Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

example: 9.6-2
Obtain the Fourier series of 
v t
t n
n
( )
(
. )
=
− −
=−∞
∞
∑
d
0 5
shown in Fig. 9.6-2.
1
–1
1
2
3
4
Time(s)
(
t
)
v
–2
–3
–4
Fig. 9.6-2 
Waveform 
v 
(
t
) for example: 9.6-2 
Solution
We observe by comparing Figs. 9.6-1 and 9.6-2 that this waveform is only a delayed version 
of waveform in Fig. 9.6-1. The delay involved is 0.5 s. Therefore if we delay all the sinusoidal 
components in the Fourier series of waveform in Fig. 9.6-1, we should get the Fourier series of 
waveform in Fig. 9.6-2. Delaying a sinusoid by t
d
s amounts to adding a phase delay of 
w
t
d
rad 
to its argument, where 
w
is its radian frequency. That is, 
w
t has to be replaced by (
w
t 
-
f
) where 
f
 
= 
w
t
d
.
n
th
harmonic component in the exponential Fourier series of any waveform is denoted by 

v e
n
jn t
o
w
, where 
w
o
is the fundamental radian frequency. Delaying this component by t
d
s results in 

---

Properties of Fourier Series and Some examples 
9.17


v e
v e
e
n
j n t n t
n
jn t
jn t
o
o d
o d
o
(
)
[
]
.
w
w
w
w
−
−
=
Therefore, the Fourier series of a time-shifted waveform can be 
expressed in terms of Fourier series of the original version as follows.
If 

v
n
are the coefficients of exponential Fourier series of a periodic waveform v(t), then the 
coefficients of exponential Fourier series of the time-shifted periodic waveform v(t 
- 
t
d
) are given by 

v e
n
jn t
o d
-
w
. This is called the ‘Time-shifting property of Fourier series expansion’.
In this example, the time delay is 0.5 s and that value is half the period. Therefore, 
e
e
n
n
jn t
j n
o d
−
−
=
=
−
w
p
1
1
for even and 
for odd .
Therefore, Fourier series of v(t) is as shown below as follows:
∴
− −
=
= +
−
−
=
∞
=−∞
∞
=−∞
∞
∑
∑
∑
d
p
p
p
(
. )
( )
cos
t k
e
e
nt
j n j
nt
n
n
k
k
0 5
1
1 2
2
2
1
(9.6-2)
There is no change in the amplitude of cosine waves, but the phase of waves alternate between 0 
and 180
°
with harmonic order n.
example: 9.6-3
Find the exponential Fourier series and trigonometric Fourier series of waveforms in Examples 9.6-1
and 9.6-2 if the magnitude of each impulse in both cases is V units and the period is T s.
Solution
We remember that there is a 1T factor in the Fourier series analysis equation. It happened to be 1 
when T was set to 1 s. We have to bring that factor back. Similarly, since 
w
p
o
T
=
2
we have to bring 
in 1T in the index of exponential in exponential Fourier series and in the argument of trigonometric 
functions in trigonometric Fourier series.
Hence the required Fourier series for waveform in Example 9.6-1 is
∴
−
=
= +
=
∞
=−∞
∞
=−∞
∞
∑
∑
∑
V
t kT
V
T
e
V
T
V
T
T
nt
j
T
nt
n
n
k
d
p
p
(
)
cos
2
1
2
2
and the required Fourier series for waveform in Example 9.6-2 is
∴
−
−
=




= +
−
−
=
∞
∑
V
t kT
T
V
T
e
e
V
T
V
T
T
nt
j n
j
T
nt
n
n
d
p
p
p
(
)
( ) cos
2
2
1
2
2
1
nn
k
=−∞
∞
=−∞
∞
∑
∑
example: 9.6-4
Let v
1
(t) be a periodic waveform same as the one used in Example 9.6-1 and v
2
(t) be a periodic waveform 
same as the waveform used in Example 9.6-2. Then, find v(t) 
= 
v
1
(t) 
-
 v
2
(t) and its Fourier series.
Solution
The waveform v(t) is constructed and shown in Fig. 9.6-3.

---

9.18
Dynamic Circuits with Periodic Inputs – Analysis by Fourier Series
We construct the Fourier series of this waveform by using Eqns. 9.6-1 and 9.6-2 and property of 
linearity of Fourier series.
∴
=
−
=
=
−
=−∞
∞
=−∞
∞
∑
∑
v t
e
e
e
e
nt
j
nt
jn
j
nt
n
j
nt
n
odd n
n
( )
cos
2
2
2
2
4
2
p
p
p
p
p
==
∞
=−∞
∞
∑
∑
1
odd n
n
It contains only odd harmonics of cosine format. Its average value is zero. It should be so since this 
waveform has even symmetry and half-wave symmetry.
1
–1
1
2
3
4
Time(s)
(
t
)
v
–2
–3
–4
Fig. 9.6-3 
Waveform of 
v
(
t
) in example: 9.6-4 
example: 9.6-5
v(t) is a periodic waveform with same waveform as that of Fig. 9.6-3, but with 2 units of magnitude in 
each impulse. Find and plot 
v t
v t dt
i
t
( )
( )
=
−∞
∫
and obtain Fourier series of v
i
(t).
Solution
Integration usually results in an arbitrary constant. This constant will become a DC component in 
v
i
(t). Since there is no way to find out this arbitrary constant we choose to ignore it and we qualify the 
Fourier series of v
i
(t) by adding a clause that a DC term can always come in the series without upsetting 
other coefficients. The waveform of v
i
(t), ignoring a possible DC term, is shown in Fig. 9.6-4.
1
–1
–1
1
2
3
4 Time (s)
(
t
)
v
i
–2
–3
–4
Fig. 9.6-4 
Waveform of 
v
i
(
t
) in example: 9.6-5 
The DC content was ignored. Therefore v
i
(t) must be pure AC. At t 
= 
0 its value will change by the 
area content of impulse at that point. This area content is 2 units. Hence the value of v
i
(t) must change 
by 2 units at t 
= 
0. Then, it must remain constant. At t 
= 
0.5 s the negative impulse of 2 unit magnitude 
will again change v
i
(t) by –2 units. With the assumption of AC output, v
i
(t) must then be alternating 
between 
+
1 and –1 with a period of 1 s. Hence v
i
(t) is a symmetric square wave of amplitude 1 unit 
and period 1 s.

---

Properties of Fourier Series and Some examples 
9.19
We wish to find the Fourier series of v
i
(t). There are two ways to do it. The first method is to 
employ the analysis equation of Fourier series, Eqn. 9.2-2, and carry out the required integration to 
determine the Fourier series coefficients. The second method relies upon the fact the waveform v
i
(t) is 
the integral of v(t) and that we already know the Fourier series of v(t). That leads us to the question – 
what is the relationship between Fourier series coefficients for a periodic waveform and Fourier series 
coefficients for its integral?
v t
v t dt
v e
dt
v e
dt
v
jn
e
i
i
n
jn t
n
jn t
t
n
o
jn t
n
o
o
o
( )
( )
=
=
=
=
−∞
=−
∫



w
w
w
w
∞
∞
∞
=−∞
∞
=−∞
∞
−∞
−∞
∑
∑
∑
∫
∫
n
n
t
t
∴
=


v
v
jn
in
n
o
w
(9.6-3)
Note that v(t) should not have an average value, i.e. 

v
a
0
0
0
=
=
.
If it has a non-zero average value, 
then its integral will have a linearly increasing term which makes it aperiodic and unbounded. There 
is no Fourier series for such a waveform.
We have arrived at the Integration in Time property of Fourier series. 
If 
v 
(
t
) is a zero-average periodic waveform with 

v
n
as its exponential Fourier series 
coefficients, then the exponential Fourier series coefficients of 
v t dt
t
( )
−∞
∫
is 

v
jn
n
o
w
.
The Fourier series coefficients of half of v(t) in this example was derived in Example 9.6-4.
v t
e
nt
v t
j
n
e
j
nt
n
odd n
n
odd n
i
j
nt
( )
cos
( )
=
=
∴
=
=
∞
=−∞
∞
∑
∑
4
8
2
4
2
2
1
2
p
p
p
p
==
+
−
=
−
=
∞
=
∞
∑
2
2
4 1
2
2
2
1
1
j n
e
j
n
e
n
nt
j
nt
j
n t
n
odd n
n
odd n
p
p
p
p
p
p
(
)
sin
(
)
∑
∑
∑
=−∞
∞
n
odd n
A unit amplitude symmetric square wave of period T will also have same coefficients, i.e. 
4
p
n
.
The 
factor of 
1
T
gets cancelled by a similar factor that is included in 
w
o 
in Eqn. 9.6-3. 
A symmetric square wave contains only odd harmonics and the harmonic amplitude 
=
4
p
n


in trigonometric Fourier series


decreases in inverse proportion to harmonic order 
n
.