Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

9.1.1 
the sinusoidal steady-state frequency response function
In Chapter 7 we noted that the most general differential equation governing a chosen response variable 
for a linear circuit is
d y
dt
a
d
y
dt
a
dy
dt
a y
b
d x
dt
b
d
x
dt
n
n
n
n
n
m
m
m
m
m
m
+
+ +
+
=
+
−
−
−
−
−
−
1
1
1
1
0
1
1
1

++ +
+

b
dx
dt
b x
1
0

---

Periodic Waveforms in Circuit Analysis 
9.5
where y is the chosen response variable, x is the input function and a’s and b’s are real positive 
constants decided by circuit parameters. The order of differential equation, n¸ is equal to the number 
of independent energy storage elements in the circuit.
If x 
= 
e 
j
w
t
, then y 
= 
A e 
j
w
t
, where A is a complex amplitude decided by the equation
A j
a
j
a j
a
b
j
b
j
b
n
n
n
m
m
m
m
[(
)
(
)
(
)
]
(
)
(
)
(
w
w
w
w
w
+
+ +
+
=
+
+ +
−
−
−
−
1
1
1
0
1
1
1


jj
b
w
)
+
0
Therefore, the complex amplitude of steady-state response to a unit amplitude complex exponential 
input comes out as a ratio of rational polynomials in the variable j
w
. 
The desired steady-state response is obtained as 
y
b j
j
a j
e
k
k
k
m
n
i
i
i
n
j t
=
+
=
=
−
∑
∑
(
)
(
)
(
)
.
w
w
w
w
0
0
1
the Sinusoidal Steady-State Frequency response Function of a linear time-invariant 
circuit is defined as the complex valued gain offered by the circuit to a signal 
e 
j
w
t
. It is 
usually represented by 
H
(
j
w
).
∴
=
+
=
=
−
∑
∑
H j
b j
j
a j
k
k
k
m
n
i
i
i
n
(
)
(
)
(
)
(
)
w
w
w
w
0
0
1
Let the input to a linear time-invariant circuit be 
x t
A
t
u t
( )
cos(
) ( ).
=
+
w q
Then the output after 
the circuit attains steady-state (if it can reach a steady-state) is 
y t
H j
A
t
( ) | (
)| cos(
)
=
+ +
w
w q f
where 
| (
)|
H j
w
is the gain magnitude of frequency response function and 
f
w
= ∠
H j
(
)
is the phase 
of frequency response function. Similarly if the input is 
x t
A
t
u t
( )
sin(
) ( ),
=
+
w q
the output after the 
circuit achieves steady-state is 
y t
H j
A
t
( ) | (
)| sin(
).
=
+ +
w
w q f
the magnitude part of frequency response function 
H
(
j
w
) gives the ratio of ‘output 
sinusoid amplitude’ to ‘input sinusoid amplitude’. the angle part of frequency response 
function gives the phase angle by which the output sinusoid leads the input sinusoid.
Note that we did not specify the nature of x and y in the definition of frequency response function. 
If both x and y are voltage (or current) variables, the frequency response function is a transfer gain 
function. If x is a voltage variable and y is the current variable at the same terminals where the voltage 
x is present, the frequency response function is a driving point admittance function. If x is a current 
variable and y is the voltage variable at the same terminals, the frequency response function is a 
driving point impedance function.
It is not necessary to derive the differential equation describing the circuit to obtain its frequency 
response function. H(j
w
) can be found from the phasor equivalent circuit by circuit reduction or 
by nodal/mesh analysis. For instance, the frequency response function connecting the circuit 

---

9.6
Dynamic Circuits with Periodic Inputs – Analysis by Fourier Series
current to applied voltage in a series RL circuit can be obtained from its phasor equivalent circuit as 
=
+
=
+
=
+
∠ −
−
1
1
1
1
1
2
1
R
j L
R
j L R
R
L R
L
R
w
w
w
w
(
/ )
tan
.
Similarly, the frequency response function describing the transfer gain between the voltage across 
capacitor and input voltage in a series RC circuit can be determined from its phasor equivalent circuit 
by applying voltage division rule in a series circuit. 
H j
j C
R
j C
j RC
RC
RC
(
)
(
)
tan
w
w
w
w
w
w
=
+
=
+
=
+
∠ −
1
1
1
1
1
1
2
1
Many circuits do reach periodic steady-state after the switch-on transient and operate for long 
periods before they are switched off. Electrical power system and the loads that draw power from it 
constitute an important example of this kind of operation. Many features of such steady-state operation 
are of considerable practical interest to the electrical engineer. Therefore, one reason why we study the 
forced response of circuits to periodic waveforms is that we are interested in steady-state response of 
circuits when such waveforms are switched on to them at t 
= 
0. Expanding the periodic waveform in 
terms of sinusoidal components and using frequency response function and superposition principle 
will help us to get the required steady-state solution with relative ease. However, if we want to get the 
complete response we have to bring in the natural response terms too and arrange for compliance with 
initial conditions.
Consider an example. An RL circuit with no initial energy storage, R 
= 
1 
W
and L
= 
1 H is connected 
to a source v(t) 
= 
sin t 
+ 
0.3 sin 3t 
+ 
0.2 sin 5t V, at t 
= 
0. Subsequently it is removed from the supply 
at t 
= 
100 s and the circuit is kept shorted after that. We wish to find the total response of the circuit 
current. 
Let v
S
(t) stand for the voltage applied to the circuit. What is v
S
(t)? 
v t
t
t
t u t
u t
S
( ) (sin
. sin
. sin )[ ( )
(
)]
=
+
+
−
−
0 3
3
0 2
5
100
The forcing function is a sum of three sinusoids. Forced response component for each sinusoid 
may be obtained using frequency response data. The relevant frequency response function is 
1
1
1
1
2
1
2
1
R
L R
L
R
+
∠ −
=
+
∠ −
−
−
(
/ )
tan
tan
.
w
w
w
w
Forced response components for the three sinusoids may be obtained by using corresponding 
values of 
w
in this function. For instance, when 
w
 
= 
1 rad/s, the gain magnitude is 0.707 
W
-
1
and lead 
phase angle is –45
°
. The individual steady-state components may be combined to obtain the total 
steady-state response. 
The result will be 
= 
0.707 sin (t – 45
°
) 
+ 
0.0945 sin(3t – 71.6
°
) 
+ 
0.04 sin(5t – 78.7
°
) A.
This solution predicts that the current at t 
= 
0
+ 
is –0.63 A. But the initial condition for current 
at t 
= 
0
+ 
is zero. Therefore the natural response component has to be –0.63 e
-
t
A. Therefore the 
solution for current for t 
≥
0
+ 
is i(t) 
= 
–0.63 e
-
t
+ 
0.707 sin (t – 45
°
) 
+ 
0.0945 sin(3t – 71.6
°
) 
+ 
0.04 
sin(5t – 78.7
°
) A.
The circuit reaches periodic steady-state after about 5 time constants, i.e. at about 5 s. It operates 
under periodic steady-state till 100 s. At that point, the applied voltage in the circuit goes to zero 
suddenly. Therefore, the solution we determined above is valid only up to 100 s. The value of current 

---

the exponential Fourier Series 
9.7
at t
= 
100
-
s can be obtained by evaluating i(t) with t 
= 
100 s. The value is 0.66 A. This current 
decreases exponentially with a time constant of 1 s since the circuit is kept shorted after 100 s.
Therefore the complete solution for current in the circuit is
i t
e
t
t
t
t
( )
.
. sin(
)
.
sin(
. )
.
sin(
=
−
+
− ° +
−
° +
−
−
0 6
0 7
45
0 095
3
71 6
0 04
5
788 7
0
100
0 66
100
100
. ) ;
.
(
)
°
≤ ≤
≥




+
−
− −
+
A
A, for 
t
e
t
t
We could arrive at the solution only because we could solve for the forced response component 
using frequency response analysis. We could do that since the statement of the problem already 
expressed the applied source as a sum of sinusoids. Fourier series helps us to get this sum for a wide 
class of periodic waveforms.
If a waveform that is applied to a circuit can be expressed as a gated periodic waveform – i.e. if 
v
S
(t) 
= 
v(t) 
×
[u(t)
-
u(t
-
t
0
)], where v
S
(t) is the applied waveform, v(t) is the underlying periodic 
waveform and t
0
is the duration for which the underlying periodic waveform is passed on to the circuit – 
then, the forced response component in the output can be obtained by using sinusoidal expansion of 
the underlying waveform along with frequency response data for the circuit. Fourier series is needed 
and will be of great help in this kind of circuit problems.