Help with Inverse Laplace transform problem

guitarguy12387

Joined Apr 10, 2008
359
Any ideas how to find the inverse laplace of this:

4s/(s+4)^2

If i try using partial fractions, i only get two of the same equations. I am thinking convolution will be the best method. Any help would be great! Thanks.

Mark44

Joined Nov 26, 2007
628
Yeah, you need to do something different in partial fraction decomposition when you have repeated factors. Try a decomposition like this:
A/(s + 4) + B/(s + 4)^2 = 4s/(s + 4)^2

Mark

Mark44

Joined Nov 26, 2007
628
To generalize this, if you have p(s)/(s + a)$$^{n}$$, where p(s) is a polynomial of degree < n, the decomposition looks like this:

A$$_{1}$$/(s + a) + A$$_{2}$$/(s + a))$$^{2}$$ + ... +A$$_{n}$$/(s + a)$$^{n}$$

You would then set this expression equal to p(s)/(s + a)$$^{n}$$ to find the constants A$$_{1}$$, A$$_{2}$$, ... , A$$_{n}$$.

Mark

guitarguy12387

Joined Apr 10, 2008
359
Hey thanks for the help Mark. I did figure out how to do that correctly (i think). However, i'm still not getting the correct answer for the whole problem... :/

The original problem:

x" + 8x' +16x = delta(t) - delta(t-3), x(0) = 4, x'(0) = 1

I am supposed to solve this IVP using Laplace methods.

So... L[x] = X(s), L[x'] = sX(s) - 4, L[x"] = s^2*X(s) - 4s - 1, L[delta(t)] = 1, L[delta(t-3)] = e^(-3s)

then, plugging those into the original equation...

s^2*X(s) - 4s - 1 + 8s*X(s) - 24 + 16X(s) = 1 - e^(-3s)
=> X(s)*(s^2 +8s +16) = 1-e^(-3s)+4s+25
=> X(s) = (26+4s-e^(-3s))/(s+4)^2

Ok... hopefully i haven't lost ya yet

So then i take the inverse laplace's of this:

L^-1[26/(s+4)^2] = 26te^(-4t)
L^-1[e^(-3s)/(s+4)^2]
L^-1[4s/(s+4)^2] = 4e^(-4t)-4te^(-4t)

The last inverse transform there i got from the partial fraction method you described.

So my final answer for x(t) should be the sum of those three inverse laplace transforms. However, this is not the correct answer and i'm not sure where i went wrong. I know this one's kinda long winded/hard to read so it's cool if you don't want to read it all... just thought i'd give it a shot. Thanks a lot in advance!

Mark44

Joined Nov 26, 2007
628
It's getting late my time, and I don't see anything obviously wrong, so I'll have to wait until tomorrow to dive into this. It looks to me like you have all the basic ideas, so if you're not getting the right answer, it might be that there's an algebraic error somewhere. I'll take a look tomorrow...

Mark44

Joined Nov 26, 2007
628
x" + 8x' +16x = delta(t) - delta(t-3), x(0) = 4, x'(0) = 1

s^2*X(s) - 4s - 1 + 8s*X(s) - 24 + 16X(s) = 1 - e^(-3s)
=> X(s)*(s^2 +8s +16) = 1-e^(-3s)+4s+25
=> X(s) = (26+4s-e^(-3s))/(s+4)^2
The number in bold should be -32 ( from -8 * x(0) = -8 * 4 ). I haven't checked the rest, but if you redo your work from there, that will give you a good start.
Mark

guitarguy12387

Joined Apr 10, 2008
359
hah wow. Good call. I'll try that and see how it works. Thanks again, Mark!

guitarguy12387

Joined Apr 10, 2008
359
hmm... so i redo that work and end up with 34 as the coefficient in front of the first inverse laplace transform instead of 26, but the online grader still isn't liking my answer.

I also think i made an error on the second transform. I believe (although i dont have my book on me) it should be:

L^-1[e^(-3s)/(s+4)^2] = step(t-3)(t+3)*e^(-4(t+3))

neither of which worked. I am going to recheck that transform when i get back home though.

scubasteve_911

Joined Dec 27, 2007
1,203
Just a hint for you if you didn't already know, you can use MATLAB to do inverse laplace. Sometimes you'll get a whacky time domain answer with cosh, etc, but sometimes it works well.

Steve

Mark44

Joined Nov 26, 2007
628
I believe (although i dont have my book on me) it should be:

L^-1[e^(-3s)/(s+4)^2] = step(t-3)(t+3)*e^(-4(t+3))

neither of which worked.
I like your initial answer, at least for L$$^{-1}$$ (e^(-3s))/(s + 4)$$^{2}$$. What happened to the other terms? I don't have the work I did in front of me, but I remember the left side looked something like this: X(s)*(s + 4)$$^{2}$$ + other terms in s. And I think I remember the right side had two terms. Have you accounted for all of these?
Mark

payenk

Joined Apr 11, 2008
3
can anyone solve my problem?it is too difficult for me...
inverse laplace for (As^2+Bs+C)/s(Ds^4+Es^3+Fs^2+Gs+H)

guitarguy12387

Joined Apr 10, 2008
359
Ya thanks scubasteve. I am trying to make sure i know this stuff for an exam coming up soon otherwise i would just use matlab. Thanks though.

I remember the left side looked something like this: X(s)*(s + 4)
$image=http://forum.allaboutcircuits.com/mimetex.cgi?%5E%7B2%7D&hash=04aa9b4f4516e2aaf0faa3829caf329a$

This is what i ended up with for X(s)

=> X(s)*(s^2 +8s +16) = 1-e^(-3s)+4s+33
=> X(s) = (34+4s-e^(-3s))/(s+4)^2

which i broke up into 34/(s+4)^2 + 4s/(s+4)^2 - e^(-3s)/(s+4)^2
I then took the inverse laplace of each of those individually and sum them together to get a total expression for x(t)

guitarguy12387

Joined Apr 10, 2008
359
can anyone solve my problem?it is too difficult for me...
inverse laplace for (As^2+Bs+C)/s(Ds^4+Es^3+Fs^2+Gs+H)
What have you done so far? Have you tried using partial fractions?

Mark44

Joined Nov 26, 2007
628
This is what i ended up with for X(s)

=> X(s)*(s^2 +8s +16) = 1-e^(-3s)+4s+33
=> X(s) = (34+4s-e^(-3s))/(s+4)^2

which i broke up into 34/(s+4)^2 + 4s/(s+4)^2 - e^(-3s)/(s+4)^2
I then took the inverse laplace of each of those individually and sum them together to get a total expression for x(t)
That's what I got, too, starting from the original 2nd order DE + initial conditions. If this answer doesn't agree with the "official" answer, here are some things to think about:
1. Are you working the right problem? Make sure you have exactly the same DE and initial conditions.
2. Check your solution to see if it satisfies both intial conditions and the DE. Keep in mind that $$\delta$$(t) is just a fancy way of saying 1 if t > 0 and 0 otherwise (don't remember if $$\delta$$(0) is 0 or 1), and that its derivative is zero.
3. If your solution satisfies the DE and initial conditions, then congratulations. If it's different than the solution that's the "official" one, then the "official" one might be incorrect.

Mark

guitarguy12387

Joined Apr 10, 2008
359
Well thanks for all the help, Mark. As far as i can tell, the answer seems right and i cant find any mistakes, but the online grader still won't take it. So we'll see in a few days when they post solutions what the correct answer was :/ Thanks again for your time.

Brian

payenk

Joined Apr 11, 2008
3
i do not know how to use partial fraction coz the denominator is in higher order and no values are given...
can you tell me how to do that?
tq

Mark44

Joined Nov 26, 2007
628
can anyone solve my problem?it is too difficult for me...
inverse laplace for (As^2+Bs+C)/s(Ds^4+Es^3+Fs^2+Gs+H)
I don't see that a partial fractions approach will do any good, since we don't have any of the eight coefficients. Without the coefficients of the quartic factor in the denominator, I don't see that factorization is feasible.
Can you give us more information? I'm assuming this is a homework problem or a problem in a textbook. Are there any hints that go with this problem? Any other information?

With what you've provided I don't see any way of finding the inverse Laplace transform.
Mark

Mark44

Joined Nov 26, 2007
628
As far as i can tell, the answer seems right and i cant find any mistakes, but the online grader still won't take it.
I guess you can't argue with an online grader

Don't forget, though, that you can check pretty easily that what you found is a solution if x(0) and x'(0) satisfy the initial conditions, and if x(t), x'(t), and x''(t) satisfy the DE. IOW, x''(t) + 8x'(t) + 16x(t) should add to 0 for t < 0, to 1 for 0 <= t <= 3, and to 0 for t > 3.

payenk

Joined Apr 11, 2008
3
thanx 4 the reply mark...but this is the only equation that given to me...no hint provided...