Why isn't the Fourier transform of a single sine wave cycle a single bar?

Question

I have tried different Fourier transform codes out there on single sine waves, and all of them produce a distributed spectrum with a resonance at the signal frequency when they should theoretically display a single bar.

The sampling frequency has little effect (10kHz here), however the number of cycles does:

One cycle:

100 cycles:

100000 cycles:

It looks like the fourier transform converges only for an infinite number of cycles, why is that? Shouldn't a time window of exactly one cycle bring the same results as that of N cycles?

Application: This is both out of curiosity and also because I want to get how much the step response of a first order system will be exciting the resonance of a mechanical assembly. Therefore I need an accurate Fourier transform of the response... Which I don't trust anymore. What could I do to improve the accuracy then, based on the "sine wave" case?

P.S: These particular screenshots are based on the code here.

Answer 1

This is a windowing artifact.

The linked code pads out a 10,000 sample signal with zeroes so that the length is a power of two.

%% Author :- Embedded Laboratory

%%This Project shows how to apply FFT on a signal and its physical 
% significance.

fSampling = 10000;          %Sampling Frequency
tSampling = 1/fSampling;    %Sampling Time
L = 10000;                  %Length of Signal
t = (0:L-1)*tSampling;      %Time Vector
F = 100;                    %Frequency of Signal

%% Signal Without Noise
xsig = sin(2*pi*F*t);
...

%%Frequency Transform of above Signal
subplot(2,1,2)
NFFT = 2^nextpow2(L);
Xsig = fft(xsig,NFFT)/L;
...

Note that in the above code, the FFT is taken with the FFT size NFFT which is the next power of 2 bigger than the signal length (in this case, 16,384.) From the Mathworks fft() documentation:

Y = fft(X,n) returns the n-point DFT. fft(X) is equivalent to fft(X, n) where n is the size of X in the first nonsingleton dimension. If the length of X is less than n, X is padded with trailing zeros to length n. If the length of X is greater than n, the sequence X is truncated. When X is a matrix, the length of the columns are adjusted in the same manner.

This means that you are not actually taking a FFT of a 'pure sine wave' - you are taking the FFT of a sine wave with a flat signal after it.

This is equivalent to taking the FFT of a sine wave multiplied with a square window function. The FFT spectrum is then the convolution of the sine wave frequency spectrum (an impulse function) with the square wave frequency spectrum (sinc(f).)

If you change L = 16,384 so that there is no zero-padding of the signal, you will observe a perfect FFT.

Further search keywords: "Spectral Leakage", "Window Function", "Hamming Window".

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Edit: I cleaned up some material I wrote on this topic back in university, which goes into substantially more detail. I've posted that on my blog.