L76 - Spring 2006 Math 308-505 7 Laplace Transforms 7.6...

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Spring 2006 Math 308-505 7 Laplace Transforms 7.6 Transforms of Discontinuous and Periodic Functions Fri, 03/Mar c ± 2006, Art Belmonte Summary Heaviside function The Heaviside unit step function is a piecewise continuous function defined by u ( t ) = ± 0 , for t < 0 ; 1 , for t 0 . For c R (typically c > 0), the translate of u is defined by u c ( t ) = u ( t - c ) = ± 0 , for t < c ; 1 , for t c . The interval function is defined by u ab ( t ) = u ( t - a ) - u ( t - b ) = 0 , for t < a ; 1 , for a t < b ; 0 , for t b . The Heaviside function is easy to implement numerically in MATLAB in a vectorized fashion. To learn all about it, read Chapter 6 (“Advanced use of dfield7”) in your lab manual! Also read about Polking’s sqw square wave function M-file. (It’s available to you on CalcLab and is easily transferred to your Windows XP LABS account or home.) Finally, study the MATLAB Examples in this lecture handout. Even better, as of Fall 2003 I have learned how to fully implement the Heaviside function in a symbolic fashion for both Laplace transform and graphing purposes. Accordingly, your work will be much easier than the folks had it in the Spring! Periodic Functions If f ( t + T ) = f ( t ) for all t ,then f is periodic with period T (or T -periodic). Its window is defined as f T ( t ) = ± f ( t ), 0 t < T ; 0 , t T . More Laplace transforms In the following, a , b , c are constants, p is a nonnegative constant, L { f ( t ) } = F ( s ) is the Laplace transform of f , and the Laplace transform of the window g T of the piecewise continuous T -periodic function g is given by L { g T ( t ) } = G T ( s ) . L { u ( t - c ) } = e - cs s L { u ( t ) } = e - as - e - bs s L { f ( t - p ) u ( t - p ) } = e - ps F ( s ) L { f ( t ) u ( t - p ) } = e - ps L { f ( t + p ) } ( s ) L { g ( t ) } = G T ( s ) 1 - e - sT = ² T 0 g ( t ) e - st dt 1 - e - Hand Examples Example A Find the Laplace transform of e 2 ( t - 1 ) u ( t - 1 ) . Solution Let f ( t ) = e 2 t .Then F ( s ) = L { f ( t ) }= 1 s - 2 . Therefore L ³ e 2 ( t - 1 ) u ( t - 1 ) ´ = L { f ( t - 1 ) u ( t - 1 ) } = e - s F ( s ) = e - s s - 2 . Example B Compute the Laplace transform of e - t u ( t - 1 ) . Solution Since L { u ( t - 1 ) } = e - s s ,wehave L µ e - t u ( t - 1 ) = e - ( s + 1 ) s + 1 . Example C Determine the Laplace transform of u ( t - π 2 ) cos 3 t . Solution We have cos ( 3 ( t + π 2 )) = cos ( 3 t + 3 2 π) = sin 3 t via trig. So L µ u ( t - π 2 ) cos 3 t = e - π 2 s L µ cos ( 3 ( t + π 2 )) = 3 e - π 2 s s 2 + 9 . Example D Let f ( t ) = 0 , t < 0 t , 0 t < 3 3 , t 3 .Define f in terms of the Heaviside function, then compute its Laplace transform. Solution f ( t ) = t ( u ( t ) - u ( t - 3 )) + 3 u ( t - 3 ) = tu ( t ) - ( t - 3 ) u ( t - 3 ) implies L { f ( t ) } = e - 0 s s 2 - e - 3 s s 2 = 1 - e - 3 s s 2 . 1
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Example E Find the inverse Laplace transform of e - 2 s s + 3 . Solution Let F ( s ) = 1 s + 3 .Then f ( t ) = L - 1 { F ( s ) }= e - 3 t . Thus L - 1 ± e - 2 s 1 s + 3 ² = u ( t - 2 ) f ( t - 2 ) = u ( t - 2 ) e - 3 ( t - 2 ) = ± 0 , t < 2 ; e - 3 ( t - 2 ) , t 2 .
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L76 - Spring 2006 Math 308-505 7 Laplace Transforms 7.6...

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