TEST BANK FOR Principles of Communication Systems, Modulation and Noise 5th Ed By R. E. Ziemer

Problem 2.1
For the single-sided spectra, write the signal in terms of cosines:
x(t) = 10cos(4πt + π/8) + 6 sin(8πt + 3π/4)
= 10cos(4πt + π/8) + 6 cos(8πt + 3π/4 − π/2)
= 10cos(4πt + π/8) + 6 cos(8πt + π/4)
For the double-sided spectra, write the signal in terms of complex exponentials using Euler’s
theorem:
x(t) = 5exp[(4πt + π/8)] + 5 exp[−j(4πt + π/8)]
+3 exp[j(8πt + 3π/4)] + 3 exp[−j(8πt + 3π/4)]
The two sets of spectra are plotted in Figures 2.1 and 2.2.
Problem 2.2
The result is
x(t) = 4ej(8πt+π/2) +4e−j(8πt+π/2) + 2ej(4πt−π/4) + 2e−j(4πt−π/4)
= 8cos(8πt + π/2) + 4 cos (4πt − π/4)
= −8 sin(8πt) + 4cos (4πt − π/4)
1
2 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY
f, Hz f, Hz
0 2 4 6 0 2 4 6
10
5
π/4
π/8
Single-sided amplitude Single-sided phase, rad.
Figure 2.1:
Problem 2.3
(a) Not periodic.
(b) Periodic. To find the period, note that
6π
2π
= n1f0 and
20π
2π
= n2f0
Therefore
10
3
=
n2
n1
Hence, take n1 = 3, n2 = 10, and f0 = 1 Hz.
(c) Periodic. Using a similar procedure as used in (b), we find that n1 = 2, n2 = 7, and
f0 = 1 Hz.
(d) Periodic. Using a similar procedure as used in (b), we find that n1 = 2, n2 = 3, n3 = 11,
and f0 = 1 Hz.
Problem 2.4
(a) The single-sided amplitude spectrum consists of a single line of height 5 at frequency 6
Hz, and the phase spectrum consists of a single line of height -π/6 radians at frequency 6
Hz. The double-sided amplitude spectrum consists of lines of height 2.5 at frequencies of
6 and -6 Hz, and the double-sided phase spectrum consists of a line of height -π/6 radians
at frequency 6 Hz and a line of height π/6 at frequency -6 radians Hz.
(b) Write the signal as
xb(t) = 3cos(12πt − π/2) + 4 cos(16πt)
From this it is seen that the single-sided amplitude spectrum consists of lines of height 3
and 4 at frequencies 6 and 8 Hz, respectively, and the single-sided phase spectrum consists
2.1. PROBLEM SOLUTIONS 3
f, Hz
0 2 4 6
π/4
π/8
Double-sided phase, rad.
f, Hz
-6 -4 -2 0 2 4 6
5
Double-sided amplitude
-π/8
-π/4
-6 -4 -2
Figure 2.2:
4 CHAPTER 2. SIGNAL AND LINEAR SYSTEM THEORY
of a line of height -π/2 radians at frequency 6 Hz. The double-sided amplitude spectrum
consists of lines of height 1.5 and 2 at frequencies of 6 and 8 Hz, respectively, and lines of
height 1.5 and 2 at frequencies -6 and -8 Hz, respectively. The double-sided phase spectrum
consists of a line of height -π/2 radians at frequency 6 Hz and a line of height π/2 radians
at frequency -6 Hz.
Problem 2.5
(a) This function has area
Area =
Z∞
−∞
²−1
·
sin(πt/²)
(πt/²)
¸2
dt
=
Z∞
−∞
·
sin(πu)
(πu)
¸2
du = 1
A sketch shows that no matter how small ² is, the area is still 1. With ² → 0, the central
lobe of the function becomes narrower and higher. Thus, in the limit, it approximates a
delta function.
(b) The area for the function is
Area =
Z∞
−∞
1
²
exp(−t/²)u (t) dt =
Z∞
0
exp(−u)du = 1
A sketch shows that no matter how small ² is, the area is still 1. With ² → 0, the function
becomes narrower and higher. Thus, in the limit, it approximates a delta function.
(c) Area =
R ²
−²
1
² (1 − |t| /²) dt =
R 1
−1 Λ (t) dt = 1. As ² → 0, the function becomes narrower
and higher, so it approximates a delta function in the limit.
Problem 2.6
(a) 513; (b) 183; (c) 0; (d) 95,583.8; (e) -157.9.
Problem 2.7
(a), (c), (e), and (f) are periodic. Their periods are 1 s, 4 s, 3 s, and 2/7 s, respectively.
The waveform of part (c) is a periodic train of impulses extending from -∞ to ∞ spaced
by 4 s. The waveform of part (a) is a complex sum of sinusoids that repeats (plot). The
waveform of part (e) is a doubly-infinite train of square pulses, each of which is one unit
high and one unit wide, centered at · · ·, −6, −3, 0, 3, 6, · · ·. Waveform(f) is a raised
cosine of minimum and maximum amplitudes 0 and 2, respectively.
2.1.