Solution Manual for Signals and Systems: A Primer with MATLAB®, 2nd Edition by Matthew N.O. Sadiku

2


Table of Contents
Chapter 1 1
Chapter 2 41
Chapter 3 77
Chapter 4 117
Chapter 5 144
Chapter 6 182
Chapter 7 205




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CHAPTER 1

P. P. 1.1
2 2
The period is T = = =1
 2
x(t + T ) = A cos((2 (t +1) + 0.1 )
= A cos(2 t + 2 + 0.1 )
= A cos(2 t + 0.1 )
= x(t)
Hence x(t) is periodic.

P.P. 1.2

(a) x(t) = t, 0 < t < 
T /2 T /2  T / 2 3
E = lim
T →
 T →

| x(t) |2dt = limt 2
dt = lim 2   =
−T / 2 −T / 2 T →  3 

1
 | x(t) |2dt = lim 1  t 2dt = lim 2  T / 2  = 
T / 2 T / 2 3
P = lim
T → T T → T T → T  3 
−T / 2 −T / 2



i.e. x(t) is neither an energy nor a power signal.

(b)
T /2 a
E = lim  | x(t) |2dt = lim  A dt = 2aA
2 2
T → T →
−T / 2 −a


i.e. x(t) is an energy signal.

(c) | x(n) |= 5 | e− j4n |= 5
 | x[n] |2 = lim  52
N N
1 1
P = lim
N → 2N +1 N → 2N +1
n=− N n=− N
1
= lim 25(2N +1) = 25 
N → 2N +1



i.e x[n] is a power signal.

P.P. 1.3
(a) ze = t 2 −10, zo = 4t




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1
h (t) = [u(t +1) − u(t −1)]
e 2
(b)
1
h (t) = [−u(t +1) + 2u(t) − u(t −1)]
o
2

These are sketched below. ho
he

½ 1/2



t -1 0 1 t
-1 0 1
-1/2


P. P. 1.4


(a)
−
 sin(t3 +  / 2)  (t)dt = sin(t3 +  / 2) t = 0 = sin( / 2) = 1
10

(b)
 (t 2 + 4t − 2) (t −1)dt = (t 2 + 4t − 2) t = 1 = 1+ 4 − 2 = 3
0


P. P. 1.5
 0, t 0

i(t) =  10, 0t 2

 −10, 2t 4
i(t) = 10u(t) − u(t − 2) −10u(t − 2) − u(t − 4)
= 10[u(t) − 2u(t − 2) + u(t − 4)]

t

Let I=  idt
−
For t < 0, I = 0.
t

For 0 < t < 2, I = 10dt = 10t
0
2 t
t
For 2 < t < 4, I = 10dt − 10dt = 20 −10t = 40 −10t
0 2
2
4
For t > 4, I = 20 −10t =0
2
Thus,




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 0, t  0
 10t,
I =  0t 2
40 −10t, 2t 4
 0, t  4
Or
I = 10[ r(t) – 2t(t-2) + r(t-4) ]
which is sketched below.



 idt

20




0 2 4 t


P.P. 1.6

(a) x(t) is sketched in (a).



x(t) y(t)



5 3



-1 0 1 2 3 4 5 t
0 1 2 3 4 t

(a) (b)

To confirm this, replace every t in eq. (1.35) with t-2 and set =4.




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