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,

Let's call this expression f(t); then

~ f

(p)

=ò₀^¥ 3 e^{-p t} e^-4tsin2t dt

=ò₀^¥ 3e^-(p+4)tsin2t dt

=3

~ g

(p+4)

where

~ g

(p)=ò0¥ e-p tsin2t dt

, ie it is the Laplace transform of g(t)=sin 2t.

This you can now do from tables: we have

~ g

(p)=2/(p2+4)

so

~ f

(p)=6/((p+4)2+4)=6/(p2+8p+20)

is the Laplace transform of f(t)=3e^-4tsin2t.

This just illustrates the effect of multiplying f by e^{-a t}: if h(t)=f(t)e^{-a t},

~ h

(p)=

~ f

(p+a)

.

David

Thankyou, Thankyou, Thankyou.
George.