Fourier transform of sin

In summary, the Fourier transform of sin(t) involves the Dirac delta function and the function is zero outside -pi<t<pi.f
  • #1
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Homework Statement



Hey guys.
I need to find the Fourier transform of sin, is this right?

http://img156.imageshack.us/img156/5531/scan0004r.jpg [Broken]

I searched the internet but all I could find is the answer with the dirac delta and I don't need that.

Thanks.


Homework Equations





The Attempt at a Solution

 
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  • #2
The Fourier transform of sin(t) involves the Dirac delta function. What do you mean by "I don't need that"? And why did you change the limits from -∞ to ∞ to -π to π in your integral?
 
  • #3
The Fourier transform of sin(t) involves the Dirac delta function. What do you mean by "I don't need that"? And why did you change the limits from -∞ to ∞ to -π to π in your integral?

Oh, sorry, I need to find it from -pi to pi.
Is there something wrong with what I did?

Thanks.
 
  • #4
I didn't read your whole solution, but there is a mistake in your first step. The Fourier transform integral goes from -∞ to ∞. Why did you change those limits?
 
  • #5
I didn't read your whole solution, but there is a mistake in your first step. The Fourier transform integral goes from -∞ to ∞. Why did you change those limits?

Yeah, I need to find it from -pi to pi.
Is that way it doesn't involves Dirac function?

Thanks.
 
  • #6
No! It's not from -pi to pi. It's -∞ to ∞.
 
  • #7
No! It's not from -pi to pi. It's -∞ to ∞.

:smile:

But that is the question.
Find Fourier transform of sin in -pi<t<pi.

What do you mean?

Thanks.
 
  • #8
Your question is to transform the function [tex]f(t) = \left\{ \begin{matrix} \sin t & \mathrm{if}\; -\pi < t < \pi \\ 0 & \mathrm{otherwise} \end{matrix} \right[/tex] ?
 
  • #9
Your question is to transform the function [tex]f(t) = \left\{ \begin{matrix} \sin t & \mathrm{if} -\pi < t < \pi \\ 0 & \mathrm{otherwise} \end{matrix} \right[/tex] ?

Yeah, sorry for the misconfusion.
 
  • #10
Then your approach is correct since the function is zero outside -pi<t<pi anyway so you may as well integrate from -pi to pi.
 
  • #11
Ah, now it makes sense! Thanks Cyosis!
 
  • #12
You're welcome.
 

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