Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

example: 9.9-3
A linear electrical element draws only sinusoidal current from a pure sinusoidal source. However, 
a non-linear load can draw a non-sinusoidal current that is rich in harmonics from a voltage source 
of sinusoidal nature. Various industrial electronic equipment such as AC–DC converters, inverter 
fed induction motor drives, thyristor controlled industrial heaters, converter fed DC motor drives, 
uninterruptible power supplies, fluorescent lamps etc. belong to this category. 

---

Analysis of Periodic Steady-State Using Fourier Series 
9.33
The voltage generated in synchronous generators located at various generating stations in a power 
system is generally a pure single frequency (50 Hz in India) sinusoid. However this voltage reaches 
individual customer’s electrical installation through transmission and distribution network. Hence, the 
voltage received by a customer will be a superposition of generated voltage and the voltage drops in 
various system impedances like transformer series reactance, line resistances and reactance and so 
on. Harmonic currents drawn by non-linear loads flow through these system impedances producing 
harmonic voltage drops in them. These harmonic voltage drops produce distorted voltage waveform 
everywhere in the power system despite the pure sinusoidal voltage generated at the generating stations.
Distorted voltage results in mal-functioning of electronic equipment, inefficient operation of 
motors, increased losses everywhere in the system etc. It can also cause damage to the power capacitors 
used in a power system for various purposes.
Harmonic current filters are used to prevent the harmonic currents generated by a load from getting 
into the power system. Both passive filters and filters using power semi conductor devices are in 
use. A passive harmonic filter is essentially a tuned LC series combination connected in parallel to 
the offending load. The LC circuit is tuned to the harmonic frequency to be eliminated. The tuned 
combination will have zero impedance at that frequency and the currents at that frequency will flow 
into the filter instead of getting into the power system. Obviously more than one such combination will 
have to be used in parallel if elimination of multiple harmonics is desired.
i
v
o
(
t
)
v
S
(
t
)
C
f
L
f
R
f
C
p
L
s
+
–
+
–
(
t
)
i
(
t
) (A)
100
–100
2
5
6
π
1
6
π
π
ω
π
t
Fig. 9.9-4 
Circuit diagram for the harmonic filtering context in example: 9.9-3 
L
S
in the circuit in Fig. 9.9-4 represents the Thevenin’s impedance of the power system at a point 
where a non-linear load is connected. The current drawn by the load is modelled as a current source 
i(t). The waveform of load current is also given in the same figure. L
f
and C
f
are the harmonic filter 
components. They are tuned to 250 Hz, i.e. it is a fifth harmonic current filter. R
f
is the series resistance 
of the inductor. The component values are L 
S
= 
0.3 mH, L
f
= 
1.84 mH, C
f
= 
0.22 mF and R
f
= 
0.072 
W
. 
C
p
is a power factor correction capacitor that is employed to draw leading current in order to cancel 
the lagging current drawn by another load which is not active in the current context. Its value is 0.47 
mF. The source voltage is a 320V amplitude, zero phase sinusoid at 50 Hz.
Solve for the three currents and the output voltage under steady-state conditions and plot them.
Solution
Step 1: Fourier series of i(t). 
Refer Example 9.7-1 for Fourier series of this waveform. It is
i t
j
n
n
n
e
n
n
n
n
j
nt
o
( )
sin
sin
sin
sin
sin
−
=




200
2
3
400
2
3
p
p
p
p
p
p
w
w
oo
n
n
n
n
t
=−∞
∞
=−∞
∞
∑
∑
odd
odd

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9.34
Dynamic Circuits with Periodic Inputs – Analysis by Fourier Series
All harmonics of 3 and its multiples will have amplitude of zero.
Step 2: Solve for steady-state response for each harmonic.
The phasor equivalent circuit for fundamental component is shown in circuit of Fig. 9.9-5 (a). The 
value of 
w
o
is 
≈
314 rad/s.
Peak values of voltage and current sources are marked instead of the rms value.It is a two-source 
problem and the solution is obtained by applying superposition principle.
Solution for (c)
Total effective impedance 
= 
j0.0942
+ 
(
-
j13.9)/(
-
j6.78) 
= 
j0.0942
-
 j4.56 
= -
j4.463 
W
\
I
S
= 
320
∠
0 
÷
-
j4.463 
= 
71.7 
∠
90
°
A
V
O
= 
320
∠
0 
×
(
-
 j4.56 
÷
-
j4.463) 
= 
327
∠
0 V
I
f
= 
71.7 
∠
90
°
×
(
-
j6.78 
÷
-
j20.68) 
= 
23.5
∠
90
°
A
I
c
= 
71.7 
∠
90
°
×
(
-
j13.9 
÷
-
j20.68) 
= 
48.2
∠
90
°
A
j
0.0942
j
0.58
approximately
–
j
13.9
–
j
14.47
–
j
6.78
0.072
320
∠
0
110
∠
0
I
s