Laplace Transforms of 3exp(-4t)sin(2t)

By Georgina Horton on Sunday, May 19, 2002 - 04:56 pm:

Hello Everyone,
Could anyone provide me with the method for discovering the Laplace Transform of:
3exp(-4t)sin(2t)
and explain the "Exponential Multiplier Rule" to me.
Thanks
George,

By David Loeffler on Sunday, May 19, 2002 - 05:14 pm:

Using the notation

\tilde{f} (p)

for the Laplace transform of

f (t)

, where

p

is the parameter in the integral:

Let's call this expression $f (t)$ ; then $\tilde{f} (p)$

$= \int_{0}^{\infty} 3 e^{- p t} e^{- 4 t} \sin 2 t dt$

$= \int_{0}^{\infty} 3 e^{- (p + 4) t} \sin 2 t dt$

$= 3 \tilde{g} (p + 4)$ where $\tilde{g} (p) = \int_{0}^{\infty} e^{- p t} \sin 2 t dt$ , ie it is the Laplace transform of $g (t) = \sin 2 t$ .

This you can now do from tables: we have

$\tilde{g} (p) = 2 / (p^{2} + 4)$

so $\tilde{f} (p) = 6 / ((p + 4)^{2} + 4) = 6 / (p^{2} + 8 p + 20)$

is the Laplace transform of $f (t) = 3 e^{- 4 t} \sin 2 t$ .

This just illustrates the effect of multiplying $f$ by $e^{- a t}$ : if $h (t) = f (t) e^{- a t}$ , $\tilde{h} (p) = \tilde{f} (p + a)$ .

David

By Georgina Horton on Sunday, May 19, 2002 - 05:23 pm:

Thankyou, Thankyou, Thankyou.
George.