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
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
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