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Topic review (newest first)

bobbym
2013-05-13 08:58:08

Hi;

Helsaint wrote:

Sin^2(t) = 1/2 - cos2t

That is not correct.

Helsaint
2013-05-13 08:41:54

Shouldn't this be
Sin^2(t) = 1/2 - cos2t
L{sin^2(t)} = 1/2s - s/(s^2+4)

bobbym
2012-10-26 15:38:46

Hi;

Welcome to the forum. That is not correct.

JamesRook
2012-10-26 09:18:26

to be clear I am talking about (sin(t))^2

JamesRook
2012-10-26 09:17:22

Hello,

I get that the laplace transform of sin^2t = -(sin^2te^-st)/s + 2/s^3+4s evaluated from 0...infinity. 

when I evaluate the limit from 0..infinity I get that the transform to equal 0.  Did I evaluate that right?

bobbym
2011-08-10 22:52:21

Wolfram is going to put it in terms of the Dirac delta function, which I think is a step function.

There are diiferent definitions for a fourier transform, that page will partly explain that.

zetafunc.
2011-08-10 22:50:01

Thanks...

I just tried the Fourier transform of f(x) = 1 and got

... is that correct? I'll check on Wolfram.

bobbym
2011-08-10 22:43:50

Hi;

There are FFT and DFT's. Wikipedia can be a horror story at times. To me that is exactly what that is saying.

I have never seen their notation. They are using small f with a cap ( borrowed from statistics)

zetafunc.
2011-08-10 22:43:40

Also what is the notation for a Fourier transform? For Laplace it's a fancy L, is it a fancy F for Fourier transforms?

zetafunc
2011-08-10 22:38:37

Sorry if it's a bother but do you know how to compute Fourier transforms? I'm trying to learn how, I've seen the Wikipedia article and saw this:

for every real number ξ.

Does this mean that if I put in some function of x, such as sin(x), I'll get f(ξ) where ξ is a real number? Not sure, I'll post my working in a second. Sorry if I sound stupid...

bobbym
2011-08-10 22:21:32

Hi;

Zeroing the LHS will leave you with just the Laplace term. That should be your answer.

I was just asking to see what you thought about it. Since t approaches infinity it will drown out s no matter how small as long as s > 0.

That is nice, spotting the Laplace Transform there.

In addition zetafunc, welcome to the forum! Why not consider becoming a member here?

zetafunc.
2011-08-10 22:20:29

I wasn't given an interval for s, sorry. I am just waiting for my GCSE results (I turn 16 in August) and I'm just trying to extend my knowledge of calculus. I want to learn about Fourier transforms too hopefully but I need some practice with that.

zetafunc.
2011-08-10 22:18:46

Hi,

What I meant is that I get



Then evaluate RHS at 0 and subtract that from the evaluation at infinity. I got 0... so then we have



Therefore
assuming s > 0.

I also tried the Laplace transformation for sina(t) and got
.

bobbym
2011-08-10 22:11:21

Hi;

I am glad to help but we are not done yet.



The LHS has to be evaluated at infinity and then you subtract the evaluation of it at 0. The RHS is untouched.

How are you getting 0 for the LHS?
If s is very small then the LHS is not zero. Were you given some interval for s?

zetafunc.
2011-08-10 22:03:07

Hi,

Thanks for the response again and confirming that my IBP was correct -- I think I get it now - subtract the rightmost term from both sides to get y(s) - 2/(s3 + 4s), evaluate the RHS at 0 and infinity to get 0 (0 - 0 = 0), then add 2/(s3 + 4s) to both sides to get the completed Laplace transform? Is that correct? Phew, thanks for your help. smile

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