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Fourier time shift

  1. Oct 24, 2009 #1
    The time shift property of the Fourier transform is defined as follows:

    [itex]x(n - n_o ) \Leftrightarrow e^{ - j\omega n_o } X(e^{j\omega } )[/itex]

    I am confused by this notation....what does [itex]X(e^{j\omega } )[/itex] mean? I know that [itex]X(\omega)[/itex] is the value of the Fourier transform at a given angular frequency but I'm confused why it has been put in a complex exponent.

    It is also sometimes written as:

    [itex]h(x) = f(x - x_0)[/itex]
    [itex]\hat{h}(\xi)= e^{-2\pi i x_0\xi }\hat{f}(\xi)[/itex]

    This notation I think I understand...it's saying that, if I have the DFT of [itex]f(x)[/itex], then I can get the DFT for [itex]f(x - x_0)[/itex] by multiplying each point in the DFT by a scale factor that depends on the frequency. Specifically, from Euler's relation, I should scale the real/complex part by,

    [itex]\cos(-2 \pi x_0 \xi ) + i \sin(2 \pi x_0 \xi )[/itex].

    I tested this out by constructing a DFT that has a single spike, then taking the inverse FFT to reconstruct a time signal...and scaling the DFT spike and reconstructing again to see if I got a time shifted signal. I did not. When I thought about it more, I realized that this doesn't make sense, because some time shifts would result in multiplication by zero, which means that a shift followed by a negative shift could result in the signal being destroyed.

    What have I got wrong?
    Last edited: Oct 25, 2009
  2. jcsd
  3. Oct 27, 2009 #2


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    Looks like a typo to me, I think it should be simply X(ω). Here's another link, they do have F(ω) (different notation):

    Not sure what is going on with your test. You might try starting in the time domain, with both a spike at t=0 and also a time-shifted spike. Take the FFT of both and compare.

    BTW, if you want to represent purely real time-domain signals, the frequency-domain should have the property

    X(-ω) = X*(ω),

    where * denotes the complex conjugate. So the only way to have a single spike in the frequency domain is when that spike is at ω=0.
  4. Oct 27, 2009 #3
    Alright, I found out where my confusion was...everything I said in my above post was correct, except for the place where I said "I tried this and it didn't work," because the reason it didn't work is I was doing the complex multiplication pointwise, and complex multiplication actually involves some addition! Now it all makes sense.

    Oh I was only referring to a single spike in the positive frequencies...because obviously the negative frequency range is just a mirror of the positive data as you point out.

    I'm still confused about the X(e^jw) notation though...I've seen it in many different places so I don't think it's just a typo..
    Last edited: Oct 27, 2009
  5. Oct 27, 2009 #4


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    That is bizarre. If X is the signal in the frequency domain, the argument must be a real number ... this is the Fourier Transform, not Laplace, after all. I am equally baffled.
  6. Oct 16, 2010 #5
    It's only a notation criterium.

    As usual in Fourier Analysis in function of the application area, differents conventions appear.

    the notation X(w) it's usual for physicist but the notation X(expiw) it's more usual for electronic engineering.
    The second notation has conection with the bilateral Laplace transform for cotninuous signal and Z transform for discrete signals.
    Esentially it means:
    a) for cotinuos signal, the Fourier Transform is a particular case of laplace transform, where the complex number s, is restricted to unit circle.
    b) for discrete signals, the same idea but with Z Transforms.
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