1. ELECTRIC CIRCUITS EIGHTH EDITION JAMES W. NILSSON & SUSAN A. RIEDEL

2. CHAPTER 13 THE LAPLACE TRANSFORM IN CIRCUIT ANALYSIS © 2008 Pearson Education

4. 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 © 2008 Pearson Education

5. CONTENTS 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 © 2008 Pearson Education

6. 6 surge: abrupt change of vtg./ct. due to switching. we will answer that problem using Lasplace transform.

7. where

9. Now solving for ct. I,

11. Now solving for vtg. V, when V o = 0,

12. 13.1 Circuit Elements in the s Domain   We can represent each of the circuit elements as an s-domain equivalent circuit by Laplace-transforming the voltage- current equation for each elements: Resistor: V = RI Inductor: V = s LI – LI 0 Capacitor: V = (1/s C)I + V 0 /s © 2008 Pearson Education

13. 13.1 Circuit Elements in the s Domain In these equations, I 0 = initial current through the inductor, V 0 = initial voltage across the capacitor. © 2008 Pearson Education V = L { v}, I = L { i)

14. 13.1 Circuit Elements in the s Domain © 2008 Pearson Education  The resistance element. Time domain Frequency domain

15. 13.1 Circuit Elements in the s Domain © 2008 Pearson Education An inductor of L henrys carrying an initial current of I 0 amperes

16. 13.1 Circuit Elements in the s Domain © 2008 Pearson Education The series equivalent circuit for an inductor of L henrys carrying an initial current of I 0 amperes

17. 13.1 Circuit Elements in the s Domain © 2008 Pearson Education The parallel equivalent circuit for an inductor of L henrys carrying an initial current of I 0 amperes

18. 13.1 Circuit Elements in the s Domain © 2008 Pearson Education The s-domain circuit for an inductor when the initial current is zero

19. 13.1 Circuit Elements in the s Domain © 2008 Pearson Education A capacitor of C farads initially charged to V 0 volts

20. 13.1 Circuit Elements in the s Domain © 2008 Pearson Education The parallel equivalent circuit for a capacitor initially charged to V 0 volts

21. 13.1 Circuit Elements in the s Domain © 2008 Pearson Education The series equivalent circuit for a capacitor initially charged to V 0 volts

22. 13.1 Circuit Elements in the s Domain © 2008 Pearson Education The s-domain circuit for a capacitor when the initial voltage is zero

23. 13.1 Circuit Elements in the s Domain © 2008 Pearson Education  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.

24. 24

25. 25

26. 13.2 Circuit Analysis in the s Domain   Circuit analysis can be performed in the s domain by replacing each circuit element with its s-domain equivalent circuit. Ohm’s Law in the s-domain © 2008 Pearson Education

27. 13.3 Applications   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. © 2008 Pearson Education

29. Solving for V, Cf:

30. 13.3 Applications © 2008 Pearson Education The capacitor discharge circuit An s-domain equivalent circuit  The Natural Response of an RC Circuit

31. 13.3 Applications © 2008 Pearson Education The step response of a parallel RLC circuit  The Step Response of a Parallel Circuit An s-domain equivalent circuit

32. Solving for V, Substituting this expression, into

33. Final value of i L

35.  New ct. S. See:

38. 13.3 Applications © 2008 Pearson Education The multiple-mesh RL circuit  The Step Response of a Multiple Mesh Circuit An s-domain equivalent circuit

42. 13.3 Applications © 2008 Pearson Education A circuit to be analyzed using Thévenin’s equivalent in the s domain  The Use of Thévenin’s Equivalent An s-domain model of the circuit A simplified version of the circuit, using a Thévenin’s equivalent

44. See :

45. v 1 = L 1 di 1 /dt + M di 2 /dt = (L 1 – M)di 1 /dt + M(di 1 /dt + di 2 /dt) v 2 = L 2 di 2 /dt + M di 1 /dt = (L 2 – M)di 2 /dt + M(di 2 /dt + di 1 /dt)

46. M(di 1 /dt + di 2 /dt)  M((sI 1 (s) – i 1 (0 - )) + (sI 2 (s) – i 2 (0 - )) = 2(sI 1 (s) – 5 + sI 2 (s)) = 2s(I 1 (s) + I 2 (s)) -10 Laplace transform

47. © 2008 Pearson Education A circuit showing the use of superposition in s-domain analysis  The Use of Superposition The s-domain equivalent for the above circuit See :

48. © 2008 Pearson Education The circuit with V g acting alone

49.  See :

50. The circuit with I g acting alone where

51. © 2008 Pearson Education The circuit with energized inductor acting alone

52. The circuit with energized capacitor acting alone

54. 13.4 The Transfer Function   The transfer function is the s-domain ratio of a circuit’s output to its input. It is represented as Y(s) is the Laplace transform of the output signal, X(s) is the Laplace transform of the input signal. © 2008 Pearson Education

56. a)Transfer fn.? b)Poles, zeros? a) KCL

57. b) Poles: Zeros:

58. 13.5 The Transfer Function in Partial Fraction Expansions   The partial fraction expansion of 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; X(s) terms correspond to the steady-state component. © 2008 Pearson Education

59. S. : v g = 50tu(t) a) find v o using transfer fn. b) transient resp. component? c) steady state resp.? d) a) V g (s) = 50/s 2 From above example, Therefore,

62. 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.

63.  If a circuit is driven by a unit impulse, x(t) = δ (t), then the response of the circuit equals the inverse Laplace transform of the transfer function, y(t) = L -1 {H(s)} = h(t)  Impulse response

64. 13.6 The Transfer Function and the Convolution Integral   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): © 2008 Pearson Education

65. Why we have interest in Convolution Integral ? 1 st, we can work in the time domain. when x(t), h(t) are known only through experimental data, transform is very difficult. 2 nd, introduces concept of memory & weighting fn. into analysis. 3 rd, can predict, to some degree, how closely output waveform replicates input waform. Finally, provides a formal procedure for finding the inverse transform of products of Laplace transforms.

66. Derivation of convolution integral: Assume ckt is linear & time invariant. Linear  principale of superposition Time invariant  amount of resp delay = that of input delay x(t) : excitation signal h(t) : impulse response y(t) : desired output signal

68.  λ i+1 - λ i = Δλ

69. h(t) : y(t) when x(t) = δ(t). So Summation  continuous integral

71. © 2008 Pearson Education 13.6 The Transfer Function and the Convolution Integral

72. © 2008 Pearson Education 13.6 The Transfer Function and the Convolution Integral

73. 

76. h(t) : See: Output quickly approaches present value of input because present value of input receives more weight than past values.

77. Multication of gives rise to weighting fn. Weighting fn., in turn, determines how much memory ckt. has. Memory is the extent to which ckt.’s resp. matches its input. For example,

78. input & output. If See:

79. This fig. shows distortion between input & output that has some memory.

80. 13.7 The Transfer Function and the Steady-State Sinusoidal Response   We can use the transfer function of a circuit to compute its steady-state response to a sinusoidal source.   To make the substitution s = j ω in H(s) and represent the resulting complex number as a magnitude and phase angle. © 2008 Pearson Education

81. Assume excitation signal to find output signal.

82. So,

83. Therefore, steady state response is

86.   If x(t) = A cos(ωt + ø), H(jω) = |H(jω)|e jθ(ω) then Steady-state sinusoidal response computed using a transfer function © 2008 Pearson Education 13.7 The Transfer Function and the Steady-State Sinusoidal Response

87. 13.8 The Impulse Function in Circuit Analysis   Laplace transform analysis correctly predicts impulsive currents and voltages arising from switching and impulsive sources.   The s-domain equivalent circuits are based on initial conditions at t = 0 -, that is, prior to the switching. © 2008 Pearson Education

88. A circuit showing the creation of an impulsive current The s-domain equivalent circuit

89. See :

90.  Physical interpretation is the following :

91. © 2008 Pearson Education A circuit showing the creation of an impulsive voltage The s-domain equivalent circuit

95.  See : eq.13.132

97. Using Laplace transform,

99. As in the

102. EE141102 Home work Prob. 13.7 13.8 13.9 13.13 13.17 13.19 13.22 13.25 13.26 13.27 13.36 13.44 13.54 13.56 13.64 13.68 13.84 제출기한 : - 다음 요일 수업시간 까지 - 제출기일을 지키지않는 레포트는 사정에서 제외함

103. THE END © 2008 Pearson Education