Inverse Fourier Transform

  • Thread starter PiRho31416
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[Solved] Inverse Fourier Transform

Homework Statement


If
[tex]F(\omega)=e^{-|\omega|\alpha}\,(\alpha>0)[/tex],
determine the inverse Fourier transform of [tex]F(\omega)[/tex]. The answer is [tex]\frac{2\alpha}{x^{2}+\alpha^{2}}[/tex]


Homework Equations


Inverse Fourier Transform is defined as:
[tex]f(x)=\int_{-\infty}^{\infty}F(\omega)e^{-i\omega x}\, d\omega[/tex]


The Attempt at a Solution


So I broke up the equation into two different integrals.
[tex] F(\omega)=e^{-|\omega|\alpha}=\int_{-\infty}^{\infty}e^{-|\omega|\alpha}e^{-i\omega x}d\omega=\int_{-\infty}^{0}e^{\omega(\alpha-ix)}+\int_{0}^{\infty}e^{-\omega(\alpha+ix)}d\omega [/tex]

[tex]\frac{e^{\omega(\alpha-ix)}}{\alpha-ix}\bigg|_{\omega=-\infty}^{\omega=0}+\frac{e^{-\omega(\alpha-ix)}}{\alpha-ix}\bigg|_{\omega=0}^{\omega=\infty}=\frac{1}{\alpha-ix}+\frac{1}{\alpha+ix}=\frac{\alpha+ix+\alpha-ix}{\alpha^{2}+x^{2}}=\frac{2\alpha}{\alpha^{2}+x^{2}}[/tex]
 
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Answers and Replies

  • #2
vela
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You got the sign of the second term wrong.
 
  • #3
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I don't see where the sign is wrong.

I broke it up the absolute value so it reads like follows:

[tex]f(\omega)=\begin{cases}
e^{\omega\alpha} & \mbox{if }\omega<0\\
e^{-\omega\alpha} & \mbox{if }\omega>0\end{cases}[/tex]
 
  • #4
vela
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I'm referring to the second integral. Once you integrated and before you plugged in the limits, you should have a minus sign in front of the second term.
 
  • #5
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Got it! Dang those minus signs :-p Thanks!
 

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