FFT Example - Done Using Matlab

        In this note we are going to analyze a triangle signal using the FFT. Here is the signal.

• The signal has 4000 samples, and the length of the record is 2 milliseconds.  (It goes from -.001 seconds to +.001 seconds.)

• Since the length of the record is 2 milliseconds, if we compute the FFT for the entire record (which we would normally do), then the fundamental frequency in the computed results is going to be 1/.002 = 500 Hz.  We will refer to this as the fundamental frequency of the data record.

• Within the data record is the triangle signal, and it has a period of 1 millisecond, so it has a frequency of 1 KHz.

• Data=dlmread('5VoltSqWvDataInclT.txt','\t');

• Time = Data(:,1);

• VData = Data(:,2);

• plot(Time,VData)

• pause

• SigFFT = fft(VData);

• plot(abs(SigFFT))

Now, if we look at the plot of the absolute value of the SigFFT array, we get a plot like the one below.

        Now, the fundamental frequency of the data record is 500 Hz.  You need to be able to get from that to the actual frequency components of the signal.  Here is what you need to use.

• The fundamental frequency of the data record is the reciprocal of the length (in seconds) of the data record.

• In Matlab, the indices start from one.  So, a_o, is going to be placed in SigFFT(1) in our m-file above.  Here is a short table of frequencies, indices, etc.

        Now, we can get at the frequencies in the FFT plot.  Notice the following.

• The first "spike" in the FFT plot is at SigFFT(3).  That corresponds to 1000 Hz which is the fundamental of the signal.  It is not the fundamental of the data record.  It is, however, related to the fact that there are exactly two cycles of the signal in the length of the data record.

• The second "spike" in the FFT plot is at SigFFT(7).  That corresponds to 3000 Hz which is the third harmonic of the signal.

• and so on. . .

        With that, you should be able to interpret the horizontal scale of the FFT plot  - at least the plots that Matlab produces.

        Now, we need to address the vertical scale.  First, you should realize that the vertical plot is the absolute value of the c's in the Fourier expansion.  If you need to understand what the c's are, check these links.

        When we did the calculation, we found that the third element in the SigFFT array (That's the one that is the first, large value - somewhere over 8000 on the plot!.) has a value of 8,103.  That's not a number that we would expect from a triangle wave that has an amplitude of 5 volts.

        The explantion for the seemingly ridiculous value of over 8000 is this:

• The calculation does not include the 2/N term.  Type "help FFT" (without the quote marks!) in the Matlab workspace for the explanation of what it actually calculates.  What is boils down to is that there is a missing 2/N term.  You have to compensate for that.  In a five volt triangle wave with 4000 data points, the first harmonic of the triangle wave should be 8A/(p²) (where A= 5 for our signal).Here are some links to pages with the expressions for the Fourier Series of a triangle wave signal.  That's where we got that amplitude for the first harmonic.

  • A page from Wolfram Research (the folks who make Mathematica).  There is a lot of information on this one web page.  You may want to print it out.

  • A Java Simulator

  • A nice derivation of the coefficients

• The actual first harmonic (i.e., the fundamental) of a triangle wave of amplitude A is 8A/p², and that would calculate out to about 4.053.  To get from 8,103 to 4.053 you need to multiply 8,103 by 2/4000 (that's 2/N).  Try it, and the result is very close to 4.053

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What do you conclude from this?

        From the material above, you should be able to determine the actual Fourier Series components if you have a signal in a file.  You should be able to distinguish between the fundamental frequency of the data record and the fundamental frequency of the signal embedded in the data record - if that signal is periodic.  And, you should be able to determine the frequencies present in the signal, as well as the amplitudes.

 

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What if the signal doesn't have an integral number of periods in the data record?

        Here is an example of a signal with 2.5 periods in the data record.  In this signal, we only have 2000 points, so we have to factor that in.

You can see that there is not an integral number of periods in the data record.  Now, let's see what happens when we FFT this data record.  That's shown below.

Generally, this can only be described as a mess.  There are no clear-cut lines in this FFT spectrum.  However, note the following table.
 

Index	Harmonic of the fundamental frequency of the data record	The actual frequency
SigFFT(1)	0	0 Hz
SigFFT(2)	1	500 Hz
SigFFT(3)	2	1000 Hz (1 KHz)
SigFFT(4)	3	1500 Hz
SigFFT(5)	4	2000 Hz
SigFFT(6)	5	2500 Hz
SigFFT(7)	6	3000 Hz

        Now, we can get at the frequencies in the FFT plot.  Notice the following.

• The fundamental frequency of the signal embedded in the data record is 1250 Hz.  That is not a frequency found in the table.  Rather, the harmonics of the fundamental frequency of the data record are 1000 Hz and 1500 Hz, and they appear at indices of 2 and 3.

• The largest spikes appear at indices of 2 and 3.

        All of that is well and good, but the third harmonic of the embedded signal is at 3750 Hz and that would appear between indices 8 and 9.