Solutions Manual for Medical Imaging Signals and Systems 2nd Edition by Prince IBSN 9780132145183
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Solutions Manual for Medical Imaging Signals and Systems 2nd Edition by Prince IBSN 9780132145183
Full download: http://downloadlink.org/p/solutionsmanualformedicalimagingsignalsandsystems2ndeditionbyprinceibsn9780132145183/
2
Signals and S
ys
t
e
ms
S
IGNALS AND THEIR
PROPERTIES
Solution 2.1(a)
δ
s
(x, y)
=
P
∞
−∞
separable signal.
P
∞
n
=
−∞
δ
(x
−
m, y
−
n)
=
P
∞
δ
(x
−
m)
·
P
∞
δ
(y
−
n), therefore it is a
(b)
δ
l
(x, y) is separable if sin(2
θ
)
=
0. In this case, either sin
θ
=
0 or cos
θ
=
0,
δ
l
(x, y) is a product of a constant function in one axis and a 1D delta function in another. But in general,
δ
l
(x, y) is not separable. (c) e(x, y)
=
exp[
j
2
π
(
u
0
x
+
v
0
y)]
=
exp(
j
2
π
u
0
x
)
·
exp(
j
2
π
v
0
y)
=
e
1D
(x; u
0
)
·
e
1D
(y;
v
0
), where e
1D
(t;
ω
)
=
exp(
j
2
π
ω
t
)
.
Therefore, e(x, y) is a separable
signal.
(d) s(x, y) is a separable signal when u
0
v
0
=
0. For example, if u
0
=
0, s(x, y)
=
sin(2
π
v
0
y) is the product of a constant signal in x and a 1D sinusoidal signal in y. But in general, when both u
0
and v
0
are nonzero, s(x, y) is not separable. Solution 2.2 (a) Not periodic.
δ
(x, y) is nonzero only when x
=
y
=
0. (b) Periodic. By definition
∞
comb(x, y)
=
X
∞
X
δ
(x
−
m, y
−
n)
.
For arbitrary integers M and N , we have
m=
−∞
n
=
−∞
comb(x
+
M, y
+
N )
=
∞ ∞
X X
δ
(x
−
m
+
M, y
−
n
+
N
)
m=
−∞
n
=
−∞
∞ ∞
=
X X
δ
(x
−
p, y
−
q)
[let
p
=
m
−
M, q
=
n
−
N
]
p
=
−∞
q
=
−∞
=
comb(x, y)
.
2
3 3
CHAPTER
2
:
SIGNALS
AND SYSTEMS
x
x
s
So the smallest period is 1 in both x and y directions. (c) Periodic. Let
f
(x
+
T
x
, y)
=
f
(x, y), we have sin(2
π
x) cos(4
πy
)
=
sin(2
π
(x
+
T
x
)) cos(4
πy
)
.
Solving the above equation, we have
2
π
T
x
=
2k
π
for arbitrary integer k. So the smallest period for x is
T
x0
=
1. Similarly, we find that the smallest period for y is T
y0
=
1
/
2.
(d) Periodic. Let
f
(x
+
T
x
, y)
=
f
(x, y), we have sin(2
π
(x
+
y))
=
sin(2
π
(x
+ T
x
+
y
))
.
So the smallest period for x is
T
x0
=
1 and the smallest period for y is T
y0
=
1. (e) Not periodic. We can see this by contradiction. Suppose
f
(x, y)
= sin(2
π
(x
2
+
y
2
)) is periodic; then there exists some
T
x
such that
f
(x
+
T
x
, y)
=
f
(x, y), and
sin(2
π
(x
2
+
y
2
))
= sin(2
π
((x +
T
x
)
2
+
y
2
))
= sin(2
π
(x
2
+
y
2
+ 2xT
x
+
T
2
))
.
In order for the above equation to hold, we must have that
2xT
x
+
T
2
=
k for some integer k. The solution for
T
x
depends on x. So
f
(x, y)
= sin(2
π
(x
2
+
y
2
)) is not periodic. (f) Periodic. Let
f
d
(m
+
M, n)
=
f
d
(m, n). Then sin
π
m cos
π
n
=
sin
π
(m
+
M ) cos
π
n
.
5 5 5 5 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
f
d
(m
+
M, n)
=
f
d
(m, n), and then
sin
1
m
5
1 cos
n
5
= sin
1
(m
+
M
)
5
1
cos n
.
5
The solution for M is M
=
10k
π
. Since
f
d
(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
=
10k
π
for some M . So,
f
d
(m, n)
=
sin
1
m cos
1
n is not periodic.
5
5
Solution 2.3 (a) We have
E
∞
(
δ
s
)
=
Z
∞
Z
∞
δ
2
(x, y) dx dy
−∞
=
lim
−∞
lim Z
X
Z
Y
∞
∞
X
δ
(x
−
m, y
−
n) dx dy
X
→∞
Y
→∞
−
X
−
Y
m=
−∞
n
=
−∞
=
lim lim
(2
b
X
c
+
1)(2
b
Y
c
+ 1)
X
→∞
Y
→∞
=
∞
,
4 4
CHAPTER
2
:
SIGNALS
AND SYSTEMS
s
2
where
b
X
c
is the greatest integer that is smaller than or equal to X . We also have
P
∞
(
δ
s
)
= lim
lim 1 Z
X
Z
Y
δ
2
(x, y) dx dy
X
→∞
Y
→∞
4X
Y
−
X
−
Y
=
lim lim 1 Z
X
Z
Y
∞
∞
X
δ
(x
−
m, y
−
n) dx dy
X
→∞
Y
→∞
4X
Y
−
X
−
Y
m=
−∞
n
=
−∞
=
lim lim
(2
b
X
c
+
1)(2
b
Y
c
+ 1)
X
→∞
Y
→∞
4X
Y
4
b
X
cb
Y
c
2
b
X
c
+
2
b
Y
c
1
=
lim lim
+ +
X
→∞
Y
→∞
=
1
.
4X
Y
4X
Y
4X
Y
(b) We have
E
∞
(
δ
l
)
=
Z
∞
Z
∞

δ
(x cos
θ
+
y sin
θ
−
l
)

2
dx dy
−∞
−∞
Z
∞
Z
∞
=
δ
(x cos
θ
+
y sin
θ
−
l) dx dy
−∞
−∞
Z
∞
1
dx, sin
θ
=
0

sin
θ

1
−∞
=
Z
∞
1
E
∞
(
δ
l
)
=
∞
.

cos
θ

dy, cos
θ
=
0
−∞
Equality
1
comes from the scaling property of the point impulse. The 1D version of Eq. (2.8) in the text is
δ
(ax)
=
1
δ
(x). Suppose cos
θ
=
0. Then

a

Therefore,
δ
(x cos
θ
+
y sin
θ
−
l)
=
Z
∞
1
δ

cos
θ

x
+
y sin
θ
cos
θ
1
l
−
cos
θ
.
We also have
δ
(x cos
θ
+
y sin
θ
−
l)dx
=
−∞
.

cos
θ

P
∞
(
δ
l
)
= lim
lim 1 Z
X
Z
Y

δ
(x cos
θ
+
y sin
θ
−
l)

dx dy
X
→∞
Y
→∞
4X
Y
−
X
−
Y
=
lim lim 1 Z
X
Z
Y
δ
(x cos
θ
+
y sin
θ
−
l)dx dy
.
X
→∞
Y
→∞
4X
Y
−
X
−
Y
Without loss of generality, assume
θ
=
0 and l
=
0, so that we have sin
θ
=
0 and cos
θ
=
1. Then it follows