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Depok, October, 2009 Laplace Transform Electric Circuit Circuit Applications of Laplace Transform Electric Power & Energy Studies (EPES) Department of.

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Presentation on theme: "Depok, October, 2009 Laplace Transform Electric Circuit Circuit Applications of Laplace Transform Electric Power & Energy Studies (EPES) Department of."— Presentation transcript:

1 Depok, October, 2009 Laplace Transform Electric Circuit Circuit Applications of Laplace Transform Electric Power & Energy Studies (EPES) Department of Electrical Engineering University of Indonesia http://www.ee.ui.ac.id/epes Chairul Hudaya, ST, M.Sc Depok, October, 2009 Electric Circuit

2 Depok, October, 2009 Laplace Transform Electric Circuit Circuit applications 1.Transfer functions 2.Convolution integrals 3.RLC circuit with initial conditions

3 Depok, October, 2009 Laplace Transform Electric Circuit Transfer function h(t)h(t) y(t)y(t)x(t)x(t) In s-domain, In time domain, Network System

4 Depok, October, 2009 Laplace Transform Electric Circuit Example 1 For the following circuit, find H(s)=V o (s)/V i (s). Assume zero initial conditions.

5 Depok, October, 2009 Laplace Transform Electric Circuit Solution Transform the circuit into s-domain with zero i.c.:

6 Depok, October, 2009 Laplace Transform Electric Circuit Using voltage divider

7 Depok, October, 2009 Laplace Transform Electric Circuit Example 2 Obtain the transfer function H(s)=V o (s)/V i (s), for the following circuit.

8 Depok, October, 2009 Laplace Transform Electric Circuit Solution Transform the circuit into s-domain (We can assume zero i.c. unless stated in the question)

9 Depok, October, 2009 Laplace Transform Electric Circuit We found that

10 Depok, October, 2009 Laplace Transform Electric Circuit Example 3 Use convolution to find v o (t) in the circuit of Fig.(a) when the excitation (input) is the signal shown in Fig.(b).

11 Depok, October, 2009 Laplace Transform Electric Circuit Solution Step 1: Transform the circuit into s-domain (assume zero i.c.) Step 2: Find the TF

12 Depok, October, 2009 Laplace Transform Electric Circuit Step 3: Find v o (t) For t < 0 For t > 0

13 Depok, October, 2009 Laplace Transform Electric Circuit Circuit element models  Apart from the transformations we must model the s-domain equivalents of the circuit elements when there is involving initial condition (i.c.)  Unlike resistor, both inductor and capacitor are able to store energy

14 Depok, October, 2009 Laplace Transform Electric Circuit  Therefore, it is important to consider the initial current of an inductor and the initial voltage of a capacitor  For an inductor : –Taking the Laplace transform on both sides of eqn gives or

15 Depok, October, 2009 Laplace Transform Electric Circuit

16  For a capacitor  Taking the Laplace transform on both sides of eqn gives or

17 Depok, October, 2009 Laplace Transform Electric Circuit

18 Example 4 Consider the parallel RLC circuit of the following. Find v(t) and i(t) given that v(0) = 5 V and i(0) = −2 A.

19 Depok, October, 2009 Laplace Transform Electric Circuit Solution Transform the circuit into s-domain (use the given i.c. to get the equivalents of L and C)

20 Depok, October, 2009 Laplace Transform Electric Circuit Then, using nodal analysis

21 Depok, October, 2009 Laplace Transform Electric Circuit Since the denominator cannot be factorized, we may write it as a completion of square: Finding i(t),

22 Depok, October, 2009 Laplace Transform Electric Circuit Using partial fractions, It can be shown that Hence,

23 Depok, October, 2009 Laplace Transform Electric Circuit Example 5 The switch in the following circuit moves from position a to position b at t = 0 second. Compute i o (t) for t > 0.

24 Depok, October, 2009 Laplace Transform Electric Circuit Solution The i.c. are not given directly. Hence, at first we need to find the i.c. by analyzing the circuit when t ≤ 0 :

25 Depok, October, 2009 Laplace Transform Electric Circuit Then, we can analyze the circuit for t > 0 by considering the i.c. Let

26 Depok, October, 2009 Laplace Transform Electric Circuit Using current divider rule, we find that Using partial fraction we have


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