ECEN 314: Signals and Systems
Solutions to HW 2
Problem 1.21
x(t−1)
x(2t+1)
x(2−t)
2
2
2
1
1
1
−1
0
1
2
3
0
1
2
3
−1.5
4
x(4−t/2)
−0.5
0
0.5
−1
−1
−1
−1
[x(t)+x(−t)]u(t)
2
3
1
1
4
6
8
10
12
0
1
2
−1.5
1.5
−1
−0.5
−0.5
Problem 1.23
x (t)
x o(t)
e
1/2
1/2
1/2
−2
−2
−1
0
1
0
2
(a)
x (t)
e
1
0
1
2
1
2
−1/2
o
1/2
1/2
−1
1
x (t)
1
1/2
−2
−1
−2
−1
0
2
−1/2
(b)
x (t)
x o(t)
e
−3t/2
3t/2
0
(c)
1
−t/2
Problem 1.25
(a) x(t) = 3 cos(4t + π3 ) is periodic with period: π2 . To see why this is true, note that we
need to find a T so that
x(t + T ) = x(t)
¡
¢
Note that x(t) = 3 cos(4t + π3 ) = 3 cos 4(t + π2 ) + π3 , which yields the desired answer.
(b) x(t) = ej(πt−1) is periodic with period =
(c) x(t) = [cos(2t − π3 )]2
[cos(2t −
2π
π
= 2.
π 2 1 + cos(4t −
)] =
3
2
2π
)
3
which is periodic with period T = π2 .
(d)
x(t) = E(cos(4πt)u(t))
cos(4πt)u(t) + cos(4πt)u(−t)
=
2
1
=
cos(4πt)
−∞<t≤∞
2
which is periodic with period 12 . (Here, E is the even part).
(e)
x(t) = E(sin(4πt)u(t))
sin(4πt)u(t) − sin(4πt)u(−t)
=
2
which is not periodic.
(f) x(t) =
P∞
−∞
e−(2t−n) u(2t − n) is not periodic.
Problem 1.26
(a) x[n] = sin(6πn/7 + 1) is periodic with period=7.
(b) x[n] = cos(n/8 − π) is not periodic.
(c) x[n] = cos(π/8 n2 ) is periodic with period=8.
2
(d) x[n] = cos(π/2 n) cos(π/4 n). cos(π/2 n) is periodic with period=4, and cos(π/4 n)
is periodic with period=8. So, x[n] is periodic with period=8.
(e) x[n] = 2 cos(π/4 n) + sin(π/8 n) − 2 cos(π/2 n + π/6) is periodic with period=16.
Problem 1.34
(a) First, note that
∞
X
∞
X
x[n] = x[0] +
{x[n] + x[−n]}
n=−∞
n=1
If x[n] is odd then it implies that x[n] + x[−n] = 0, which further implies that x[0] = 0.
Hence the above summation is zero.
(b) Let, y[n] = x1 [n]x2 [n] . Then,
y[−n] = x1 [−n]x2 [−n] = −x1 [n]x2 [n] = −y[n].
(c) Consider
∞
X
∞
X
2
x [n] =
n=−∞
(xe [n] + xo [n])2
n=−∞
∞
X
=
∞
X
x2e [n] +
n=−∞
∞
X
x2o [n] + 2
n=−∞
xe [n]xo [n]
n=−∞
Using the result of part (b), we conclude xe [n]xo [n] = 0,
xe [n]xo [n] is an odd signal.
Pbecause
∞
Therefore, using the result of part (a), we conclude 2 n=−∞ xe [n]xo [n] = 0. Therefore,
∞
X
2
x [n] =
n=−∞
(d) Consider
Z∞
∞
X
−∞
x2o [n].
n=−∞
Z∞
Z∞
x2e (t)dt
x (t)dt =
∞
X
+
n=−∞
Z∞
2
x2e [n]
x2o (t)dt
+
−∞
Since xe (t)xo (t) is an odd signal, we have 2
−∞
R∞
+2
xe (t)xo (t)dt
−∞
xe (t)xo (t)dt = 0 and hence we have
−∞
Z∞
x2e (t)dt
x (t)dt =
−∞
Z∞
Z∞
2
+
−∞
−∞
3
x2o (t)dt.