Signal Processing: DFT spectrum of sinusoid signals

[Master1022]

TL;DR Summary

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.

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.

 

[BvU]

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)

Excel | Data | Data Analysis | Fourier transform

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If you have Excel ( ) or some other (F)FT software, play with it ! It's fun !

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[Baluncore]

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.

 

[Master1022]

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!