1. ## Laplace Transforms

find the laplace transform F(s) of f(t) = sin wt

the w is actually supposed to be omega

what i got was

integeral from 0 to infinity of: (sin wt)*(e^-st) dt

then i integrated by parts and obtained a large expression
which involves the limit to infinity of sin and cos ... how do i solve this limit
can someone post a step by step explanation starting from the limit part

2. L{sin(wt)} = w/ (s^2 + w^2)

It's a formula.. I didn't do the integration.

3. really its a formula
but i think i have to show it because i haven't yet learned that formula
is it hard to show or is there some easy way to show it???

4. Originally Posted by razorfever
really its a formula
but i think i have to show it because i haven't yet learned that formula
is it hard to show or is there some easy way to show it???
$\displaystyle \int_0^{\infty}e^{-st}\sin\!\left(\omega t\right)\,dt$

By parts, let $\displaystyle u=\sin\!\left(\omega t\right)$ and $\displaystyle \,dv=e^{-st}\,dt$. Thus, $\displaystyle \,du=\omega\cos\left(\omega t\right)\,dt$ and $\displaystyle v=-\frac{1}{s}e^{-st}$

Thus, we have $\displaystyle \int_0^{\infty}e^{-st}\sin\!\left(\omega t\right)\,dt=\left.\left[-\frac{1}{s}e^{-st}\sin\!\left(\omega t\right)\right]\right|_0^{\infty}+\frac{\omega}{s}\int_0^{\infty} e^{-st}\cos\!\left(\omega t\right)\,dt$ $\displaystyle =\frac{\omega}{s}\int_0^{\infty}e^{-st}\cos\!\left(\omega t\right)\,dt$, since $\displaystyle \lim_{t\to\infty}e^{-st}\sin\!\left(\omega t\right)=0$ (You can take note that sine oscilates bewteen -1 and 1, but $\displaystyle e^{-st}$ approaches zero very quicly. Thus its the term we are most concerned with when evaluating the limit.)

Now, let $\displaystyle u=\cos\!\left(\omega t\right)$ and $\displaystyle \,dv=e^{-st}\,dt$. Thus, $\displaystyle \,du=-\omega\sin\!\left(\omega t\right)\,dt$ and $\displaystyle v=-\frac{1}{s}e^{-st}$

Thus, we now see that $\displaystyle \int_0^{\infty}e^{-st}\sin\!\left(\omega t\right)\,dt=\frac{\omega}{s}\left[\left.\left[-\frac{1}{s}e^{-st}\cos\!\left(\omega t\right)\right]\right|_0^{\infty}-\frac{\omega}{s}\int_0^{\infty}e^{-st}\sin\!\left(\omega t\right)\,dt\right]$ $\displaystyle \implies \left(1+\frac{\omega^2}{s^2}\right)\int_0^{\infty} e^{-st}\sin\!\left(\omega t\right)\,dt = \frac{\omega}{s^2}$ since $\displaystyle \lim_{t\to{\infty}}e^{-sy}\cos\!\left(\omega t\right)=0$ by similar explanation above. Also note that $\displaystyle \lim_{t\to0}e^{-st}\cos\!\left(\omega t\right)=1$.

Thus, it is now evident that

$\displaystyle \left(1+\frac{\omega^2}{s^2}\right)\int_0^{\infty} e^{-st}\sin\!\left(\omega t\right)\,dt = \frac{\omega}{s^2}\implies\frac{s^2+\omega^2}{s^2} \int_0^{\infty}e^{-st}\sin\!\left(\omega t\right)\,dt = \frac{\omega}{s^2}$ $\displaystyle \implies \int_0^{\infty}e^{-st}\sin\!\left(\omega t\right)\,dt = \frac{s^2}{\omega^2+s^2}\cdot\frac{\omega}{s^2}$ $\displaystyle \implies \color{red}\boxed{\mathcal{L}\left\{\sin\!\left(\o mega t\right)\right\}=\int_0^{\infty}e^{-st}\sin\!\left(\omega t\right)\,dt=\frac{\omega}{s^2+\omega^2}}$

Keep in mind this is only true when $\displaystyle s>0$.

Does this make sense?

5. thanks for that explanation i almost made it to the end but i just had a hard time evaluating the limit for sin

the second part is:

show that

L[e^at * sin wt] = F(s-a)

how do i get started on this

and how do i get the answer in the form of F()???

6. Originally Posted by razorfever
thanks for that explanation i almost made it to the end but i just had a hard time evaluating the limit for sin

the second part is:

show that

L[e^at * sin wt] = F(s-a)

how do i get started on this

and how do i get the answer in the form of F()???
Recall that $\displaystyle \mathcal{L}\left\{f\!\left(t\right)\right\}=\int_0 ^{\infty}e^{-st}f\!\left(t\right)\,dt=F\!\left(s\right)$

In general, if you are asked to find $\displaystyle \mathcal{L}\left\{e^{at}f\!\left(t\right)\right\}$, you should see that $\displaystyle \mathcal{L}\left\{e^{at}f\!\left(t\right)\right\}= \int_0^{\infty}e^{-st}e^{at}f\!\left(t\right)\,dt=\int_0^{\infty}e^{\ left(a-s\right)t} f\!\left(t\right)\,dt$ $\displaystyle =\int_0^{\infty}e^{-\left(s-a\right)t}f\left(t\right)\,dt=F\!\left(s-a\right)$.

All you're really doing is manipilating the definition of the Laplace Transform.

Does this make sense?

7. i don't undertand the last step ... can you go into more detail ...
i really want to undertand this and my prof didn't go into much detail over this last week

8. Originally Posted by razorfever
i don't undertand the last step ... can you go into more detail ...
i really want to undertand this and my prof didn't go into much detail over this last week
The last step is by definition of the Laplace Transform.

If what I'm answering is not what you're asking then it would help if you quoted what it is that you're refering to.

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# laplace transforms =sin(wt)

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