Solutions of Chapter 3
Part 1/2
Problem 3.1-1 Find the energy of the signals depicted in Figs.P3.1-1.
Figure 1: Fig3.1-1
Solution:
2
2
2
2
2
2
(a) Ex = ∑∞
n=−∞ |x[n]| = 1 + 2 + 3 + 2 + 1 = 19
2
2
2
2
2
2
(b) Ex = ∑∞
n=−∞ |x[n]| = 1 + 2 + 3 + 2 + 1 = 19
2
2
2
2
2
2
2
2
(c) Ex = ∑∞
n=−∞ |x[n]| = | − 9| + | − 6| + | − 3| + | − 9| + |3| + |6| + |9| = 252
2
2
2
2
2
(d) Ex = ∑∞
n=−∞ |x[n]| = 4 + 2 + 2 + 4 = 40
Page 1 of 7
Problem 3.1-3 Show that a power of a signal De j(2π /N0 )n is |D|2 . Hence, show that a power of a signal
N0 −1
0 −1
x[n] = ∑r=0
Dr e jr(2π /N0 )n is Px = ∑Nr=0
|Dr |2 . Use the fact that
N0 −1
∑
e j(r−m)2π k/N0 =
{
k=0
N0 , r = m;
0, otherwise.
Solution:
(1) Let x[n] = De j(2π /N0 )n = D(cos(2π /N0 )n + j sin(2π /N0 )n), we see that x[n] is periodic with N0 its
period. For a complex number |z|2 = Z ∗Z ∗ , so we have |De j(2π /N0 )n |2 = (De j(2π /N0 )n )(D ∗ e j(2π /N0 )n ) = |D|2 .
Thus, its power is given by
Px =
1 N0 −1
1 N0 −1
|De j(2π /N0 )n |2 =
∑
∑ |D|2 = |D|2 .
N0 n=0
N0 n=0
(2) Clearly, the signal x[n] is periodic, its period is N0 . Thus, the power can be written as
1 N0 −1
1 N0 −1
Px =
|x[n]|2 =
∑
∑
N0 n=0
N0 n=0
Since
N0 −1
∑
2
Dr e
jr(2π /N0 )n
N0 −1
=
∑
N0 −1
∑
Dr e
.
r=0
Dr e jr(2π /N0 )n
r=0
r=0
2
jr(2π /N0 )n
N0 −1
∑
Dm∗ e− jm(2π /N0 )n ,
m=0
the above equation can be written as follows by interchanging the order of summation.
[
]
N0 −1
1 N0 −1 N0 −1
Px =
∑ ∑ Dr Dm∗ ∑ e j(r−m)(2π /N0 )n .
N0 r=0
n=0
m=0
The summation within square brackets is N0 when r = m and 0 otherwise. Therefore, we have
N0 −1
Px =
∑
Dr Dr∗ =
r=0
N0 −1
∑
|Dr |2 .
r=0
Problem 3.2-1 If the energy of a signal x[n] is Ex , then find the energy of the following:
(a) x[−n]
(b) x[n − m]
(c) x[m − n]
(d) Kx[n] (m integer and K constant)
Solution:
(a) The energy of x[−n] is given by
∞
E1 =
∑
|x[−n]|2 .
n=−∞
Page 2 of 7
Let m = −n, we have
−∞
∑
E1 =
∞
|x[m]|2 =
m=∞
∑
|x[m]|2 = Ex .
m=−∞
(b) The energy of x[n − m] is given by
∞
∑
E2 =
|x[n − m]|2 =
n=−∞
∞
∑
|x[r]|2 = Ex .
r=−∞
(c) The energy of x[m − n] is given by
∞
E3 =
∑
|x[m − n]|2 =
n=−∞
−∞
∑ |x[r]|2 =
r=∞
∞
∑
|x[r]|2 = Ex .
r=−∞
(d) The energy of Kx[n] is given by
∞
E4 =
∑
n=−∞
|Kx[n]|2 = K 2
∞
∑
|x[n]|2 = K 2 Ex .
n=−∞
Problem 3.2-3 and 3.2-4 For the signal depicted in Fig. P3.1-1 (a) and (c), sketch the following signals:
(a) x[−n] (b) x[n + 6] (c) x[n − 6]
(d) x[3n] (e) x[n/3]
(f) x[3 − n]
Solution:
Figure 2: Fig3.2-4
Page 3 of 7
Problem 3.3-4 Describe each of the signals in Fig.P3.1-1 by a signal expression valid for all n.
Solution:
There are many different ways of viewing x[n]. Here is just one possible expression.
(a) x[n] = x1 [n] + x2 [n] = (n + 3)(u[n + 3] − u[n]) + (−n + 3)(u[n] − u[n − 4])
Other possible solutions:
x[n] = (n + 3)(u[n + 3] − u[n]) + (−n + 3)(u[n] − u[n − 3])
x[n] = (n + 3)(u[n + 2] − u[n]) + (−n + 3)(u[n] − u[n − 3])
x[n] = (n + 3)(u[n + 3] − u[n − 1]) + (−n + 3)(u[n − 1] − u[n − 4])
(b) x[n] = n(u[n] − u[n − 4]) + (−n + 6)(u[n − 4] − u[n − 7])
(c) x[n] = n(u[n + 3] − u[n − 4])
(d) x[n] = −2n(u[n + 2] − u[n]) + 2n(u[n] − u[n − 3])
Problem 3.4-3 A moving average is used to detect a trend of a rapidly fluctuating variable such as the stock
market average. A variable may fluctuate (up and down) daily, masking its long-term (secular) trend. We
can discern the long-term trend by smoothing or averaging the past N values of the variable. For the stock
market average, we may consider a 5-day moving average y[n] to be the mean of the past 5 days’ market
closing values x[n], x[n − 1], · · · , x[n − 4].
(a) Write the difference equation relating y[n] to the input x[n].
(b) Use time-delay elements to realize the 5-day moving-average filter.
Solution:
(a) y[n] = 15 (x[n] + x[n − 1] + x[n − 2] + x[n − 3] + x[n − 4])
(b) Let ’D’ represent unit delay, the realization is as follows.
X[n]
D
D
D
D
ě
1/5
Y[n]
Figure 3: Fig3.4-3
Page 4 of 7
Problem 3.4-5 Approximate the following second-order differential equation with a difference equation.
d2y
dy
+ a1 + a0 y(t) = x(t)
dt 2
dt
Solution:
Let x[n] and y[n] represent the samples T seconds apart of the signals x(t) and y(t), respectively. i.e.
x[n] = x(nT ), y[n] = y(nT )
Suppose T is small enough so that the assumption T → 0 may be made. Then, we have
y(t) = y[n]
dy
y[n] − y[n − 1]
≃
dt
T
y[n]−y[n−1]
− y[n−1]−y[n−2]
y[n] − 2y[n − 1] + y[n − 2]
d2y
T
T
=
≃
dt 2
T
T2
Substituting the above expressions into the given differential equation, we can obtain the following
approximated difference equation
A1 y[n] + A2 y[n − 1] + A3 y[n − 2] = Bx[n]
where
A1 = 1 + a1 T + a0 T 2
A2 = −(2 + a1 T )
A3 = 1
B = T2
Page 5 of 7
Problem 3.4-8 A linear, time-invariant system produces output y1 [n] in response to input x1 [n], as shown
in Fig.P3.4-8. Determine and sketch the output y2 [n] that results when input x2 [n] is applied to the same
system.
Figure 4: Fig3.4-8
Solution:
The output y1 [n] can be expressed as y1 [n] = −δ [n] + δ [n − 1] + 2δ [n − 2]. The input x2 [n] can be
expressed as x2 [n] = x1 [n − 1] − 2x1 [n − 2].
Since the system is LTI, from the property of superposition, we have
ě
y2 [n] = y1 [n − 1] − 2y1 [n − 2] = −δ [n − 1] + 3δ [n − 2] − 4δ [n − 4]
The output is sketched as follows.
4
y2[n]
2
0
-2
-4
-2
n
-1
0
1
2
3
4
5
6
7
8
Figure 5: Fig3.4-8a
Page 6 of 7
Problem 3.4-9 A system is described by
y[n] =
1 ∞
∑ x[k](δ [n − k] + δ [n + k])
2 k=−∞
(a) Explain what this system does.
(b) Is the system BIBO stable? Justify your answer.
(c) Is the system linear? Justify your answer.
(d) Is the system memoryless? Justify your answer.
(e) Is the system causal? Justify your answer.
(f) Is the system time invariant? Justify your answer.
Solution:
In the summation, n is a constant, the signal x[k] is sampled at k = n and k = −n. Thus, the system can
be rewriten as
1
1
y[n] = (x[n]δ [n − k] + x[−n]δ [n + k]) = (x[n] + x[−n])
2
2
(a) Each signal can be expressed as a sum of an even component and an odd component as
1
1
x[n] = (x[n] + x[−n]) + (x[n] − x[−n]).
2
2
The system extracts the even portion of the input.
(b) The system is BIBO stable. If the input is bounded, then the output is necessarily bounded. Assume
input x[n] is bounded as |x[n]| ≤ Mx < ∞, then its reverse x[−n] is also bounded and |x[−n]| ≤ Mx < ∞. Thus,
|y[n]| = | 12 (x[n] + x[−n])| ≤ 12 (x[n] + x[−n]) ≤ Mx < ∞.
(c) Yes. The system is linear. Let
1
1
y1 [n] = (x1 [n] + x1 [−n]), y2 [n] = (x2 [n] + x2 [−n])
2
2
Applying x[n] = a1 x1 [n] + a2 x2 [n] to the system yields y[n] = 12 (x[n] + x[−n]) = 12 (a1 x1 [n] + a2 x2 [n] +
(a1 x1 [−n] + a2 x2 [−n])) = a1 y1 [n] + a2 y2 [n].
(d) No, The system is not memoryless since the output does not only depend on its current input. For
example, at time n = 2, the output y[2] = 21 (x[2] + x[−2]) depends on a past value of input x[−2].
(e) No, The system is not causal. For example, at time n = −1, the output y[−1] = 21 (x[−1] + x[1])
depends on a future value of input x[1].
(f) No, The system is not time invariant. For example, let the input be x1 [n] = u[n+10]−u[n−11]. Since
the input is already even, the output equals to the input as y1 [n] = x1 [n] = u[n + 10] − u[n − 11]. Shifting by
a non-zero integer N, then the signal x2 [n] = x1 [n − N] is not even; the output y2 [n] ̸= y1 [n − N] = x1 [n − N].
Therefore, the system is time varying.
Page 7 of 7