I have a single tone sinusoid, and its fft in scilab is giving me double spikes, why is that?
Please explain me the concept of why is it happening.
clf(); fs=1000; t=0:1/fs:0.2; L=length(t); f=((0:1:L-1)*fs)/L; l=length(f); fm=40; s=sin(2*%pi*fm*t); plot(t,s) S=fft(s) S_abs=abs(S) figure(2) plot(f(1:l),S_abs) 
Welcome!
Actually, in your case windowing only affects the width of your pulse (as you see, your peaks are not impulsive). You can see the difference by just increasing the temporal length of your signal (equivalent to increase the width of your window, assuming that the original signal is the infinite length sinusoid) and taking the fft. The longer your signal, the narrower your pulse.
What causes the double peaks is that the fft function returns the double-sided transform of the signal you are considering. This means that the first half of S contains the positive bandwidth (from 0 to fs/2), while the second half is just its mirrored version (mathematically, this is equivalent to the spectrum from -fs/2 to 0).
Details are given in the fft documentation by Mathworks.
fftshift() function to rotate the zero frequency to the middle of the array. $\endgroup$ You're experiencing a basic property of the DFT of a real input signal, that it is conjugate symmetric:
Let $x[n], \, n = 0\cdots N-1$ be a real-valued sequence, then the DFT coefficients are conjugate symmetric: $$X[k] = \overline{X[N-k]},\,k = 0\cdots N-1$$
For analysis, we usually discard half of the spectrum because the information is therefore redundant:
figure(2) S_abs = S_abs(1:(l+1)/2); f = (0:(L-1)/2)*fs/L; plot(f,S_abs) See also this answer for correct scaling of even and odd length DFT.
Proof of conjugate symmetry:
\begin{align} X[N-k] &= \sum_{n=0}^{N-1}x[n]e^{-j2\pi \frac{N-k}{N}n}\\ &= \sum_{n=0}^{N-1}x[n]e^{+j2\pi \frac{k-N}{N}n}\\ &= \sum_{n=0}^{N-1}x[n]e^{+j2\pi \frac{k}{N}n}e^{-j2\pi n}\quad \left(e^{-j2\pi n} = 1\right)\\ &= \sum_{n=0}^{N-1}x[n]\overline{e^{-j2\pi \frac{k}{N}n}}\\ &= \sum_{n=0}^{N-1}\overline{x[n]e^{-j2\pi \frac{k}{N}n}}\\ &= \overline{X[k]} \end{align}
Where the last step stems from the fact that $x[n]$ is real so $x[n] = \overline{x[n]}$
figure(2) f=(-L/2:L/2-1)*fs/L; plot(f, fftshift(S_abs)); $\endgroup$