Signal Processing: DFT spectrum of sinusoid signals

AI Thread Summary
The discussion centers on the discrete Fourier transform (DFT) and its application to sinusoidal signals, specifically addressing why two peaks appear in the spectrum at frequencies m and N - m. It is noted that the peak at m corresponds to the frequency of the sinusoid, while the peak at N - m arises due to the symmetry in the DFT, reflecting the periodic nature of the transform. The mathematical reasoning is explored, indicating that reversing the order of samples in time leads to conjugate phase changes in the frequency domain. Additionally, participants suggest experimenting with software like Excel to visualize the DFT results. Understanding these peaks is crucial for analyzing frequency components in sampled signals.
Master1022
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Why does the DTFT of a sinusoidal signal with an integral number of cycles in ## N ## samples yield a spectrum with two peaks
Hi,

I was recently reading about the discrete Fourier transform and its application to a basic sinusoidal signal. If we know that it has an integer number of cycles in ## N ## samples (and thus no leakage), why would there be two peaks in the spectrum: one at ## m ## and another at ## N - m## (as shown in the image below)? I am guessing that ## m ## is the number of samples per cycle. It makes sense that there is a peak at ## m ##, but it isn't immediately clear why there should be one at ## N - m##. I can see the apparent symmetry of the term, but cannot intuitively reason why it should be present. I would appreciate any help or guidance as to why this is the case.

Screen Shot 2021-03-24 at 8.55.03 AM.png
By looking at the mathematical form of the DFT, we have:
F(n) = \sum_{k = 0}^{N - 1} f[k] e^{-j\frac{2\pi n}{N} k}

so would the following be correct?
F(m) = \sum_{k = 0}^{N - 1} f[k] e^{-j\frac{2\pi m}{N} k}
F(N - m) = \sum_{k = 0}^{N - 1} f[k] e^{-j\frac{2\pi (N - m)}{N} k} = \sum_{k = 0}^{N - 1} f[k] e^{-j2 \pi k(1 - \frac{m}{N})}
\rightarrow \sum_{k = 0}^{N - 1} f[k] e^{j2 \pi k \frac{m}{N}} e^{-j2 \pi k} \rightarrow \sum_{k = 0}^{N - 1} f[k] e^{j2 \pi k \frac{m}{N}}
because ## e^{-j2 \pi k} = 1 ## for an integer ## k ##. Then somehow due to symmetry, this causes the peak at ## N - m## ( I am not really sure on the exact logic for this last part).

Thanks in advance.
 
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What you plotted is the absolute value of the complex FT of ##\displaystyle {\sin mx = {1\over 2i} \Bigl (e^{imx} - e^{-imx}\Bigr )} ##.
You get one peak at m and one at -m . That last one shows up at ## N - m ##

Example: 16 samples of sin(x)
1616597463812.png
Excel | Data | Data Analysis | Fourier transform

1616597601356.png
##\quad##
1616597524545.png
If you have Excel ( ?:) :smile: ) or some other (F)FT software, play with it ! It's fun !

##\ ##
 
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Master1022 said:
It makes sense that there is a peak at m, but it isn't immediately clear why there should be one at N−m. I can see the apparent symmetry of the term, but cannot intuitively reason why it should be present.
You are analysing the time window from 0 to 2π = N.
Consider instead shifting the time window to be -π to +π.
The second peak will then have the negative frequency, -m.
You cannot separate the positive and negative frequencies, but they will have conjugate phase.
Reversing the order of samples in time is equivalent to taking the conjugate of the phase in the frequency domain.
 
Thank you very much @BvU and @Baluncore for your replies! Yes that is correct, I usually just plot the magnitude of the spectrum to avoid worrying about the +- signs...

Does the ## N - m ## peak also have to do with the periodic nature of the DFT?

Also, thanks for the heads up about Excel @BvU - I didn't know it was possible to do that there. Will definitely go give it a go!
 
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