Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

v
(
t
)
t
(s)
1
1
–1
–1
–0.5
0.5
–
–
–
Fig. 9.13-7 
17. Find the trigonometric Fourier series of the waveform v(t) in Fig. 9.13-8 and plot its spectrum.
v
(
t
)
t
(s)
1
0.5
–0.5
–0.5 –0.25
–0.75
–1
0.5 0.75
0.25
–1
Fig. 9.13-8 
18. v(t) is a cosine wave and v
1
(t) is a square wave in Fig. 9.13-9. (i) Find v
2
(t) 
= 
v(t) v
2
(t) and plot it. 
(ii) Find the trigonometric Fourier series of v
3
(t) from Fourier series of v(t) and v
1
(t) and plot its 
spectrum.
(
t
)
v
1
1
1
0.5
0.25
0.75
1.25
0.5
–0.5
–1
v
(
t
)
t
(s)
Fig. 9.13-9 
19. v(t) is a sine wave and v
1
(t) is a square wave in Fig. 9.13-10. (i) Find v
2
(t) 
= 
v(t)v
1
(t) and plot it. 
(ii) Find the trigonometric Fourier series of v
3
(t) from Fourier series of v(t) and v
1
(t) in terms of 
a
. 
(iii)Plot its spectrum for 
a
 
= 
p
/6.
(
t
)
v
1
t
(s)
1
0.5
–0.5
–1
/2 –
v
(
t
)
π
/2
3
π
2
π
π
α
α
–
π
2
π
α
+
π α
Fig. 9.13-10 

---

Problems 
9.51
20. Positive half-cycle of v(t) with a period of 2 s is shown in Fig. 9.13-11. The waveform has odd 
symmetry. Find the exponential and trigonometric Fourier series of this waveform and plot its 
one-sided spectrum. If this waveform is used as an approximation to a sine wave find its THD.
v
(
t
)
t
(s)
1.5
1.0
0.5
0.25 0.5
0.75 1.0
Fig. 9.13-11 
21. v
S
(t) 
= 
5|sin
w
o
t| V with 
w
o
= 
100
p
rad/s in Fig. 9.13-12. Assume ideal Opamp and find the output 
voltage v
o
(t). Draw its one-sided spectrum. What function does this circuit perform?
5 
µ
F
5 k
10 k
20 k
+
–
+
–
+
–
v
O
(
t
)
v
S
(
t
)
Fig. 9.13-12 
22. The input voltage applied to the Opamp circuit in Fig. 9.13-13 is a symmetric triangle periodic 
waveform moving between 
+
5 V and –5 V with a period of 1 ms. Find the plot the output voltage 
as a function of time. What function does this circuit perform?
10 nF
10 k
10 k
10 k
+
–
+
–
+
–
v
S
(
t
)
v
O
(
t
)
Fig. 9.13-13 
23. The circuit in Fig. 9.13-14 is a practical differentiator circuit using an Opamp. The components 
C and R are sufficient to carry out differentiation. However, the non-ideal frequency response 
of the Opamp makes the circuit highly under-damped usually and the additional component, 
R
d
, imparts damping to the circuit. But, with R
d
present, the circuit is no more a differentiator at 
high frequencies. The input voltage applied to the practical differentiator circuit using Opamp in 
Fig. 9.13-14 is a 
±
1V symmetric triangular periodic waveform at 2.5 kHz. Obtain and plot the 
output voltage waveform. What is the expected output from a good differentiator for this input 
waveform? How does the calculated output compare with it?

---

9.52
Dynamic Circuits with Periodic Inputs – Analysis by Fourier Series
10 nF
C
10 k
100 k
R
+
–
+
–
+
–
v
S
(
t
)
R
d
v
O
(
t
)
Fig. 9.13-14 
24. The circuit in Fig. 9.13-15 is a practical integrator using an Opamp. The resistor R
off
is needed to 
control the DC offset at output terminals. However, R
off
makes the circuit an imperfect integrator. 
The input to this integrator is the waveform shown in Fig. 9.13-6. Find and plot the output taking 
the first five non-zero harmonics of input into account.
10 k
100 
µ
F
100 k
R
C
off
+
–
+
–
+
–
v
S
(
t
)
R
v
O
(
t
)
Fig. 9.13-15 
25. (i) Predict the DC content in current through 6 
W
and in voltage across the parallel combination 
without finding out Fourier series coefficients in the circuit in Fig. 9.13-16. (ii) Find the output 
voltage v
o
(t) and plot its one-sided Fourier spectrum. (iii) Find the rms value of current through 
0.3 
W
and the power dissipated in it.
t
(
µ
s)
16 A
12.5
50
62.5
6 
Ω
+
–
0.3 
Ω
200 
µ
F
i
S
(
t
)
i
S
(
t
)
v
O
(
t
)
Fig. 9.13-16 
26. Find the output voltage v
o
(t) in the circuit in Fig. 9.13-17 considering the DC component and first 
two non-zero harmonics in the input current source.
0.159 mH
0.159 mF
10 
Ω
5 A
–
+
–1
–0.2 0.2
1
i
S
(
t
)
v
O
(
t
)
i
S
(
t
)
t
(ms)
Fig. 9.13-17 

---

Problems 
9.53
27. R 
= 
1k 
W
and C 
= 
1 
m
F in the circuit in Fig. 9.13-18. The source voltage is a periodic impulse 
train given by v
S
(t) 
= 
d
(
)
t n
n
− ×
−
=−∞
∞
∑
10
3
V.
Find and plot the two-sided discrete power spectrum 
of v
o
(t).
R
C
C
R
v
S
(
t
)
v
O
(
t
)
+
+
–
–
Fig. 9.13-18 
28. The output voltage of a Power Electronic Inverter Circuit is related to the DC voltage used in 
the inverter by the equation 
v t
V
m
t
o
dc
( )
sin
=
×
100
p
, where m is the so-called modulation index. 
Assume that V
dc
is not a pure DC source and it contains AC components. Let V
dc
= 
400 +
20cos200
p
t
-
10cos400
p
t V and m 
= 
0.8. (i) Find and plot the output v
o
(t) of the Inverter. (ii) Find the THD and 
rms value of Inverter output. (iii) Plot the two-sided power spectrum of output voltage.
29. The switch S in the circuit in Fig. 9.13-19 operates periodically with a frequency of 10 kHz, 
spending 27 
m
s in position-1 and 73 
m
s in position-2. (i) Find the average charging current in the 
12 V battery under periodic steady-state operation. (ii) Find the exponential Fourier series of i(t) 
under steady-state operation and plot its power spectrum. (iii) Find the rms value of i(t) and the 
power dissipated in the resistor.
+
12 V
i
(
t
)
2
1
15 mH
0.096 
Ω
S
48 V
–
+
–
Fig. 9.13-19 
30. The switch S in the circuit in Fig. 9.13-20 operates periodically with a frequency of 10 kHz, 
spending 77 
m
s in position-1 and 23 
m
s in position-2. (i) Find the average current delivered by 
the 12V battery under periodic steady-state operation. (ii) Find the exponential Fourier series 
of i(t) and plot its power spectrum. (iii) Find the rms value of i(t), the power dissipated in the 
resistor and the power delivered by the 12V battery. (iv) Find the average charging current in 48V 
battery. 
+
12 V
i
(
t
)
S
1
2
15 mH 0.096 
Ω
48 V
–
+
–
Fig. 9.13-20 

---

9.54
Dynamic Circuits with Periodic Inputs – Analysis by Fourier Series
31. The applied voltage v
S
(t) in the circuit in Fig. 9.13-21 is (320 sin100
p
t – 40 sin300
p
t – 20sin 500t) V. 
(i) Find the rms value of applied voltage. (ii) Find the current delivered by the source as a function 
of time. (iii) Find the power delivered by the source and the VA delivered by it.
+
–
1 H
20 mH
100 
Ω
v
S
(
t
)
Fig. 9.13-21 
32. The exponential Fourier series coefficients of i(t) in the circuit in Fig. 9.13
-
22 are 

i
o
= 
1 A, 

i
1
= 
1
-
j1 A, 

i
-
1
= 
1
+ 
j1, 

i
3
= 
0.3
+ 
j0.2 and 

i
-
3
= 
0.3
-
j0.2. The value of L is 10 mH and value of R is 
100 
w
. The period of v
S
(t) is 50 ms. Find the Fourier series of v
S
(t).
+
–
R
R
L
L
i
(
t
)
v
S
(
t
)
Fig. 9.13-22 

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Introduction 
10.1
F i r s t - O r d e r
RL
C i r c u i t s
CHAPTER OBJECTIVES
• To develop the differential equations for series and parallel RL circuits
• Initial condition, its need and interpretation
• Complementary function, particular integral and their interpretation
• Natural response, transient response and forced response in an RL circuit 
• Interpretation of various response components
• Nature and details of step response of RL circuit and time-domain specifications based on it
• Time constant and various interpretations for it
• Steady-state response versus forced response and various kinds of steady-state response 
• Zero-input response and zero-state response and their interpretation
• Linearity and superposition principle as applied to various response components
• Impulse response and its importance
• Equivalence between impulse forcing function and non-zero initial condition
• Zero-state response for other inputs from impulse response
• Relations between 
d
(t), u(t) and r(t) and corresponding responses
• Zero-state response of RL circuit for exponential and sinusoidal inputs
• Frequency response of RL circuit 
• General analysis procedure for single time constant RL circuits