# Laplace transform

Discussion in 'Homework Help' started by ihaveaquestion, Oct 12, 2009.

1. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
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everything included here:

Last edited: Oct 12, 2009
2. ### steveb Senior Member

Jul 3, 2008
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It looks like you are trying to find the Laplace transform of a function g(t) equal to another function f(t) divided by time: i.e. L{g(t)}=L{f(t)/t}.

Since you know the Laplace transform of f(t), you are using the known integration property; which is, division by time, in the time domain, corresponds to integration over s, in the s-domain.

The only thing is that the final answer does not look right to me. Did you mean: arctan(s/B) instead of arctan(B/s)? I only looked very quickly, so perhaps I'm wrong.

3. ### notxjack New Member

Sep 7, 2009
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Are you asking us how to integrate that last term?

How familiar are you with complex analysis?

4. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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steveb, I'm pretty sure that's the answer (I found it in another laplace table online also).

notxjack, no I'm not familiar with complex analysis... only how to do normal complex number algebra stuff.

5. ### steveb Senior Member

Jul 3, 2008
2,433
469
Ah ok, I think I see why it is correct. So which part is not making sense to you? Is the method you showed your own work and are you just trying to get to the final answer?

Using an integration table for the last step, I get the following.

pi/2-arctan(s/B)

however there is a trig identity which shows this to be equal to

arctan(B/s)

Is this the confusion you are having? If not, please explain what the issue is.

6. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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Yes, I just had no idea how to integrate that last term I wrote manually... not sure if that's common knowledge though.

7. ### gregcoll New Member

Oct 11, 2009
21
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See if you can follow this, it seems more difficult than it is:
1. take a B from the denominator and you will have 1/B*int[1/(1+(S/B)^2)*dS]
2. make the substitution-> S/B = tan (x), then dS = B*sec^2(x)*dx
3. using the identity -> 1+tan^2 = sec^2, you should get simply 1*int[1*dx] (don't worry about the limits for now, always change back to your original variables when doing this to avoid missing any stray solutions!)
4. the answer to this integral is simply -> x
5. now, since S/B = tan (x) -> x = arctan(S/B)
6. THIS is evaluated at the original limits of s to positive infinity which gives the previously given solution of pi/2 - arctan(S/B)m\, or -> arctan(B/S)
NOTE: (arctan(infinity) must mean that the denominator of the tangent function has a value of zero and since tangent is sin/cos, the arc with cos=0 is at pi/2 and -pi/2 ergo arctan(infinity) is at the arclength of pi/2)
7. use the fact that if a = arctan(x) then x = tan(a) and sin(-x) = -sin(x) and cos(-x) = cos(x) (and maybe some other identities) if you want to show the proof of the solution
8. end

8. ### gregcoll New Member

Oct 11, 2009
21
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sorry, after looking over the steps I meant to say: take out a B^2 from the denominator. (and I don't know what that 'm\' is in step six

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
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This is how I did it ...

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10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Hi gregcoll,

In case you were wondering what's going on with this thread, ihaveaquestion was asking for the Laplace Transform of the function, f(t)=sin(Bt)/t. The method for setting up the LT shown in the original working by the OP was incorrect and the integration shown there was not a requirement for a correct solution.

Rgds,

t_n_k

11. ### steveb Senior Member

Jul 3, 2008
2,433
469
t_n_k,

Why do you say that the method for setting up the LT was incorrect? Are you sure about that? His method does lead to the correct answer, as I mentioned above.

I do agree that the integration shown is not a requirement, but it seems to be a valid approach.

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Yes - you are absolutely correct. One of my not-uncommon brain "farts" I'm afraid.

Apologies to all - "ihaveaquestion" especially.

Rgds,

t_n_k

13. ### notxjack New Member

Sep 7, 2009
20
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It's a complex integral over a line. It's a relatively straightforward process, though the steps will make no sense if you are unfamiliar with complex integration.

14. ### gregcoll New Member

Oct 11, 2009
21
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You are exactly correct, a complex integration using the residue method is more thorough but unnecessary in this case as you know it will lead to the same result. I did not include these methods as it seemed that ihaveaquestion might not be familiar and it is better (I think) to not scare people, especially scientists and engineers, from math and this method might seem more familiar. Sorry for any ambiguity.