ECEN 314: Signals and Systems
Solutions to HW 5
Problem 2.11
(a) Since h(t) is non-zero only for 0 ≤ t ≤ ∞, we have
y(t) = x(t) ∗ h(t)
Z ∞
¡
¢
=
e−3τ u(t − τ − 3) − u(t − τ − 5) dτ
τ =0
¡
¢
Note that u(t − τ − 3) − u(t − τ − 5) is non-zero only for (t − 5) < τ < (t − 3). Therefore,
for t ≤ 3, the above integral evaluates to zero. For 3 < t ≤ 5, the above integral is:
Z t−3
1 − e−3(t−3)
y(t) =
e−3τ dτ =
.
3
0
For t > 5, we have:
Z
t−5
y(t) =
e−3τ dτ =
t−3
(1 − e−6 )e−3(t−5)
.
3
Putting the above together, we get


0
−∞ < t ≤ 3

1−e−3(t−3)
,
3<t≤5
y(t) =
3

 (1−e−6 )e−3(t−5) 5 < t ≤ ∞
3
(b) Differentiating x(t) we get
g(t) =
dx(t)
dt
= δ(t − 3) − δ(t − 5). Therefore
dx(t)
∗ h(t) = e−3(t−3) u(t − 3) − e−3(t−5) u(t − 5)
dt
(c) From (a), we compute the derivative of y(t) to be

0
−∞ < t ≤ 3
dy(t) 
−3(t−3)
e
,
3<t≤5
=
 −6
dt
−3(t−5)
(e − 1)e
5<t≤∞
Problem 2.17
(a) Note that y(t) is the sum of the particular and homogenous solution to the given differential equation. We first determine the particular solution yp (t) by using the method specified
in Example 2.14. Since the input x(t) = e(−1+3j)t u(t) for t > 0, we hypothesize that
yp (t) = Ke(−1+3j)t
1
Substituting for x(t) and y(t) in the given differential equation yields:
(−1 + 3j)Ke(−1+3j)t + 4Ke(−1+3j)t = e(−1+3j)t
This gives
(−1 + 3j)K + 4K = 1
Therefore
yp (t) =
⇒K=
1
3(1 + j)
1
e(−1+3j)t
3(1 + j)
In order to determine the homogenous solution, we hypothesize that
yh (t) = Aest
Since the homogenous solution has to satisfy the differential equation
dyh (t)
+ 4yh (t) = 0,
dt
we obtain
Asest + 4Aest = Aest (s + 4) = 0,
This implies that s = −4 for any A. The overall solution to the differential equation now
becomes
1
y(t) = Ae−4t +
e(−1+3j)t ,
t>0
3(1 + j)
To determine the constant A, we use the fact that y(0) = 0 (condition of initial rest). Hence
A+
1
=0
3(1 + j)
Therefore for t > 0,
y(t) =
⇒A=−
1
3(1 + j)
£
¤
1
− e−4t + e(−1+3j)t
3(1 + j)
Since the system satisfies the condition of initial rest we have y(t) = 0 for t < 0.
(b) The output will now be the real part of the answer obtained in part (a).
y(t) =
¤
1 £ −t
e cos(3t) + e−t sin(3t) − e−4t
6
Problem 2.21
2
(a)
y[n] = x[n] ∗ h[n]
∞
X
=
x[k]h[n − k]
−∞
= β
=
(b) From (a), y[n] = αn
£ Pn
k=0
n
£β
n
X
(α/β)k
n≥0
k=0
n+1
− αn+1 ¤
u[n]
β−α
¤
1 u[n] = (n + 1)αn u[n]
(c) For n ≤ 6,
¾
½X
3
∞
X
k
k
(−1/8)
(−1/8) −
y[n] = 4
n
k=0
k=0
For n > 6,
¾
½X
n−1
∞
X
k
k
(−1/8)
(−1/8) −
y[n] = 4
n
k=0
k=0
Therefore,
½
y[n] =
(8/9)(−1/8)4 4n , n ≤ 6
(8/9)(−1/2)n , n > 6
(d) The convolution is shown in the figure below:
5
y[n]
4
3
2
1
0
n
1
Problem 2.23
3
−5
−4 −3 −1
T=2
T=4
1
0
1
1
3
4
t
5
−3
−2 −1
0
1
T=3/2
2
3
t
T=1
1
1
1/2
−3/2
3/2
t
t
Problem 2.27
Z
∞
Ay =
y(t)dt
Z ∞
=
x(τ )h(t − τ )dτ dt
−∞ −∞
Z ∞
Z ∞
=
x(τ )
h(t − τ )dtdτ
−∞
Z−∞
∞
=
x(τ )Ah dτ
Z−∞
∞
−∞
= Ax Ah
Problem 7
(a)
µ
¶
1
−d
1
X(s) =
+
s+2
ds s + 3
1
1
+
=
s + 2 (s + 3)2
The ROC for the first term is Re{s} > −2 and for the second term it is Re{s} > −3 and
hence the ROC of the sum is Re{s} > −2.
(b)
X(s) =
1
e−s e−2s
−2
+
s
s
s
4
with ROC being Re{s} > 0.
Problem 8
(a) Note that
s+1
(s + 2)(s + 4)
3
1
=
−
2(s + 4) 2(s + 2)
X(s) =
Hence x(t) = 32 e−4t − 12 e−2t .
(b) Note that
s+4
(s + 2)2 + 4
s+2
2
=
+
2
2
(s + 2) + 2
(s + 2)2 + 22
X(s) =
Hence x(t) = e−2t cos(2t) u(t) + e−2t sin(2t) u(t).
5