How to verify a Laplace transform?

tibetan-knight · #1 $Old$ November 4th, 2009, 01:43 PM

$\mathcal{L}\left\{e^{at}\cos\!\left(kt\right)\right\}=\frac{s-a}{\left(s-a\right)^2+k^2}$

What is the proof of this? Or rather, could somebody get me started and point me in the right direction for this?

Thanks! Any help is appreciated.

chisigma · #2 $Old$ November 4th, 2009, 01:50 PM

One possible start is the computation of the integral...

$\mathcal{L} \{\cos kt\} = \int_{0}^{\infty} \cos kt\cdot e^{-st}\cdot dt$

... that can be done by parts...

Kind regards

$\chi$ $\sigma$

tibetan-knight · #3 $Old$ November 4th, 2009, 02:15 PM

But what do you do with the e^at? Does that fit somewhere in the integral or no?

TheEmptySet · #4 $Old$ November 4th, 2009, 03:48 PM

Quote:

Originally Posted by tibetan-knight $View Post$

But what do you do with the e^at? Does that fit somewhere in the integral or no?

yes...

$\int_{0}^{\infty}e^{at}e^{-st}cos(kt)dt=\int_{0}^{\infty}e^{-(s-a)t}cos(kt)dt$

Now just use integration by parts on this

Or note in general that

$\mathcal{L}(e^{at}f(t))=\int_{0}^{\infty}e^{at}e^{-st}f(t)dt$

$\mathcal{L}(e^{at}f(t))=\int_{0}^{\infty}e^{-(s-a)t}f(t)dt$

Let $s-a=m$ then we get

$\mathcal{L}(e^{at}f(t))=\int_{0}^{\infty}e^{-mt}f(t)dt=F(m)$
The above integral is the form of the laplace transform so we get
Where $F(m)=F(s-a)$

This is commonly know as the s-axis shift theorem.

So the above would give

$\mathcal{L}(e^{at}cos(kt))=\mathcal{L}(\cos(kt))\bigg|_{s\to s-a}=\frac{s}{s^2+k^2}\bigg|_{s \to s-a}=\frac{s-a}{(s-a)^2+k^2}$

tibetan-knight · #5 $Old$ November 4th, 2009, 05:02 PM

Ooh. Shift theorem is helpful. Thank you.

But, to integrate the former by parts, which would be the 'u' part and which would be the 'dv' part?