1. The Laplace Transform in Circuit Analysis
CHAPTER 13
The Laplace Transform
in Circuit Analysis

2. CHAPTER CONTENTS 13.1 Circuit Elements in the s Domain
13.2 Circuit Analysis in the s Domain
13.3 Applications
13.4 The Transfer Function
13.5 The Transfer Function in Partial Fraction Expansions
13.6 The Transfer Function and the Convolution Integral
13.7 The Transfer Function and the Steady-State Sinusoidal Response
13.8 The Impulse Function in Circuit Analysis

3. CHAPTER OBJECTIVES
Be able to transform a circuit into the s domain using Laplace transforms; be sure you understand how to represent the initial conditions on energy-storage elements in the s domain.
Know how to analyze a circuit in the s-domain and be able to transform an s-domain solution back to the time domain.
Understand the definition and significance of the transfer function and be able to calculate the transfer function for a circuit using s-domain techniques.
Know how to use a circuit’s transfer function to calculate the circuit’s unit impulse response, its unit step response, and its steady-state response to a sinusoidal input.

4. 13.1 Circuit Elements in the s Domain
A voltage-to-current ratio in the s domain carries the dimension of volts per ampere. An impedance in the s domain is measured in ohms, and an admittance is measured in siemens.

5. A Resistor in the s Domain
We begin with the resistance element. From Ohm’s law, Because R is a constant, the Laplace transform of Eq is where

6. Figure 13.1 The resistance element.
(a) Time domain. (b) Frequency domain.

7. An Inductor in the s Domain
Figure An inductor of L henrys carrying an initial current of I0 amperes.

8. Figure The series equivalent circuit for an inductor of L henrys carrying an initial current of I0 amperes.

9. Figure The parallel equivalent circuit for an inductor of L henrys carrying an initial current of I0 amperes.

10. Figure 13.5 The s-domain circuit for an inductor when the initial current is zero.

11. A Capacitor in the s Domain
Figure A capacitor of farads initially charged to V0 volts.

12. Figure 13.7 The parallel equivalent circuit for a capacitor initially charged to V0 volts.

13. Figure 13.8 The series equivalent circuit for a capacitor initially charged to V0 volts.

14. Figure 13.9 The s-domain circuit for a capacitor when the initial voltage is zero.

16. 13.2 Circuit Analysis in the s Domain
Ohm’s law in the s-domain
Kirchhoff’s laws
Thévenin-Norton equivalents are all valid techniques, even when energy is stored initially in the inductors and capacitors.

18. 13.3 Applications The Natural Response of an RC Circuit
Figure The capacitor discharge circuit.
Figure An s-domain equivalent circuit for the circuit shown in Fig

19. Summing the voltages around the mesh generates the expression Solving Eq. 13.12 for I yields
Note that the expression for I is a proper rational function of s and can be inverse-transformed by inspection: which is equivalent to the expression for the current derived by the classical methods discussed in Chapter 7.

20. After we have found i, the easiest way to determine v is simply to apply Ohm’s law; that is, from the circuit,
Figure An s-domain equivalent circuit for the circuit shown in Fig

21. The node-voltage equation that describes the new circuit is Solving Eq
The node-voltage equation that describes the new circuit is Solving Eq for V gives Inverse-transforming Eq leads to the same expression for given by Eq , namely,

23. The Step Response of a Parallel Circuit
The parallel RLC circuit, shown in Fig , that we first analyzed in Example 8.7.
Figure The step response of a parallel RLC circuit.
Figure The s-domain equivalent circuit for the circuit shown in Fig

24. We first solve for V and then use to establish the s-domain expression for IL. Summing the currents away from the top node generates the expression
Solving Eq for V gives Substituting Eq into Eq gives

25. In Example 8.7, we factor the quadratic term in the denominator:
The limit of sIL as s → ∞is

26. We now proceed with the partial fraction expansion of Eq. 13.24:
The partial fraction coefficients are

27. Substituting the numerical values of K1 and K2 into Eq. 13
Substituting the numerical values of K1 and K2 into Eq and inverse transforming the resulting expression yields
The answer given by Eq is equivalent to the answer given for Example 8.7 because

29. The Transient Response of a Parallel RLC Circuit
Figure The step response of a parallel RLC circuit.
where Im = 24 mA and w = 40,000 rad/s.

30. The s-domain expression for the source current is
The voltage across the parallel elements is
Substituting Eq into Eq results in from which

31. Substituting the numerical values of Im, w, R, L, and C into Eq. 13
Substituting the numerical values of Im, w, R, L, and C into Eq gives We now write the denominator in factored form: where w = 40,000, a = 32,000, and b = 24,000.

32. When we expand Eq. 13.36 into a sum of partial fractions, we generate the equation

33. The numerical values of the coefficients K1 and K2 are

34. Substituting the numerical values from Eqs. 13. 38 and 13. 39 into Eq
Substituting the numerical values from Eqs and into Eq and inverse-transforming the resulting expression yields
For t = 0 Eq predicts zero initial current, which agrees with the initial energy of zero in the circuit. Equation also predicts a steady-state current of

35. The Step Response of a Multiple Mesh Circuit
Figure A multiple-mesh RL circuit.
Figure The s-domain equivalent circuit for the circuit shown in Fig

36. The two mesh-current equations are
Using Cramer’s method to solve for I1 and I2 we obtain

41. The use of Thévenin’s Equivalent
Figure A circuit to be analyzed using Thévenin’s equivalent in the s domain.

42. Figure 13.18 The s-domain model of the circuit shown in Fig. 13.17.

43. The Thévenin voltage is the open-circuit voltage across terminals a, b
The Thévenin voltage is the open-circuit voltage across terminals a, b. Under open-circuit conditions, there is no voltage across the 60 Ω resistor. Hence
The Thévenin impedance seen from terminals a and b equals the 60 Ω resistor in series with the parallel combination of the 20 Ω resistor and the 2 mH inductor. Thus

44. Figure 13. 19 A simplified version of the circuit shown in Fig. 13
Figure A simplified version of the circuit shown in Fig , using a Thévenin equivalent.

45. Using the Thévenin equivalent, we reduce the circuit shown in Fig. 13
Using the Thévenin equivalent, we reduce the circuit shown in Fig to the one shown in Fig It indicates that the capacitor current IC equals the Thévenin voltage divided by the total series impedance. Thus,
We simplify Eq to

46. A partial fraction expansion of Eq. 13
A partial fraction expansion of Eq generates the inverse transform of which is
This result agrees with the initial current in the capacitor, as calculated from the circuit in Fig

47. Let’s assume that the voltage drop across the capacitor vC is also of interest. Once we know iC, we find vC by integration in the time domain; that is,

49. A Circuit with Mutual Inductance
With the switch in position b and the magnetically coupled coils replaced with a T-equivalent circuit. Figure shows the new circuit.
Figure A circuit containing magnetically coupled coils.

50. Figure 13. 21 The circuit shown in Fig. 13
Figure The circuit shown in Fig , with the magnetically coupled coils replaced by a T-equivalent circuit.
This source appears in the vertical leg of the tee to account for the initial value of the current in the 2 H inductor of i1(0–) + i2 (0–), or 5A. The branch carrying i1 has no voltage source because L1 – M = 0.

51. The two s-domain mesh equations that describe the circuit in Fig. 13
The two s-domain mesh equations that describe the circuit in Fig are
Figure The s-domain equivalent circuit for the circuit shown in Fig

52. Solving for I2 yields
Expanding Eq into a sum of partial fractions generates
Then,

53. Figure 13. 23 The plot of i2 versus t for the circuit shown in Fig. 13

55. The use of Superposition
Figure A circuit showing the use of superposition in s-domain analysis.

56. Figure 13.25 The s-domain equivalent for the circuit of Fig. 13.24.

57. Figure 13.26 The circuit shown in Fig. 13.25 with Vg acting alone.

58. The two equations that describe the circuit in Fig. 13.26 are
For convenience, we introduce the notation

59. Substituting Eqs. 13.75–13.77 into Eqs. 13.73 and 13.74 gives
Solving Eqs and for gives

60. Solving Eqs. 13.81 and 13.82 for yields
Figure The circuit shown in Fig , with Ig acting alone.
the two node-voltage equations that describe the circuit in Fig are and
Solving Eqs and for yields

61. Figure 13. 28 The circuit shown in Fig. 13
Figure The circuit shown in Fig , with the energized inductor acting alone.

62. The node-voltage equations describing this circuit are
Figure The circuit shown in Fig , with the energized capacitor acting alone.
The node-voltage equations describing this circuit are

63. The expression for V2 is

64. We can find V2 without using superposition by solving the two node-voltage equations that describe the circuit shown in Fig Thus

66. 13.4 The Transfer Function
The transfer function is defined as the s-domain ratio of the Laplace transform of the output (response) to the Laplace transform of the input (source).
Definition of a transfer function where Y(s) is the Laplace transform of the output signal, and X(s) is the Laplace transform of the input signal. Note that the transfer function depends on what is defined as the output signal.

67. Figure 13.30 A series RLC circuit.

68. Example 13.1
The voltage source vg drives the circuit shown in Fig The response signal is the voltage across the capacitor, vo.
a) Calculate the numerical expression for the transfer function.
b) Calculate the numerical values for the poles and zeros of the transfer function.
Figure The circuit for Example 13.1.

69. Example 13.1
Figure The s-domain equivalent circuit for the circuit shown in Fig

70. Example 13.1

72. The Location of Poles and Zeros of H(s)
For linear lumped-parameter circuits, H(s) is always a rational function of s. Complex poles and zeros always appear in conjugate pairs. The poles of H(s) must lie in the left half of the s plane if the response to a bounded source is to be bounded.
H(s) plays in determining the response function.

73. 13.5 The Transfer Function in Partial Fraction Expansions
Example 13.2
The circuit in Example 13.1 (Fig ) is driven by a voltage source whose voltage increases linearly with time, namely, vg = 50 tu(t).
a) Use the transfer function to find vo.
b) Identify the transient component of the response.
c) Identify the steady-state component of the response.
d) Sketch vo versus t for 0 ≤ t ≤ 1.5 ms.

74. Example 13.2

75. Example 13.2

76. Example 13.2

77. Example 13.2
Figure The graph of υo versus t for Example 13.2.

79. Observations on the Use of H(s) in Circuit Analysis
If the input is delayed by a seconds,
If then, from Eq
Delaying the input by a seconds simply delays the response function by a seconds. A circuit that exhibits this characteristic is said to be time invariant.

80. A unit impulse source drives the circuit

81. 13.6 The Transfer Function and the Convolution Integral
The circuit and the circuit’s impulse response h(t).
We are interested in the convolution integral for several reasons.
First, it allows us to work entirely in the time domain.
Second, the convolution integral introduces the concepts of memory and the weighting function into analysis.
Finally, the convolution integral provides a formal procedure for finding the inverse transform of products of Laplace transforms

82. Figure 13.34 A block diagram of a general circuit.

83. Figure 13.35 The excitation signal of x(t)
(a) A general excitation signal.
(b) Approximating x(t) with a series of pulses.

84. Figure 13.35 The excitation signal of x(t)
(c) Approximating x(t) with a series of impulses.

85. Figure 13.36 The approximation of y(t).
The impulse response of the box shown in Fig
Summing the impulse responses.

86. As Dl → 0, the summation

87. whereas x(t) * h(t) is read as “x(t) is convolved with h(t)” and implies that

88. Figure 13.37 A graphic interpretation of the convolution integral (λ)x(t − λ) dλ.
The impulse response.
The excitation function.
The folded excitation function.
The folded excitation function displaced t units.
The product h(λ)x(t − λ).

89. Figure 13.38 A graphic interpretation of the convolution integral (t − λ)x(λ)dλ
The impulse response.
The excitation function.
The folded impulse response.
The folded impulse response displaced t units.
The product h(t − λ)x(λ).

90. Example 13.3
The excitation voltage vi for the circuit shown in Fig (a) is shown in Fig (b).
a) Use the convolution integral to find vo.
b) Plot vo over the range of 0 ≤ t ≤ 15 s.
Figure The circuit and excitation voltage for Example (a) The circuit. (b) The excitation voltage.

91. Example 13.3

92. Example 13.3

93. Example 13.3
Figure The impulse response and the folded excitation function for Example 13.3.

94. Example 13.3
Figure The displacement of υi(t − λ) for three different time intervals.

95. Example 13.3

96. Example 13.3
Figure The voltage response versus time for Example 13.3.

98. The Concepts of Memory and the Weighting Function
We can view the folding and sliding of the excitation function on a timescale characterized as past, present, and future. The vertical axis, over which the excitation function x(t) is folded, represents the present value; past values of x(t) lie to the right of the vertical axis, and future values lie to the left.
Figure The past, present, and future values of the excitation function.

99. The multiplication of x(t – λ)by h(λ)gives rise to the practice of referring to the impulse response as the circuit weighting function. The weighting function, in turn, determines how much memory the circuit has.
Memory is the extent to which the circuit’s response matches its input.

100. Figure 13.44 Weighting functions. (a) Perfect memory. (b) No memory.

101. Figure 13.45 The input and output waveforms for Example 13.3.

102. 13.7 The Transfer Function and the Steady-State Sinusoidal Response
Use the transfer function to relate the steady state response to the excitation source.

104. Steady-state sinusoidal response computed using a transfer function

105. Example 13.4
The circuit from Example 13.1 is shown in Fig The sinusoidal source voltage is 120 cos(5000t + 30°) V. Find the steady-state expression for vo.
Figure The circuit for Example 13.4.

106. Example 13.4

108. 13.8 The Impulse Function in Circuit Analysis
Impulse functions occur in circuit analysis either because of a switching operation or because a circuit is excited by an impulsive source.

109. Switching Operations Capacitor Circuit
Figure A circuit showing the creation of an impulsive current.
Figure The s-domain equivalent circuit for the circuit shown in Fig

110. Figure 13.49 The plot of i(t) versus t for two different values of R.

112. Series Inductor Circuit
Figure A circuit showing the creation of an impulsive voltage.
Figure The s-domain equivalent circuit for the circuit shown in Fig

114. Does this solution make sense?
Verify

115. Figure 13.52 The inductor currents versus t for the circuit shown in Fig. 13.50.

117. Impulsive Sources
Impulse functions can occur in sources as well as responses; such sources are called impulsive sources.
Figure An RL circuit excited by an impulsive voltage source.

118. The current is Given that the integral of d(t) over any interval that includes zero is 1, we find that Eq yields
Thus, in an infinitesimal moment, the impulsive voltage source has stored in the inductor.
The current now decays to zero in accordance with the natural response of the circuit; that is, where t = L/R.

119. Figure 13.54 The s-domain equivalent circuit for the circuit shown in Fig. 13.53.

120. Internally generated impulses and externally applied impulses occur simultaneously.
Figure The circuit shown in Fig with an impulsive voltage source added in series with the 100 V source.

121. Figure 13.56 The s-domain equivalent circuit for the circuit shown in Fig. 13.55.

122. At t = 0–, i1(0–) = 10 A and i2(0–) = 0A
At t = 0–, i1(0–) = 10 A and i2(0–) = 0A. The Laplace transform of 50d(t) = 50.
The expression for I is from which

123. The expression for V0 is from which

124. Figure 13.57 The inductor currents versus t for the circuit shown in Fig. 13.55.

126. Figure 13.58 The derivative of i1 and i2.

127. Summary
We can represent each of the circuit elements as an s-domain equivalent circuit by Laplace-transforming the voltage-current equation for each element:
Resistor: V = RI
Inductor: V = sLI – LI0
Capacitor: V = (1/sC)I + Vo/s
In these equations, I0 is the initial current through the inductor, and V0 is the initial voltage across the capacitor.

128. Summary
We can perform circuit analysis in the s domain by replacing each circuit element with its s-domain equivalent circuit. The resulting equivalent circuit is solved by writing algebraic equations using the circuit analysis techniques from resistive circuits. Table 13.1 summarizes the equivalent circuits for resistors, inductors, and capacitors in the s domain.

129. Summary
Circuit analysis in the s domain is particularly advantageous for solving transient response problems in linear lumped parameter circuits when initial conditions are known. It is also useful for problems involving multiple simultaneous mesh-current or node-voltage equations, because it reduces problems to algebraic rather than differential equations.

130. Summary
The transfer function is the s-domain ratio of a circuit’s output to its input. It is represented as where Y(s) is the Laplace transform of the output signal, and X(s) is the Laplace transform of the input signal.

131. Summary
The partial fraction expansion of the product H(s)X(s) yields a term for each pole of H(s) and X(s). The H(s) terms correspond to the transient component of the total response; the X(s) terms correspond to the steady-state component.
If a circuit is driven by a unit impulse, x(t) = d(t) then the response of the circuit equals the inverse Laplace transform of the transfer function, = h(t).

132. Summary
A time-invariant circuit is one for which, if the input is delayed by a seconds, the response function is also delayed by a seconds.
The output of a circuit, y(t) can be computed by convolving the input, x(t) with the impulse response of the circuit, h(t): A graphical interpretation of the convolution integral often provides an easier computational method to generate y(t).

133. Summary
We can use the transfer function of a circuit to compute its steady-state response to a sinusoidal source. To do so, make the substitution s = jw in H(s) and represent the resulting complex number as a magnitude and phase angle. If then

134. Summary
Laplace transform analysis correctly predicts impulsive currents and voltages arising from switching and impulsive sources. You must ensure that the s-domain equivalent circuits are based on initial conditions at that is, prior to the switching.