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Fundamentals Of Communication Systems Solution Manual Proakis J.G., Salehi M.
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Solution ManualFundamentals of Communication Systems
John G. Proakis Masoud Salehi
Second Edition
2013
, Chapter 2
Problem 2.1
1. Π (2t + 5) = Π 2 t + 52 . This indicates first we have to plot Π(2t) and then shift it to left by
5
2. A plot is shown below:
Π (2t + 5)
6
1
- t
11 9
− 4 −4
P∞
2. n=0 Λ(t − n) is a sum of shifted triangular pulses. Note that the sum of the left and right side
of triangular pulses that are displaced by one unit of time is equal to 1, The plot is given below
x2 (t)
✻
1
✲ t
−1
3. It is obvious from the definition of sgn(t) that sgn(2t) = sgn(t). Therefore x3 (t) = 0.
4. x4 (t) is sinc(t) contracted by a factor of 10.
1
0.8
0.6
0.4
0.2
0
−0.2
−0.4
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
3
,Problem 2.2
1. x[n] = sinc(3n/9) = sinc(n/3).
1
0.8
0.6
0.4
0.2
0
−0.2
−0.4
−20 −15 −10 −5 0 5 10 15 20
n n
4 −1 4 −1
2. x[n] = Π 3 . If − 21 ≤ 3 ≤ 12 , i.e., −2 ≤ n ≤ 10, we have x[n] = 1.
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
−20 −15 −10 −5 0 5 10 15 20
n n n
3. x[n] = 4 u−1 (n/4) − ( 4 − 1)u−1 (n/4 − 1). For n < 0, x[n] = 0, for 0 ≤ n ≤ 3, x[n] = 4 and
n n
for n ≥ 4, x[n] = 4 − 4 + 1 = 1.
4
, 1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
−5 0 5 10 15 20
Problem 2.3
x1 [n] = 1 and x2 [n] = cos(2π n) = 1, for all n. This shows that two signals can be different but
their sampled versions be the same.
Problem 2.4
Let x1 [n] and x2 [n] be two periodic signals with periods N1 and N2 , respectively, and let N =
LCM(N1 , N2 ), and define x[n] = x1 [n] + x2 [n]. Then obviously x1 [n + N] = x1 [n] and x2 [n + N] =
x2 [n], and hence x[n] = x[n + N], i.e., x[n] is periodic with period N.
For continuous-time signals x1 (t) and x2 (t) with periods T1 and T2 respectively, in general we
cannot find a T such that T = k1 T1 = k2 T2 for integers k1 and k2 . This is obvious for instance if
T1 = 1 and T2 = π . The necessary and sufficient condition for the sum to be periodic is that TT12 be a
rational number.
Problem 2.5
Using the result of problem 2.4 we have:
1. The frequencies are 2000 and 5500, their ratio (and therefore the ratio of the periods) is
rational, hence the sum is periodic.
5500
2. The frequencies are 2000 and π . Their ratio is not rational, hence the sum is not periodic.
3. The sum of two periodic discrete-time signal is periodic.
4. The fist signal is periodic but cos[11000n] is not periodic, since there is no N such that
cos[11000(n + N)] = cos(11000n) for all n. Therefore the sum cannot be periodic.
5
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