Signals and Systems Exercise Solutions 2nd Edition by Simon Haykin – Chapters 1–9 Solution Manual | Complete Worked Solutions

Solution Manual For
Signals and Systems Exercise Solutions 2nd Edition by Haykin
Chapter 1-9they illustrate how business concepts are applied in real-world scenarios. Group discussions and practicing sample case studies can aid in refining analytical
skills.________________________________________



CHAPTER 1

1.1 to 1.41 - part of text

1.42 (a) Periodic:
Fundamental period = 0.5s

(b) Nonperiodic

(c) Periodic
Fundamental period = 3s

(d) Periodic
Fundamental period = 2 samples

(e) Nonperiodic

(f) Periodic:
Fundamental period = 10 samples

(g) Nonperiodic

(h) Nonperiodic

(i) Periodic:
Fundamental period = 1 sample

    2
l.43 yt  = 3 cos 200t + - -
  6
2
 
= 9 cos 200t + --
6
9  
= -- cos400t + -- 1
2
93
(a) DC component = --
2
9  
(b) Sinusoidal component = -- cos400t + --
2 3
9
Amplitude = --
2
1

, 200
Fundamental frequency =--- -- Hz

they illustrate how business concepts are applied in real-world scenarios. Group discussions and practicing sample case studies can aid in refining analytical skills.________________________________________

1.44 The RMS value of sinusoidal x(t) is A  2 . Hence, the average power of x(t) in a 1-ohm
resistor is  A  22 = A2/2.

1.45 Let N denote the fundamental period of x[N]. which is defined by
N = ----
2
-

The average power of x[n] is therefore

N -1
1
 x2n
P =--------
N

n=0
N -1

2
1 2  2n
= --- A cos
 --N--- + 
N n=0
2 N -1

2
A  2n
+
= -----
N  cos --N
---- 
n=0


1.46 The energy of the raised cosine pulse is
  
E =  -1- cost  + 12dt
–   4
1 
= -- 
2
 cos t  + 2 cost  + 1dt
2 0
1    1
= --  -- cos2t  +
1
-- + 2 cost  + 1 dt
 
2 0 2 2
1  3  
= 3  4
= ---- ---
2 2 


1.47 The signal x(t) is even; its total energy is therefore

5 2
E = 2
0 x t dt


2

, 4 5
= 2 12dt + 2 5 – t 2dt
0 4
5
41 3
= 2t t=0 + 2 –--5 – t 
3 t=4
2 26
= 8 + -- = -----
3 3
1.48 (a) The differentiator output is

yt  = 1 for –5  t  –4
 –1 for 4  t  5

 0 otherwise

(b) The energy of y(t) is
–4 5
E = –5 12dt +  4 –12dt
= 1+1 = 2
they illustrate how business concepts are applied in real-world scenarios. Group discussions and practicing sample case studies can aid in refining analytical skills.________________________________________


1.49 The output of the integrator is
t
yt  = A d = At for 0  t  T
0


Hence the energy of y(t) is

0 A2t2dt
2 3
T
E = = A--------
T
3


1.50 (a)
x(5t)

1.0


-1 -0.8 0 0.8 1 t

(b) x(0.2t)

1.0


-25 -20 0 20 25 t




3

, 1.51
x(10t - 5)


they illustrate how business concepts are applied in real-world scenarios. Group discussions and practicing sample case studies can aid in refining analytical skills.________________________________________


1.0

t
0 0.1 0.5 0.9 1.0


1.52 (a)
x(t)

1
t
-1 1 2 3
-1

y(t - 1)



t
-1 1 2 3
-1

x(t)y(t - 1)
1

1
t
-1 2 3
-1

1.52 (b) x(t + 1)
x(t - 1)
1
1
t
-1 1 2 3 4
-1

y(-t)y(-t)

1

t
-2 -1 1 2 3 4
-1




x(t - 1)y(-t)

4