TEST BANK FOR Principles of Communication Systems,
Exam (elaborations)
Exam (elaborations) TEST BANK FOR Principles of Communication Systems,
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Course
TEST BANK FOR Principles of Communication Systems,
Institution
Maastricht University (UM)
Exam (elaborations) TEST BANK FOR Principles of Communication Systems,
For the single-sided spectra, write the signal in terms of cosines:
x(t) = 10 cos(4πt + π/8) + 6 sin(8πt + 3π/4)
= 10 cos(4πt + π/8) + 6 cos(8πt + 3π/4 − π/2)
= 10 cos(4πt + π/8) + 6 cos(8πt + π/4)
For the d...
Problem 2.3
(a) Not periodic.
(b) Periodic. To find the period, note that
6π 20π
= n1 f0 and = n2 f0
2π 2π
Therefore
10 n2
=
3 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) = 3 cos(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
Double-sided amplitude
5
f, Hz
-6 -4 -2 0 2 4 6
Double-sided phase, rad.
π/4
π/8
-6 -4 -2
f, Hz
0 2 4 6
-π/8
-π/4
Figure 2.2:
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