# Laplace Transform

1. May 26, 2005

### splitendz

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

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

I realise 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 :)

2. May 26, 2005

### 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.

3. May 26, 2005

### 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?

4. May 27, 2005

### dextercioby

5. May 27, 2005

### Corneo

Are you trying to compute

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

6. May 27, 2005

### splitendz

The question that i'm asking is:

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

Heaviside function definition is:

$$h(t - a)=\left\{\begin{array}{cc}0,&\mbox{ if } t< a\\1, & \mbox{ if } t\geq a\end{array}\right.$$

I hope this makes it clearer to understand...

7. May 27, 2005

### dextercioby

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

Daniel.

8. May 27, 2005

### 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 :)

9. May 27, 2005

### dextercioby

I don't see any problem

$$f(t)=\left\{\begin{array}{c}0 \ \mbox{if} \ t<0\\ -\sin 3t \ \mbox{if} \ t\in\left[0,\frac{\pi}{2}\right) \\ 0 \ \mbox{if} \ t>\frac{\pi}{2} \end{array}\right$$

Daniel.

10. May 27, 2005

### 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?

11. May 27, 2005

### 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.

12. May 28, 2005

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).

13. May 28, 2005

### dextercioby

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

Daniel.

14. May 28, 2005

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?

15. May 28, 2005

### dextercioby

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

Daniel.

16. May 29, 2005