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
π 2
l.43 y ( t ) = 3 cos 200t + ---
6
2 π
= 9 cos 200t + ---
6
π
= --- cos 400t + --- 1
9
2 3
9
(a) DC component = ---
2
π
(b) Sinusoidal component = --- cos 400t + ---
9
2 3
9
Amplitude = ---
2
1
, 200
Fundamental frequency = --------- Hz
π
1.44 The RMS value of sinusoidal x(t) is A ⁄ 2 . Hence, the average power of x(t) in a 1-ohm
2
resistor is ( A ⁄ 2 ) = A2/2.
1.45 Let N denote the fundamental period of x[N]. which is defined by
2π
N = ------
Ω
The average power of x[n] is therefore
N -1
1
P = ---- ∑ x [ n ]
2
N
n=0
N -1
2 2πn
cos ---------- + φ
1
∑A
2
= ----
N N
n=0
2 N -1 2
A 2πn
---------- + φ
= ------
N ∑ cos N
n=0
1.46 The energy of the raised cosine pulse is
π⁄ω 1
∫–π ⁄ ω --4- ( cos ( ωt ) + 1 )
2
E = dt
1 π⁄ω
= --- ∫
2
( cos ( ωt ) + 2 cos ( ωt ) + 1 ) dt
2 0
1 π ⁄ ω 1
--- cos ( 2ωt ) + --- + 2 cos ( ωt ) + 1 dt
1
= --- ∫
2 0 2 2
1 3 π
= --- --- ---- = 3π ⁄ 4ω
2 2 ω
1.47 The signal x(t) is even; its total energy is therefore
5 2
E = 2 ∫ x ( t ) dt
0
2
, 4 5
= 2 ∫ ( 1 ) dt + 2 ∫ ( 5 – t ) dt
2 2
0 4
1 5
4 3
= 2 [ t ] t=0 + 2 – --- ( 5 – t )
3 t=4
2 26
= 8 + --- = ------
3 3
1.48 (a) The differentiator output is
1 for – 5 < t < – 4
y ( t ) = –1 for 4 < t < 5
0 otherwise
(b) The energy of y(t) is
–4 5
∫–5 ( 1 ) dt + ∫ ( – 1 ) dt
2 2
E =
4
= 1+1 = 2
1.49 The output of the integrator is
t
y ( t ) = A ∫ τ dτ = At for 0 ≤ t ≤ T
0
Hence the energy of y(t) is
T 2 3
A T
∫0
2 2
E = A t dt = -------------
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)
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
4
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
π 2
l.43 y ( t ) = 3 cos 200t + ---
6
2 π
= 9 cos 200t + ---
6
π
= --- cos 400t + --- 1
9
2 3
9
(a) DC component = ---
2
π
(b) Sinusoidal component = --- cos 400t + ---
9
2 3
9
Amplitude = ---
2
1
, 200
Fundamental frequency = --------- Hz
π
1.44 The RMS value of sinusoidal x(t) is A ⁄ 2 . Hence, the average power of x(t) in a 1-ohm
2
resistor is ( A ⁄ 2 ) = A2/2.
1.45 Let N denote the fundamental period of x[N]. which is defined by
2π
N = ------
Ω
The average power of x[n] is therefore
N -1
1
P = ---- ∑ x [ n ]
2
N
n=0
N -1
2 2πn
cos ---------- + φ
1
∑A
2
= ----
N N
n=0
2 N -1 2
A 2πn
---------- + φ
= ------
N ∑ cos N
n=0
1.46 The energy of the raised cosine pulse is
π⁄ω 1
∫–π ⁄ ω --4- ( cos ( ωt ) + 1 )
2
E = dt
1 π⁄ω
= --- ∫
2
( cos ( ωt ) + 2 cos ( ωt ) + 1 ) dt
2 0
1 π ⁄ ω 1
--- cos ( 2ωt ) + --- + 2 cos ( ωt ) + 1 dt
1
= --- ∫
2 0 2 2
1 3 π
= --- --- ---- = 3π ⁄ 4ω
2 2 ω
1.47 The signal x(t) is even; its total energy is therefore
5 2
E = 2 ∫ x ( t ) dt
0
2
, 4 5
= 2 ∫ ( 1 ) dt + 2 ∫ ( 5 – t ) dt
2 2
0 4
1 5
4 3
= 2 [ t ] t=0 + 2 – --- ( 5 – t )
3 t=4
2 26
= 8 + --- = ------
3 3
1.48 (a) The differentiator output is
1 for – 5 < t < – 4
y ( t ) = –1 for 4 < t < 5
0 otherwise
(b) The energy of y(t) is
–4 5
∫–5 ( 1 ) dt + ∫ ( – 1 ) dt
2 2
E =
4
= 1+1 = 2
1.49 The output of the integrator is
t
y ( t ) = A ∫ τ dτ = At for 0 ≤ t ≤ T
0
Hence the energy of y(t) is
T 2 3
A T
∫0
2 2
E = A t dt = -------------
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)
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
4
