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 de±ned 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 Let’s check: - 1 s + 1 s + 1 s = - 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 ² EE102: Systems and Signals; Spr 09-10, Lee 4
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= r 1 L - 1 ± 1 s - λ 1 ² + · · · + r n L - 1 ± 1 s - λ n ² = 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: fnd poles λ 1 , . . . , λ n ( i.e. , Factor a ( s ) ) fnd 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 f cients: coe f 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 f 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 f cient of s 2 : b 2 = r 1 + r 2 + r 3 Now solve for r 1 ,r 2 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 ! We can get r 2 ,r 3 the same way. EE102: Systems and Signals; Spr 09-10, Lee 8
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In the general case we have the formula r k =( s - λ k ) F ( s ) | s = λ k which means: multiply F by s - λ k then cancel s - λ k from numerator and denominator then evaluate at s = λ k to get r k EE102: Systems and Signals; Spr 09-10, Lee 9
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Example: s 2 - 2 s ( s + 1)( s + 2) = r 1 s + r 2
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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_Lecture15 - UCLA Spring 2009-2010 Systems and Signals...

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