Exam (elaborations)
Solutions Manual for Principles of Communications 7th Edition Ziemer
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- Alderson-Broaddus College
Solutions Manual for Principles of Communications 7th Edition Ziemer
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SOLUTIONS MANUAL FORPRINCIPLES OF
COMMUNICATIONS 7TH
EDITION ZIEMER
,Principles of Communications 7th Edition Ziemer Solutions Manual
Chapter 2
Signal and Linear System Analysis
2.1 Problem Solutions
Problem 2.1
a. For the single-sided spectra, write the signal as
x1 (t) = 10 cos(4 t + =8) + 6 sin(8 t + 3 =4)
= 10 cos(4 t + =8) + 6 cos(8 t + 3 =4 =2)
= 10 cos(4 t + =8) + 6 cos(8 t + =4)
h i
= Re 10ej(4 t+ =8) + 6ej(8 t+ =4)
For the double-sided spectra, write the signal in terms of complex exponentials using Euler’s
theorem:
x1 (t) = 5 exp[j(4 t + =8)] + 5 exp[ j(4 t + =8)]
+3 exp[j(8 t + 3 =4)] + 3 exp[ j(8 t + 3 =4)]
The spectra are plotted in Fig. 2.1.
b. Write the given signal as
h i
x2 (t) = Re 8ej(2 t+ =3)
+ 4ej(6 t+ =4)
to plot the single-sided spectra. For the double-side spectra, write it as
x2 (t) = 4ej(2 t+ =3)
+ 4e j(2 t+ =3)
+ 2ej(6 t+ =4)
+ 2e j(6 t+ =4)
The spectra are plotted in Fig. 2.2.
1
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,2 CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
c. Change the sines to cosines by subtracting =2 from their arguments to get
x3 (t) = 2 cos (4 t + =8 =2) + 12 cos (10 t =2)
= 2 cos (4 t 3 =8) + 12 cos (10 t =2)
h i
= Re 2ej(4 t 3 =8) + 12ej(10 t =2)
= ej(4 t 3 =8)
+e j(4 t 3 =8)
+ 6ej(10 t =2)
+ 6e j(10 t =2)
Spectral plots are given in Fig. 2.3.
d. Use a trig identity to write
3 sin (18 t + =2) = 3 cos (18 t)
and get
x4 (t) = 2 cos (7 t + =4) + 3 cos (18 t)
h i
= Re 2ej(7 t+ =4) + 3ej18 t
= ej(7 t+ =4)
+e j(7 t+ =4)
+ 1:5ej18 t
+ 1:5e j18 t
From this it is seen that the singe-sided amplitude spectrum consists of lines of amplitudes
2 and 3 at frequencies of 3.5 and 9 Hz, respectively, and the phase spectrum consists of
a line of height =4 at 3.5 Hz. The double-sided amplitude spectrum consists of lines of
amplitudes 1, 1, 1.5, and 1.5 at frequencies of 3.5, -3.5, 9, and -9 Hz, respectively. The
double-sided phase spectrum consists of lines of heights =4 and =4 at frequencies 3.5
Hz and 3:5 Hz, respectively.
e. Use sin (2 t) = cos (2 t =2) to write
x5 (t) = 5 cos (2 t =2) + 4 cos (5 t + =4)
h i
= Re 5ej(2 t =2)
+ 4ej(5 t+ =4)
= 2:5ej(2 t =2)
+ 2:5e j(2 t =2)
+ 2ej(5 t+ =4)
+ 2e j(5 t+ =4)
From this it is seen that the singe-sided amplitude spectrum consists of lines of amplitudes
5 and 4 at frequencies of 1 and 2.5 Hz, respectively, and the phase spectrum consists of
lines of heights =2 and =4 at 1 and 2.5 Hz, respectively. The double-sided amplitude
spectrum consists of lines of amplitudes 2.5, 2.5, 2, and 2 at frequencies of 1, -1, 2.5, and
-2.5 Hz, respectively. The double-sided phase spectrum consists of lines of heights =2,
=2, =4, and =4 at frequencies of 1, -1, 2.5, and -2.5 Hz, respectively.
, 2.1. PROBLEM SOLUTIONS 3
Single sided Double sided
6
10
8
4
Amplitude
Amplitude
6
4 2
2
0 0
0 1 2 3 4 5 -5 0 5
f, Hz f, Hz
0.8
0.5
0.6
Phase, rad
Phase, rad
0.4 0
0.2
-0.5
0
0 1 2 3 4 5 -5 0 5
f, Hz f, Hz
f. Use sin (10 t + =6) = cos (10 t + =6 =2) = cos (10 t =3) to write
x6 (t) = 3 cos (4 t + =8) + 4 cos (10 t =3)
h i
= Re 3ej(4 t+ =8) + 4ej(10 t =3)
= 1:5ej(4 t+ =8)
+ 1:5e j(4 t+ =8)
+ 2ej10 t =3)
+ 2e j(10 t =3)
From this it is seen that the singe-sided amplitude spectrum consists of lines of amplitudes
3 and 4 at frequencies of 2 and 5 Hz, respectively, and the phase spectrum consists of
lines of heights =8 and =3 at 2 and 5 Hz, respectively. The double-sided amplitude
spectrum consists of lines of amplitudes 1.5, 1.5, 2, and 2 at frequencies of 2, -2, 5, and -5
Hz, respectively. The double-sided phase spectrum consists of lines of heights =8, =8,
=3, and =3 at frequencies of 2, -2, 5, and -5 Hz, respectively.
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