Thread: Laplace transform with heaviside step function

1. Laplace transform with heaviside step function

Hi Everyone,

I'm a little confused with this problem so I just wanted to check my solution. I have to find the Laplace Transform of the diff eq:

d^2y/dt^2+2dy/dt+2y=x(t)
where x(t) = 1, 0≤t≤1;
2, t≥2;
0 elsewhere
y(0)=0, y'(0)=1

I got the following:

Y=1/(s^2+2s+2)=1/(s+1)^2
Y=te^-t

The heaviside function is:
e^-s/s +2e^-2s/s

so would the final solution be:

te^-t=e^-s/s +2e^-2s/s??

Thanks for any help!!

2. So if $H(t)$ is the Heaviside function defined by

$\displaystyle H(t)=\begin{cases}1,\;t\ge 0\\0,\;t<0\end{cases},$ then you could write your RHS function

$x(t)=H(t-1)+H(t-2).$

(Technically, you need to use a strict inequality for one or the other of your function pieces for $t=2$ - the way I've done it assigns $x(2)=2.$)

So now your DE appears as

$y''(t)+2y'(t)+y(t)=H(t-1)+H(t-2),$ subject to $y(0)=0$ and $y'(0)=1.$

So now we take the LT of both sides:

$\displaystyle s^{2}Y(s)-sy(0)-y'(0)+2\left[sY(s)-y(0)\right]+Y(s)=\frac{e^{-s}}{s}+\frac{e^{-2s}}{s},$ or

$\displaystyle s^{2}Y(s)-1+2sY(s)+Y(s)=\frac{e^{-s}+e^{-2s}}{s}.$

I'm not sure you did your LT correctly, especially the RHS.

Does this make sense?

[EDIT] There are corrections to both the RHS and LHS below.

3. Re:Laplace transform with heaviside step function

Well I seem to have a similar transform of the RHS as you but the LHS differs. The LHS seems incomplete? Is that considered the full solution to the differential equation?

4. All I did was take the LT of both sides. I have not yet solved the new equation for $Y(s).$

Your LT and mine are actually quite different and will lead to different solutions.

Did you follow how I got my LT?

5. Yes, I follow how you got that LT. Should there be a 2 in front of the Y(s) term on the LHS?

6. No. I take it you mean the last occurrence of $Y(s)$ on the LHS, right? The one with no $s$'s of any power multiplying it, right? The reason that should not have a $2$ multiplying it is that there is no $2$ multiplying the corresponding $y(t)$ term from which we got the $Y(s)$ term via the LT. The only $2$ that shows up is the one multiplying the $y'(t)$ term, and you can see that I've accounted for that.

7. The original equation has a 2 in front of the y term as well as the y' term. My understanding is that this will stay through the Laplace transform like the 2 in front of the y' term. Once I do that I end up with 1/(s^2+2s+2)=1/(s+1)^2 which there does not seem to be an obvious transform for in the table. It is this part of which I'm a little unsure.

8. Hmm. I didn't notice that 2 there before. The LT is linear, so any constant multiplying the y(t) will end up multiplying the Y(s), as you say.

Your LT is not correct. You have to have some exponentials in there, due to the Heaviside step functions.

My (corrected) LT of the ODE is now

$\displaystyle s^{2}Y(s)-1+2sY(s)+2Y(s)=\frac{e^{-s}+e^{-2s}}{s}.$ Therefore,

$\displaystyle Y(s)(s^{2}+2s+2)=1+\frac{e^{-s}+e^{-2s}}{s}.$ Hence,

$\displaystyle Y(s)=\frac{s+e^{-s}+e^{-2s}}{s(s^{2}+2s+2)}.$

[EDIT]: RHS is incorrect. See below for corrections.

9. Well yes I understand I still need the exponentials from the RHS I was just trying to get the LHS correct. Do I now turn to the Laplace tables to find the transform of this function? Stupid question, I know, but this is where I get confused. I would generally transform the 's' functions but a transform of the exponential functions will only take me back to a heaviside function.

10. I would probably factor the denominator, and then use theorems about the inverse LT in order to find the inverse. I would also split up the numerator so that you get three separate fractions. What do you get when you do that?

11. Apologies for butting in on what is almost a private conversation, but before you go any further could you check on the expression on the RHS ?
Shouldn't the step function representation be 1 - H(t-1) + H(t-2) ?

12. From an example in my book I can see that the e terms drop away in the inverse...it appears that way anyway. If that's the case I can break it up into the following fractions:

L^-1{1/s} + L^-1{1/(s+1)}+L^-1{s/s+1}
=1*e^-t*cos(-t) (or something close to this)

13. Reply to BobP @ 11:

Your function would be $1$ for $t < 0$, $0$ for $0 \le t < 2$, and $1$ for $t \ge 2$. That is not the function described in the OP. You're correct, however, in pointing out that my function doesn't work, either. The correct function is

$x(t)=H(t)+H(t-1).$ Its plot is here. Your function is plotted here.

So this changes the LT as follows:

$\displaystyle s^{2}Y(s)-1+2sY(s)+2Y(s)=\frac{1}{s}+
\frac{e^{-s}}{s}.$

14. So the inverse would now be:
L^-1{1/s^2} + L^-1{1/(s+1)}+L^-1{s/s+1}

= t*e^-t*(cos(-t)?)

it appears i was missing the last term in my original solution for the inverse

15. I have no idea how you arrived at your line:

L^-1{1/s^2} + L^-1{1/(s+1)}+L^-1{s/s+1}.

You skipped way too many steps there, and your final answer is incorrect. I don't think the intermediate step was correct, either. Exponentials don't just "go away" in inverse LT's. They turn into unit step functions that ought to be present in your final answer.

Take it from here, and work step-by-step, and show your work:

$\displaystyle Y(s)(s^{2}+2s+2)=1+\frac{1}{s}+
\frac{e^{-s}}{s}.$

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