An FFT Example - 75% Pulse Signal

        In this note we are going to analyze a pulse signal using the FFT. Here is the signal.  (NOTE:  If you are used to Matlab, the indices here start at zero.)

• 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.

Now, we need to examine what happens when we FFT this signal.  We will use the Mathcad file this link takes you to.  (You may have to change the file name for the data file that is loaded in this workspace.)  That file assumes that the data file has two columns (Time in the first column, data in the second column.  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.

• Notice that there is a DC component in this signal.
  • Since the pulse is 5 volts and is high for 75% of the time (at +5v) and is low for 25% of the time) at -5v), the DC level is 2.5 volts.  However, the Mathcad CFFT function multiplies by 1/N (not 2/N) so the computed result is the correct value for the DC component.  (Remember that the absolute value of the coefficients is plotted.)
• The fundamental frequency of the data record is the reciprocal of the length (in seconds) of the data record.  Since the data record is 2 milliseconds long (i.e. .002 sec), the fundamental frequency of the data record is 500 Hz.
• In Mathcad, the indices start from zero.  So, a_o, is going to be placed in SigFFT₀ in our workspace above.  Here is a short table of frequencies, indices, etc.  In this table, the indices in the FFT array (SigFFT) are the number of the harmonics of the fundamental frequency of the data arecord.

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

• All of the odd components are zero.
  • They are components at 500, 1500, 2500 Hz, etc.  We would expect them to be zero in a signal that has a 1000 Hz fundamental.
• The first "spike" in the FFT plot is at SigFFT₂.  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.
  • Since the "2" in the 2/N factor is missing, the actual value of the fundamental is around 2x2.3 = 4.6.
• The second "spike" in the FFT plot is at SigFFT₄.  That corresponds to 2000 Hz which is the second harmonic of the signal.
  • The actual value of the second harmonic is around 2x1.55 = 3.1.
  • The same computations can be used for higher harmonics.
• There are some signal harmonics that are missing.  For example, SigFFT₈ is missing (in this case, maybe just very small).  That means the fourth harmonic of the signal is zero!  Examining the plot we also see that the 16^th, 24^th and 32^nd, etc. are missing, so that the 8^th, 12^th and 16^th harmonics are missing.