1
S
OLUTIONS
TO
ECE2025 P
ROBLEM
S
ET
#13, A
PRIL
24, 2009
PROBLEM 13.1.
(a)
Since

j
ω

=

ω

, we get the following:
(b)
Since
v
(
t
) =
x
(
t
)cos(66
π
t
)
, the modulation property of Table 113 tells us that
V
(
j
ω
)
can be found by shifting
X
(
j
ω
)
up and down by
66
π
, then multiplying by
1/2
, so that:
V
(
j
ω
) =
X
(
j
(
ω
–
66
π
)) +
X
(
j
(
ω
+ 66
π
)):
Similarly, since
w
(
t
) =
v
(
t
)cos(15
π
t
)
we get:
W
(
j
ω
) =
V
(
j
(
ω
–
15
π
)) +
V
(
j
(
ω
+ 15
π
)):
Finally, since
z
(
t
) =
w
(
t
)cos(51
π
t
)
we get:
Z
(
j
ω
) =
W
(
j
(
ω
–
51
π
)) +
W
(
j
(
ω
+ 51
π
)):
(c)
The picture above shows that
z
(
t
)
=
x
(
t
)/4 +
highfrequency terms, so we can recover
x
(
t
)
by applying
z
(
t
)
to a lowpass filter which passes –
7
π
<
ω
<
7
π
with a gain of
4
,
and its stop band must start at
23
π
, so that the frequency response is zero for

ω

>
23
π
:
7
π
–
7
π
7
π

X
(
j
ω
)

ω


73
π
7
π
/2

V
(
j
ω
)

ω
7
π
/2
59
π
–
73
π
–
59
π
66
π
–
66
π
0


58
π

W
(
j
ω
)

ω
44
π
51
π
–
81
π
88
π
7
π
/4
7
π
/4
74
π
81
π
–
51
π
0



Z
(
j
ω
)

ω
102
π
–
30
π
7
π
/8
7
π
/4
132
π
0 7
π
–
132
π
7
π
/8
7
π
/8
30
π
–
102
π
–
7
π
23
π
H
(
j
ω
)
ω
–
23
π
0 7
π
–
7
π
23
π
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2
PROBLEM 13.2.
From Table 112 we see that the Fourier transform of the pulse train
p
(
t
)
=
∑
n
δ
(
t
–
nT
s
)
with period
T
s
is another pulse train
P
(
j
ω
) =
ω
s
∑
k
δ
(
ω
–
k
ω
s
),
where
ω
s
= 2
π
/
T
s
. Since
x
s
(
t
) =
x
(
t
)
p
(
t
)
, the multiplication property of Table 113 tells us that the Fourier
transform
X
s
(
j
ω
)
can be found by convolving the triangle
X
(
j
ω
)
with the pulse train
P
(
j
ω
)
. This places a copy of the triangle at every integer multiple of the sampling
frequency, yielding:
X
s
(
j
ω
) =
∑
k
X
(
ω
–
k
ω
s
).
(a)
We must avoid overlapping triangles if we want
x
r
(
t
)
=
x
(
t
)
. Therefore, the smallest
sampling rate is twice the maximum frequency in
x
(
t
)
, namely
ω
s
=
2
ω
max
=
160
π
. The
choice
ω
s
=
2
ω
max
is known as the Nyquist rate. With this choice, the Fourier transform
of
x
s
(
t
)
looks like this:
(b)
Putting a copy of the triangle at every integer multiple of
150
π
will result in overlapping
triangles, i.e. aliasing, as shown below:
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 Spring '08
 JUANG
 Digital Signal Processing, Signal Processing, Lowpass filter, jω, LPF

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