Solution Manual for Medical Imaging Signals and Systems, 2nd Edition

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Medical Imaging
Signals and Systems (2e):
Solutions Manual
Version 1.4 (December 21, 2016)
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

2
Signals and Systems
SIGNALS AND THEIR PROPERTIES
Solution 2.1
(a) δs(x, y) = ∑∞
m=−∞
∑∞
n=−∞ δ(x − m, y − n) = ∑∞
m=−∞ δ(x − m) · ∑∞
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.
(c) e(x, y) = exp[j2π(u0x+v0y)] = exp(j2πu0x)·exp(j2πv0y) = e1D(x; u0)·e1D(y; v0), where e1D(t; ω) =
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 x and a 1-D sinusoidal signal in y. But in general, when both u0 and v0 are 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 M and 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) .
2

3
So the smallest period is 1 in both x and y directions.
(c) Periodic. Let f (x + Tx, y) = f (x, y), we have
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 x is
Tx0 = 1. Similarly, we find that the smallest period for y is 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 x is Tx0 = 1 and the smallest period for y is 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 Tx such that f (x + Tx, y) = f (x, y), and
sin(2π(x2 + y2)) = sin(2π((x + Tx)2 + y2))
= sin(2π(x2 + y2 + 2xTx + T 2
x )) .
In order for the above equation to hold, we must have that 2xTx + T 2
x = k for some integer k. The solution
for Tx depends 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
5 m
)
cos
( π
5 n
)
= sin
( π
5 (m + M )
)
cos
( π
5 n
)
.
Solving for M , we find that M = 10k for any integer k. The smallest period for both m and n is 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
5 m
)
cos
( 1
5 n
)
= sin
( 1
5 (m + M )
)
cos
( 1
5 n
)
.
The solution for M is 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 k that solves the equality for M = 10kπ for some M . So,
fd(m, n) = sin ( 1
5 m) cos ( 1
5 n) is not periodic.
Solution 2.3
(a) We have
E∞(δs) =
∫ ∞
−∞
∫ ∞
−∞
δ2
s (x, y) dx dy
= lim
X→∞ lim
Y →∞
∫ X
−X
∫ Y
−Y
∞∑
m=−∞
∞∑
n=−∞
δ(x − m, y − n) dx dy
= lim
X→∞ lim
Y →∞(2bXc + 1)(2bY c + 1)
= ∞ ,

4 CHAPTER 2: SIGNALS AND SYSTEMS
where bXc is the greatest integer that is smaller than or equal to X. We also have
P∞ (δs) = lim
X→∞ lim
Y →∞
1
4XY
∫ X
−X
∫ Y
−Y
δ2
s (x, y) dx dy
= lim
X→∞ lim
Y →∞
1
4XY
∫ X
−X
∫ Y
−Y
∞∑
m=−∞
∞∑
n=−∞
δ(x − m, y − n) dx dy
= lim
X→∞ lim
Y →∞
(2bXc + 1)(2bY c + 1)
4XY
= lim
X→∞ lim
Y →∞
{ 4bXcbY c
4XY + 2bXc + 2bY c
4XY + 1
4XY
}
=

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5
that
P∞(δl) = lim
X→∞ lim
Y →∞
1
4XY
∫ X
−X
∫ Y
−Y
δ(x) dx dy
= lim
X→∞ lim
Y →∞
1
4XY
∫ Y
−Y
{∫ X
−X
δ(x)dx
}
dy
= lim
X→∞ lim
Y →∞
1
4XY
∫ Y
−Y
1dx
= lim
X→∞ lim
Y →∞
2Y
4XY
= lim
X→∞
1
2X
= 0 .
(c) We have
E∞(e) =
∫ ∞
−∞
∫ ∞
−∞
|exp [j2π(u0x + v0y)]|2 dx dy
=
∫ ∞
−∞
∫ ∞
−∞
1 dx dy
= ∞ .
And also
P∞(e) = lim
X→∞ lim
Y →∞
1
4XY
∫ X
−X
∫ Y
−Y
| exp[j2π(u0x + v0y)]|2 dx dy
= lim
X→∞ lim
Y →∞
1
4XY
∫ X
−X
∫ Y
−Y
1 dx dy
= 1 .
(d) We have
E∞(s) =
∫ ∞
−∞
∫ ∞
−∞
sin2[2π(u0x + v0y)] dx dy
©2
=
∫ ∞
−∞
∫ ∞
−∞
1 − cos[4π(u0x + v0y)]
2 dx dy
=
∫ ∞
−∞
∫ ∞
−∞
1
2 dx dy −
∫ ∞
−∞
∫

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8 CHAPTER 2: SIGNALS AND SYSTEMS
Solution 2.5
1. Suppose the PSF of an LSI system is absolutely integrable.
∫ ∞
−∞
∫ ∞
−∞
|h(x, y)| dx dy ≤ C < ∞ (S2.1)
where C is 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)|
=
∣
∣
∣
∣
∫ ∞
−∞
∫ ∞
−∞
h(x − ξ, y − η)f (ξ, η)dξdη
∣
∣
∣
∣
≤
∫ ∞
−∞
∫ ∞
−∞
|h(x − ξ, y − η)| · |f (ξ, η)| dξdη
≤ B
∫ ∞
−∞
∫ ∞
−∞
|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,
∫ ∞
−∞
∫ ∞
−∞
|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)| =
∫ ∞
−∞
∫ ∞
−∞
|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 g′(x, y

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12 CHAPTER 2: SIGNALS AND SYSTEMS
(d)
δ(x − 1, y − 2) ∗ f (x + 1, y + 2) ©3
=
∫ ∞
−∞
∫ ∞
−∞
δ(x − ξ − 1, y − η − 2)f (ξ + 1, η + 2)dξ dη
©4
=
∫ ∞
−∞
∫ ∞
−∞
δ(x − ξ − 1, y − η − 2)f ((x − 1) + 1, (y − 2) + 2)dξ dη
=
∫ ∞
−∞
∫ ∞
−∞
δ(x − ξ − 1, y − η − 2)f (x, y)dξ dη
©5
= f (x, y) = x + y2
©3 comes from the definition of convolution; ©4 comes from the Eq. (2.7) in text; ©5 is the same as

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13
Solution 2.12
g(x, y) = f (x, y) ∗ h(x, y)
=
∫ ∞
−∞
∫ ∞
−∞
f (x − ξ, y − η)h(ξ, η)dξdη
=
∫ ∞
−∞
∫ ∞
−∞
(x − ξ + y − η)exp{−(ξ2 + η2)}dξdη
= (x + y)
∫ ∞
−∞
∫ ∞
−∞
e−ξ2−η2
dξdη −
∫ ∞
−∞
∫ ∞
−∞
ξe−ξ2−η2
dξdη −
∫ ∞
−∞
∫ ∞
−∞
ηe−ξ2−η2
dξdη
= (x + y)
[∫ ∞
−∞
e−ξ2
dξ
]2
−
∫ ∞
−∞
e−η2
[∫ ∞
−∞

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