The window function specifies values of f t over a single period This can be

# The window function specifies values of f t over a

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The window function specifies values of f ( t ) over a single period This can be replicated k periods to the right as f T ( t - kT ) u kT ( t ) = f ( t - kT ) , kT t ( k + 1) T 0 , otherwise Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (22/35) Subscribe to view the full document.

Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function Laplace for Periodic Functions By summing n time shifted replications of the window function , f T ( t - kT ) u kT ( t ), k = 0,..., n - 1, gives f nT ( t ), the periodic extension of f T ( t ) to the interval [0 , nT ], f nT ( t ) = n - 1 X k =0 f T ( t - kT ) u kT ( t ) Theorem If f is periodic with period T and is piecewise continuous on [0 , T ] , then L [ f ( t )] = F T ( s ) 1 - e - sT = R T 0 e - st f ( t ) dt 1 - e - sT . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (23/35) Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function Laplace for Periodic Functions Proof: From our earlier theorem, we have for each k 0, L [ f T ( t - kT ) u kT ( t )] = e - kT s L [ f T ( t )] = e - kT s F T ( s ) . By linearity of L , the Laplace transform of f nT is F nT ( s ) = Z nT 0 e - st f ( t ) dt = n - 1 X k =0 L [ f T ( t - kT ) u kT ( t )] = n - 1 X k =0 e - kT s F T ( s ) = F T ( s ) n - 1 X k =0 ( e - T s ) k = F T ( s ) 1 - ( e - T s ) n 1 - e - sT . The last term comes from summing a geometric series. With e - sT < 1, F ( s ) = lim n - > Z nT 0 e - st dt = lim n - > F T ( s ) 1 - ( e - T s ) n 1 - e - sT = F T ( s ) 1 - e - sT Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (24/35) Subscribe to view the full document.

Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function Sawtooth Function Return to sawtooth waveform f ( t ) = t, 0 t < 1 , 0 , 1 t < 2 . and f ( t ) has period 2 The theorem for the Laplace transform of periodic function gives L [ f ( t )] = R 2 0 e - st f ( t ) dt 1 - e - 2 s But Z 2 0 e - st f ( t ) dt = Z 1 0 te - st dt = 1 - se - s - e - s s 2 , so L [ f ( t )] = 1 - se - s - e - s s 2 (1 - e - 2 s ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (25/35) Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function IVP with Periodic Forcing Function 1 Example: Consider the following initial value problem: y 00 + 4 y = f ( t ) , y (0) = 0 , y 0 (0) = 0 , with the square waveform as the periodic forcing function : f ( t ) = 1 , 0 t < 1 , 0 , 1 t < 2 . and f ( t ) has period 2 The theorem for the Laplace transform of square waveform gives L [ f ( t )] = R 2 0 e - st f ( t ) dt 1 - e - 2 s But Z 2 0 e - st f ( t ) dt = Z 1 0 e - st dt = 1 - e - s s , so L [ f ( t )] = 1 - e - s s (1 - e - 2 s ) = 1 s (1 + e - s ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part B — (26/35) Subscribe to view the full document.

Inverse Laplace Transforms Special Functions Heaviside or Step function Periodic functions Impulse or δ Function IVP with Periodic Forcing Function 2 Example: Taking the Laplace transform of the IVP with L [ y ( t )] = Y ( s ), we have: s 2 Y ( s ) - sy (0) - y 0 (0) + 4 Y ( s ) = 1 s (1 + e - s ) Thus, Y ( s ) = 1 s ( s 2 + 4) (1 + e - s ) Partial fractions decomposition gives 1 s ( s 2 + 4) = 1 / 4 s - s/ 4 s 2 + 4 , while 1 1 + e - s = 1 1 - ( - e - s ) = 1 - e - s + e - 2 s - ... + ( - 1) n e - ns + Joseph M. Mahaffy, h [email protected] i  • Fall '08
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