Laplace Transform help needed

  • #1
Hi,

I've been asked to find the Laplace transform of a function and I have not the slightest clue where to begin. My professor derived the basic Laplace transforms in class(sin, cos, delta function, step function, etc), all of which I understood perfectly. However, he never really gave us an example of how to use those to find other Laplace transforms, let alone discuss how to approach the homework problem.

rc34wp.png


i made a few attempts. The first was by saying that the transform of a summation is the summation of the transforms. And then trying to take the transform of that, which seems ugly. I also tried graphing it and re-writing it as a series of step functions to get an idea, but I obviously can't do that until the end of time, so I'm stuck. Could somebody lead me in the right direction? Thanks!
 

Answers and Replies

  • #2
35,292
7,149
Hi,

I've been asked to find the Laplace transform of a function and I have not the slightest clue where to begin. My professor derived the basic Laplace transforms in class(sin, cos, delta function, step function, etc), all of which I understood perfectly. However, he never really gave us an example of how to use those to find other Laplace transforms, let alone discuss how to approach the homework problem.

rc34wp.png


i made a few attempts. The first was by saying that the transform of a summation is the summation of the transforms. And then trying to take the transform of that, which seems ugly. I also tried graphing it and re-writing it as a series of step functions to get an idea, but I obviously can't do that until the end of time, so I'm stuck. Could somebody lead me in the right direction? Thanks!

Try sketching a graph of the summation - it's just the sum of some step functions, with each multiplied by either 1 or -1.
 
  • #3
Try sketching a graph of the summation - it's just the sum of some step functions, with each multiplied by either 1 or -1.

Right, I've done that already, but how do I get the Laplace transform from the graph? My graph starts at 1 when n=0, drops to -1 a n=1, goes to 1 at n=2, -1 at n=3, etc.

So the series I computed from that was u(t)-2u(t-1)+2u(t-2)-2u(t-3)+2u(t-4)...

But now what?
 
  • #4
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7,149
Where are the 2's coming from?

If you expand the series, don't you get just u(t) - u(t - 1) + u(t - 2) -+ ... + (-1)nu(t - n) + ... ?

Now, what's the Laplace transform of u(t - a)?
 
  • #5
Where are the 2's coming from?

If you expand the series, don't you get just u(t) - u(t - 1) + u(t - 2) -+ ... + (-1)nu(t - n) + ... ?

Now, what's the Laplace transform of u(t - a)?

Well isn't a step function minus another step function zero (1-1=0)? so I need to subtract it by another step function, 2u(t-n), to make it reach -1.

And the laplace transform of u(t-a) = e^(-as)/s
 
  • #6
35,292
7,149
Well isn't a step function minus another step function zero (1-1=0)?
Yes
so I need to subtract it by another step function, 2u(t-n), to make it reach -1.
???
Why do you think you need to reach -1?

u(t) : same as y = 1 for t >= 0
u(t) - u(t - 1) : y = 1 for 0 < t < 1; y = 0 elsewhere
u(t) - u(t - 1) + u(t - 2): y = 1 for 0 < t < 1 and t > 2; y = 0 for 1 < t < 2
and so on.
And the laplace transform of u(t-a) = e^(-as)/s
Yes.
 
  • #7
Yes
???
Why do you think you need to reach -1?

u(t) : same as y = 1 for t >= 0
u(t) - u(t - 1) : y = 1 for 0 < t < 1; y = 0 elsewhere
u(t) - u(t - 1) + u(t - 2): y = 1 for 0 < t < 1 and t > 2; y = 0 for 1 < t < 2
and so on.
Yes.

I must've been mixing myself up with a plot from the book.

Anyway, now that I have the Laplace transform of u(t-a), then is the answer I'm looking for the summation of that Laplace transform times (-1)^n from 0 to infinity?
 
  • #9
Now is there any way to simplify that even further, like get rid of the summation?
 
  • #12
wish you were able to get it worked out!http://www.infoocean.info/avatar1.jpg [Broken]
 
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  • #13
Ray Vickson
Science Advisor
Homework Helper
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[QUOTYE=audifanatic51;3870693]Now is there any way to simplify that even further, like get rid of the summation?[/QUOTE]

Yes. You just have a geometric series.

RGV
 

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