TEST BANK FOR Fundamentals Of Communication Systems 2nd Edition By Proakis J.G., Salehi M

, Solution Manual



Fundamentals 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