This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1 ME261  Solved Examples for Homework #3 Problem 1: For each circuit in Figure 1., derive the transfer function using equivalent impedance method and impulse responses using partial fraction expansion and inverse Laplace transform. Figure 1. Solution: Z C = 1 Cs ; Z L = Ls; Z R = R Circuit 1: loop equations: (1) Y ( s ) = I & 1 sC ¡ (2) X ( s ) = Z R I + Z C I = IR + I 1 Cs  {z } Y ( s ) = I 1 Cs CsR + I 1 Cs = Y ( s ) ( RCs + 1) transfer function: H ( s ) = Y ( s ) X ( s ) = 1 RCs +1 = 1 RC h 1 s + 1 RC i de&ne: & = RC , impulse response: h ( t ) = L & 1 f H ( s ) g = 1 & & e & t=& ¡ u ( t ) (see Laplace Transform Table 11.1 Ambardar) Circuit 2: loop equations: (1) X ( s ) = i 1 Z R 1 + ( i 1 & i 2 ) Z R 2 (2) Y ( s ) = i 2 Z c (3) Y ( s ) + Z R 2 ( i 2 & i 1 ) = 0 i 2 = Y ( s ) Z c and ( i 1 + i 2 ) = Y ( s ) Z R ) i 1 = Y ( s ) ¢ 1 Z R & 1 Z c £ now sub i 1 and i 2 into X ( s ) equation: X ( s ) = Y ( s ) ¢¢ 1 + Z R Z c £ + 1 £ transfer function: H ( s ) = Y ( s ) X ( s ) = 1 2+ Z R Zc = 1 2+ RCs = 1 RC ¢ 1 2 RC + s £ impulse response: h ( t ) = L & 1 f H ( s ) g = 1 RC & e & 2 t=RC ¡ u ( t ) (see table) Circuit 3: loop equations: (1) X ( s ) = i 1 Z R 1 + ( i 1 & i 2 ) Z R 2 (2) Y ( s ) = i 2 Z L = ( i 1 & i 2 ) Z R (3) i 2 = Y ( s ) Z L (4) ( i 1 & i 2 ) = Y ( s ) Z R now sub i 1 and...
View
Full
Document
This note was uploaded on 04/18/2010 for the course ME 360 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Staff
 Laplace

Click to edit the document details