# Considering the FFT of Real & Complex Signals

I've been implementing a website to perform the FFT of various signals, real & complex.

Examining the first example, a real signal $x[n] = 10 cos(2\pi\times4n)$, I got the following FFT:

Which was exactly what I expected - two nice peaks of half amplitude at $\pm 4$

So I then extended to the FFT of the complex signal $x[n] = 10e^{j2\pi\times4n}$ (or equivalently $x[n] = 10cos(2\pi\times4n) + j10sin(2\pi\times4n)$). This is shown in the time domain as; However, this time I got the following as the FFT, however, it now only has a single peak at $+4$, rather than the $\pm4$ mirrored peak I was expecting, and received in the real signal.

After reading through a wide number of articles relating to the conjugate symmetric property of the FFT with regard to real, even signals that states $x[n] => F*(\omega) = F(-\omega)$.

However, this doesn't help me understand what the FFT of a complex signal should look like - i.e. if my figure is correct? How should an FFT for a real/complex/imaginary signal appear in terms of mirroring and symmetry?

NB: My own way of justifying this so far is that for a real signal, $cos(2\pi ft) = \frac{1}{2}(e^{2\pi ft}+e^{-2\pi ft})$ which creates negative frequencies - hence the mirroring, whilst the FFT of $e^{2\pi ft}$ directly does not...however I may be completely wrong!

• Sample $F(t)= sin(at)+sin(bt)$ at different sampling rate, perform FFT on the sampling rates. Please assist me in solving this problem Jan 26, 2022 at 17:51

the cosine has two peaks, one at +f, the other at -f. That's because Euler's formula actually says $\cos x = \frac12\left(e^{ix}+e^{-ix}\right)$. You can see the two complex sinusoids that lead to your two peaks.
Now, doing the same for the sine, we see that $i \cdot\sin x =i\frac1{2i}\left(e^{ix}-e^{-ix}\right)$. That means, the sine kills one of the two peaks of the cosine, if you will so.