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Introduction to Signal Processing
Copyright © by Hayder Radha
 1 
ECE 366
Homework Set #11
Due Friday, April 24, 2009 (In class)
ECE 366 – Introduction to Signal Processing
Spring 2009
Michigan State University
Department of Electrical and Computer Engineering
Please remember to follow the rules and policies outlined in the ECE 366 Syllabus
[1]
Problem 8.11
The highest frequency for
()
1
x
t
is:
5
1
10
B
=
Hz.
Therefore, the Nyquist rate is:
1
5
1
22
1
0
s
fB
==
×
samples/second.
The highest frequency for
1
x
t
is:
5
2
1.5 10
B
=×
Hz.
Therefore, the Nyquist rate is:
2
5
2
23
1
0
s
×
samples/second.
For
2
1
x
t
, this is equivalent to multiplying
( )
1
x
t
by itself. Since:
(1)
Multiplication in the time domain is equivalent to convolution in the frequency
domain.
(2)
The width of the convolution of two functions is the sum of the widths,
Then the highest bandwidth of
( )
2
1
x
t
is:
2
1
5
21
0
xt
B
Hz.
Therefore, the Nyquist rate is:
11
5
24
1
0
×
samples/second.
Similarly, for
3
2
x
t
:
The highest bandwidth of
3
2
x
t
is:
2
2
555
5
1.5 10
1.5 10
1.5 10
4.5 10
B
+×
Hz.
Therefore, the Nyquist rate is:
5
29
1
0
×
samples/second.
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View Full Document Introduction to Signal Processing
Copyright © by Hayder Radha
 2 
ECE 366
For
() ()
12
x tx t
:
The highest bandwidth is:
55
5
10
1.5 10
2.5 10
BB
+=
+×
=
×
Hz.
Therefore, the Nyquist rate is:
22
.
51
0 51
0
xtx t
f
=× × =×
samples/second.
Introduction to Signal Processing
Copyright © by Hayder Radha
 3 
ECE 366
[2]
Problem 8.15
As can be seen from the following two figures,
The minimum sampling frequency required is
22
At this sampling rate, the uncorrupted part of the signal can still be recovered.
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This note was uploaded on 06/08/2009 for the course ECE 366 taught by Professor Staff during the Spring '08 term at Michigan State University.
 Spring '08
 STAFF
 Signal Processing

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