A Note About the FFT (Fast Fourier Transform)

        We assume that you know about Fourier Series.  The FFT algorithm does the following.

• Assume that the Fourier Series - in complex form - is given by the expressions below.  Note that you can use a complex form to represent the coefficients (a_k + jb_k).  It's the same information, but it turns out that it is easier to compute the complex form when you program this.  It's easier, and - most importantly - it is faster, and these computations can use a lot of compute power.
  

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• The FFT calculates an approximation to a_k and b_k, by doing a sum.  Assume that  we have a set of N samples of f(t) - spaced Dt seconds apart  (so that T = NDt.).  In essence, we are assuming that the record of data we have is one cycle of a periodic function - even though we have only the one record of length T seconds.  In any event, we can approximate the integral with this sum.

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• The FFT algorithm is a fast way to compute the sum.

  • CAUTION:  Although the FFT algorithm is available in Mathcad, Matlab and other analysis programs, that pesky 2/N factor is not always handled the same way in those computations.  In Matlab, the fft function does not put that 2/N factor in the final result it gives you.  More later.

• When you use the FFT algorithm, you will have to remember that frequency information will be lost if the data is simply stored in an array in a computer program.  Instead of f(nDt), you will have only a sequence, f_n, that has no reference to the specific times at which the samples were taken.  You will have to keep track of the length of the record and the interval between samples.  If you have done that, then you should find the following.

  • The fundamental frequency is 1/T, so that the k^th computed coefficient (assuming that the array starts at k = 0) is the coefficient of the k^th harmonic.

Example 1

        Let us assume that you take data for two seconds and record that data in a file.  Let's also assume that you get 4000 data points in that two second period.  Here are the conclusions that you can draw from those two facts.

• Since the length of the data record is two seconds, if you take the FFT you are getting a Fourier Series for a signal that has a period of two seconds.

  • With a period of two seconds, the fundamental frequency for the data in the data record is 0.5 Hz.

• Now assume you want to look for a frequency component in the recorded data, and you want to see if there is a signal at 100 Hz.

  • A signal at 100 Hz is at a multiple of 200 times the fundamental frequency of 0.5 Hz.  In other words, a 100 Hz signal is the 200^th harmonic of 0.5 Hz.

  • The 200^th harmonic will be in the 201^st position in the array returned by the Matlab function.

  • You would look in the 201^st position in the array to determine the magnitude of the signal at 100 Hz.

Example 2

        Let use imagine you have a periodic signal.  You do the following

• You take data on the signal (using an oscilloscope, for example) and store the data in a file.

• The length of the data record is T seconds.

  • If the data record is T seconds long, you can imagine that record repeating over and over again, and you can imagine that the data record is one period of a periodic signal.  (We know that's not true, but if you do that you can use Fourier techniques to determine frequencies of signals in the data in the record.)

• If there is a periodic signal T seconds long, that signal would have a fundamental of 1/T Hz and 2p

  /T rad/sec (angular frequency).

• If there is a sine wave buried in the data record, it will not have the same frequency as the data record (unless you record exactly one period of the sine!).  Let's assume that the frequency of the sine wave is "f".)

• At this point we can consider a specific numerical example.

  • Assume that the data record is 2 seconds long (i.e. T = 2.)

  • The fundamental frequency of the record is 0.5 Hz (i.e. 1/T = 1/2 = 0.5)

  • Assume that the sine wave signal in the record is at 7 Hz.  In other words, there would be exactly 14 periods in 2 seconds - the length of the data record.

  • We would expect to find a peak in the computed Fourier coefficients at the 14^th harmonic of the data record.

  • The data for the 14^th harmonic would be stored in the 15^th array element if you are using Matlab - since the DC (constant) term is stored in the first array element.  So, you would look in the 15^th array element to find data for a 7 Hz sine wave.

• The next issue you need to address is how big the 7 Hz sine wave component is.  You need to note the following.

  • The algorithm programmed into Matlab is fft(FileData).  That algorithm does not compute the sum above, given again below.
    

  • The algorithm does not have the 2/N factor in the computed result, and you have to correct for that.  If you have an oscilloscope that returns 4000 data points, you would have to divide every computed a and b by 2000 (i.e. 2/4000).

• Now you are ready to determine how big the 7 Hz sine wave component is.  Let's say that you found that the computation gave you 10,000 at the 15^th array element.  Dividing by 2000, you should find that you have a sine wave with an amplitude of 5.

• If you have a more complex signal (square wave, triangular wave, short pulse, etc.) those signals would have harmonics.  If you have a 7 Hz signal - not a sine - you would have harmonics at 14 Hz, 21 Hz, etc. and you would need to look in the appropriate array element to determine what the size of those components are.