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

This preview shows page 6 - 8 out of 9 pages.

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) 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)
Image of page 6

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) 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
Image of page 7
Image of page 8
  • Fall '08
  • staff

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Ask Expert Tutors You can ask You can ask ( soon) You can ask (will expire )
Answers in as fast as 15 minutes