Fourier transform of sin(3pix/L)

  • #1

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So we want sine in terms of the exponentials when we take the fourier transform [tex]F(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx}dx[/tex] where [itex]f(x)=\sin(3\pi x/L)[/itex]. Let a=3pi/L. Then [tex]\sin(ax)=\frac{e^{iax}-e^{-iax}}{2i}[/tex].
(Is this correct?)
Then we can take the fourier transform:
[tex]F(k)=\int_{-\infty}^{\infty}\frac{e^{iax}-e^{-iax}}{2i}e^{-ikx}dx[/tex]. Rearranging gives [tex]\frac{1}{2i}[\delta(K+a)-\delta(K-a)][/tex]. But my notes says there is [itex]\sqrt{2\pi}[/itex] in front and I'm not sure where it came from?
Any help will be appreciated.
 

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  • #2
Ray Vickson
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Homework Statement




Homework Equations




The Attempt at a Solution


So we want sine in terms of the exponentials when we take the fourier transform [tex]F(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx}dx[/tex] where [itex]f(x)=\sin(3\pi x/L)[/itex]. Let a=3pi/L. Then [tex]\sin(ax)=\frac{e^{iax}-e^{-iax}}{2i}[/tex].
(Is this correct?)
Then we can take the fourier transform:
[tex]F(k)=\int_{-\infty}^{\infty}\frac{e^{iax}-e^{-iax}}{2i}e^{-ikx}dx[/tex]. Rearranging gives [tex]\frac{1}{2i}[\delta(K+a)-\delta(K-a)][/tex]. But my notes says there is [itex]\sqrt{2\pi}[/itex] in front and I'm not sure where it came from?
Any help will be appreciated.
Google is your friend; look up "Dirac delta function".
 
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  • #3
Google is your friend; look up "Dirac delta function".
I don't understand how to apply the dirac delta function here? I just used the integral representation of delta to get to the last line.
 
  • #4
Oh I am missing 2pi when I integrate. But where is the square root coming from?
 
  • #5
vela
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You're probably using a different convention for the Fourier transform compared to what was done in your notes (or your notes are wrong). I think you also made a sign error.
 

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