Electronic – Aperiod vs Period waveform Fourier Transform: How does nature understand which is the case

fourierharmonicssignal-theoryspectrum

Today I reviewed the theory behind the Fourier Transform, and I asked myself a question that I couldn't answer to in the process.

Theory: A periodic waveform has a Fourier series which can be seen as the sampling of the corresponding aperiodic waveform (i.e. the series of a square wave is the sampling of the transformation of a single rectangle).
So, a rectangle and a square wave have different harmonic components.

And now my question (square wave is just an example, the question is applicable to any aperiodic function): A digital driver (a clock generator if you want), drives a square wave at frequency f. It starts with a single "rectangle", its spectrum is that of the Fourier transform of a rectangle? Then it continues with many other – infinite – rectangles and you can say that it is indeed a periodic waveform (or at least it's been until that point), is its spectrum that of a square wave?

Of course, if I watch the signal with a spectrum analyzer I'll see the odd harmonics typical of a square wave… And it's absurd to say that the spectrum of the signal changes (initially that of a rectangle, and then that of a square wave). So what does really happen there?

I hope you see the incongruency that I'm trying to point out, and I would say this is not useless thinking as the spectrum of a signal says how the signal will propagate through the medium.

EDIT

From the comments it seems the question was not clear enough, let me add this:

"are you asking why you don't observe the spectrum changing in your analyzer between the 1st square wave and every subsequent square wave as it passes through?" –> My question is if this change really happens, and how it can make sense. If that really happened, it's clear why the spectrum analyzer would not show it.

EDIT 2: Clarify the meaning of rectangle and square wave

Also, when I refer to a "rectangle" or "single rectangle" I mean, basically, a single period of the square wave. To be more precise, the rising edge, high level and then falling edge are enough. Referring to it as a "square" didn't sound right to me. So I see a "square wave" as a series of "rectangles".

Best Answer

The Fourier transform is a linear transform. That means the transform of a long series of pulses is identical to adding the full complex transform of each individual pulse. If you try this, carefully respecting the time offset of each pulse, and the different FT phase result that that implies, you will see something interesting. What happens is that the spectra at the harmonic frequencies will reinforce (due to the phase at that frequency of all the individual FTs being identical), and all the other frequencies will start to destructively interfere (due to non-alignment of phase of each pulses FT from all others), and thus will tend towards zero.

This is because the phases of any frequency will only all line up when adding pulses that are spaced an identical amount equal to an exact integer multiple of the period. e.g. periodic. The phases of all other frequencies will be “scrambled”. And more scrambled the more you add them together. Eventually summing to near zero (relative to the peaks).

That’s how the change happens, by constructive and destructive interference of the Fourier spectra as the pulses are added, gradually changing a single Sinc-like spectra into something approaching the spectra of a long periodic signal, which is a comb of harmonics.

Note that to do this addition, you need the full complex transform (including phase), not just a magnitude spectra. A spectrum analyzer's long FFT window does this.

Added: You can't see the change because the spectrum analyzer has already added up multiple pulses before displaying anything. (long FFT window)

Related Solutions

Electronic – Why isn’t the Fourier transform of a single sine wave cycle a single bar

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".

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.

Electrical – Fourier transform to understand filtering on PWM

The Fourier transform is used to analyze PWM signals. The result depends on the type of PWM like type of carrier and kind of modulation.

A typical result looks like shown in the picture below. The baseband signal is present and a number of intermodulation products around the carrier frequency.

The two spectra are for different amplitudes of the modulating signal.

In order to sufficiently suppress the unwanted components most often a higher order filter is used.

Best Answer

Related Solutions

Electronic – Why isn’t the Fourier transform of a single sine wave cycle a single bar

Electrical – Fourier transform to understand filtering on PWM

Related Topic