Math Is Fun Forum / Laplace Transform of sin^2t

zetafunc. · 2011-08-10 20:59:17

Hi all,

I have started learning about Laplace transforms and done some easy examples such as proving that

and

Here's my working:

Using integration by parts we get

then I noticed that

is the Laplace transform of since by the double-angle rule sin(2t) = 2sintcost, and we can separate the 1/2 since it's a constant.

So if that bit is the Laplace transform of , then we can use our identity for Laplace transform of sin(at) to rewrite the integral as;

So;

But when I evaluate the RHS at t = and t = , I end up with y(s) = 0... is that really correct? I haven't checked the answer online anywhere because I don't want the answer spoiled for me...

I also tried a different method using integration by parts 3 times before noticing a repeat. Not sure what to do though. Would appreciate your help.

Thanks.

bobbym · 2011-08-10 21:08:08

Hi;

That I am not getting?

zetafunc. · 2011-08-10 21:18:08

Thanks for the quick response, I don't understand why I'm not getting the same answer as you...

Using integration by parts I let u = sin²t so du = sin(2t)dt

And dv = e^-stdt so v =

Using the formula uv - ∫vdu for integration by parts, that's how I got my answer...

And using the identity for sin(at) (where a = 2), I am getting

why is this incorrect?

Thanks.

bobbym · 2011-08-10 21:39:37

Sorry, I am being disconnected. I will check your IBP please hold.

Your first IBP is correct. And you set it up right.

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.

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 22:18:46

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 sin^a(t) and got .

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.

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 22:38:37

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...

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?

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 22:50:01

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 22:52:21

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.

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?

JamesRook · 2012-10-26 09:18:26

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

bobbym · 2012-10-26 15:38:46

Hi;

Welcome to the forum. 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 · 2013-05-13 08:58:08

Hi;

Helsaint wrote:
Sin^2(t) = 1/2 - cos2t

That is not correct.

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#1 2011-08-10 20:59:17

Laplace Transform of sin^2t

#2 2011-08-10 21:08:08

Re: Laplace Transform of sin^2t

#3 2011-08-10 21:18:08

Re: Laplace Transform of sin^2t

#4 2011-08-10 21:39:37

Re: Laplace Transform of sin^2t

#5 2011-08-10 22:03:07

Re: Laplace Transform of sin^2t

#6 2011-08-10 22:11:21

Re: Laplace Transform of sin^2t

#7 2011-08-10 22:18:46

Re: Laplace Transform of sin^2t

#8 2011-08-10 22:20:29

Re: Laplace Transform of sin^2t

#9 2011-08-10 22:21:32

Re: Laplace Transform of sin^2t

#10 2011-08-10 22:38:37

Re: Laplace Transform of sin^2t

#11 2011-08-10 22:43:40

Re: Laplace Transform of sin^2t

#12 2011-08-10 22:43:50

Re: Laplace Transform of sin^2t

#13 2011-08-10 22:50:01

Re: Laplace Transform of sin^2t

#14 2011-08-10 22:52:21

Re: Laplace Transform of sin^2t

#15 2012-10-26 09:17:22

Re: Laplace Transform of sin^2t

#16 2012-10-26 09:18:26

Re: Laplace Transform of sin^2t

#17 2012-10-26 15:38:46

Re: Laplace Transform of sin^2t

#18 2013-05-13 08:41:54

Re: Laplace Transform of sin^2t

#19 2013-05-13 08:58:08

Re: Laplace Transform of sin^2t

Helsaint wrote:

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