102_1_Lecture15

102_1_Lecture15 - UCLA Spring 2009-2010 Systems and Signals...

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UCLA Spring 2009-2010 Systems and Signals Lecture 15: Inversion of the Laplace Transform May 19, 2010 EE102: Systems and Signals; Spr 09-10, Lee 1
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Inversion of the Laplace transform From Lecture 14, the inverse Laplace transform is given by f ( t ) = 1 2 π j c + j c - j F ( s ) e st ds where σ is large enough that F ( s ) is defined for s c This involves a contour integral in the complex plane. Simpler approach: rewrite a rational Laplace transform into simple terms we recognize, and can invert by inspection ( partial fraction expansion ). Computationally the same operations as contour integral. EE102: Systems and Signals; Spr 09-10, Lee 2
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Partial fraction expansion F ( s ) = b ( s ) a ( s ) = b 0 + b 1 s + · · · + b m s m a 0 + a 1 s + · · · + a n s n Let’s assume (for now) no poles are repeated, i.e. , all roots of a have multiplicity one m < n then we can write F in the form F ( s ) = r 1 s - λ 1 + · · · + r n s - λ n called partial fraction expansion of F EE102: Systems and Signals; Spr 09-10, Lee 3
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λ 1 , . . . , λ n are the poles of F the numbers r 1 , . . . , r n are called the residues when λ k = λ * l , r k = r * l Example: s 2 - 2 s 3 + 3 s 2 + 2 s = - 1 s + 1 s + 1 + 1 s + 2 Let’s check: - 1 s + 1 s + 1 + 1 s + 2 = - 1( s + 1)( s + 2) + s ( s + 2) + s ( s + 1) s ( s + 1)( s + 2) In partial fraction form, inverse Laplace transform is easy: L - 1 [ F ( s )] = L - 1 r 1 s - λ 1 + · · · + r n s - λ n EE102: Systems and Signals; Spr 09-10, Lee 4
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= r 1 L - 1 1 s - λ 1 + · · · + r n L - 1 1 s - λ n = r 1 e λ 1 t + · · · + r n e λ n t This is real since whenever the poles are conjugates, the corresponding residues are also. EE102: Systems and Signals; Spr 09-10, Lee 5
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Finding the partial fraction expansion Two steps: find poles λ 1 , . . . , λ n ( i.e. , factor a ( s ) ) find residues r 1 , . . . , r n (several methods) Method 1: Clear Fractions, Solve linear equations We’ll illustrate for m = 2 , n = 3 b 0 + b 1 s + b 2 s 2 ( s - λ 1 )( s - λ 2 )( s - λ 3 ) = r 1 s - λ 1 + r 2 s - λ 2 + r 3 s - λ 3 Clear denominators: b 0 + b 1 s + b 2 s 2 = r 1 ( s - λ 2 )( s - λ 3 )+ r 2 ( s - λ 1 )( s - λ 3 )+ r 3 ( s - λ 1 )( s - λ 2 ) EE102: Systems and Signals; Spr 09-10, Lee 6
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Equate coe ffi cients: coe ffi cient of s 0 : b 0 = ( λ 2 λ 3 ) r 1 + ( λ 1 λ 3 ) r 2 + ( λ 1 λ 2 ) r 3 coe ffi cient of s 1 : b 1 = ( - λ 2 - λ 3 ) r 1 + ( - λ 1 - λ 3 ) r 2 + ( - λ 1 - λ 2 ) r 3 coe ffi cient of s 2 : b 2 = r 1 + r 2 + r 3 Now solve for r 1 , r 2 , r 3 (three equations in three variables) EE102: Systems and Signals; Spr 09-10, Lee 7
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Method 2: Heaviside “Cover-up” Procedure To get r 1 , multiply both sides by s - λ 1 to get ( s - λ 1 )( b 0 + b 1 s + b 2 s 2 ) ( s - λ 1 )( s - λ 2 )( s - λ 3 ) = r 1 + r 2 ( s - λ 1 ) s - λ 2 + r 3 ( s - λ 1 ) s - λ 3 cancel s - λ 1 term on left and set s = λ 1 : b 0 + b 1 λ 1 + b 2 λ 2 1 ( λ 1 - λ 2 )( λ 1 - λ 3 ) = r 1 an explicit formula for r 1 !
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