Solving Second-Order Circuits Electrical Circuits for Engineering Students

Solution to Second-Order Circuits

1. 6.1Finding Initial and Final Values
6.2The Source-Free Series RLC Circuit
6.3The Source-Free Parallel RLC Circuit
6.4Step Response of a Series RLC Circuit
6.5Step Response of a Parallel RLC Circuit
6.6General Second-Order Circuits
6.Second-Order Circuits

2. What is a second-order
circuit?
•A second-order circuit is
characterized by a second-order
differential equation. It consists of
resistors and the equivalent of two
energy storage elements.
•Typical examples of second-order
circuits:
a)series RLC circuit,
b)parallel RLC circuit,
c)RL circuit,
d)RC circuit

3. 6.1 Finding Initial and Final Values

4. What are the values that we
need to focus on?
•
•
•
•
•
•
•Where:
•is capacitor voltage; and
•is inductor current.

5. What are the key points to keep in mind in
determining the initial conditions?
1.As always in circuit
analysis—carefully handle
the polarity of voltage
across the capacitor and
the direction of the
current through the
inductor.
•and are defined strictly
according to the passive sign
convention.
•Carefully observe how these
are defined and apply them
accordingly.
2.Keep in mind that the capacitor voltage
is always continuous so that
> ?
and the inductor current is always continuous
so that
> ?
•Where
•
?
denotes the time just before a
switching event and
•
>
is the time just after the
switching event, assuming that the
switching event takes place at .

6. Practice Problems

7. Practice Problem
•The switch has been closed
for a long time.
•It is open at .
•Find:
>
,
>
;
>
,
>
;
, .

8. To find , :
•Since the switch has been closed for a long time
before , the circuit reached dc steady state.
•Therefore, at
?
:
?
? ?
•Since the inductor current and the capacitor voltage
cannot change abruptly:
> ?
> ?

9. To find , :
•At
>
, the switch opens. The same current flows
through both the inductor and capacitor.
•Hence,
?
> >
.
•Since
?
> >
>
?
>
•Given
>
use KVL to find
?
>
:
>
?
> >
?
>
?
>
•In the same manner that
?
> >
:
>
?
>

10. To find , :
•At
>
, the circuit will
experience transient
responses. However, as
, the circuit reaches
steady state again.
•Therefore:

11. Answers
> >
;
??4
6
??
?
?
??4
6
??
?
?

12. Practice Problem
•Given the circuit
shown, find:
, ,
;
,
,
;
, , .

13. To find , , :
•Since the switch has been closed for a long time
before , the circuit reached dc steady state.
•Therefore, at
?
:
?
?
?
?
?
?
•Since the inductor current and the capacitor voltage
cannot change abruptly:
?
>
?
?
?
>
?
?

14. To find , , :
•Use KCL and KVL to find
?
>
:
•KCL:
?
?
4
6
6
?
0?
4
6
8
•KVL:
?
>
8?
>
?
>
•Given that
?
?
:
?
>
8?
>
?
>
8?
>
8?
>
8?
>
8?
>
8?
>
?
>
8?
>

15. To find , , :
•Since
?
> >
:
>
?
>
•Given that
?
?
, use KVL on the
rightmost mesh:
?
>
?
>
?
>
?
>
?
>
?
>
•Therefore:
?
>
?
>

16. To find , , :
•Since
?
> >
:
?
>
?
>
•Given
8?
>
and
?
>
, use KCL on the
rightmost node:
8?
>
?
>
?
>
?
>
?
>
•Therefore:
?
>
?
>

17. To find , , :
•To get
?
>
, use KCL at the leftmost node:
? 8? ? 8?
? 8?
•Taking the derivative of each term when
>
:
?
>
8?
>
•Applying KVL to middle mesh:
? 8??
•Again, taking derivative of each term when
>
:
?
>
8?
>
?
>

18. To find , , :
•Substitute ????????????
?
0
>
/????????????=2 ??????/?????? to the KVL equation:
−
????????????
?
0
>
????????????
+
????????????
8?
0
>
????????????
+2=0;
????????????
?
0
>
????????????
−
????????????
8?
0
>
????????????
=2
•Given0=2
??
?
4
6
??
+
??
0?
4
6
??
, substitute:
2
????????????
?
0
>
????????????
=−
????????????
8?
0
>
????????????
;
????????????
?
0
>
????????????
−
????????????
8?
0
>
????????????
=2
????????????
?
0
>
????????????
+2
????????????
?
0
>
????????????
=2; 3
????????????
?
0
>
????????????
=2
∴
????????????
?
0
>
????????????
=
2
3
??????/??????
•Had we been looking for ????????????
?
0
>
/????????????, use ??????
?
=2??????
?
:
????????????
?
0
>
????????????
=
1
2
????????????
?
0
>
????????????
=
1
2
×
2
3
=
1
3
??????/??????

19. To find , , :
•At
>
, the circuit will experience transient
responses. However, as , the circuit reaches
steady state again. The 3-A current source is now
working on the circuit.
•Therefore:
?
?
? ?
?
?

20. Practice Problem

21. Thank You!

22. 6.1Finding Initial and Final Values
6.2The Source-Free Series RLC Circuit
6.3The Source-Free Parallel RLC Circuit
6.4Step Response of a Series RLC Circuit
6.5Step Response of a Parallel RLC Circuit
6.6General Second-Order Circuits
6.Second-Order Circuits

23. 6.2The Source-Free Series RLC Circuit

24. Why is this considered a second-
order differential equation?
•Using KVL:
?? ?
6
6
•Given
?
.
?
??
.
?
?
??
??
?
??
and letting
??
,
differentiate:
6?? ?? ??
??6
6
•This is the series RLC circuit’s characteristic
equation.

25. How can this be simplified?
•Given
6
?
?
5
??
, determine the roots:
5
?
6?
?
6?
6
5
??
6
?
6?
?
6?
6
5
??
or
5
6
4
6
6
6
4
6
•Where:
?
6?
4
•The roots
5
and
6
are the natural frequenciesof the
circuit.
•The resonant frequencyis strictly referred to as the
undamped natural frequency of the circuit.

26. What is the implication of having
two roots?
•In simple terms, there are three solutions.
•Response vs. roots of circuit’s characteristic equation:
1.If overdamped,
4
or unequal and real
2.If critically damped,
4
or equal and real
3.If underdamped,
4
or complex

27. What happens when a
circuit is overdamped?
•When
4
,
6
•Both roots
5
and
6
are negative and
real.
•The response decays and approaches
zero as increases.
5
?
-
?
6
?
.
?

28. What happens when a circuit
is critically damped?
•When
4
,
6
•Therefore,
5 6
•The natural response of the critically
damped circuit is a sum of two terms: a
negative exponential and a negative
exponential multiplied by a linear term.
6 5
??

29. What happens when a
circuit is underdamped?
•When
4
,
6
•The roots are complex numbers:
5 4
6 6
?
6 4
6 6
?
•Where:
? 4
6 6
•Therefore:
5
?>?
?
?
6
???
?
?
??
5
?
?
?
6
??
?
?

30. What happens when a
circuit is underdamped?
•Given
??
5
?
?
?
6
??
?
?
•Using Euler’s identities:
??5 ? ?
6 ? ?
??5 6 ?
5 6 ?
•Replacing
5 6 5
and
5 6 6
:
??
5 ? 6 ?
•Exponentially damped and oscillatory in
nature
•The response has and

31. Practice Problems

32. Practice Problem
•Given R = 40 Ω, L = 4 H, and C = 1∕4 F:
•Calculate the characteristic roots of the circuit.
•Is the natural response overdamped, underdamped,
or critically damped?
•Find where , and .

33. Solution
•Given R = 40 Ω, L = 4 H, and C = 1∕4 F:
•Find:
4
5,6
6
4
6 6 6
5
6
•Since
4
, the response is overdamped. Thus:
5
?9>6:?
6
?9?6:?

34. Solution
•Given
5
?9>6:?
6
?9?6:?
, obtain
5
and
6
using , and :
•At :
5
?9>6:4
6
?9?6:4
5 6
•Given R = 40 Ω, L = 4 H, and C = 1∕4 F, use KVL:
?? ? ?
?@4 ?
?@4

35. Solution
•Given
??
??
?@4
, take the derivative of :
5
?9>6:?
6
?9?6:?
5
?9>6:?
6
?9?6:?
•At :
5
4
6
4
5 6
5 6
5 6

36. Solution
•Given:
5 6 5 6
•Get
5
and
6
:
5
6
•Answer:
5
6
•Therefore:
?9>6:? ?9?6:?

37. Answers
•The characteristic roots of the circuit are:
5
6
•Since
4
, the response is overdamped.
• where , and is:
?9>6:? ?9?6:?

38. Practice Problem
•Find in the circuit.
•Assume that the circuit has reached steady state at
?
.

39. Solution
•For , the switch is closed. Thus, at
•For , the switch is opened, and the voltage
source is disconnected.
4
5,6
6
4
6 6 6
5
6
•Since
4
, the response is underdamped.

40. Solution
•Given thatresponse is underdamped, , and
?
:
??
5 ? 6 ?
?=?
5 6
•Obtain
5
and
6
using and :
•At :
?=(4)
5 6 5
•Using KVL:
?? ?
??
??
?
?@4 ?
?@4

41. Solution
•Given
5
and
??
??
?@4
, take the derivative of
?=?
6
•
??
??
?=?
6
?=?
6
•At :
?=(4)
6
?=4
6
6
6
•Therefore:
?=?

42. Thank You!

43. 6.1Finding Initial and Final Values
6.2The Source-Free Series RLC Circuit
6.3The Source-Free Parallel RLC Circuit
6.4Step Response of a Series RLC Circuit
6.5Step Response of a Parallel RLC Circuit
6.6General Second-Order Circuits
6.Second-Order Circuits

44. 6.3The Source-Free Parallel RLC Circuit

45. Why is this considered a second-
order differential equation?
•Using KCL:
???
6
6
•The parallel RLC circuit’s characteristic equation is:
6
•Therefore:
5,6
5
6??
5
6??
6
5
??
or
5,6
6
4
6
•Where:
5
6??
4

46. What is the implication of having
two roots?
•In simple terms, there are three solutions.
•Response vs. roots of circuit’s characteristic
equation:
•If overdamped,
4
or unequal and real
•If critically damped,
4
or equal and real
•If underdamped,
4
or complex

47. What happens when a
circuit is overdamped?
•When
4
,
6
•Both roots
5
and
6
are negative and
real.
•The response decays and approaches
zero as increases.
5
?
-
?
6
?
.
?

48. What happens when a circuit
is critically damped?
•When
4
,
6
•Therefore,
5 6
•The natural response of the critically
damped circuit is a sum of two terms: a
negative exponential and a negative
exponential multiplied by a linear term.
5 6
??

49. What happens when a
circuit is underdamped?
•When
4
,
6
•The roots are complex numbers:
5,6 4
6 6
?
•Where:
? 4
6 6
•Therefore:
??
5 ? 6 ?
•Exponentially damped and oscillatory in
nature
•Thus, and
?

50. Practice Problems

51. Practice Problem
•In the circuit shown, find ??????(??????)for ??????>0for the
following resistances:
•R = 1.923 Ω,
•R = 5 Ω, and
•R = 6.25 Ω.
•Assume ??????(0) = 5 ??????,??????(0) = 0,?????? = 1 ??????, and ?????? =
10 ????????????.
•What is the natural response for each case?

52. If R = 1.923 Ω:
•Given
, and
:
?7
4
?7
5,6
6
4
6 6 6
5,6
5
6
•Since
4
, the response is overdamped.
•Thus:
5
?6?
6
?94?

53. If R = 1.923 Ω:
•Given
5
?6?
6
?94?
, obtain
5
and
6
using and :
•At :
5
4
6
4
5 6
•Given , and , use KCL:
??? ?
?@4 ?
?@4 ?7
?@4

54. If R = 1.923 Ω:
•Given
??
??
?@4
and
5 6
, take the
derivative of
5
?6?
6
?94?
•At :
5
4
6
4
5 6 5 6
•Get
5
and
6
:
5
6
•Answer:
5
6
•Therefore:
?6? ?94?

55. If R = 5 Ω:
•Given
, and
:
?7
4
?7
5,6
6
4
6 6 6
5,6
•Since
4
, the response is critically damped.
•Thus:
5 6
?54?

56. If R = 5 Ω:
•Given
5 6
?54?
, obtain
5
and
6
using and :
•At :
5 6
4
5
•Given , and , use KCL:
??? ?
?@4 ?
?@4 ?7
?@4

57. If R = 5 Ω:
•Given
??
??
?@4
and
5
, take the
derivative of
6
?54?
6
?54?
6
?54?
?54?
6 6
•At :
4
6 6
6 6
•Get
5
and
6
:
5
6
•Therefore:
?54?

58. If R = 6.25 Ω:
•Given
, and
:
?7
4
?7
5,6
6
4
6 6 6
5,6
5 6
•Since
4
, the response is underdamped.
•Thus:
?<?
5 6

59. If R = 6.25 Ω:
•Given
?<?
5 6
, obtain
5
and
6
using and :
•At :
4
5 6 5
•Given , and , use KCL:
??? ?
?@4 ?
?@4 ?7
?@4

60. If R = 6.25 Ω:
•Given
??
??
?@4
and
5
, take the derivative
of
?<?
5 6
•
?<?
6
?<?
6
•At :
•
4
6
4
6
6 6
•Get
5
and
6
:
5
6
•Therefore:
?<?

61. Practice Problem
•When R = 1.923 Ω, overdamped.
????????????=−0.2083??????
?6?
+5.208??????
?94?
•When R = 5 Ω, critically damped.
????????????=5−50????????????
?54?
•When R = 6.25 Ω, underdamped.
????????????=??????
?<
5cos6??????−
20
3
sin6??????

62. Practice
Problem
Find for in the RLC
circuit.

63. Solution
•For , the switch is open. Thus, at
•For , the switch is closed, and the voltage source is
disconnected.
?:
4
?:
5,6
6 6
5
6
•Since
4
, the response is overdamped.
•Thus:
5
6946?6?
6
?6946>6?

64. Solution
•Given
5
6946?6?
6
?6946>6?
, obtain
5
and
6
using and :
•At :
5
4
6
4
5 6
•Use KCL:
4.8?94?64? 4.8?
?@4 4.8?
?@4 ?:
?@4

65. Solution
•Given
??
??
?@4
and
5 6
, take the derivative
of
5
6946?6?
6
?6946>6?
•
??
??
5
6946?6?
6
?6946>6?
•At :
5
4
6
4
5 6
5 6
•Get
5
and
6
:
5
6
•Answer:
5
6
696>69
6
69?696
6
•Therefore:
6946?6? ?6946>6?

66. Thank You!

67. 6.1Finding Initial and Final Values
6.2The Source-Free Series RLC Circuit
6.3The Source-Free Parallel RLC Circuit
6.4Step Response of a Series RLC Circuit
6.5Step Response of a Parallel RLC Circuit
6.6General Second-Order Circuits
6.Second-Order Circuits

68. 6.4Step Response of a Series RLC Circuit

69. Why is this considered a second-
order differential equation?
•Using KVL:
?? ? ?
??
??
?
•Replace every with :
?
.
?
??
.
??
??
?
6
6
?
•The series RLC circuit’s characteristic equation is still:
6
•Therefore:
5,6
?
6?
?
6?
6
5
??
or
5,6
6
4
6
•Where:
?
6?
4

70. What is the implication of having
two roots?
•In simple terms, there are three solutions.
•Response vs. roots of circuit’s characteristic
equation:
•If overdamped,
4
or unequal and real
•If critically damped,
4
or equal and real
•If underdamped,
4
or complex

71. What is the implication of having
a power source?
•The solution to
6
6
?
•has two components:
•the transient response
?
and
•the steady-state response
??
.
? ??
•Steady-state response
??
is the final value of
•Transient response
?
dies out with time.

72. How is the steady-state response
obtained?
•Steady-state response
??
is the final value of .
•In the circuit shown, the final value of the capacitor
voltage is the same as the source voltage
?
.
•Therefore:
?? ?

73. How is the transient response
obtained?
•Transient response is the solution for the source-
free circuit.
•Thus, the transient response for the overdamped,
underdamped, and critically damped cases are:
•Overdamped:
? 5
?
-
?
6
?
.
?
•Critically damped:
? 5 6
??
•Underdamped:
?
??
5 ? 6 ?

74. How is the complete response
obtained?
•Given that
?? ?
, the complete
solution for the overdamped, underdamped, and
critically damped cases are:
•Overdamped:
? 5
?
-
?
6
?
.
?
•Critically damped:
? 5 6
??
•Underdamped:
?
?
5 ? 6 ?

75. Practice Problems

76. Practice Problem
•Given the circuit
shown, find and
for .
•Consider these cases:
•R = 5 Ω
•R = 4 Ω, and
•R = 1 Ω.

77. Answers
•When R = 5 Ω:
????????????=24−
64
3
??????
??
+
4
3
??????
?8?
??????
????????????=
16
3
??????
??
−
1
3
??????
?8?
A
•When R = 4 Ω:
?6?
?6?
•When R = 1 Ω:
?
5
6
?
?
5
6
?
•All for .

78. If R = 5 Ω:
•For , the switch is closed. Thus, at
•For , the switch is opened, and the 1-ohm resistor
is disconnected.
4
5,6
6 6
5
6
•Since
4
, the response is overdamped. Thus:
??5
??
6
?8?

79. If R = 5 Ω:
•Given
??5
??
6
?8?
obtain
?
,
5
and
6
using and :
•At
>
:
??
•At :
5
4
6
4
5 6
•Find
??
??
?@4
:
?@4
•Take the derivative of :
5
??
6
?8?
•At :
5
4
6
4
5 6

80. If R = 5 Ω:
•Given
5 6
and
5 6
:
•Get
5
and
6
:
5
6
•Answer:
5
6
•Therefore:
?? ?8?
•Find :
?? ?8? ?? ?8?
?? ?8?

81. If R = 4 Ω:
•For , the switch is closed. Thus, at
•For , the switch is opened, and the 1-ohm resistor
is disconnected.
4
5,6
6 6
5 6
•Since
4
, the response is critically damped. Thus:
? 5 6
?6?

82. If R = 4 Ω:
•Given
? 5 6
?6?
obtain
?
,
5
and
6
using and :
•At
>
:
??
•At :
5 6
4
5
•Find
??
??
?@4
:
?@4
•Take the derivative of :
6
?6?
6
?6?
•At :
4
6
4
6

83. If R = 4 Ω:
•Given
5
6
•Therefore:
?6?
?6?
•Find :
?6?
?6? ?6
?6?
?6?
?6?

84. If R = 1 Ω:
•For , the switch is closed. Thus, at
•For , the switch is opened, and the 1-ohm resistor
is disconnected.
4
5,6
6
6
•Since
4
, the response is underdamped. Thus:
?
?
5
6
?
5 6

85. If R = 1 Ω:
•Given
?
?
-
.
?
5
59
6
6
59
6
obtain
?
,
5
and
6
using and :
•At
>
:
??
•At :
4
5 6 5
•Find
??
??
?@4
:
?@4

86. If R = 1 Ω:
•Given
??
??
?@4
and
5
, take the derivative of
?
?
-
.
?
5
59
6
6
59
6
:
?
5
6
?
6
?
5
6
?
6
•At :
4
6
4
6
6
6
6

87. If R = 1 Ω:
•Given
??????
5
??????
6
=
−12
6<59
9
=
−12
21.69
•Therefore: ∴????????????=24+??????
?
-
.
?
−12cos
59
6
??????+
6<59
9
sin
59
6
??????
•Find ????????????: ????????????=?????? ????????????/????????????
????????????=0.25
??????
????????????
24+??????
?
5
6
?
−12cos
15
2
??????+
2815
5
sin
15
2
??????
????????????=
1
4
−
1
2
??????
?
5
6
?
−12cos
15
2
??????+
2815
5
sin
15
2
??????
+
15
2
??????
?
5
6
?
12sin
15
2
??????+
2815
5
cos
15
2
??????
????????????=
1
4
×
1
2
??????
?
5
6
?
1215−
2815
5
sin
15
2
??????+12+84cos
15
2
??????
????????????=
1
4
×
1
2
??????
?
5
6
?
3215
5
sin
15
2
??????+96cos
15
2
??????
∴????????????=??????
?
5
6
?
415
5
sin
15
2
??????+12cos
15
2
??????A

88. Answers
•When R = 5 Ω:
????????????=24−
64
3
??????
??
+
4
3
??????
?8?
??????
????????????=
16
3
??????
??
−
1
3
??????
?8?
A
•When R = 4 Ω:
?6?
?6?
•When R = 1 Ω:
?
5
6
?
?
5
6
?
•All for .

89. Thank You!

90. 6.1Finding Initial and Final Values
6.2The Source-Free Series RLC Circuit
6.3The Source-Free Parallel RLC Circuit
6.4Step Response of a Series RLC Circuit
6.5Step Response of a Parallel RLC Circuit
6.6General Second-Order Circuits
6.Second-Order Circuits

91. 6.5Step Response of a Parallel RLC Circuit

92. Why is this considered a second-
order differential equation?
•Using KCL:
???? ?
•Replace every with :
?
.
?
??
.
?
?
??
??
?
6
6
?
•The parallel RLC circuit’s characteristic equation is still:
6
•Therefore:
5,6
5
6?
5
6??
6
5
??
or
5,6
6
4
6
•Where:
5
6??
4

93. What is the implication of having
two roots?
•In simple terms, there are three solutions.
•Response vs. roots of circuit’s characteristic
equation:
•If overdamped,
4
or unequal and real
•If critically damped,
4
or equal and real
•If underdamped,
4
or complex

94. What is the implication of having
a power source?
•The solution to
6
6
?
•has two components:
•the transient response
?
and
•the steady-state response
??
.
?? ?
•Transient response
?
dies out with time.
•Steady-state response
??
is the final value of

95. How is the steady-state response
obtained?
•Steady-state response
??
is the final value of
.
•In the circuit shown, the final value of the inductor
current is the same as the current source
?
.
•Therefore:
?? ?

96. How is the transient response
obtained?
•Transient response is the solution for the source-
free circuit.
•Thus, the transient response for the overdamped,
underdamped, and critically damped cases are:
•Overdamped:
? 5
?
-
?
6
?
.
?
•Critically damped:
? 5 6
??
•Underdamped:
?
??
5 ? 6 ?

97. How is the complete response
obtained?
•Given that
?? ?
, the complete solution
for the overdamped, underdamped, and critically
damped cases are:
•Overdamped:
? 5
?
-
?
6
?
.
?
•Critically damped:
? 5 6
??
•Underdamped:
?
?
5 ? 6 ?

98. Practice Problems

99. Practice
Problem
Given the circuit shown, find
and
?
for .

100. Solution
•For , the switch is open and the 30-V source is on.
Thus, at
? ?
•For , the switch is closed and 30-V source is turned
off (shorted).
?7
4
?7
5,6
6 6
5
6
•Since
4
, the response is overdamped. Thus:
?? 5
?4.965<
6
?55.=;<?

101. Solution
•Given
??5
?4.965<?
6
?55.=;<?
obtain
?
,
5
and
6
using
?
and :
•At
>
:
??
•At :
5
4
6
4
5 6
•Find
??
??
?@4
:
??
??
?@4
?4
?
59
64
7
8
•Take the derivative of :
5
?4.965<?
6
?55.=;<?
•At :
5
4
6
4
5 6
5 6

102. Solution
•Given ??????
5
=−??????
6
and 521−25??????
5
−521+25??????
6
=3:
•Get ??????
5
and ??????
6
:
1 1
521−25−5 21−25
??????
5
??????
6
=
0
3
•Answer:
??????
5
??????
6
=
21/70
−21/70
=
0.0655
−0.0655
•Thus: ∴????????????=4+
65
;4
??????
1.-7.
0
?
−??????
?
1.-6.
0
?
??????
•Find ??????
?
(??????): ??????
?
??????=
??
64
=
?
64
??
??
=
64
64
?
??
4+
65
;4
??????
1.-7.1
0
?
−??????
?
1.-6.
0
?
??????
?
??????=
21
70
??????
????????????
??????
965?6
8
?
−??????
?
965>69
8
?
??????
?
(??????)=
21
70
521−25
4
??????
965?6
8
?
+
521+25
4
??????
?
965>69
8
?
??????
?
(??????)=
21
56
21−5??????
965?6
8
?
+21+5??????
?
965>6
8
?
??????
∴??????
?
(??????)=−0.03415??????
?4.965<?
+0.7842??????
?55.=;<?
??????

103. Answers
All for .

104. Thank You!

105. 6.1Finding Initial and Final Values
6.2The Source-Free Series RLC Circuit
6.3The Source-Free Parallel RLC Circuit
6.4Step Response of a Series RLC Circuit
6.5Step Response of a Parallel RLC Circuit
6.6General Second-Order Circuits
6.Second-Order Circuits

106. 6.6General Second-Order Circuits

107. Rules to Remember:
1.Determine the initial conditionsand and final value
2.Turn off the independent sources and find the form of the transient response
?
by
applying KCL and KVL.
Once a second-order differential equation is obtained, determine its characteristic roots.
Depending on whether the response is overdamped, critically damped, or underdamped, get
?
with two unknown constants as in the previous sections.
3.Obtain the steady-state response
??
4.Given that the total response is now found as the sum of the transient response and
steady-state response
?? ?
, determine the constants associated with the
transient response by imposing the initial conditions and found at (1).

108. Practice Problems

109. Practice
Problem
Find the complete response
and then for in the
circuit shown.

110. 1. Determine the initial and final
values
For initial values:
•At
?
: ;
•At
>
: ; ;
?
•Using KCL:
6? ?
? ?
?
For final values at
>
:

111. 2. Find the form of the transient
response
Turn off the independent sources:
1.Apply KCL:
6??
2.Apply KVL:
8??6?
•Substitute (1) to (2):

112. 2. Find the form of the transient
response
•Determine first then :
6
6
6
6
5,6
6 6
5,6
5 5
•Since roots are unequal and real, overdamped.

113. 2-3. Find transient response and
obtain steady-state response
•Given
5
and
6
:
?
?6? ?7?
•At
>
:
??
•Thus:
???
?6? ?7?

114. 4. Determine the constants
associated with transient response
Given
??4
??
and
?6? ?7?
:
•Find A and B at
>
:
4 4
•Take the derivative of :
?6? ?7?
?6? ?7?
•Find
??
??
?@4
:

115. 4. Determine the constants
associated with transient response
•Given and :
•Get and :
•Answer:
•Thus:
?6 ?7?

116. 4. Determine the constants
associated with transient response
Given
?6??7?
:
•Find using KCL:
?6??7? ?6 ?7?
?6 ?7? ?6? ?7?
?6 ?7?
?6 ?7?
?6??7?

117. Answers
?6??7?
?6??7?
All for

118. Practice
Problem
Find
4
for in the
circuit shown.
A second-order circuit with two
inductors.

119. 1. Determine the initial
conditions
For initial values:
•At
?
, 7-volt source is off. Thus:
5 6
•At
>
, the 7-volt source is turned on but the
inductors do not change instantaneously.
•Using KVL:
5 ?
-
4 ?
-
?
-
5
5 5 ?
-
5
?
.
6
6 6 ?
.
6
For final values at
>
:
5 6

120. 2. Find the form of the transient
response
Turn off the independent sources and apply KVL:
1.For
5
:
7?4?
-
5 56
5
56
5
6
5
5
2.For
6
:
4?
.
56
6 6
65
•Substitute (1) to (2):
5
5
5
55

121. 2. Find the form of the transient
response
•Determine first then :
5
5
5
55
6
5
6
5 5
5
6
5
6
5
5
5,6
6 6
5,6
5 5
•Since roots are unequal and real, overdamped.

122. 2-3. Find transient response and
obtain steady-state response
•Given
5
and
6
10:
5?
?7? ?54?
•At
>
:
5??5
•Thus:
5 5??5?
5
?7 ?54?

123. 4. Determine the constants
associated with transient response
Given
5
??
-
4
??
and
5
;
7
?7
?54?
:
•Find A and B at
>
:
4 4
•Take the derivative of :
5
?7? ?54?
5
?7? ?54
•Find
??
-
??
?@4
:

124. 4. Determine the constants
associated with transient response
•Given and :
•Get and :
•Answer:
•Thus:
5
?7??54?

125. 4. Determine the constants
associated with transient response
Given
5
;
7
8
7
?7??54?
:
•Find
6
using KVL on
5
:
5
5
6
??
-
??
56
6 5
5
6
?7??54? ?7??54?
6
?7??54? ?7 ?54?
6
?7? ?54?
6
?7??54?

126. Find for :
Given:
5
?7??54?
6
?7??54?
•Find
4
:
4 5 6
4
?7??54? ?7??54?
4
?7? ?54?
4
?7??54?
4
?7??54?

127. Thank You!