EC204: Networks & Systems
Problem Set 4
1. A signal x(t) can be expressed as the sum of even and odd components as x(t) = x e (t) +
xo (t). (a) If x(t) X ( ), show that for real x(t), xe (t) Re[X ( )] and xo (t)
j Im[X ( )]. (b) Verify these results for
EC204: Networks & Systems
Problem Set 5
1. x1 (t) = 104 rect(104 t) and x2 (t) = (t) are
applied as inputs to LTI systems with frequency responses H1 ( ) = rect(/40000 ) and
H2 ( ) = rect(/20000 ).
x1 (t)
y1 (t)
H1 ( )
y (t)
(a) Sketch Y1 ( ) and Y2 ( ).
EC204: Networks & Systems
Solutions to Problem Set 4
1. (a) x(t) is real x(t) = x (t) X ( ) = X ( ) X ( ) = X ( )
x (t ) + x ( t )
2
x (t ) x ( t )
x o (t ) =
2
X ( ) + X ( )
xe (t)
2
X ( ) X ( )
xo (t)
2
x e (t ) =
x (t ) + x ( t )
2
x (t ) + x ( t )
=
EC204: Networks & Systems
Problem Set 6
1. Find the Laplace transform of x(t) and y (t) shown below.
x(t)
y (t)
4
0
1
2
4
t
T /2
0
T
3T /2 2T
5T /2
2. Consider the partial fraction expansions shown below.
1
K1
K2
K3
) = s + a + s p + s p
(s + a)(s p1 )(s
EC204: Networks & Systems
Problem Set 7
1. Determine i(t) for t 0 in the circuit shown below given that i(0 ) = . Identify the
steady state and transient components of the response.
R
L
t=0
+
i (t )
10 sin t
2. Determine vC (t) for t 0 in the circuit show
EC204: Networks & Systems
Problem Set 2
1. u(t) is the step function. Find u(t) u(t).
2. Find y (t) = x(t) h(t).
x(t)
h(t)
1
1
0
2
3
-3
t
0
3
t
Figure 1:
3. Find x(t) h(t), where h(t) = (et + 2e2t )u(t) and x(t) = 10e3t u(t).
4. Express f1 (t) and f2 (t)
EC204: Networks & Systems
Problem Set 3
1. (a) Determine the coecients of the Fourier series (in exponential form) of the periodic
signal x(t) shown below. Sketch the magnitude and phase spectrum.
x(t)
A
3
2
1
0
1
2
3
t
(b) Using the Fourier series for x(
EC204: Networks & Systems
Problem Set 1
1. For the signal x(t) illustrated in Fig. 1, sketch (a) x(t 4), (b) x(t/1.5), (c) x(t), (d)
x(2t 4), and (e) x(2 t).
x(t)
4
2
4
0
2
t
Figure 1:
2. Consider the signal y (t) = (1/5)x(2t 3) shown in Fig. 2. Determine
EC204: Networks & Systems
Problem Set 9
1. Find i(t) for t 0 using (a) Thevenins theorem, and (b) substitution and superposition
theorems.
1H
t=0
1
+
1V
1
i(t)
1F
1
2. The galvanometer current Ig is zero if RX = 600. If RX varies between 570 and 630,
then
EC204: Networks & Systems
Solutions to Problem Set 2
1.
u (t ) u (t ) =
u( )u(t )d =
0
if t < 0
t
d = t if t 0
0
2.
y (t) = x(t) h(t)
1
-1
0
5
Figure 1: Solution to problem 2
1
6
t
3.
+
x (t ) h (t ) =
h( )x(t )d
+
(e + 2e2 )u( )10e3(t ) u(t )d
=
t
(e + 2
EC204: Networks & Systems
Solutions to Problem Set 8
1. We have
Y (s) = K
(s a)(s b)
(s + 1 j 1)(s + 1 + j 1)
I (s) = V (s)Y (s)
Y (s)|s=0 = 0 =
Kab
2
= a = 0 or b = 0
Without loss of generality, we take a = 0.
According to the second condition specied, t
EC204: Networks & Systems
Solution to Problem Set 10
1. Here,
A=
0
1
2 3
1
s 1
2 s+3
10
C= 1 1
02
10
11
B=
and
(sI A)1 =
00
D= 1 0
01
1
(s + 1)(s + 2)
s
(s + 1)(s + 2)
s+3
(s + 1)(s + 2)
=
2
(s + 1)(s + 2)
Hence, the transfer function H(s) is given by
EC204: Networks & Systems
Solutions to Problem Set 9
1. (a) To solve for i(t), t 0 using Thevenins theorem, we rst transform the given
network to the Laplace domain, as in gure (1).
We obtain values for the Thevenin impedance Z0 (s) and voltage source Voc
EC204: Networks & Systems
Solution to Problem Set 7
1. The transformed network is shown below.
I (s)
+
L
Ls
R
+
10
s 2 + 2
I (s) can be determined as follows.
10
+ L
10/L
s2 + 2
I (s) =
+
=
R
2 + 2 ) s +
R + Ls
s+ L
(s
R
L
The second term of I (s) can be
EC204: Networks & Systems
Problem Set 10
1. Consider the system with the state equation
x1
x2
=
0
1
2 3
x1
x2
+
10
11
u1
u2
and the output equation
10
y1
y2 = 1 1
02
y3
x1
x2
00
+ 1 0
01
u1
u2
.
Find H32 (s).
2. Find the state vector x for the system w
EC204: Networks & Systems
Solutions to Problem Set 6
1. (i) x(t) = 2r(t) 4r(t 2) + 2r(t 4)
X (s) =
4
2
2
2
2 e2s + 2 e4s = 2 [1 e2s + e2s ]
2
s
s
s
s
(ii) Dene y1 (t) as
y1(t)
1
t
T
2
0
Figure 1: Problem 1
T
2
sT
1
= (1 e 2 )
s
y 1 (t ) = u (t ) u t
= Y
EC204: Networks & Systems
Solutions to Problem Set 3
cn ej 2nt . The d.c. value c0 =
1. (a) The period of x(t) is 1. Therefore, x(t) =
n=
A/2. Now, let the rst derivative of x(t) (shown in gure below) be expressed as
dx(t)
=
dn ej 2nt .
dt
n=
dx(t)
dt
A
3
EC204: Networks & Systems
Problem Set 8
1. The admittance function Y (s) has poles at s = 1 j 1 and two zeros.
i(t)
+
LTI
1port
Network
v (t)
Y (s)
The steady state current to a 6V dc input and a sinusoidal input sin t is given below.
v (t)
6V dc
sin t V