# Calculate a Kind of Equivalent Baseband Signal by Examining the Zero Crossing

I have a device (whose behaviour I can't change) which does the following signal processing:

The incomming signal contains two frequencies $f_1$ and $f_2$. One has a huge amplitude, the other a smaller one:

$x_1 = X_{01} \cdot cos(2\pi f_1 \cdot t)$

$x_2 = X_{02} \cdot cos(2\pi f_2 \cdot t)$

with (for example):

$f_1 = 20.02e6, f_2 = 20e6, X_{01} >> X_{02}$ (Actually $f_2$ is an interfering signal).

An example for this signal (only the beating envelope is visible) is depicted below.

At first, phase durations are calculated by "measuring" the duration between two zero crossings. Then the phase duration of $f_1$ is subtracted - so if $f_2$ wouldn't be there, the phase duration signal would always be zero. But since there is $f_2$ also, the phase duration signal contains a new periodic signal, beeing periodic with $f = f_1-f_2 = 20kHz$. I get the following signal: With zoom

My problem is: This signal is not a pure sine wave but it contains some additional frequencies - as already visible in the zoomed phase signal. The DFT shows the containing frequencies (which are harmonics). They are depicted below.

Does anyone know an equation for the amplitudes of those frequencies as a function of the amplitude and frequency of $X_{01}$ and $X_{02}$?

To make it more complex, I unfortunaly have a second interfering signal at $f_3=19.08e6$. Is there an equation for this case, too?

• Why are you doing the zero-crossing math? Is your aim to suppress the interference, measure its characteristics, etc.? – Jason R Nov 12 '13 at 16:58
• This zero-crossing math is done in the measurement card of an heterodyne michelson interferometer.The interferometer measures the speed of a measuremnt object which does influence $f_1$ (via Doppler effect) - and so the displacement is shown in the phase of the $f_1$-signal. So the aim is to measure the change in phase which is done via zero-crossing. In my example above there is no change in phase because I want to keep it as simple as possible. The interference signal $f_2$ results of improper optics (which can't be changed). One problem I have to deal with is $f_2$ and its harmonics. – Semjon Mössinger Nov 13 '13 at 8:32