laplace transforms are causing me trouble

**action259** · August 28th 2006, 05:49 PM

I'm having a little trouble with these laplace transforms, don't know why... they look pretty simple from first glance was wondering if someone could give me some direction solving these.

1. $exp(-at+b)$

2. $sin(wt+b)$

3. $cosh^23t$

**ThePerfectHacker** · August 28th 2006, 05:52 PM

Originally Posted by action259

$exp(-at+b)$

I assume $a>0$ .
---
$e^{-at+b}$
Now there is no rule for that. But you need to change it.
You can write.
$e^{-at}e^b$ because the product of these is adding exponents.

Remember that,
$\mathcal{L}e^{-at}=\frac{1}{s+a}$
And,
$\mathcal{L}\{\kappa f(x)\}=\kappa \mathcal{L}\{f(x)\}$

Since, $e^b$ is a KoNsTaNt.
We have,
$\frac{e^b}{s+a}$

**ThePerfectHacker** · August 28th 2006, 06:00 PM

Originally Posted by action259

$sin(wt+b)$

Did you Botox it?
$\sin \omega t\cos b+\cos \omega t\sin b$
And, $b$ is a KoNsTaNt.
Apply the rules,
$\mathcal{L}\{ \sin at\}=\frac{a}{s^2+a^2}$
$\mathcal{L}\{ \cos at\}=\frac{s}{s^2+a^2}$
Thus,
$\frac{\omega \cos b}{s^2+\omega^2}+\frac{s\sin b}{s^2+\omega^2}$

**ThePerfectHacker** · August 28th 2006, 06:21 PM

Originally Posted by action259

$cosh^23t$

The problem is that you have a square. However, you can eliminate it.
Use the following identity,
$\cosh^2 x=\frac{1+\cosh 2x}{2}$
This will eliminate the square and turn it into one of the forms on the table.
(If you want I can prove that identity. The reason why I am saying this is because my Calculus book does not contain this identity on its treatise on hyperbolic functions. Thus, I assumed might be unfamiliar).

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