1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fourier Transform of integral of a signal

  1. Nov 5, 2007 #1
    Hi. I have a question regarding the continuous time Fourier Transform of an input signal:

    [tex]x(t) \rightarrow X(j\omega)[/tex]

    then

    [tex]\int_{-\infty}^{t}x(\tau)d\tau \rightarrow \frac{X(j\omega)}{j\omega} + \pi X(0)\delta(\omega)[/tex]

    but if I want to write it in terms of [itex]f = \frac{\omega}{2\pi}[/itex], should it be:

    [tex]\int_{-\infty}^{t}x(\tau)d\tau \rightarrow \frac{X(j\omega)}{j\omega} + \frac{1}{2}X(0)\delta(f)[/tex]

    How does the [itex]\pi[/itex] get replaced by [itex]\frac{1}{2}[/itex] here?
     
  2. jcsd
  3. Nov 5, 2007 #2

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    How does the [itex]\pi[/itex] get replaced by [itex]\frac{1}{2}[/itex] here?[/QUOTE]

    For [itex]a>0[/itex], what does [itex]\delta (ax ) [/itex] equal? Why?
     
  4. Nov 5, 2007 #3
    Oh ok, so

    [itex]\delta(\omega) = \delta(2\pi f) = \frac{1}{2\pi}\delta(f)[/itex]

    Thanks :cool:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Fourier Transform of integral of a signal
  1. Fourier transform (Replies: 7)

  2. Signal transformation (Replies: 1)

Loading...