Thread: Simple Laplace Transform w/ Final Value Theorem

1. Simple Laplace Transform w/ Final Value Theorem

Help! This is the last problem on my math homework and we are starting a section on Fourier transforms which I understand, but we never learned Laplace transforms. My book doesn't give any examples of how do a problem similar to this one and I can't find any close solved examples on the internet. I only have to do three of these so I'm posting the one I could learn the most from:

about the only think I found on the internet is that the final value theorem is lim y(t)=lim Sy(s) which at this point doesn't help because I don' know how to do the rest of the problem.

Any steps to this problem are greatly appreciated,
Evan

2. Originally Posted by 360modina
Help! This is the last problem on my math homework and we are starting a section on Fourier transforms which I understand, but we never learned Laplace transforms. My book doesn't give any examples of how do a problem similar to this one and I can't find any close solved examples on the internet. I only have to do three of these so I'm posting the one I could learn the most from:

about the only think I found on the internet is that the final value theorem is lim y(t)=lim Sy(s) which at this point doesn't help because I don' know how to do the rest of the problem.

Any steps to this problem are greatly appreciated,
Evan
Why not use the transfer function?

Remember that:

$\mathcal L({f''}) = s^{2}F(s) - sf(0) - f'(0)$
$\mathcal L({f'}) = sF(s) - f(0)$

The transfer function, $Q(S)$, is a solution of the equation and it is given by:

$\frac{Y(s)}{R(s)}$

EDIT: In fact, with your initial conditions, you don't need to consider the transfer function, you just need the above! ^^^

3. L[y''] = L [(y')'] = p L [y'] - y'(o) = p^2 L[y] - p y(o) - y'(0)
Then you're plugging it in and solve for y using L^-1

4. I was trying to solve this problem, as I've studied this before and need to reivew for the final. I also got stuck.

on the right hand side:
L [2r(t)] = 2 L [ r(t) ] = 2 L [ y/ sint ] can some body help? L [ y/sint ] = d/dp L [ (sint)^-1] = ???

5. Originally Posted by tennis14321
I was trying to solve this problem, as I've studied this before and need to reivew for the final. I also got stuck.

on the right hand side:
L [2r(t)] = 2 L [ r(t) ] = 2 L [ y/ sint ] can some body help? L [ y/sint ] = d/dp L [ (sint)^-1] = ???
Where are you getting sine from?!

$\mathcal L (\frac{d^2y}{dt^2}+4\frac{dy}{dt}+3y)=\mathcal L (2r(t))$

$s^{2}Y(s)-sy(0)-y'(0) +4sY(s)-y(0)+3Y(s)=2R(s)$

We know that $y(0) = 1, y'(0) = 0, r(t) = 1.$

If $r(t) = 1$, then $R(s) = \mathcal L (1) = \frac{1}{s}$. Plug these all in:

$s^{2}Y(s)-s+4sY(s)-1+3Y(s)=2(\frac{1}{s})$

$Y(s)(s^{2}+4s+3)-s-1=\frac{2}{s}$

$Y(s)(s^{2}+4s+3)=\frac{2}{s} + s +1$

$Y(s)(s^{2}+4s+3)=\frac{2+s^2+s}{s}$

$Y(s)=\frac{s^2 +2+s}{s}\times \frac{1}{s^{2}+4s+3}$

Multiply it out and use partial fractions to get it into a recognisable form...

6. Originally Posted by Mush

$Y(s)=\frac{s^2 +2+s}{s}\times \frac{1}{s^{2}+4s+3}$

Multiply it out and use partial fractions to get it into a recognisable form...
$Y(s)=\frac{1}{s^2+4s+3} + \frac{2}{s(s^2+4s+3)} + \frac{s}{s^2+4s+3}$

$Y(s)=\frac{1}{(s+1)(s+3)} + \frac{2}{s(s+1)(s+3)} + \frac{s}{(s+1)(s+3)}$

That was extremely helpful, do you mind taking all the way through the final value theorem, Mush you are good at this.

7. Originally Posted by 360modina
$Y(s)=\frac{1}{s^2+4s+3} + \frac{2}{s(s^2+4s+3)} + \frac{s}{s^2+4s+3}$

That was extremely helpful, do you mind taking all the way through the final value theorem, Mush you are good at this.
Final value theorem states:

$\lim_{s \to 0} sF(s) = \lim_{t \to \infty} f(t)$

Or, in our case:

$\lim_{s \to 0} sY(s) = \lim_{t \to \infty} y(t)$

$Y(s)=\frac{1}{s^2+4s+3} + \frac{2}{s(s^2+4s+3)} + \frac{s}{s^2+4s+3}$

Multiply this by s to get sY(s):

$sY(s)=\frac{s}{s^2+4s+3} + \frac{2}{(s^2+4s+3)} + \frac{s^2}{s^2+4s+3}$

As s tends towards zero, the first and third terms of this equation disappear, because there is s and s^2 on the numerator. However, the 2nd term does not disappear! The s^2 and 4s in the denominator disappear leaving only 3 in the denominator, and the 2 in the numerator remains.

Hence:

$\lim_{s \to 0} sY(s) = \lim_{t \to \infty} y(t) = \frac{2}{3}$

8. ohh,
r(t) = 1
I was thinking y = r sin (t)

Ok. it's easier than i thought