Laplace Transform

**Maccaman** · August 27th 2009, 07:29 AM

Could someone please help with the following:

Find the laplace transform of the following:

$f(t) = e^{4t} \ \sin^2(3t)$

I dont know if my method is correct, but here is what I started with.

f(t) becomes

$f(t) = \frac{1}{2} e^{4t} - \frac{1}{2} e^{4t} \cos(6t)$

Let $g(t) = e^{4t}, ... h(t) = e^{4t} \cos(6t)$

and
$A = \frac{1}{2}, B = -\frac{1}{2}$

Then $F(s) = \mathcal{L}(f(t)) = \int_0^{\infty} e^{-st} (A(g(t)) + B(h(t) )$

skipping a few lines of calculation....

$= \frac{1}{2} \int_0^{\infty} e^{(4-s)t} dt + (\frac{-1}{2} ) \ \mathcal{L} (\cos(6t) e^{4t})$

$= \frac{1}{2s-8} + \frac{-1}{2} \times \frac{s-4}{(s-4)^2 + 6^2}$

but I am not really sure. Im lost

**Opalg** · August 27th 2009, 07:57 AM

Originally Posted by Maccaman

Find the Laplace transform of the following:

$f(t) = e^{4t} \ \sin^2(3t)$

.....

$= \frac{1}{2s-8} + \frac{-1}{2} \times \frac{s-4}{(s-4)^2 + 6^2}$

but I am not really sure. Im lost

Your answer looks correct to me.

**DeMath** · August 27th 2009, 08:34 AM

Originally Posted by Maccaman

Could someone please help with the following:

Find the laplace transform of the following:

$f(t) = e^{4t} \ \sin^2(3t)$

I dont know if my method is correct, but here is what I started with.

f(t) becomes

$f(t) = \frac{1}{2} e^{4t} - \frac{1}{2} e^{4t} \cos(6t)$

Let $g(t) = e^{4t}, ... h(t) = e^{4t} \cos(6t)$

and
$A = \frac{1}{2}, B = -\frac{1}{2}$

Then $F(s) = \mathcal{L}(f(t)) = \int_0^{\infty} e^{-st} (A(g(t)) + B(h(t) )$

skipping a few lines of calculation....

$= \frac{1}{2} \int_0^{\infty} e^{(4-s)t} dt + (\frac{-1}{2} ) \ \mathcal{L} (\cos(6t) e^{4t})$

$= \frac{1}{2s-8} + \frac{-1}{2} \times \frac{s-4}{(s-4)^2 + 6^2}$

but I am not really sure. Im lost

You can just use Frequency shifting $\mathcal{L}\left\{ {{e^{\alpha t}}f\left( t \right)} \right\} = F\left( {s - \alpha } \right)$

So $g\left( t \right) = {e^{4t}}{\sin ^2}3t$

Then $f\left( t \right) = {\sin ^2}3t$

$\mathcal{L}\left\{ {{{\sin }^2}3t} \right\} = \mathcal{L}\left\{ {\frac{1}{2}\left( {1 - \cos 6t} \right)} \right\} = \frac{1}{2}\left( {\frac{1}{s} - \frac{s}{{{s^2} + 36}}} \right)$

Finally $\mathcal{L}\left\{ {{e^{4t}}{{\sin }^2}3t} \right\} = \frac{1}{2}\left( {\frac{1}{{s - 4}} - \frac{{s - 4}}{{{{\left( {s - 4} \right)}^2} + 36}}} \right)$

Thread: Laplace Transform

LinkBack

Thread Tools

Display

Laplace Transform

Similar Math Help Forum Discussions

Laplace transform and Fourier transform what is the different?

Laplace transform

Laplace transform

Laplace Transform Help

LaPlace Transform

Search Tags