Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

example: 9.6-6
v(t) is a 
±
4 unit symmetric square wave with T 
= 
1 s. Find and plot 
v t
v t dt
i
t
( )
( )
=
−∞
∫
and find its Fourier 
series.

---

9.20
Dynamic Circuits with Periodic Inputs – Analysis by Fourier Series
v(t) and v
i
(t) are shown in Fig. 9.6-5.
4
–4
–1
1
2
Time(s)
Time(s)
(
t
)
v
i
(
t
)
v
–2
1
–1
–1
1
2
–2
Fig. 9.6-5 
Waveforms of 
v 
(
t
) and its integral, 
v
i 
(
t
), in example: 9.6-6 
Using the result from the previous example, we get, 
v t
J n
e
n
nt
j
nt
n
n
n
n
( )
sin
.
=
=
=
∞
=−∞
∞
∑
∑
8
16
2
2
1
p
p
p
p
odd
odd
The exponential Fourier series coefficients get divided by jn
w
o
when the waveform gets integrated 
in time.
∴
=
=
−
+
=
−
v t
j
n
ej
nt
n
ej
nt e
n
nt
i
j
nt
( )
(
)
(
)
cos
16
2
2
4
2
8
2
2
2 2
2
2 2
p
p
p
p
p
p
p
nn
n
n
n
n
n
=
∞
=
∞
=−∞
∞
∑
∑
∑
1
1
odd
odd
odd
The triangle wave in Fig. 9.6-5 has 
±
1 unit amplitude and has even half-wave symmetry. Hence it 
contains odd cosine harmonics.
A symmetric triangle wave contains only odd harmonics and the harmonic amplitude 
=
8
2 2
p
n


in trigonometric Fourier series


decreases in inverse proportion to square of 
harmonic order 
n
.
example: 9.6-7
Let v(t) be a periodic waveform with period T and let v
C
(t) be defined as 
v t
v t
C
( )
( )
=
a
, where 
a
> 0. 
Show that v(t) and v
C
(t) have same coefficients in their Fourier series.

---

Properties of Fourier Series and Some examples 
9.21
Solution
v
C
(t) will be a time-compressed version of v(t) for 
a
>1 and time-expanded version of v(t) for 
a
< 1. 
Therefore period of v
C
(t) will be 
T
a
. Let 


v
v
cn
n
and 
be the exponential Fourier series coefficients of 
v
C
(t) and v(t) respectively.

v
T
v t e
dt
T
v t e
dt
cn
c
j
T
nt
j
T
n
t
T
T
T
=
=
−
−
∫
a
a
a
ap
p a
a
a
a
( )
( )
(
)
2
2
1
2
1
2
1
2
1
22
1
2
1
2
2
2
1
1
a
a
a
a
a
a
a
a
p a
p a
T
T
T
T
v t e
d t
T
v t e
j
T
n
t
j
T
n
t
∫
∫
=
=
−
( )
( )
( )
(
)
(
)
dd t
v
T
T
n
( )
a
a
a
−
∫
=
1
2
1
2

Compression or expansion of a periodic waveform in time does not change its Fourier 
series coefficients. but fundamental frequency and harmonic frequency values will 
change. this property is called 
time-scaling property of Fourier series
. 
Note that time-scaling is not the same as a simple change in T alone. Let v(t) a periodic train of 
rectangular pulses of unit amplitude and 0.1 s width repeating with a period of 1 s. Then v(0.5t) is 
a periodic train of rectangular pulses of unit amplitude and 0.2 s repeating with a period of 2 s. Its 
Fourier series coefficients will be same as that of v(t), but its fundamental frequency will be 
p
rad/s 
whereas that of v(t) will be 2
p
rad/s.
However, consider another periodic train of rectangular pulses of unit height and 0.1 s width 
repeating with a period of 2 s. This is not the same as v(0.5t). Its Fourier series coefficients will be 
different from that of v(t). In fact, all Fourier series coefficients will get multiplied by 
1
2 .
example: 9.6-8
Find the Fourier series of the periodic rectangular pulse train shown in Fig. 9.6-6.
Solution

v
T
v t e
dt
T
e
dt
jn T
e
e
n
jn t
jn t
o
j
j
o
o
n o
n o
=
=
= −
−
−
−
−
−
1
1
1
2
2
( )
[
]
w
w
t
t
w
w
w
tt
t
w t
w t
w t
/
/
/
/
/
/
sin
2
2
2
2
2
2
2
2
∫
∫
−
−
−
= −
T
T
jn
jn
n
e
e
j
o
o
o
, by Euler’s Formula
∴ =
=
= 




v
T
n
n
T
n
n
T
n
o
o
o
o
t
w t
w Tt
t
w t
w t
t
2
2
2
2
sin
sin
sin
xx
x
x
n
o




=
, where 
w t
2
Fig. 9.6-6 
Waveform for 
example: 9.6-8
1
v
(
t
)
t
2 2
τ
τ
τ
τ
2
2
–
–

---

9.22
Dynamic Circuits with Periodic Inputs – Analysis by Fourier Series
The DC content of the waveform is 
t
T
. The Fourier series contains only cosine terms since 
exponential Fourier series coefficients are real.
example: 9.6-9
Figure 9.6-7 shows the output of an absolute value 
circuit when input is a sinusoidal wave. This is 
the shape of output voltage of a full-bridge diode 
rectifier routinely used in AC–DC conversion 
applications. We wish to obtain the Fourier series 
for this waveshape. We bring out an important 
property of Fourier series first and use that property 
to obtain the required Fourier series in this example.
Solution
The waveform v(t) is visualised first as the product of two waveforms as in Fig. 9.6-8.
The product of v
2
(t) which is a unit amplitude square 
wave with v
1
(t) which is a pure sine wave results in v(t), 
the full-wave rectified waveform.
That raises the question – what are the exponential 
Fourier series coefficients of a product of two waveforms 
with same period in terms of exponential Fourier series 
coefficients of the constituent waveforms?
v t
v e
v t
v e
v t
v t
n
jn t
n
n
jn t
n
o
o
1
1
2
2
1
( )
( )
( )
(
=
=
=
=−∞
∞
=−∞
∞
∑
∑


w
w
and 
)) ( )
v t
v e
k
jk
t
n
o
2
=
=−∞
∞
∑

w
Consider kth
component in the exponential Fourier series of v(t). The waveform contributed to v(t) 
by this component is 

v e
k
jk
t
o
w
. Since v(t) is the product of v
1
(t) and v
2
(t), this contribution can come up 
in v(t) due to products of n
th
contribution 

v e
n
jn t
o
1
w
in v
1
(t) and (k 
- 
n)th contribution 

v
e
k n
j k n
t
o
2(
)
(
)
-
-
w
with n varying from 
-∞
to 
+∞
.
∴ =
−
=−∞
∞
∑

 
v
v v
k
n
k n
n
1
2(
)
This is the so-called Multiplication in Time property of Fourier series.
If 
v
1
(
t
) and 
v
2
(
t
) are two periodic waveforms with same period and 
v
(
t
) 
= 
v
1
(
t
) 
×
v
2
(
t
), then the 
exponential Fourier series coefficients of 
v 
(
t
) is given by 

 
v
v v
k
k
n
k n
n
=
− ∞ < < ∞
−
=−∞
∞
∑
1
2(
)
for 
,
where 


v
v
n
n
1
2
and 
are the exponential Fourier series coefficients of 
v
1
(
t
) and 
v
2
(
t
), 
respectively.
Fig. 9.6-7 
Full-wave rectified waveform 
in example: 9.6-9
v
(
t
)
t
1
0.5
–1
1
0.5
–0.5
Fig. 9.6-8 
two waveforms in a 
product results in 
v 
(
t
) 
of example: 9.6-9
2
1
–1
1
0.5
0.5
–0.5
–0.5
v
(
t
)
1
v
(
t
)
t

---

Discrete Magnitude and Phase Spectrum 
9.23
v
1
(t) is a sine wave in this example.
v t
t
e
e
j
v
j
v
j
t
j
t
1
2
2
1
2
2
1
2
( ) sin
,
=
=
−
∴ =
−
p
p
p
(By Euler’s Formula)


−−
=
−
=
1
1
2
0
j
v
n
n
and 
for all other values of 

,
v
2
(t) is a unit square wave. Its Fourier series was obtained in Example 9.6-5 as

v
j n
e
n
j
nt
n
n
2
2
2
=
=−∞
∞
∑
p
p
odd
∴ =
= −
+
+
−
−
=−∞

 
v
v v
j j
k
j j
k
k
k
n
k n
n
1
2
1
2
2
1
1
2
2
1
(
)
(
)
(
)
p
p
, for even 
∞
∞
∑
=
−
−
2
1
2
p
(
)
k
k
, for even 
Since n takes only two values –1 and 1, and square wave has only odd harmonics, (k
+ 
1) and (k 
- 
1) 
have to be odd for 

v
k
to be non-zero.
∴
=
−
−
=
−
−
=
∞
=−∞
∑
v t
k
e
k
kt
j
kt
k
k
k
( )
(
)
(
)
cos
2
1
2
4
1
1
2
2
2
2
1
p
p p
p
p
even
evenn
k
∞
∑
is the Fourier series of a full-wave 
rectified sinusoid of unit amplitude.
The average value of rectified waveform is 
2
p
or 
2
V
m
p
in general, where V
m
is the peak value of 
sine wave undergoing rectification. It has only even cosine harmonics in it. What kind of symmetry 
is responsible for this? It is even function and hence only cosines are expected. Notice that the 
fundamental period of the waveform is 0.5 s and not 1 s. This 1 s period comes up only in the square 
waveform and sine waveform we used to form the product. Hence, there is no component at odd 
harmonic order in the rectified output. If we treat it as a waveform with 0.5-s period, all odd and even 
harmonic components of its fundamental frequency are present in it. Its fundamental frequency will 
be double the fundamental frequency of the sinusoid that underwent rectification.