (b), but it was a good check to do it out anyway.) Using past results we can ﬁnd the phase for the ﬁrst

waveform as follows:

φ1(T)

γ= 0 + GvT 2

4+ZT/2

(−G)1

2aτ2dτ +ZT

T/2

2aτ2dτ

=GvT 2

4+−Ga

τ3

3

T/2

+Ga

τ3

3

T/2

=GvT 2

4+−Ga

(T/2)3

3+Ga

3−Ga

(T/2)3

=GvT 2

4+GaT 3

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221

For the second waveform, we get

φ2(2T)

γ= 0 + 0 + ZT/2

(−G)1

2aτ2dτ +Z3T /2

T/2

2aτ2dτ +Z2T

3T/2

(−G)1

2aτ2dτ

=−GaT 3

48 +Ga

2τ33

3T/2

T/2

+−Ga

2τ33

3T/2

=−GaT 3

48 +Ga(3T/2)3

6−Ga(T/2)3

6+−Ga(2T)3

6−−Ga(3T/2)3

=−GaT 3

48 +27GaT 3

48 −GaT 3

48 −64GaT 3

48 +27GaT 3

= (−1 + 27 −1−64 + 27)GaT 3

=−GaT 3

We see that the ﬁrst gradient waveform has phase effects from both velocity and acceleration, while the

second only has phase effects from acceleration.

(d) We use the same “trick” that took us from part (a) to part (b). We invert the pulse sequence in part (b)

and replay it, as shown in Figure S13.1. The static position term and the velocity is still nulled because the

integrals evaluated in the second phase are still zero. The acceleration term is now nulled as well, since the

integral will be negated.

Figure S13.1 See Problem 13.8.

(e) It is possible. Now we need to show it. Consider the general gradient pulse of height Aand duration T

starting at t0. Suppose a particle has the position r(t) = x+vt +at2/2. What is the phase accumulation

after this pulse? We have gone through this exercise above, but in not quite this general way. Here we do it

again.

γA =Zt0+T

(x+vτ +aτ2

2)dτ

=xT +v

2(T2+ 2t0T) + a

6(3t2

0T+ 2T2t0+T3),

where the second equation follows after some algebra. Now consider a sequence of three pulses starting at

t= 0, with heights A,B, and C, and each of duration T. What is the phase accumulation after this sequence?

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222 CHAPTER 13: MAGNETIC RESONANCE IMAGING

We apply the previous result three times and add the result.

γ=xAT +Av

2(T2) + Aa

6(T3)

+BxT +Bv

2(3T2) + Ba

6(7T3)

+CxT +Cv

2(5T2) + Ca

6(19T3),

where the second equation follows after some algebra. Now the sequence should be independent of position,

which implies

A+B+C= 0 .

And the phase should be dependent on velocity, so the sum of the coefﬁcients multiplying vshould not be

zero, for convenience, we make that sum equal to T2/2, which implies

A+ 3B+ 5C= 1 .

And ﬁnally, there should be no dependence on acceleration, which implies

A+ 7B+ 19C= 0 .

Solving these equations for A,B, and Cyields

A=−1,

B= 1.5,

C=−0.5.

The implied pulse sequence is shown in Figure S13.2.

Figure S13.2 See Problem 13.8.

Solution 13.9

(a) The sampling interval is

T=10 ms

256 = 39.1µs

The FOV in the x-direction is

FOVx=1

42.58 MHz/T ×1G/cm ×39.1µs= 6 cm

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223

(b) The (nominal) pixel size is

∆x=FOVx

256 = 0.234 mm

The spatial extent of the cube is 5 cm, so the number of pixels across the object is 50 mm/0.234 mm = 213.3.

same doubles T, which in turn halves the FOV. Thus, there is no net effect on the FOV by making these

collective changes.

(d) This pulse sequence will require reassembling Fourier space by conjugate symmetry. It will scan higher

spatial frequencies than the previous pulse sequence. Nevertheless, the answers to (a) to (c) are the same.

Solution 13.10

(a) Use the following algebraic manipulation:

f=AM0sin αe−TE/T21−e−TR/T1

1−cos αe−TR/T1

f(1 −cos αe−TR/T1) = AM0sin αe−TE/T2(1 −e−TR/T1)

f−fcos αe−TR/T1

sin α=AM0e−TE/T2(1 −e−TR/T1)

sin α=e−TR/T1f

tan α+AM0e−TE/T2(1 −e−TR/T1)

(b) Let

x=f

tan α,

y=f

sin α.

Then the equation proven in part (a) becomes

y=e−TR/T1x+AM0e−TE/T2(1 −e−TR/T1),

which is the equation of a line with slope

m=e−TR/T1

and y-intercept

b=AM0e−TE/T2(1 −e−TR/T1).

Since only xand ywill vary when we change α, the computed points from the three acquisitions will form

different points on the same line.

m=y2−y1

x2−x1

=f2/sin α2−f1/sin α1

f2/tan α2−f1/tan α1

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224 CHAPTER 13: MAGNETIC RESONANCE IMAGING

Since m=e−TR/T1, we have

T1=−TR

ln m

=−TR

ln f2/sin α2−f1/sin α1

f2/tan α2−f1/tan α1

=TR

ln(f2/tan α2−f1/tan α1)−ln(f2/sin α2−f1/sin α1).

SAMPLING THE FREQUENCY SPACE

Solution 13.11

The basic relationship between the pulse sequence parameters and the Fourier frequencies is

u=γ–Gxt ,

v=γ–Ay.

There are four regions to consider.

(1) 0≤t≤1ms: The ucomponent is given by

u=γ–Gxt

= 4,258 Hz/G ×10 G/cm t

= 4,258 mm−1s−1t .

The vcomponent requires a determination of the area of the ygradient as a function of time. Since

Gy(t) = 10 G/cm

1ms t

= 1 ×103G

mm s−1t .

the area is

Ay(t) = Zt

Gy(τ)dτ

= 1 ×103G

mm s−1t2

Therefore,

v=γ–Ay(t)

= 4,258 Hz/G ×1×103G

mm sec−1t2

= 2.129 ×106mm−1sec−2t2.

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225

Writing vin terms of uyields:

v= 0.117u2,

both in units of mm−1. Thus, the ﬁrst segment of the Fourier trajectory is a parabola, starting at u=v= 0

and ending at (u, v) = (4.258,2.129) mm−1.

(2) 1ms ≤t≤2ms: In this interval, uis the same as in the previous interval. However, although vcontinues to

increase, its rate of increase is decreasing. We have

Gy(t) = 20 G/cm −10 G/cm

1ms t

= 2 G

mm −1×103G

mm s−1t .

Ay(t) = Zt

1ms Gy(τ)dτ

= 2τ−1×103τ2

2

0.001 s

=−500t2+ 2t−0.001 .

The vertical spatial frequency is then

v=γ–Ay(t)

= 4,258(−500t2+ 2t−0.001)

=−2.129 ×106t2+ 8,516t−4.258 .

This is a quadratic function that peaks at t= 2 ms. The ﬁnal value (at 2 ms) is v= 4.258 mm−1.

(3) 2ms ≤t≤3ms: This range can be worked out in a fashion similar to the work in intervals (1) and (2).

However, it is not necessary to do this since there is symmetry in the gradient pulses. Since both gradients

are negative in this range, their Fourier trajectories will be decreasing. Since Gxis constant, uwill decrease

with uniform speed. Since Gyis a linear function with negative slope, it will cause the vcomponent of the

trajectory to behave parabolically, as in intervals (1) and (2). For a given increment in time (and equivalently

a decrement in u), the drop in area in this time interval starts off small and gets larger over time. Therefore,

by appealing to symmetry, we see that the trajectory will trace over that of interval (2).

(4) 3ms ≤t≤4ms: Using a similar argument to that in interval (3), we see that this trajectory is identical to that

in interval (1), only traveling toward the origin.

A sketch of the resulting Fourier trajectory is shown in Figure S13.3. Since the net area of the two gradients is zero,

this pulse sequence ends exactly where it starts—at the origin.

Solution 13.12

(a) In this interval, the Fourier trajectory goes from the origin to the point (−0.25,−0.5) mm−1. Using the

relations:

u=γ–Gxt ,

v=γ–Gyt ,

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226 CHAPTER 13: MAGNETIC RESONANCE IMAGING

Figure S13.3 [Problem 13.11]

leads to

Gx=u

γ–t=−0.25 mm−1

4,258 Hz/G 0.0001 s=−0.587 G/mm .

Gy=v

γ–t=−0.5mm−1

4,258 Hz/G 0.0001 s=−1.174 G/mm .

(b) A similar argument as in (a) leads to

Gx=∆u

γ–Ts

=0.5mm−1

4,258 Hz/G 0.01 s= 11.7mG/mm ,

Gy=∆v

γ–Ts

=1.0mm−1

4,258 Hz/G 0.01 s= 23.5mG/mm .

The sampling rate is

fs=128

10 ms = 12.8kHz .

Solution 13.13

The required timing diagram is shown in Figure S13.4. The timings and the amplitudes of the gradients:

Figure S13.4 See Problem 13.13.

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227

kx=γ–ZGxdt, ky=γ–ZGydt

are determined by

−1mm−1=−1,000 m−1= 42.6×106×(−G1)×t1⇒G1×t1= 23.5×10−6T·s

m

0.5mm−1= 500 m−1= 42.6×106×G2×t1⇒G2×t1= 11.7×10−6T·s

m

2mm−1= 2000 m−1= 42.6×106×G1×t2⇒G1×t2= 47 ×10−6T·s

m

0.1mm−1= 100 m−1= 42.6×106×G3×t3⇒G3×t3= 2.3×10−6T·sec

m

−2mm−1=−2,000 m−1= 42.6×106×(−G1)×t4⇒G1×t4= 47 ×10−6T·s

m

Now, let G1= 10 mT/m, G2= 5 mT/m, and G3= 1 mT/m. Then, t1= 2.3ms, t2= 4.7ms, t3= 2.3ms, and

t4= 4.7ms.

Solution 13.14

Suppose the objects are at positions yoand y1. Also, let’s suppose that the amplitude of the signal from the point

samples is A0and A1respectively. If we acquire data from a pulse sequence, such as that shown in Figure 13.15,

with Gy= G0

y, we can apply a Fourier transform to the signal to yield a proﬁle So(x). Because the two point objects

are at the same xcoordinate, we are only interested in the value of the proﬁle at So(xo). We can write the equation

for the amplitude and phase of this value as

S0(xo) = A0e−iγ(G0

ytyyo)+A1e−iγ(G0

ytyy1).(S13.1)

In a separate acquisition, if we acquire data with Gy= G1

y, the amplitude and phase of this point in the proﬁle

becomes

S1(xo) = A0e−iγ(G1

ytyyo)+A1e−iγ(G1

ytyy1).(S13.2)

Because the Si(xo)are complex, the above equations represent four equations in four unknowns (Ao, A1, yo, y1);

therefore, if Ao, A1are real numbers, we can determine the ycoordinates of the objects exactly.

Solution 13.15

All three pulses are concerned with the phase of the precessing transverse magnetization. The refocusing lobe is

correcting the linear phase that is produced by the slice selection pulse. This is done by applying a negative gradient

to that of the slice selection for a duration half of that of the slice selection gradient. The phase-encoding gradient

is deliberately applying a linear phase in the phase encode (y) direction, so that Fourier space position is encoded.

The gradient echo formation lobe can be called a prefocusing lobe, since it essentially corrects the phase prior to

the readout gradient so that the spins are in phase at the center of the readout gradient.

Since all these phase corrections are accomplished by either increasing or decreasing the Larmor frequency in a

spatially encoded fashion, they can all be combined. The net effect will be that the overall phase of each point will

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228 CHAPTER 13: MAGNETIC RESONANCE IMAGING

arrive at its correct ﬁnal value faster than if each of these corrections had been done sequentially.

MR IMAGE RECONSTRUCTION

Solution 13.16

(a) The baseband signal is given by

s0(t) = Z∞

−∞Z∞

−∞

f(x, y)e−j2πγ–Gxxte−j2πγ–Gyyt dx dy

=Z∞

−∞Z∞

−∞

f(x, y)e−j2πγ–Gcos θ xte−j2πγ–Gsin θ yt dx dy

=Z∞

−∞Z∞

−∞

f(x, y)e−j2πγ–Gt(xcos θ+ysin θ)dx dy .

(b) The Fourier transform of f(x, y)is

F(u, v) = Ae−j2πu +Be+j2πv .

The baseband signal is sampling the Fourier transform according to the following formulas

u=γ–Gt cos θ ,

v=γ–Gt sin θ .

When θ= 0, we have

u=γ–Gt ,

v= 0 ,

and therefore

s0(t)|θ=0◦=F(γ–Gt, 0) = Ae−j2πγ–Gt +B .

This is a complex signal having two components, as shown in Figure S13.5(a). Similar reasoning yields the

following baseband signal for θ= 90◦

s0(t)|θ=90◦=F(0, γ–Gt) = A+Be+j2πγ–Gt ,

which is shown in Figure S13.5(b).

cross section), one needs to apply this basic pulse sequence for different values of θranging over 180 degrees,

say 0≤θ < π. This will cover half of Fourier space. The remainder is ﬁlled in using conjugate symmetry:

F(u, v) = F∗(−u, −v).

Since data are acquired along rays passing through the origin (and conjugate symmetry assures us that the

data are deﬁned in both directions), we can use convolution backprojection to do the reconstruction.

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229

Figure S13.5 Baseband signals. See Problem 13.16.

Solution 13.17

First, it is useful to see what happens during standard rectilinear scanning. All aspects of imaging are linear. So,

the object can be decomposed and analyzed separately:

f(x, y) = f1(x, y) + f2(x, y),

where f2(x, y)corresponds to the point object in the perturbed ﬁeld. By linearity, the resultant baseband signal is

the sum

s0(t) = s1(x, y) + s2(x, y).

During phase encoding, a phase equal to 2πγ–∆BTp, where Tpis the duration of the phase encode pulse, is added

to the signal arising from f2(x, y). During readout, a phase equal to −2πγ–∆Bt accumulates. Putting both terms

together yields the baseband signal equation for f2(x, y)

s2(t) = Z∞

−∞Z∞

−∞

f2(x, y)ej2πγ–∆BTpe−j2πγ–∆Bte−j2πγ–Gxxte−j2πγ–GyTpydx dy .

Using the usual equivalences

u=γ–Gxt ,

v=γ–GyTp,

yields

F(u, v) = ej2πγ–∆BTpe−j2π(∆B/Gx)uZ∞

−∞Z∞

−∞

f2(x, y)e−j2π(ux+vy)dx dy .

The inverse Fourier transform of ˆ

F(u, v)is

f(x, y) = ej2πγ–∆BTpf(x−(∆B/Gx), y).

The leading phase term is irrelevant, since the complex modulus is typically displayed. The second phase term,

since it was a linear phase term in the x-direction causes a shift of f2in the x-direction by ∆B/Gx. Thus, the

relative positions of f1(x, y)(which will be reconstructed correctly) and f2(x, y)will be altered in the x-direction,

but otherwise, the two point functions will be reconstructed correctly as point functions.

Reconstruction using the 2-D projection (polar scanning) method leads to a different—less desirable—result,

primarily because the readout direction is different with each excitation pulse. With no phase encoding in this pulse

sequence, there is no leading phase constant. However, both x- and y-gradients are on at the same time, in general.

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230 CHAPTER 13: MAGNETIC RESONANCE IMAGING

This gives the baseband signal for f2(x, y)as follows

s2(t) = Z∞

−∞Z∞

−∞

f2(x, y)e−j2πγ–∆Bte−j2πγ–(Gxx+Gyy)tdx dy .

Let the gradients be given by

Gx=¯

Gcos θ ,

Gy=¯

Gsin θ ,

and make the spatial frequency associations

u=γ–Gxt ,

v=γ–Gyt .

Then it follows that

%=pu2+v2=γ–t¯

G ,

and the observed signal becomes

G(%, θ) = s0%

γ–¯

G

=Z∞

−∞Z∞

−∞

f2(x, y)e−j2π(∆B/ ¯

G)%e−j2π(x% cos θ/ ¯

G+y% sin θ/ ¯

G)dx dy

=e−j2π(∆B/ ¯

G)%Z∞

−∞Z∞

−∞

f2(x, y)e−j2π(x% cos θ/ ¯

G+y% sin θ/ ¯

G)dx dy

=e−j2π(∆B/ ¯

G)%F2(%cos θ/ ¯

G, % sin θ/ ¯

G).

The inverse Fourier transform of this gives a projection of f2(x, y)shifted by ∆B/ ¯

G. Each projection is shifted by

the same amount on the `axis. For example, if f2(x, y) = δ(x, y), whose Radon transform is g(`, θ) = δ(`), the

observed “projection” would be ˆg=δ(`−∆B/ ¯

G).

Unfortunately, ˆgis not a Radon transform, which can be proven by showing that there is no object whose center

of mass agrees with the centers of mass of the collection of projections. This makes it difﬁcult to determine the

precise effect that this term will have on the resulting reconstruction. It should be apparent, however, that shifting

each projection of an impulse function by ∆B/ ¯

Gcreates a disk with radius ∆B/ ¯

Gin the object domain. Thus,

although the precise details are not developed here, we can conclude that rather than being shifted, as in the case of

rectilinear scanning, 2-D projection imaging will blur objects when the Larmor frequency varies across the FOV.

Solution 13.18

We have

s0(t) = Z∞

−∞Z∞

−∞

f(x, y)e−j2πγ–Gxxt dx dy

=Z∞

−∞Z∞

−∞

AMxy(x, y; 0+)e−t/T2e−j2πγ–Gxxt dx dy

=e−t/T2Z∞

−∞Z∞

−∞

AMxy(x, y; 0+)e−j2πγ–Gxxt dx dy .

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231

The inverse Fourier transform of s0(t)is

S0(ν) = F−1{s0(t)}

=F−1{e−t/T2}∗F−1Z∞

−∞Z∞

−∞

AMxy(x, y; 0+)e−j2πγ–Gxxt dx dy

=F−1{e−t/T2} ∗ Z∞

−∞Z∞

−∞

Mxy(x, y)e−iγGxxte+j2πνt dt dx dy

=F−1{e−t/T2} ∗ Z∞

−∞Z∞

−∞

Mxy(x, y)δ(2π(ν−γ–Gxx)) dx dy .

Therefore, the amplitude of the function S0(ν)at a speciﬁc frequency ν0will be proportional to the line integral

of Mxy(x, y; 0+)along the ydirection at the xposition x0=νo/γ–Gx; hence, S(ν)is a projection of the ob-

ject Mxy(x, y; 0+)onto the x-axis. This data, however, will be convolved with F−1{e−t/T 2}, which is called a

Lorentzian function. If the data are acquired fast enough, this convolution can be ignored.

MR IMAGE QUALITY

Solution 13.19

(a) With reference to Figure 13.6, we see that the through-plane direction coincides with the x-axis, which is

therefore the slice selection gradient direction as well. There is no choice in this matter.

In a standard (axial) image, it is customary to make the +ydirection be the phase encode direction. Although

this is arbitrary, provided that the direction is within the y-zplane, it makes sense to retain the +ydirection

as the phase encode direction [see part (b)].

The standard (axial) image uses the +xdirection as the frequency encode direction. The frequency encode

direction should be orthogonal to the phase encode direction, so it could be either the −zor +zdirection.

For simplicity, we choose the +zdirection to be the readout direction.

(b) Aliasing in the phase encode direction is prevented by making sure that the object being imaged is conﬁned

to the FOV in the phase encode direction,

FOVy=1

∆v,

where ∆vare the phase encode increments in the frequency domain.

With reference to Figure 13.6, it becomes a bit more clear, why the phase encode direction should be in the y

direction. In this ﬁgure, the extent of the patient in the ydirection is given by distance from the back to the

chest, whereas the extent in the zdirection is the entire height of the patient. It is usually the case that the

thinnest section of the patient is chosen as the phase-encode direction in an arbitrary scan.

Aliasing is prevented in the frequency encode direction by using an anti-aliasing ﬁlter prior to sampling, as

usual.

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232 CHAPTER 13: MAGNETIC RESONANCE IMAGING

Solution 13.20

(a) The spatial extent is deﬁned by the ﬁeld-of-view in the y(phase encode) direction, which is given by

FOVy=1

γ–∆Ay

Since γ– is ﬁxed, it is the change in the area of the phase encode gradient, ∆Ay, between imaging pulses

that determines the spatial extent in the ydirection. This parameter is independent of the number of phase

encodes—that is, 256 in the problem statement —that are acquired.

(b) The spatial extent in the readout direction is deﬁned by the ﬁeld-of-view in the x(readout) direction, which

is given by

FOVx=fs

γ–Gx

Thus, the sampling rate fsand the readout gradient strength Gxdetermine the spatial extent in the readout

direction. If that is the case, it must be assumed that the readout gradient is left on for a duration that is

long enough to collect 256 samples. It is also possible to view the duration of the readout interval Tsas a

parameter. In this case, the sampling rate is determined by the number of samples acquired over the interval,

fs= 256/Ts. Then the spatial extent can be written as

FOVx=256

γ–TsGx

From Section 13.4.2, we have

FWHMy=1

Nyγ–∆Ay

Here, Ny= 256, so the only parameter actually affecting resolution is ∆Ay. Increasing ∆Ayreduces

FWHMy, improving the resolution. But given the result in part (a), we see that this can only be done provided

that the spatial extent covers the object (otherwise wraparound will occur).

(d) In the readout direction, we have

FWHMx=1

γ–NxT Gx

Here, Nx= 256, so we can change either Tor Gxin order to change the spatial resolution in the readout

direction. Since the readout time is Ts=NxT, it is more fundamental to write

FWHMx=1

γ–TsGx

Now we see that the readout resolution is actually inversely proportional to the product TsGxand increasing

either Tsor Gxwill improve this resolution.

Solution 13.21

(a) The spatial extent in the diagonal direction is

D=√2×25.6cm = 36.2cm .

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233

There are 256 samples.

(b) The sampling rate in the xand ydirections are

fu= 1/∆v=FOVy= 25.6cm ,

fv= 1/∆u=FOVx= 25.6cm ,

But the sampling rate in the diagonal direction is lower, equal to

fd=25.6cm

√2= 18.1cm .

Solution 13.22

(a) The average power dissipated in the object is:

Pave =1

TZT

I2(t)Rdt ,

where Tis the period of the current and Ris the effective electrical resistance. Substitute I(t) = cos(2πν0t)

into the above equation, we have

Pave =1

TZT

cos2(2πν0t)Rdt

2πZ2π

cos2(u)du, let u= 2πν0t

Since the current in the coil is I(t) = cos(2πν0t), the magnetic ﬂux density is B1(t) = µ0NI(t). And the

induced voltage in a cylindrical shell of radius ris given by

V(t, r) = dφ

dt =d(AB1)

dt = 2π2ν0r2µ0Nsin(2πν0t),

where φis the magnetic ﬂux and A=πr2is the area subtended by the cylindrical shell.

(b) Given the differential conductance in a thin shell of radius r, the average power dissipated in the shell is:

dPave =1

TZT

V2(t, r)dGdt =1

(2π2ν0r2µ0N)2L

2πrρ dr.

The average power dissipated in the object can also be expressed using the voltage as:

Pave =Zr0

dPave =R

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234 CHAPTER 13: MAGNETIC RESONANCE IMAGING

Therefore, the effective electrical resistance is:

R= 2Pave

=Zr0

(2π2ν0r2µ0N)2L

2πrρ dr

=2π3ν2

0µ2

0N2L

ρZr0

r3dr

=π3ν2

0µ2

0N2Lr4

2ρ,

which is Equation (13.79).

Solution 13.23

Start with the imaging equation

f(x, y) = AM0sin αe−TE/T2(x,y)1−e−TR/T1

1−cos αe−TR/T1,

Using TE= 0 and α=π/2, and setting the overall gain to unity, gives

f(x, y) = 1 −e−TR/T1.

The two tissues have values

fa= 1 −e−TR/T a

1,andfb= 1 −e−TR/T b

so the image difference between these two tissues is

∆ = fa−fb

= (1 −e−TR/T a

1)−(1 −e−TR/T b

=e−TR/T b

1−e−TR/T a

Taking the derivative of this expression with respect to TRyields

d∆

dTR

(e−TR/T b

1−e−TR/T a

=−1

e−TR/T b

1−−1

e−TR/T a

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235

Now set this equal to zero and solve for ˆ

TR:

−1

e−TR/T b

1−−1

e−TR/T a

1= 0

1e−ˆ

TR/T a

1=Ta

1e−ˆ

TR/T b

ln Tb

1−ˆ

= ln Ta

1−ˆ

TR1

1−1

1= ln Ta

1−ln Tb

TR=ln Ta

1−ln Tb

1−1

Whether this expression should yield a maximum or minimum depends on the relationship between Ta

1and Tb

and hence the sign of ∆. Suppose Ta

1< T b

1; then ∆>0and our goal is to maximize ∆. Otherwise, our goal is

to minimize ∆(making the difference more negative). It is readily veriﬁed by plotting ∆as a function of TRthat

these conditions are satisﬁed by the expression we have derived for ˆ

TR.

Solution 13.24

The conventional magnet strength of whole-body scanners today is 1.5 T. Therefore, Larmor frequency is

f=γ–B0

= 42.58 MHz/T ×1.5T

= 63.84 MHz .

In air, the wavelength of radio frequency waves at this frequency is

λ=3×108m/s

63.84 ×106Hz

= 4.7m.

The Rayleigh limit is λ/2 = 2.35 meters. Clearly, MRI does not work according to conventional optical imaging

principles.

Solution 13.25

From Chapter 12, we know that the Larmor frequency of fat is

ν0(fat) = ν0(water)(1 −ς).

where ς= 3.35 ×10−6. The frequency difference is

∆ν0=−ςγ–B0,

which at 1.5 T is −214 Hz. But the above analysis is in a static ﬁeld. When a readout gradient is turned on, the

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236 CHAPTER 13: MAGNETIC RESONANCE IMAGING

Larmor frequencies of water and fat will be

ν0(water) = γ–(B0+Gxx),

ν0(fat) = γ–(B0+Gxx)−ςγ–(B0+Gxx).

After demodulation to baseband (assuming that the Larmor frequency of water is used for demodulation), the

following frequencies are encoded during the readout interval

ν(water) = γ–Gxx ,

ν(fat) = γ–Gxx−ςγ–(B0+Gxx).

Therefore, during frequency encoding (the readout interval), the fat signal will be slightly mispositioned in the

readout direction relative to that of water.

To suppress the fat signal, we can apply a 180-degree (so-called inversion) pulse and then wait for both compo-

nents of longitudinal magnetization to recover. Since the fat T1is so much shorter than the water T1, it will recover

faster, passing the Mz= 0 point at a predictable time. At that time, one can begin imaging with the application of

a standard π/2RF pulse. Because the fat signal has Mz= 0 at that time, it will not contribute to the transverse

signal, and only water will be imaged. This type of sequence, called inversion recovery sequence, can be tuned to

suppress the water signal as well by waiting for the longitudinal magnetization recovery of water to cross zero.

Solution 13.26

(a) When Gx→0.5Gx, the FOV doubles, that is, FOVx→2FOVx. Since image size remains unchanged, the

pixel size must double as well, that is, Vs→2Vs. Therefore, the SNR also doubles, that is,

SNR →2SNR .

(b) Ny→2Nymeans to double the number of phase encodes. If everything else is to remain unchanged, this

implies that these phase encodes either repeat the ﬁrst set of phase encodes or add on to those already acquired

but at a different vlocations in Fourier space. In either case, the net effect is to double the scan time; therefore,

the SNR will improve, but only by a √2factor. In other words,

SNR →√2SNR .

time has remained constant, that is, TA→TA. The only possible remaining factor affecting SNR is voxel

size. If the image size is assumed to follow J=Nx, then the image size has doubled, J→2J. But since

FOVxis inversely proportional to T, we also have that FOVx→2FOVx. Under this assumption, voxel size

is constant and SNR is constant

SNR →SNR ,assuming J=Nx.

(d) It is reasonable to make the assumption from the problem statement that the image size should not change,

that is, J→J. In this case, since the FOVxhas doubled, the voxel size would also double, Vs→2Vsand

SNR →2SNR ,assuming J→J .

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237

Figure S13.6 See Problem 13.27(a).

APPLICATIONS, EXTENSIONS, AND ADVANCED TOPICS

Solution 13.27

(a) The 2D function f(x, y)is

f(x, y) = rect x−5

10 ,y−5

10 .

It is sketched in Figure S13.6. The Fourier transform of f(x, y)is

F(u, v) = 10 sinc(10u)×10 sinc(10v)e−j2π(5u)e−j2π(5v),

and

|F(u, 0)|= 100 sinc(10u).

(b) We have

γ–Gxt= 42.58 kHz/G ·0.5G/mm ·t= 0.4cm−1,

and after solving yields

t= 18.788 µs.

(d) Perfect g(`, 0◦)needs perfect G(u, 0). We collect only partial information of G(u, 0), since we do not collect

data outside of −0.4cm−1≤u≤0.4cm−1. Therefore, we cannot get a perfect reconstruction of g(`, 0◦).

Solution 13.28

(a) We have

ω0=γB0= 2π×4,258 (rad/s)/G ×1.5T×104G/T = 4.01 ×108rad/s ,

f0= 6.39 ×107Hz .

The tip angle can be computed as

α=γZ1×10−3

−1×10−3

1(t)dt = 2π×4,258 Z1×10−3

−1×10−3

2×4.258 ×104sinc(4.258 ×104t)dt .

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238 CHAPTER 13: MAGNETIC RESONANCE IMAGING

Figure S13.7 See Problem 13.28(b).

Figure S13.8 See Problem 13.28(d).

(b) Carry out the following manipulations

B1(t) = A∆fsinc(∆ft)ejω0t,

F{B1(t)}=Arect f−f0

∆f,

= 2 rect f−6.39 ×107

4.258 ×104,

∆ω= 2π∆f= 267,538 rad/s .

ωc=γ(B0+Gz·zc) = ω0⇒zc= 0

∆ω=γGz·∆z⇒∆z=∆ω

γGz

=267,538

1π·4,258 ·2= 5 cm

(d) We have

c=zc+ ∆z= 5 cm ,

ω0

c=γ(B0+Gz·z0

c) = ω0+γGz·z0

c=ω0+ 2π·4,258 ·2·5 = ω0+ 2.68 ×105rad/s .

So in order to select the adjacent slice, we need

1(t) = A∆fsinc(∆ft)ej(ω0+ω0

c)t.

(e) Carry out these steps:

ω1=γ(B0+Gzz1),

ω2=γ(B0+Gzz2),

∆ω=γGz∆z ,

∆φ= ∆ω·τp

2=γGz∆zτp

2= 2π·4,258 ·2·5·1×10−3= 267.5rad .

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239

Figure S13.9 See Problem 13.28(f).

In order to rephase the spins, we need to use a refocussing lobe on Gz.

(f) We have

kx=γ

2πZt1

Gx(τ)dτ =γ

2πGx·t1

ky=γ

2πZt2

Gy(τ)dτ =γ

2πGy·t2

t1=kx

2πGx

= 2 ms

t2=ky

2πGy

= 0.5ms

The pulse sequence is shown in Figure S13.9.

(g) We have

FOVx=2π

∆kx

=2π

γGxT=2πfs

−γ·Gx

fs=FOVx· −γ·Gx

2π=50 cm ·4,258 ·2π·2.5

2π= 5.32 ×105Hz .

Solution 13.29

(a) We will use phase encoding in both the yand zdirections. The pulse sequence shown in Figure S13.10, a

modiﬁcation of Figure 13.16, shows the general idea (though this is not a realistic pulse sequence).

(b) The following baseband signal is the same as the standard 2D gradient echo pulse sequence, except that there

is a new term that depends on the zphase encoding:

s0(t) = Z∞

−∞Z∞

−∞

f(x, y)e−j2πγ–Gxxte−j2πγ–GyTpye−j2πγ–GzTpzdx dy .

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240 CHAPTER 13: MAGNETIC RESONANCE IMAGING

Figure S13.10 See Problem 13.29.

above expression to the 3-D Fourier transform yields the following identiﬁcations:

u=γ–Gxt ,

v=γ–GyTp,

w=γ–GzTp.

Therefore, the 3D Fourier transform F(u, v, w)of f(x, y, z)is built up by successive imaging pulses as

follows

F(u, γ–GyTp, γ–GzTp) = s0u

γ–Gx0≤u≤γ–GxTs.

where Gyand Gzmust take on a series of different values in order to cover 3D Fourier space.

Solution 13.30

In general, if we have N points, we can write the set of signal equations as

Sm(xo) =

j=N

j=0

Aje−iγ(Gm

ytyyj),(S13.3)

which is 2N equations with 2N unknowns. Figure S13.11 shows one possible scheme for changing the value of the

phase-encoding gradient amplitudes. For each acquisition, we have Gm

y=m∆Gy.

It is important to note that while the data acquisition is similar to that used in spin-warp imaging, the procedure

described here for reconstructing the MR image is not what is usually done. Usually, the positions of the objects are

assumed to be known; that is, we reconstruct the signal amplitudes (and phases) of a set of decaying oscillators that

are sitting on a ﬁxed coordinate grid by using the Fast Fourier Transform (FFT). This coordinate grid is the pixel

array in the image. If the assumption is not true (which in most cases it is not), image artifacts from Gibbs ringing

occur.

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241

Figure S13.11 Pulse sequence for phase encoding. See Problem 13.30.

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Medical Imaging Signals And Systems Solutions Manual

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