FFT on triangular and square waves.

Question

Hi,
I have been computing FFTs on a some sinusoidal waves and have been doing fine, gett the right Y axis values in the units Vrms^2.  For example a 10V amplitude (20v peak to peak) wave at 100 htz would give a peak at 100 htz and 49Vrms^2 since the RMS value fo 10 is 7 and 7^2 is 49.  Now when I input a Square wave with the same characteristics I no longer get the same scaling instead getting a peak at a higher y value.  With a triangular wave I get peaks at lower values.  Can anyone explain to me why this is happening?
Thanks.
Intern NSWCCD Carderock.

Answer 1

Actually RMS (Root Mean Square) value of periodic signal f(t) is defined as a square root of squared signal averaged over the period T, mathematically:
f_rms = sqrt ( aver (f^2) ),
where aver (f^2) = [integral of f^2 over the period T] / T
It is easy to show that for pure sine wave f(t) = A*sin(2*pi*t / T), f_rms = A / sqrt(2) ~ 0.707*A – this is well-known result anyone can find in almost every textbook. It is not difficult to calculate exact value of f_rms for square and triangular waves. Results are as follows:
f_rms (square) = A, where square wave f(t) = A, if 0<t<c*T and f(t) = -A, if c*T<t<T (here 0<c<1 and c=0.5 for 50% duty cycle wave)
and
f_rms (triangle) = A / srqt(3) ~ 0.577*A, where I assumed triangle wave that raises from 0 to +A over 0 < t <T/4, goes down to –A over T/4 < t < 3T/4 and returns to ZERO in the last quarter of the period T
Parseval’s theorem (http://en.wikipedia.org/wiki/Parseval's_theorem) of Fourier analysis states that energy of periodic signal (=integral of f^2 over the period T) must equal to energy of its Fourier transform ( sum of squared amplitude of all harmonics for periodic signal). That’s why Fourier analysis in Labview is capable of producing signal’s RMS value in addition to its spectrum.
Now, main harmonic of square and triangular waves carries majority of signal energy, but not all of it because of presence of other harmonics. That means that amplitude of main harmonic as reported by Labview will be close to but slightly less than signal’s RMS value. This explains the results you reported in your very first post.