Fourier Transform of Heaviside function

  • #1
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Homework Statement


Find the Fourier transform of [tex] H(x-a)e^{-bx}, [/tex] where H(x) is the Heaviside function.

Homework Equations


[tex] \mathcal{F}[f(t)]=\frac{1}{2 \pi} \int_{- \infty}^{\infty} f(t) \cdot e^{-i \omega t} dt [/tex]
Convolution theory equations that might be relevant:
[tex] \mathcal{F}[f(t) \cdot g(t)] = \mathcal{F}[f(t)] * \mathcal{F}[g(t)] [/tex]
[tex] \mathcal{F}[e^{\alpha t} \cdot f(t)] = \mathcal{F}[f(t+\alpha t)] [/tex]
Derivative of the Heaviside function:
[tex] H'(t) = \delta(t) [/tex]
where [itex] \delta(t) [/itex] is the Dirac Delta function.

The Attempt at a Solution


Using the first relevant equation, and assuming the Heaviside function simply changed the boundary conditions from a to infinity:
[tex] \mathcal{F}[H(x-a)e^{-bx}]=\frac{1}{2 \pi} \int_{a}^{\infty} e^{-bx} \cdot e^{-i \omega t} dt [/tex]
[tex] = - \frac{1}{2 \pi} \left[\frac{i e^{-bx-i \omega t}}{\omega} \right]_{a}^{\infty} = - \frac{i e^{-bx-i \omega a}}{\sqrt{2\pi} \omega} [/tex]

This was not right so I tried to take the integral of the Heaviside function via integration by parts:
[tex] \mathcal{F}[f(t)]=\frac{1}{2 \pi} \int_{- \infty}^{\infty}H(x-a)e^{-bx-i \omega t} dt [/tex]
[tex] =\frac{1}{2 \pi} \left[\frac{-H(x-a)ie^{-bx-i \omega t}}{\omega} \right]_{-\infty}^{\infty} - \int_{\infty}^{\infty} \delta(x-a)e^{-bx-i \omega t} [/tex]
At this point I could go on with the integration however I'm unsure if it'll lead to the right answer or if what I'm doing makes sense. I have considered using the last equation in part of the convolution theory section, to get an integral of perhaps [itex] H(x-a-i \omega t) [/itex] but that doesn't seem right either.. I am unsure how to take the Fourier transformation of the Heaviside function, any help will be gladly appreciated.
 

Answers and Replies

  • #2
Orodruin
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You are mixing variables. Either use x or t, for this problem it is quite obvious that x is the variable and you need to set up your Fourier transform accordingly. (If it was in t, then your function would be a constant and the Heaviside would not change the integration domain.)
 
  • #3
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Would it then just be the integral from a to infinity of [itex] H(x-a)e^{-bx-i \omega x} dx [/itex] or the integral from a to infinity of [itex] e^{-bx-i \omega x} [/itex] ?
 
  • #4
Ray Vickson
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Would it then just be the integral from a to infinity of [itex] H(x-a)e^{-bx-i \omega x} dx [/itex] or the integral from a to infinity of [itex] e^{-bx-i \omega x} [/itex] ?
Aren't these exactly the same thing?
 

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