Solving Laplace Transform of sin[3t] ( h[t - pi/2] - h[t] )

[splitendz]

Hi. I'm trying to find the Laplace transform of:

sin[3t] ( h[t - pi/2] - h[t] )

I realize that this function is only valid for 0 <= t <= pi/2.

Since the limit is not from zero to infinity (0 to pi/2 instead) how can I use the laplace tranform definition to solve this?

Also, does anyone know how to insert mathematical symbols in your post?

Thanks :)

 

[dextercioby]

Yes,use the LaTex code.Type the code using [ tex ] [ /tex ] (without the spacings) and the compiler will do the rest.

That "h" is Heaviside's distribution,right...?

Daniel.

 

[splitendz]

Yes, the h is the heaviside function. The correct answer to the problem is -1/(s^2 + 9) * (3 + se^( (pi * s) / 2 ). I can get -3/(s^2 + 9) but not sure how they get (3 + se^( (pi * s) / 2 ) for the next term.

Do you know where I can get a list of LaTex commands?

 

[dextercioby]

Check out this thread and write everything again. https://www.physicsforums.com/showthread.php?t=8997

Daniel.

 

[Corneo]

Are you trying to compute

\mathcal{L}\left\{ \sin (3t) [u_{t - \pi}(t) - u_0(t)] \right\}?

 

[splitendz]

The question that I'm asking is:

\mathcal{L}\left\{ \sin (3t) [ h(t - \pi/2) - h(t) ] \right\}

Heaviside function definition is:

h(t - a)=\left\{\begin{array}{cc}0,&amp;\mbox{ if }<br /> t&lt; a\\1, &amp; \mbox{ if } t\geq a\end{array}\right.

I hope this makes it clearer to understand...

 

[dextercioby]

Isn't that \mathcal{L}\left(-\sin 3t\right) with t\in\left[0,\frac{\pi}{2}\left)...?

Daniel.

 

[splitendz]

Yes, it is but it's the next step that confuses me.

Laplace transform is defined as

\int_{0}^{\infty} e^{-st} f(t) dt


Since we have a different limit how do we transform the function \mathcal{L}\left(-\sin 3t\right) for t\in\left[0,\frac{\pi}{2}\left)? I've only been learning laplace for 2 days so I'm a bit vauge, sorry if I'm asking a boring question :)

 

[dextercioby]

I don't see any problem

f(t)=\left\{\begin{array}{c}0 \ \mbox{if} \ t&lt;0\\ -\sin 3t \ \mbox{if} \ t\in\left[0,\frac{\pi}{2}\right) \\ 0 \ \mbox{if} \ t&gt;\frac{\pi}{2} \end{array}\right

Daniel.

 

[splitendz]

I understand that part. It's the actual integral that I need to evaluate that I'm not sure about.

-\int_{0}^{\pi / 2} e^{-st} \sin 3t dt Can i just do that or must the integration limits be between zero and inifinity to transform the function?

 

[dextercioby]

The branch function is defined on R and splitting the integral on R into 3 integrals on each interval,you'll run into that integral.


Daniel.

 

[LeBrad]

Laplace transform is defined as

\int_{0}^{\infty} e^{-st}f(t)dt

If you really want to integrate from 0 to inf, I think you could rewrite

\sin(3t)u(t-\frac{\pi}{2})=-\cos(3(t-\frac{\pi}{2})u(t-\frac{\pi}{2})

Then use the time shifting property which says that

\mathcal{L}(f(t-t_0))=e^{-st_0}\mathcal{L}(f(t))

Now you just have to transform sin(3t) and cos(3t).

You could probably also use the differentiation property and time scaling so you would only have to actually transform sin(t).

 

[dextercioby]

I'm afraid it doesn't work,unless that "3" was missing from the argument of cosine.

Daniel.

 

[LeBrad]

Oops, my mistake.

Why is that though? It doesn't seem like messing with the time scaling of the step function should matter provided they both turn on at the same time. Isn't u(5t) the same as u(t). Actually, does u(5t) even make any sense?

 

[dextercioby]

Yes,i guess you're right.The function should be the same,as i depicted in post #9.

Daniel.

P.S.Just wondering,what's your brother's name...?

 

[splitendz]

Thanks for your help :]

 

[LeBrad]

P.S.Just wondering,what's your brother's name...?

When pronounced incorrectly, his name is similar to that of a certain throat lozenge.