Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

example: 9.11-1
Fourier analysis of the AC voltage received at the substation of an industry that used uses a large 
number of variable-speed drives shows that the received phase voltage can be represented by the 
following truncated Fourier series approximately.
v t
t
t
t
( )
.
sin(
)
.
sin(
.
)
.
sin(
=
+
−
+
−
6 35 2
100
0 2 2
500
0 5
0 15 2
700
p
p
p
rad
11rad kV
)
The industry draws a distorted current from this supply due to the power electronics involved in 
variable speed drives. Fourier analysis of current drawn shows that the following Fourier series can 
represent it approximately.
i t
t
t
( )
sin(
.
)
sin(
.
)
sin(
=
−
+
−
+
110 2
100
0 6
22 2
500
1 5
15 2
700
p
p
rad
rad
pp
p
p
t
t
t
−
+
−
+
−
1
10 2
1100
0 5
7 2
1300
0 5
rad
rad
rad A
)
sin(
.
)
sin(
.
)
These waveforms are shown in Fig. 9.11-1.
v
(
t
)
2
2
–2
–4
–6
–8
–10
4
6
8
10
4
6
(kV)
(rad)
t
ω
–50
50
100
150
200
–100
–150
–200
2
4
6
(rad)
i
(
t
) (A)
t
ω
Fig. 9.11-1 
Phase voltage and line current waveforms in example: 9.11-1 
Find the active power, VA, power factor, fundamental power factor and THD of voltage and current 
assuming balanced three-phase operation.

---

Summary 
9.45
Solution
v t
t
t
t
( )
.
sin(
)
.
sin(
.
)
.
sin(
=
+
−
+
−
6 35 2
100
0 2 2
500
0 5
0 15 2
700
p
p
p
rad
11rad kV
)
i t
t
t
( )
sin(
.
)
sin(
.
)
sin(
=
−
+
−
+
100 2
100
0 6
22 2
500
1 5
15 2
700
p
p
rad
rad
pp
p
p
t
t
t
−
+
−
+
−
1
10 2
1100
0 5
7 2
1300
0 5
rad
rad
rad A
)
sin(
.
)
sin(
.
)
∴ =
×
×
+
×
×
−
+
× ×
P
6 35 110
0 6
0 2 22
1 5 0 5
0 15 15
.
cos( .
)
.
cos( .
.
)
.
cos
rad
rad
((
)
.
.
1 1
581 1
1 743
−
=
=
rad
kW per phase
MW (three phase)
V
rms
kV (phase)
kV (line)
=
+
+
=
=
6 35
0 2
0 15
6 355
11
2
2
2
.
.
.
.
I
S V I
rms
rms rms
A
kVA (per ph
=
+
+
+
+
=
∴ =
=
110
22
15
10
7
113 8
723
2
2
2
2
2
.
aase)
MW (three-phase)
=
2 17
.
Power factor
Fundamental power factor
=
=
=
=
P
S
1 74
2 17
0 8
0
.
.
.
cos( ..
)
.
.
6
0 825
0 2
2
rad
lag
Total harmonic distortion in voltage
=
=
++
×
=
=
0 15
6 35
100 3 94
2
.
.
. %
Total harmonic distortion in current
222
15
10
7
110
100 26 63
2
2
2
2
+
+
+
×
=
. %
9.12 
summary
• Almost all periodic signals employed in circuits can be expressed as the sum of infinite number of 
sinusoids. Given a periodic waveform v(t) with period T, its exponential Fourier series is given by 
v t
v e
jn t
n
( )
=
−
=−∞
∞
∑

n
o
w
, where 
w
o
= 
2
p
T
and 

v
n
are the exponential Fourier series coefficients of v(t).
• The exponential Fourier series coefficients 

v
n
are found by 

v
T
v t e
dt
jn t
T
T
n
o
=
−
−
∫
1
2
2
( )
.
w
• The Fourier series may also be expressed in trigonometric form as follows:
∴
=
+
+
=
∞
=
∞
∑
∑
v t
a
a
n t
b
n t
n
n
( )
cos
sin
o
n
o
n
o
w
w
1
1
where 
o
o
n
n
o
a
v
T
v t dt
a
v
T
v t
n t dt
T
T
T
=
=
=
=
−
−
∫


1
2
2
2
2
2
( ) ,
Re( )
( ) cos
w
T
T
T
T
b
v
T
v t
n t dt
2
2
2
2
2
∫
∫
= −
=
−
and 
n
n
o
Im( )
( )sin

w

---

9.46
Dynamic Circuits with Periodic Inputs – Analysis by Fourier Series
• Another form of trigonometric Fourier series is given in the following:
∴
=
+ ∑
−
=
=
+
=
=
∞
v t
c
c
n t
c
v
c
a
b
v v
n
n
n
( )
cos(
)
,
o
n
o
n
o
o
n
n
where 
1
2
2
2
w
f

 
nn
v
b
a
v
n
∗
−
=
=
= −∠
=
2
1 2 3
1


n
n
n
n
n
and 
of , for 
f
tan
, , ...
• Following properties are exhibited by the Fourier series for a real v(t). 
table 9.12-1
Properties of exponential Fourier Series for a real 
v
(
t
) 
Aperiodic Signal Property
Fourier Series
v(t) 

v a
b c
n
n
n
n
n
and 
and 
,
,
f
-
Definition
a v
1
(t) 
+ 
b v
2
(t) 
av
bv
n
n


1
2
+
-
Linearity
v(t
-
t
d
) 
e
v
j t
d
-
w

n
– Time-shifting
e
jk
t
w
o
v(t) 

v
n k
-
– Frequency-shifting
v(
-
t) 

v
-
n
– Time reversal
v(
a
t), 
a
> 0

v
n
– Time-scaling
v
1
(t) v
2
(t) 
∑
=−∞
∞
−
k
k
n k
v v
 
1
2(
)
dv t
dt
( )
jn
v
w
o n

-
Time-domain differentiation
v t dt
t
( )
−∞
∫
(v(t) has zero DC content)
v
jn
n
o
w
– Time-domain integration
v(t) real


v
v
−
∗
=
n
n
– Conjugate symmetry
Re( ) Re(
)
Im( )
Im(
)




v
v
v
v
n
n
n
n
and
=
= −
−
−
|

v
n
| 
= 
|

v
-
n
| and 
∠ = −∠
−


v
v
n
n
v t
v t
T
t
( )
(
),
= −
±
1
2
for all 
-half-wave 
symmetry

v
n
= 
0 for even n
v t
v t
( )
( )
− =
, even symmetry
Im( )

v
n
= 
0, for all n. Only cosines in trigonometric Fourier 
series. 
v t
v t
( )
( )
− = −
, odd symmetry
Re( )

v
n
= 
0, Only sine terms in trigonometric Fourier series.
v t
v t
v t
e
( )
. [ ( )
( )]
=
+ −
0 5
Re( )

v
n
v t
v t
v t
o
( )
. [ ( )
( )]
=
− −
0 5
Im( )

v
n
Parseval’s power relation 
P
v t
dt
v
r
r
n
n
n
=
= ∑
−
=−∞
∞
∫
[ ( )]
2
2
2
2


---

Problems 
9.47
• The magnitude of exponential Fourier series coefficients plotted against frequency in a two-sided 
line plot is called the discrete magnitude spectrum of the waveform. A similar plot of phase of 
exponential Fourier series coefficients is the discrete phase spectrum. 
• The two components at 
±
n
w
o
in the discrete spectrum form a pair that cannot be separated in any 
analysis. They always go together to form a real sinusoidal component.
• The steady-state response of a circuit to a periodic input can be found in four steps. (i) Find the 
Fourier series of the input. (ii) Find out the frequency response function between the desired 
output variable and the input variable by using phasor equivalent circuit. (iii) Obtain the steady-
state response in time-domain for each sinusoidal component in the input. (iv) Decide how many 
terms in the output Fourier series are needed to represent the output waveform with the desired 
degree of accuracy. 
• Normalised power P 
n
of v(t) is defined as the average power that will be dissipated in a 1
W
resistance 
if this waveform is applied to it as a voltage. Parseval’s theorem states that the normalised power 
of a periodic waveform v(t), P
n
, is given by 
P
v v
v v
v
v
v
n
n
n
n
n
= ∑
= ∑
= ∑
=
+ ∑
=
=−∞
∞
−
=−∞
∞
∗
=−∞
∞
=
∞
 
 



n
n
n n
n
o
n
2
2
1
2
2
aa
a
b
c
c
n
n
n
n
n
o
o
2
1
2
2
2
1
2
2
2
+ ∑
+
=
+ ∑
=
∞
=
∞
(
)
• This is equivalent to 
P
n
n
n
th
= ∑
=
∞
0
2
(r.m.s value of 
harmonic component)
• The rms value of a waveform v(t) containing harmonics can be found by
rms value of 
(r.m.s value of 
harmonic compon
v t
n
n
th
( )
= ∑
=
∞
0
eent)
2