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 ff erential equations with Constant coe ffi cients Initial conditions Input signals Solution procedure: Laplace transform converts di ff 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
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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 + 5 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 + 1 s + 4 . Solving for Y ( s ) , Y ( s ) = 2 s + 11 s 2 + 5 s + 6 from init. cond. + s + 1 ( s + 4)( s 2 + 5 s + 6) from input . The first 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
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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
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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 u ( t ) Zero-State and Zero-Input Solutions: This solution combines the e ff 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 + 5 s + 6) = 2 s + 11 ( s + 2)( s + 3) zero input + s + 1 ( s + 2)( s + 3)( s + 4) zero state . EE102: Systems and Signals; Spr 09-10, Lee 6
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The zero input component is 2 s + 11 ( s + 2)( s + 3) = r 1 s + 2 + r 2 s + 3 = 7 s + 2 - 5 s + 3 . which corresponds to the signal y zi ( t ) = 7 e - 2 t - 5 e - 3 t u ( t ) The zero state component is s + 1 ( s + 2)( s + 3)( s + 4) = r 1 s + 2 + r 2 s + 3 + r 3 s + 4 = - 1 / 2 s + 2 + 2 s + 3 - 3 / 2 s + 4 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 u ( t ) The complete solution is then y ( t ) = y zi ( t ) + y zs ( t ) = 7 e - 2 t - 5 e - 3 t u ( t ) + - 1 2 e - 2 t + 2 e - 3 t - 3 2 e - 4 t u ( t ) = 13 2 e - 2 t - 3 e - 3 t - 3 2 e - 4 t u ( t ) EE102: Systems and Signals; Spr 09-10, Lee 8
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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 ) for an input x ( t ) is y ( t ) = ( h * x )( t ) .
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