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Discrete Fourier transform mirrored? 
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#1
Dec2912, 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?



#2
Dec2912, 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. 


#3
Dec2912, 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?



#4
Dec2912, 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.



#5
Dec3112, 11:54 AM

P: 5

Thanks for your help!



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