Chapter 9: Sampling and Quantization
Problem 9.1
(a)
(
)
1
1 200
200
( )
5sinc 200
sinc
40
t
x t
t
π
π
π
π
×
⎛
⎞
=
=
⎜
⎟
⎝
⎠
. Therefore
(
)
1
400
1/ 40
200
1
(
)=
rect
0
200 .
40
X
ω
π
ω
π
ω
ω
π
⎧
≤
=
⎨
>
⎩
The maximum frequency is given by 100 Hz. Based on the Nyquist theorem, the maximum
sampling period is
1
5
200
s
T
s
ms
=
=
.
(b)
(
)
(
)
2
max
100
max
50
max
100
( )
5sinc 200
8sin 100
freq
Hz
freq
Hz
freq
Hz
x
t
t
t
π
=
=
=
=
+
±²³²´
±²²³²²´
±²²²²³²²²²´
.
Therefore, maximum sampling period is given by
1
5
200
s
T
s
ms
=
=
.
Fig. S9.1:
A
(
f
) denotes the spectrum of sinc(200
t
) and
B
(
f
) denotes the spectrum of sin(100
π
t
).
X
3
(
f
)
is the sum of two shifted replicas of
A
(
f
), and is non-zero within the band [
−
150,150] Hz.
(c)
Since
x
3
(
t
) is a product of sinc(200
t
) and sin(100
π
t
), the spectrum of
x
3
(
t
) can be obtained by
convolving the spectrums of sinc(200
t
) and sin(100
π
t
). From the theory of CT convolution, it can
easily be seen that (see Fig. S9.1), the maximum frequency present in
x
3
(
t
) is 150 Hz. Therefore, the
maximum sampling period is given by

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