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Lec 16 - UCLA Spring 2009-2010 Systems and Signals Lecture...

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Unformatted text preview: UCLA Spring 2009-2010 Systems and Signals Lecture 16: Applications of the Laplace Transform May 24, 2010 EE102: Systems and Signals; Spr 09-10, Lee 1 Solutions of LCCODEs The Laplace transforms provide an easy solution for linear differential equations with Constant coefficients Initial conditions Input signals Solution procedure: Laplace transform converts differential equation, initial conditions, and input signals into an algebraic equation Solve for the Laplace transform of the output Inverse Laplace transform provides the solution EE102: Systems and Signals; Spr 09-10, Lee 2 Example Solve the LCCODE y 00 ( t ) + 5 y ( t ) + 6 y ( t ) = f ( t ) + f ( t ) where y ( t ) is the output signal, and initial conditions are y (0- ) = 2 , and y (0) = 1 . Assume that the input signal is f ( t ) = e- 4 t u ( t ) . Solution: Taking the Laplace transform of the left side of the equation is s 2 Y ( s )- sy (0- )- y (0- ) + 5( sY ( s )- y (0- )) + 6 Y ( s ) = Y ( s )( s 2 + 5 s + 6)- 2 s- 1- 10 = Y ( s )( s 2 + 5 s + 6)- (2 s + 11) EE102: Systems and Signals; Spr 09-10, Lee 3 The Laplace transform of the right side is sF ( s )- f (0- ) + F ( s ) = F ( s )( s + 1) = s + 1 s + 4 where we have used the fact that that L [ f ( t )] = L [ e- 4 t u ( t )] = 1 s +4 . The Laplace transform of the equation is then Y ( s )( s 2 + 5 s + 6)- (2 s + 11) = s + 1 s + 4 . Solving for Y ( s ) , Y ( s ) = 2 s + 11 s 2 + 5 s + 6 | {z } from init. cond. + s + 1 ( s + 4)( s 2 + 5 s + 6) | {z } from input . The first part can be traced back to the initial conditions, and the second part is due to the input. Well return to this in a few pages. EE102: Systems and Signals; Spr 09-10, Lee 4 Combining these terms we get Y ( s ) = (2 s + 11)( s + 4) + ( s + 1) ( s + 4)( s 2 + 5 s + 6) = 2 s 2 + 20 s + 45 ( s + 4)( s + 3)( s + 2) To find the inverse Laplace transform we first find the partial expansion, 2 s 2 + 20 s + 45 ( s + 2)( s + 3)( s + 4) = r 1 s + 2 + r 2 s + 3 + r 3 s + 4 . Using the cover up algorithm, r 1 = 2(- 2) 2 + 20(- 2) + 45 (- 2 + 4)(- 2 + 3) = 8- 40 + 45 (2)(1) = 13 2 r 2 = 2(- 3) 2 + 20(- 3) + 45 (- 3 + 4)(- 3 + 2) = 18- 60 + 45 (1)(- 1) =- 3 r 3 = 2(- 4) 2 + 20(- 4) + 45 (- 4 + 3)(- 4 + 2) = 32- 80 + 45 (- 1)(- 2) =- 3 2 EE102: Systems and Signals; Spr 09-10, Lee 5 Then Y ( s ) = 13 / 2 s + 2- 3 s + 3- 3 / 2 s + 4 . The solution y ( t ) is then found by taking the term-by-term inverse Laplace transform y ( t ) = 13 2 e- 2 t- 3 e- 3 t- 3 2 e- 4 t u ( t ) Zero-State and Zero-Input Solutions: This solution combines the effect of the initial conditions, and the input. We can also keep them separate, and solve for the zero-input and zero-state signals, Y ( s ) = 2 s + 11 s 2 + 5 s + 6 + s + 1 ( s + 4)( s 2 + 5 s + 6) = 2 s + 11 ( s + 2)( s + 3) | {z } zero input + s + 1 ( s + 2)( s + 3)( s + 4) | {z } zero state ....
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This note was uploaded on 10/21/2010 for the course EE ee102 taught by Professor Levan during the Spring '09 term at UCLA.

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Lec 16 - UCLA Spring 2009-2010 Systems and Signals Lecture...

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