Homework Help: A very basic laplace transform problem (picture included)

1. Sep 4, 2011

saiparcha

Hi..new here. I'm having trouble finishing off a very basic laplace transform problem. I'd greatly appreciate it if someone could show me the way.

The problem is #5 from this picture: http://www.flickr.com/photos/67208468@N08/6111476312/

What I did was use the Euler's formula for sin t and I got to this: 1/i * (-s2e-pi*s+i+e-pi*s)/s2+1

I always kinda get confused typing math/equations, so I apologize if what i described wasn't clear. Thanks for any help.

2. Sep 4, 2011

lanedance

i would probably just write it as a combination of sin, say u(t) is the unit step function
$$f(t) = u(t) sin(t) - u(t-\pi)sin(t)$$

most laplace transform tables will have entries for all these terms, or do you need to perform the integral yourself?

3. Sep 4, 2011

saiparcha

We haven't been taught the step function stuff yet, so I've never seen that equation before but I'm sure the professor meant for us to solve this without tables. It's just been one lecture so far, so we haven't seen the tables you're talking about either.

We have to do the integral, yes. Thanks for the response.

4. Sep 4, 2011

lanedance

ok so you just need to perform

$$F(s) = L\{f(t)\} = \int_0^{\infty}e^{-st}f(t)dt = \int_0^{\pi}e^{-st}sin(t)dt$$

can you show your steps for the integration and we'll try and find where things are falling over

5. Sep 4, 2011

saiparcha

Last edited by a moderator: Apr 26, 2017
6. Sep 4, 2011

TheBestMilke

You may have figured it out by now, but in case you haven't this might help a little. The case you have is a little bit tricky, and you have to solve it similarly to the situation e^t*sint back in calc 2.

F(t) = $\int$$^{\pi}_{0}$e-stsin(t)dt = $\frac{-1}{s}$e-stsin(t) + $\frac{1}{s}$$\int$$^{\pi}_{0}$e-stcos(t)dt
In that we used integration by parts with u = sin(t) and dv = e-st. Now, we see that the last term is $\int^{\pi}_{0}$e-stcos(t)dt. If we tried to integrate that we would come back to our original equation:
F2(t) = $\int$$^{\pi}_{0}$e-stcos(t)dt = $\frac{-1}{s}$e-stcos(t) - $\frac{1}{s}$$\int$$^{\pi}_{0}$e-stsin(t)dt

From Here we can make a substitution:

F(t) = $\int$$^{\pi}_{0}$e-stsin(t)dt = $\frac{-1}{s}$e-stsin(t) + $\frac{1}{s}$($\frac{-1}{s}$e-stcos(t) - $\frac{1}{s}$$\int$$^{\pi}_{0}$e-stsin(t)dt)

(1+$\frac{1}{s^{2}}$)$\int^{\pi}_{0}$e-stsin(t)dt = $\frac{-1}{s}$e-stsin(t) - $\frac{1}{s^{2}}$e-stcos(t))

$\int^{\pi}_{0}$e-stsin(t)dt = ($1/(1+\frac{1}{s^{2}})$)($\frac{-1}{s}$e-stsin(t) - $\frac{1}{s^{2}}$e-stcos(t))

You'll have to include the bounds and solve, of course, but that's the general breakdown of how you'd want to tackle that type of problem. Good luck!

7. Sep 4, 2011

saiparcha

BestMilke, appreciate it a lot. If no one can do it using the Euler's formula, then I'll do it the way you did.

8. Sep 4, 2011

Ray Vickson

Use brackets around the denominator: (1+ exp(-s*pi))/(s^2+1). I got the correct answer, so I cannot tell what you did wrong.

RGV

9. Sep 4, 2011