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Discrete Fourier transform mirrored?

by lordchaos
Tags: discrete, fourier, mirrored, transform
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lordchaos
#1
Dec29-12, 11:07 AM
P: 5
Why does a discrete Fourier transform seems to produce two peaks for a single sine wave? It seems to be the case that the spectrum ends halfway through the transform and then reappears as a mirror image; why is that? And what is the use of this mirror image? If I want to recover the frequency, phase and magnitude of an oscillation, do I need to use any data from this mirror image?
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rbj
#2
Dec29-12, 11:23 AM
P: 2,251
because

[tex] \cos(\omega t + \phi) = \frac{1}{2} \left( e^{+i \omega t} + e^{-i \omega t} \right) [/tex]

so there is a frequency component at [itex]+\omega[/itex] and at [itex]-\omega[/itex].

because of aliasing due to sampling, negative frequencies are displayed in the upper half of the output of the DFT.
lordchaos
#3
Dec29-12, 12:22 PM
P: 5
Thanks for that. Does this affect how I should extract the magnitude & phase from the transform? Or is it OK to throw the second half away for that purpose?

rbj
#4
Dec29-12, 01:16 PM
P: 2,251
Discrete Fourier transform mirrored?

if your input to the DFT is real (i.e. they are complex numbers, but the imaginary part is zero), then yes, the second half is a mirror image of the first half. the real part (or the magnitude) of the DFT output has even symmetry and the imaginary part (or the phase) has odd symmetry.
lordchaos
#5
Dec31-12, 11:54 AM
P: 5
Thanks for your help!


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