1. ## Laplace Transform

How do we prove that L{f(2t)} = 1/2F(S/2).
I know to prove L{f(t)} = 1/S

Thanks,

2. ## Re: Laplace Transform

$F(s)=\int f(t) e^{-s t}~dt$

$G(s)=\int f(2t) e^{-s t}~dt$

$u=2t, du=2dt,~t=u/2~,dt=du/2$

$G(s)=\frac 1 2 \int f(u) e^{-s u/2}~du=\frac 1 2 \in f(u) e^{-(s/2)u}~du=\frac 1 2 F(s/2)$

3. ## Re: Laplace Transform

Uh, you know that "L(f)= 1/s" is only true for the specific function, f(t)= 1 for all t, don't you? Saying that "L{f(t)}= 1/s" is meaningless.

I hope, at least, that you know that $L(f)= \int_0^\infty f(t)e^{-st}dt$.
So that $L{f(2t)}= \int_0^\infty f(2t)e^{-st}dt$. Now make the substitution u= 2x in that integral.

4. ## Re: Laplace Transform

Originally Posted by romsek
$F(s)=\int f(t) e^{-s t}~dt$

$G(s)=\int f(2t) e^{-s t}~dt$

$u=2t, du=2dt,~t=u/2~,dt=du/2$

$G(s)=\frac 1 2 \int f(u) e^{-s u/2}~du=\frac 1 2 \in f(u) e^{-(s/2)u}~du=\frac 1 2 F(s/2)$
this should be

$G(s)=\frac 1 2 \int f(u) e^{-s u/2}~du=\frac 1 2 \int f(u) e^{-(s/2)u}~du=\frac 1 2 F(s/2)$

5. ## Re: Laplace Transform

Thank you Romsek