If you use Mathcad to get the FFT you could use a workspace like the one below.  Here is a link to the complete workspace.  This workspace does the following:

• Data := READPRN("5VoltSqWvData.txt")
• Brings the data from the file 5VoltSqWvData.txt into the Mathcad workspace.
• The file is "tab-delimited".
• The file should be a pure text file, and that's the kind of file many data acquisition programs generate, even if they use other extensions like ".dat".
• The file is assumed to have only one column, so if you generated the data with a program like Benchlink, you need to strip out the header and the time information, and just have a column of data that starts in the first row.
• SigFFT := CFFT(Data)
• This uses the CFFT algorithm which will FFT an arbitray length vector (which can have complex elements).
• If you are lucky enough to have a vector that has a length that is exactly a power of 2 (something like 512 elements, or 1024 or 4048, etc.) use FFT - which will compute faster.

• Mathcad indices start at 0, and we start with a coefficient, a_o.  The first coefficient is a_o and it will be stored in the first element of the array (SigFFT in the example m-file), i.e, SigFFT(0).  Unlike Matlab, the indices work out in Mathcad.  For example, SigFFT(11) will contain the 11^th harmonic.
• The calculation will be done in terms of the harmonics of the fundamental frequency of the data set.  If you have a square wave that shows two periods in the data set, the first harmonic of the square wave will be the second harmonic of the fundamental frequency of the data set, and - given the first point just above - will appear in the third element of the FFT array.
• When you do the calculation, you will often have data that has many periods in one data set.  Those frequencies - in the periodice signal - will be at exact multiples of the fundamental frequency of the data set (a sharp line in the computed FFT) or near a multiple (a spread-out line).  You shouldn't expect the fundamental of a periodice signal to be the fundamental frequency of the data set unless you have exactly one period of your periodic signal in the data set.
• Example:  You have a data set that is 2 milliseconds long.  The fundamental frequency of the data set is 0.5 kHz (= 500 Hz).  If you have a one kHz signal embedded in that data, it will be at the second harmonic of the 500 Hz.  (In Mathcad, that will be at the index = 2.)
• The calculations includes a 1/N term.  You have to compensate for that since you will probably want it to be 2/N.  In a five volt square wave with 4000 data points, the first harmonic of the square wave will calculate out at a magnitude of 3.2 or thereabouts.  The actual first harmonic (i.e., the fundamental) of a square wave of amplitude A is 4A/p, and that would calculate out to about 6.4.  To get from 3.2 to 6.4 you need to multiply 3.2 by 2.
• There are other options in Mathcad.  They include the following - which have different normalizing factors and other peculiarities:
• FFT - if you have exactly 2^m data points, where m is an integer.
• cfft - which gives results that need to be multiplied by 2*SQRT(N).
• fft - which gives results that need to be multiplied by 2*SQRT(N) - if you have exactly 2^m data points, where m is an integer.
• Note that there are also inverse transform functions available if you need to go from FFT data back to time data.