ECE
314 Signals
Systems
ECE
314 and
– Signals
and Communications
Fall 2012
Fall/2004
2
Solutions to Homework 1
Problem 1.44
Consider the sinusoidal signal
x(t) = A cos(ωt + φ).
Determine the average power of x(t).
Solution: As discussed in the solutions to the previous homework, the average
power of a periodic signal is given by the ratio of the signal energy of one
period Eo , over the length of the period To , that is
Eo
P =
,
To
where
Z To /2
Eo =
|x(t)|2 dt, and
−To /2
2π
To =
.
ω
So, let’s compute the energy Eo :
Z To /2
|x(t)|2 dt
Eo =
−To /2
π/ω
Z
=
|A cos(ωt + φ)|2 dt
−π/ω
π/ω
Z
=
A2 cos2 (ωt + φ)dt
−π/ω
π/ω
1 + cos(2ωt + 2φ)
=
A
dt
2
−π/ω
"Z
#
Z π/ω
π/ω
A2
=
dt +
cos(2ωt + 2φ)dt
2
−π/ω
−π/ω
(
π/ω )
A2 2π
sin(2ωt + 2φ)
=
+
2
ω
2ω
−π/ω
Z
=
2
A2 To
.
2
1
The null value of the second integral is due to the fact that sin(2π + 2φ) =
sin(−2π + 2φ).
As a conclusion, we get that
A2
Eo
=
P =
To
2
Problem 1.47
defined by
The trapezoidal pulse x(t) shown in the graph below is

5 − t,



1,
x(t) =
t + 5,



0,
4 ≤ t ≤ 5,
−4 ≤ t ≤ t4,
−5 ≤ t ≤ −4,
otherwise.
Determine the total energy of x(t).
Solution: By definition
Z ∞
E =
x2 (t)dt
−∞
Z 5
=
x2 (t)dt
−5
Since x2 (t) is even, we may write
Z 5
E = 2
x2 (t)dt
0
Z
Z 5
4
2
= 2
dt +
(5 − t) dt
0
4
(
t=5 )
26
(5 − t)3
=
= 8.666
= 2 4−
3
3
t=4
2
y(t)
1
0 1/10
t
9/10
Problem 1.52 Let x(t) and y(t) be given in Fig. below. Carefully sketch
the following signals:
−1
x(t)
y(t)
1
1
−1
1
2
t
3
−2
−1
(a) x(t)y(t − 1)
Solution:
x (t)
1
t
−1
y (t−1)
1
t
−1
x (t)y(t−1)
1
t
−1
0
1
2
3
4
1
2
t
(g) x(4 − t)y(t)
Solution:
x(4−t)
1
−1
y(t)
1
−1
x(4−t)y(t)
1
−2 −1
0
1
2
3
4
5
Problem 1.54 Sketch the wave forms of the following signals:
(a) x(t) = u(t) − u(t − 2)
Solution:
u(t)
1
−u(t−2)
−1
1
x(t) = u(t) − u(t−2)
0
1
2
5
(b) x(t) = u(t + 1) − 2u(t) + u(t − 1)
Solution:
u(t+1)
1
−2u(t)
−1
−2
1
u(t−1)
x(t) = u(t+1) −2 u(t)+u(t−1)
1
−1
−1
0
1
Problem 1.56 Let x[n] and y[n] be given in the figure below. Carefully
sketch the following signals:
x[n]
y[n]
3
2
1
−5 −4 −3 −2 −1
1
−5 −4 −3 −2 −1
0 1
2
3
4
1
−1
5
(h) x[3 − n]y[n]
Solution:
6
2
3
4
5
x[3−n]
y[n]
x[3−n]y[n]
2
1
−5 −4 −3 −2 −1
0 1
2
3
4
5
6
7
Problem 1.64 The systems that follow have input x(t) or x[n] and output
y(t) or y[n]. For each system, determine whether it is (iv) linear.
(a) y(t) = cos(x(t))
Solution: From the definition, a system is linear if, and only if, the application
of an input α1 x1 (t) + α2 x2 (t) generates the output α1 y1 (t) + α2 y2 (t), where
y1 (t) and y2 (t) are the outputs generated by the individual inputs x1 (t) and
x2 (t), respectively, no matter what those inputs may be, and for any choice
of coefficients α1 and α2 . For the system above, we can see that
cos(α1 x1 (t) + α2 x2 (t)) 6= α1 cos(x1 (t)) + α2 cos(x2 (t)),
for some choice of x1 (t), x2 (t), α1 and α2 . One such choice could be x1 (t) = π,
x2 (t) = −π, α1 = α2 = 1/2. We can see that
cos(α1 x1 (t) + α2 x2 (t)) = 1 6=
α1 cos(x1 (t)) + α2 cos(x2 (t)) = −1.
Therefore, the system is NON-LINEAR.
(b) y[n] = 2x[n]u[n]
7
Solution: For this system, we can see that
y[n] = 2 (α1 x1 [n] + α2 x2 [n])
= α1 (2x1 [n]u[n]) + α2 (2x2 [n]u[n])
= α1 y1 [n] + α2 y2 [n].
Therefore, the system IS LINEAR.
R t/2
(d) y(t) = −∞ x(τ )dτ
Solution: For this system, we can see that
t/2
Z
y(t) =
[α1 x1 (τ ) + α2 x2 (τ )] dτ
−∞
Z
t/2
= α1
Z
t/2
x1 (τ )dτ + α2
−∞
x2 (τ )dτ
−∞
= α1 y1 (t) + α2 y2 (t)
Therefore, the system IS LINEAR.
(f ) y(t) = dtd x(t)
Solution: For this system, we can see that
d
[α1 x1 (t) + α2 x2 (t)]
dt
d
d
= α1 x1 (t) + α2 x2 (t)
dt
dt
= α1 y1 (t) + α2 y2 (t)
y(t) =
Therefore, the system IS LINEAR.
(i) y(t) = x(2 − t)
Solution: For this system, we can see that
y(t) = α1 x1 (2 − t) + α2 x2 (2 − t)
= α1 y1 (t) + α2 y2 (t)
Therefore, the system IS LINEAR.
8