Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

I
f
I
c
I
V
s
+
+
320
∠
0
V
s
+
j
0.0942
–
j
13.9
–
j
6.78
320
∠
0
110
∠
0
I
s
v
o
v
o
v
o
I
f
I
c
I
V
s
+
+
j
0.0942
–
j
13.9
–
j
6.78
110
∠
0
I
s
I
f
I
c
I
+
j
0.0942
–
j
13.9
–
j
6.78
I
s
V
o
I
f
(c)
[Impedance values are in 
Ω
]
(a)
(d)
(b)
I
c
+
Fig. 9.9-5 
(a) and (b) Phasor equivalent circuit (c) and (d) Circuits 
for applying superposition principle
Solution for (d)
Total effective admittance 
= -
j10.4 S
Total effective impedance 
= 
j0.096
\
Vo 
= -
110
∠
0 
×
j0.096
= -
10.6 
∠
90
°
V
I
S
= 
10.6
∠
90
°
÷
j0.094 
= 
112.8 
∠
0
°
A

---

Analysis of Periodic Steady-State Using Fourier Series 
9.35
I
C
= 
10.6
∠
90
°
÷
(
-
j6.78) 
= 
1.54
∠
180
°
A
I
f
= 
10.6
∠
90
°
÷
(
-
j13.9) 
= 
0.74
∠
180
°
A
Solution for (c) 
+ 
(d)
V
O
= 
327
∠
0 
-
10.6 
∠
90
°
V 
= 
327.2
∠-
1.9
°
V
I
S
 
= 
71.7 
∠
90
°
+ 
112.8 
∠
0
°
A
= 
133.66
∠
32.4
°
A
I
f
= 
23.5
∠
90
°
+ 
0.74
∠
180
°
A
= 
23.5
∠
91.8
°
A
I
c
= 
48.2
∠
90
°
+ 
1.54
∠
180
°
A
= 
48.2
∠
91.8
°
A
Time-domain solution
v
o
(t) 
= 
327.2 sin(314t 
-
1.9
°
)V
i
S
(t) 
= 
133.7 sin(314t 
+ 
32.4
°
) A
i
f
(t) 
= 
23.5 sin(314t 
+ 
91.8
°
) A
i
C
(t) 
= 
48.2 sin(314t 
+ 
91.8
°
) A
The equivalent circuits for solving the harmonic currents are shown in Fig. 9.9-6. Only the first four 
non-zero harmonics in load current are considered.
The circuits can be solved by current division principle – current divides among parallel elements 
in proportion to their admittance values. Solution is illustrated in the case of n 
= 
11.
V
O
= 
10
∠
0 
÷
(j1.63
-
j0.2
-
j0.96)
= 
21.3
∠-
90
°
V
I
S
= 
10
∠
0 
×
-
j0.96 
÷
(j1.63
-
j0.2
-
j0.96) 
= -
20.4
∠
0
°
A
I
f
= 
10
∠
0 
×
-
j0.2 
÷
(j1.63
-
j0.2
-
j0.96) 
= -
4.26
∠
0
°
A
I
C
= 
10
∠
0 
×
j1.63 
÷
(j1.63
-
j0.2
-
j0.96) 
= 
34.7
∠
0
°
A
Observe that the load current had only 10 A at 550 Hz (11th harmonic). However, the source 
current has 21.3 A and the power factor correction capacitor C
p
has 34.7 A in it. If C
p
were not there, 
the source current would have been close to 10 A only and the 11th harmonic component in voltage 
would have been close to 10 V instead of its present value of 21.3 V. This amplification of 11th 
harmonic illustrates a potential problem that can exist in harmonic context in power systems. The 
admittance of source inductance and the power factor capacitor cancel each other partially at 11th 
harmonic and this results in an increase in the impedance level presented to 11th harmonic. Higher 
impedance results in higher voltage and higher currents in individual parallel paths. There can be 
dangerous harmonic resonance if the parallel resonant frequency of the circuit coincides or comes 
near one of the harmonics of supply frequency. In this case, the resonant frequency is around 424 Hz 
(decided practically by the 0.3 mH inductor and 0.47 mF capacitor) and there can be large harmonic 
amplification if the load current contains 8th of 9th harmonics.

---

9.36
Dynamic Circuits with Periodic Inputs – Analysis by Fourier Series
j
0.47 (–
j
2.13)
(
j
0.74)
j
2.9
(13.9)
0.072
–
j
2.9
–
j
1.36
0.072
–22
∠
0
I
s
I
f
I
c
I
+
[Impedance values are in 
Ω
. Admittance values (in brackets) are in siemens.]
(a)
n
= 5
j
0.66 (–
j
1.52)
(
j
1.04)
j
4.05
(–
j
0.5)
i
2
approximately
–
j
2.9
–
j
0.96
0.072
–15.7
∠
0
I
s
v
o
v
o
v
o
v
o
I
f
I
c
I
+
(b)
n
= 7
j
1.226(–
j
0.82)
(
j
1.94)
j
7.52
(
–j
0.156)
j
6.41
approximately
–
j
1.11
–
j
0.516
0.072
8.5
∠
0
I
s
I
f
I
c
I
+
(d)
n
= 13
j
1.037(–
j
0.96)
(
j
1.63)
j
6.364
(–
j
0.2)
j
5.05
approximately
–
j
1.37
–
j
0.61
0.072
10
∠
0
I
s
I
f
I
c
I
+
(c)
n
= 11
Fig. 9.9-6 
Circuit equivalents for current harmonics
The time-domain solution is obtained by using 
w
= 
11 
× 
314 rad/s.
v
o
(t) 
= 
21.3 sin (11 
× 
314t 
-
90
°
) V
i
S
(t) 
= -
20.4 sin (11 
× 
314t) A
i
f
(t) 
= -
4.26 sin (11 
× 
314t) A
i
C
(t) 
= 
34.7 sin (11 
× 
314t) A
The equivalent circuits for other harmonics also can be solved similarly.
The time-domain solution for n 
= 
5 is given in the following. Note that the tuned circuit diverts 
most of the 22 A fifth harmonic into it. The source current still contains fifth harmonic and this is due 
to the fact that the filter path has a resistance of 0.072 
W
.
v
o
(t) 
= -
1.58 sin(5 
× 
314t
+ 
5.7
°
) V
i
S
(t) 
= -
3.35 sin(5 
× 
314t 
- 
84.3
°
) A
i
f
(t) 
= -
21.9 sin(5 
× 
314t
+ 
5.7
°
) A
i
C
(t) 
= -
1.17 sin(5 
× 
314t
+ 
95.7
°
) A
The time-domain solution for n 
= 
7 is given in the following:
v
o
(t) 
= -
16.1 sin(7 
× 
314t
+ 
90
°
) V
i
S
(t) 
= -
24.4 sin(7 
× 
314t) A
i
f
(t) 
= -
8.05 sin(7 
× 
314t) A
i
C
(t) 
= 
16.7 sin(7 
× 
314t) A

---

Analysis of Periodic Steady-State Using Fourier Series 
9.37
The time-domain solution for n 
= 
13 is given in the following:
v
o
(t) 
= 
8.8 sin(13
×
314t 
-
90
°
) V
i
S
(t) 
= -
7.2 sin(13
×
314t) A
i
f
(t) 
= -
1.38 sin(13
×
314t) A
i
C
(t) 
= 
17 sin(13
×
314t) A
The total solution is obtained by combining all these solutions. We neglect the harmonics above 
13th in this example.
v
o
(t) 
=
327sin(
w
o
t
-
1.9
°
) 
-
1.6sin(5
w
o
t
+ 
5.7
°
) 
-
16.1sin(7
w
o
t
+ 
90
°
) 
+ 
21.3sin(11
w
o
t
-
90
°
) 
+ 
8.8sin(13
w
o
t
-
90
°
) V
i
S
(t) 
=
133.7 sin(
w
o
t 
+ 
32.4
°
) 
-
3.35 sin(5
w
o
t 
-
84.3
°
) 
-
24.4 sin(7
w
o
t) 
-
20.4 sin(11
w
o
t) 
-
7.2 sin(13
w
o
t) A
i
f
(t) 
=
23.5 sin(
w
o
t 
+ 
91.8
°
) 
-
21.9 sin(5
w
o
t 
+ 
5.7
°
) 
-
8.05 sin(7
w
o
t) 
-
4.26 sin(11
w
o
t) 
-
1.38 sin(13
w
o
t) A
i
C
(t) 
=
48.2 sin(
w
o
t 
+ 
91.8
°
) 
-
1.17 sin(5
w
o
t 
+ 
95.7
°
) 
+ 
16.7 sin(7
w
o
t) 
+ 
34.7sin(11
w
o
t) 
+ 
17 sin(13
w
o
t) A
These waveforms are plotted in Fig. 9.9-7. The waveforms reveal that the presence of power factor 
correction capacitor has caused deterioration in the harmonic performance of the system and that 
the harmonic filter is not very effective in this context. Observe the high frequency currents in the 
capacitor.
300
200
100
–100
–200
–300
t
v
O
(
t
)(V)
60
50
–50
40
–40
30
–30
20
–20
10
–10
t
i
f
(
t
) (A)
150
125
100
75
50
25
–25
–50
–75
–100
–125
–150
t
i
S
(
t
) (A)
125
i
c
(
t
) (A)
100
75
50
25
–25
–50
–75
–100
t
Fig. 9.9-7 
Waveforms of 
v
o
(
t
), 
i
S
(
t
), 
i
C
(
t
) and 
i
f 
(
t
) in example: 9.9-3

---

9.38
Dynamic Circuits with Periodic Inputs – Analysis by Fourier Series