102_1_Lecture16

# 102_1_Lecture16 - UCLA Spring 2009-2010 Systems and Signals...

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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

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Solutions of LCCODE’s The Laplace transforms provide an easy solution for linear di f erential equations with Constant coe F cients Initial conditions Input signals Solution procedure: Laplace transform converts di f erential 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 ± ( 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 s + 6) - (2 s + 11) EE102: Systems and Signals; Spr 09-10, Lee 3

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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 s . Solving for Y ( s ) , Y ( s 2 s + 11 s 2 s +6 ± ²³ ´ from init. cond. + s ( s + 4)( s 2 s + 6) ± ²³ ´ from input . The Frst part can be traced back to the initial conditions, and the second part is due to the input. We’ll 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 fnd the inverse Laplace transForm we frst fnd 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

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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 ² u ( t ) Zero-State and Zero-Input Solutions: This solution combines the e f ect 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 s + 6) = 2 s + 11 ( s + 2)( s + 3) ³ ´µ zero input + s ( s + 2)( s + 3)( s + 4) ³ ´µ zero state . EE102: Systems and Signals; Spr 09-10, Lee 6
The zero input component is 2 s + 11 ( s + 2)( s + 3) = r 1 s +2 + r 2 s +3 = 7 s - 5 s . which corresponds to the signal y zi ( t )= ± 7 e - 2 t - 5 e - 3 t ² u ( t ) The zero state component is s +1 ( s + 2)( s + 3)( s + 4) = r 1 s + r 2 s + r 3 s +4 = - 1 / 2 s + 2 s - 3 / 2 s EE102: Systems and Signals; Spr 09-10, Lee 7

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which corresponds to the signal y zs ( t )= ± - 3 2 e - 4 t +2 e - 3 t - 1 2 e - 2 t ² u ( t ) The complete solution is then y ( t y zi ( t )+ y ( t ) = ³ 7 e - 2 t - 5 e - 3 t ´ u ( t ± - 1 2 e - 2 t e - 3 t - 3 2 e - 4 t ² u ( t ) = ± 13 2 e - 2 t - 3 e - 3 t - 3 2 e - 4 t ² u ( t ) EE102: Systems and Signals; Spr 09-10, Lee 8
Zero-State Response: Transfer Function of LTI Systems A linear time invariant system is complete characterized by its impulse response. If the impulse response is h ( t ) , the output y ( t

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## This note was uploaded on 12/09/2010 for the course EE ee102 taught by Professor Levan during the Spring '09 term at UCLA.

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102_1_Lecture16 - UCLA Spring 2009-2010 Systems and Signals...

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