Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

example: 9.10-1
The output of a fully controlled AC–DC converter operating from a sinusoidal voltage of 320 V peak 
and 50 Hz frequency is shown in Fig. 9.10-1. (i) Find and plot its discrete power spectrum. (ii) This 
waveform is applied to RL circuit with L 
= 
150 mH and R 
= 
10 
W
. Find and plot the discrete power 
spectrum of voltage appearing across the resistor and find the power dissipation in it.
320 V
–20
–17.5
–10
–7.5
2.5
10
12.5
20
Time in ms
Fig. 9.10-1 
Waveform for example: 9.10-1
Solution-(i)
This waveform, v(t), can be expressed as the product of two waveforms – v
1
(t) 
= 
320sin100
p
t and v
2
(t) 
which is a symmetric 
±
1, 50 Hz square wave which is delayed by 2.5 ms.
v t
t
e
e
j
j
t
j
t
1
100
100
320
100
320
2
( )
sin
=
=
−
−
p
p
p
(By Euler’s Formulaa)
and 
for all other values
∴
=
=
−
=
−



v
j
v
j
v
n
11
1 1
1
160
160
0
,
,
of 
n
v
2
(t) is a unit amplitude square wave. Fourier series of a unit amplitude square wave of 1 s 
period that crosses zero at origin was obtained in Example 9.6-5 as 
2
2
j n
e
j
nt
n
n
p
p
.
=−∞
∞
∑
odd
Using time-shift 
property of exponential Fourier series we write the exponential Fourier series of the square wave in 
this example as 

v
j n
e
e
n
j n j
nt
n
n
2
4
100
2
=
−
=−∞
∞
∑
p
p
p
.
odd
Using the multiplication in time property of exponential 
Fourier series we write the exponential Fourier series coefficients of the waveform v(t) as
∴ =
= −
+
+
−
−
−
+
−
−

 
v
v v
j
e
j
k
j
e
j
k
k
n
k n
j
k
j
k
1
2
1
4
1
4
160 2
1
160 2
(
)
(
)
(
p
p
p
p
11
320
1
1
1
4
1
4
)
, for even 
k
e
k
e
k
n
j
k
j
k
=−∞
∞
−
+
−
−
∑
=
−
+
+
−







p
p
p

, for even 
k
The magnitude part of the quantity inside the brackets can be shown as 
2 1
1
2
2
(
)
.
+
−
k
k

---

9.42
Dynamic Circuits with Periodic Inputs – Analysis by Fourier Series
∴
= × 



+
−
| |
(
)
v
k
k
k
n
2
2
2
2
2
2
320
1
1
p
, for even 
The discrete power spectrum of 
| |
v
n
2
is plotted against harmonic order n in Fig. 9.10-2. The 
common factor 
2
320
2
× 



p
is normalised to unity. 88% of P
n
of this waveform is contributed by 
the DC component and the 100 Hz component (n 
= 
2). The next harmonic component at 200 Hz 
(n 
= 
4) adds another 6.3%. Thus all the harmonics above 200 Hz contribute only 5.7% of power 
to P
n
. The contribution from a spectral component to normalised power depends on the square of 
amplitude of spectral component. The sharp edge in the waveform in this example is synthesised by 
small amplitude high frequency components. But they contribute almost nothing to the normalised
power.
n
1
0.556
0.556
0.076
0.076
0.03
0.03
–9 –8 –7 –6 –5 –4 –3 –2 –1
1
2
3
4
5
6
7
8
9
Fig. 9.10-2 
Two-sided discrete power spectrum of waveform 
v
(
t
) in Example: 9.10-1 
Solution (ii)
The frequency response of resistor voltage in a series RL circuit is obtained as 
V j
V j
R
(
)
(
)
w
w
=
R
R
j L
j
+
=
+
w
wt
1
1
.
t
 
= 
150 mH/10
W
= 
15 ms and 
w
= 
100
p
n in this case.
V j
V j
j
n
n
n
R
(
)
(
)
.
.
tan
.
w
w
p
=
+
=
+
∠
−
1
1
1 5
1
1 22 21
4 71
2
1
Exponential Fourier series coefficients of output 
= 
exponential Fourier series coefficients of output 
×
value of frequency response function at the corresponding frequency.
\
Power spectral component in output 
= 
power spectral component at input 
×
square of magnitude 
of frequency response function value at the corresponding frequency.
The values of square of magnitude of frequency response function is 1 at n 
= 
0, 0.1055 at 
n 
= 
2, 0.053 at n 
= 
4 and 0.035 at n 
= 
6. Hence, the power spectral component (normalised with 
respect to maximum value) in the output at n 
= 
0 is 1, at n 
= ±
2 is 0.1055
×
0.556 
= 
0.059, at n 
= 
±
4 is 0.053
×
0.076 
= 
0.004 and at n 
= ±
6 is 0.035
×
0.03 
= 
0.001. Thus almost the entire normalised 
power in the output waveform comes from its DC component. This spectrum is plotted in
Fig. 9.10-3.

---

Power and Power Factor in AC System with Distorted Waveforms 
9.43
n
1
0.06
0.06
0.04
0.04
0.01
0.01
1
2
3
4
5
6
7
8
9
–9 –8 –7 –6 –5 –4 –3 –2 –1
Fig. 9.10-3 
Discrete power spectrum of output waveform in example: 9.10-1 
Power dissipated in the 10 
W
resistor 
= 
P
n
/10 
≈
2
320
2
× 



p
×
(1
+ 
0.06
+ 
0.06
+ 
0.04
+ 
0.04
+ 
0.01
+ 
0.01
+ .
..) /10 
= 
2.53 kW.
9.11 
PoWer and PoWer factor In ac system WIth dIstorted Waveforms
Let v(t) and i(t) be the voltage across an electrical element and current through that element under 
periodic steady-state conditions in an AC system working under distorted waveform conditions. 
Both v(t) and i(t) are expressed in the form of trigonometric Fourier series below where 
w
o
is the 
fundamental radian frequency of the system. AC system voltages and currents usually do not contain 
any DC component under steady-state and hence DC components are not included in these Fourier 
series.
v t
V
n t
i t
I
n t
n
o
vn
n
n
o
in
n
( )
cos(
)
( )
cos(
)
=
−
=
−
=
∞
=
∞
∑
∑
2
2
1
1
w
f
w
f
V
n
and 
f
vn 
represent the rms value of n
th
harmonic component of voltage and its phase. Similarly I
n
and 
f
in
represent the rms value of n
th
harmonic component of current and its phase.
The power delivered to the element, assuming passive sign convention, is given by v(t)
×
i(t). The 
average power delivered, P, is obtained by averaging this quantity over one period of the AC system.
The product v(t)i(t) contains two kinds of terms – product of cosine functions of same harmonic order 
which have the general form of 
2
V I
k
t
k
t
k k
o
vk
o
ik
cos(
) cos(
)
w
f
w
f
-
-
is the first kind. The second kind of 
terms will be of the form 
2
V I
k
t
r
t
k
r
k r
o
vk
o
ir
cos(
) cos(
);
.
w
f
w
f
−
−
≠
The first kind can be rewritten as 
V I
k
t
k k
o
vk
ik
ik
vk
[cos(
) cos(
)].
2
w
f
f
f
f
−
−
+
−
When integrated over a period and divided by the period, 
this term will result in a contribution of V I
k k
ik
vk
cos(
)
f
f
-
to the average delivered power. The second 
kind of terms can be expressed as V I
k r
t
k r
t
k
r
k k
o
vk
ir
o
ir
vk
[cos((
)
) cos((
)
)];
.
+
−
−
+
−
+
−
≠
w
f
f
w
f
f
Since k and r are integers, (k
+ 
r) and (k 
- 
r) are also integers. Therefore, the two cosine components at 
the sum and difference frequencies will have zero average value over one period of the system. Hence, 
only the first kind of terms contribute to the average power.
∴ =
−
=
∞
∑
P
V I
n n
in
vn
n
cos(
)
f
f
1
W

---

9.44
Dynamic Circuits with Periodic Inputs – Analysis by Fourier Series
Thus, the active power (i.e. the average power) delivered by each voltage and current harmonic pair 
with same harmonic order is independent of whether they are acting alone or acting along with other 
harmonic pairs. Each harmonic pair delivers an active power given by the product of rms voltage, rms 
current and power factor of that pair.
However, the volt–ampere product (VA) is given by the product of rms values of v(t) and i(t). 
We express the rms value of v(t) in terms of rms values of its harmonic components by using 
Eqn. 9.10-4 as 
V
V
n
n
rms
=
=
∞
∑
( )
2
1
and the rms value of i(t) as I
I
n
n
rms
=
=
∞
∑
( ) .
2
1
Then V-A product 
S 
= 
V I
V
I
n
n
n
n
rms rms
=
=
∞
=
∞
∑
∑
( )
( )
2
1
2
1
VA
The power factor (apparent power factor) is the ratio between P and S. It includes the effect of 
reactive power of each harmonic voltage–current pair as well as the cross-power terms resulting from 
product of voltage and current of different harmonic order. Cross-power terms do not contribute to 
active power, but they contribute to VA.
The fundamental power factor is 
cos(
)
f
f
i
v
1
1
-
and is lagging if 
(
)
f
f
i
v
1
1
-
is negative and is leading 
if 
(
)
f
f
i
v
1
1
-
is positive.