Type:
Solution Manual
Resource:
Principles of Communications
Edition:
7th Edition
Author(s):
Rodger E. Ziemer William H. Tranter
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) i h = 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
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 cos (4 t 3 =8) + 12 cos (10 t h i = Re 2ej(4 t 3 =8) + 12ej(10 t =2)
=2)
=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) i h = 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 h = Re 5ej(2 t = 2:5ej(2 t
=2) + 4 cos (5 t + =4) i =2) + 4ej(5 t+ =4)
=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
8
Amplitude
Amplitude
10
6 4
4
2
2 0
0
1
2 3 f, Hz
4
0 -5
5
0 f, Hz
5
0 f, Hz
5
0.5
0.6
Phase, rad
Phase, rad
0.8
0.4 0.2
0 -0.5
0 0
1
2 3 f, Hz
4
5
f. Use sin (10 t + =6) = cos (10 t + =6
-5
=2) = cos (10 t
x6 (t) = 3 cos (4 t + =8) + 4 cos (10 t =3) i h = Re 3ej(4 t+ =8) + 4ej(10 t =3)
= 1:5ej(4 t+ =8) + 1:5e j(4 t+ =8) + 2ej10 t
=3) to write
=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.
4
CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS
Single sided
Double sided 5
8 6
Amplitude
Amplitude
4
4 2 0
2 1
0
1
2 f, Hz
3
0 -4
4
-2
0 f, Hz
2
4
-2
0 f, Hz
2
4
1 Phase, rad
1 Phase, rad
3
0.5
0
0.5 0 -0.5 -1
0
1
2 f, Hz
3
4
-4