lecture_05 - MIT OpenCourseWare http:/ocw.mit.edu 2.161...

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MIT OpenCourseWare http://ocw.mit.edu 2.161 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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Massachusetts Institute of Technology Department of Mechanical Engineering 2.161 Signal Processing - Continuous and Discrete Fall Term 2008 Lecture 5 1 Reading: Class handout: The Laplace Transform . Class handout: Understanding Poles and Zeros . Class handout: Sinusoidal Frequency Response of Linear Systems . 1 The One-Sided Laplace Transform Consider a causal waveform x ( t ), such as the unit-step function u s ( t ), for which the Fourier integral x ( t ) e j Ω t dt = x ( t ) e j Ω t dt −∞ 0 does not converge. Clearly, for u s ( t ), the Fourier integral U s ( j Ω) = 1 .e j Ω t dt 0 does not converge. Now consider a modified function x Þ ( t ) = x ( t ) w ( t ) where w ( t ) is a weighting function with the property lim t →∞ w ( t ) = 0, chosen to ensure convergence, so that X Þ ( j Ω ,w ) = x ( t ) w ( t ) e j Ω t dt 0 may be considered an approximation to X ( j Ω). In particular, consider w ( t ) = e σt for real σ , and note that as σ 0, x ( t ) w ( t ) x ( t ), so that the Fourier transform is F ± x ( t ) e σt ² = X ( j Ω ) = x ( t ) e σt e j Ω t dt = x ( t ) e ( σ + j Ω) t dt. 0 0 If we define a complex variable s = σ + j Ω we can write x ( t ) e st dt L{ x ( t ) } = X ( s ) = 0 which defines the one-sided Laplace transform. (See the handout for the definition of the two-sided transform). 1 copyright c D.Rowell 2008 5–1
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t f ( t ) f ( t ) ± ² ± ² ³ ³ The Laplace transform may be considered as an extension of the Fourier transform (for causal functions) that includes an exponential weighting term to extend the range of functions for which the integral converges. Note that for causal waveforms X ( j Ω) = X ( s ) | s = j Ω and if x ( t ) is non-causal X ( j Ω) ± X ( s ) | = s = j Ω , for example F{ sin(Ω 0 t ) } ± L{ sin(Ω 0 t ) }| = s = j Ω , since the Laplace transform assumes x ( t ) 0 for t< 0 . 0 n o n - c a u s a l s i n u s o i d c a u s a l s i n u s o i d 0 t Example 1 The following are some simple examples of Laplace transforms: 1 u s ( t ) } = 1 .e st dt = 0 s δ ( t ) } = δ ( t ) e st dt =1 0 ´ µ 1 L e at = e ( s + a ) t dt = 0 s + a 1.1 The Derivative Property of the Laplace Transform: If a function x ( t ) has a Laplace transform X ( s ), the Laplace transform of the derivative of x ( t )is dx L = sX ( s ) x (0) . dt Using integration by parts dx dx L = e st dt dt 0 dt = ³ x ( t ) e st ³ 0 + sx ( t ) e st dt 0 = sX ( s ) x (0) . 5–2
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± ² ² ² ² ² ² ² ² ² ³ ± ± ´ µ ´ µ 2 This procedure may be repeated to find the Laplace transform of higher order derivatives, for example the Laplace transform of the second derivative is d 2 x dx ² L = s [ s L{ x ( t ) }− x (0)] dt 2 dt t =0 dx ² = s 2 X ( s ) sx (0) dt t =0 which may be generalized to ± n L d n x = s n X ( s ) · s n i d i 1 x ² dt n dt i 1 i =1 t =0 for the n derivative of x ( t ).
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lecture_05 - MIT OpenCourseWare http:/ocw.mit.edu 2.161...

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