From what I have read about signal (de)modulation I understand that multiplying two sine waves, both with the same frequency, gives a two frequency component sine wave. One at DC and one at $2\omega$. Using a low pass filter will let us remove the $2\omega$ component. Now DC amplitude is proportional to the phase difference between the two sine waves.

So: $$ A \sin(\omega t) \sin(\omega t + \theta) = \frac{A}{2}\cos(\theta) + \frac{A}{2}\cos(2\omega t + \theta) \rightarrow {\rm LPF} \rightarrow \frac{A}{2}\cos(\theta) $$ where $\omega$ is the frequency, LPF is Low Pass Filter, $A$ is the amplitude and $\theta$ is the phase difference between the sine waves.

I made a Matlab program to plot this, then I tried with a square wave multiplied with a sine wave. I can see that there is still a connection between DC and phase by doing a FFT as well as in the time domain. I would like to see the math for the relationship between DC and phase.

For sine $\times$ sine, I have read that sampling with a frequency four times the modulated signal frequency will make it possible to get the I and Q values by subtracting samples in the following order:

$$ I(t) = s(t) - s(t-2) $$ $$ Q(t) = s(t-3) - s(t-1) $$ where $$ s(t) = A\sin(\omega t + \theta) $$

This I do not quite understand yet and therefore leads me to be unable to determine if this also is valid for a square $\times$ sine. Can anyone try to explain this to me?

I will appreciate any feedback on how to effectively do IQ demodulation on a micro-controller using a square reference waveform.

Additional information:

I wish to use a micro-controller to sample the analog sine and square (reference) signal and do IQ demodulation. I almost only find information about sine $\times$ sine and therefore want to know if there are any pitfalls using a reference square wave that I am not seeing yet.


1 Answer 1


So if your sampling rate is four times the modulated signal frequency (and sampling synchronously) then $\omega= \pi/2$. So then $$ \begin{array} II(t) &=& s(t) - s(t-2)\\ &=& A\sin(\pi/2 t + \theta) - A\sin(\pi/2 (t - 2) + \theta)\\ &=& 2A\sin(\pi/2 t + \theta) \end{array} $$ and $$ \begin{array} QQ(t) &=& s(t-3) - s(t-1)\\ &=& A\sin(\pi/2 (t-3) + \theta) - A\sin(\pi/2 (t - 1) + \theta)\\ &=& A\sin(\pi/2 t + \theta - 3\pi/2) - A\sin(\pi/2 t + \theta - \pi/2)\\ &=& 2A \cos((\pi/2 t + \theta - 3\pi/2 + \pi/2 t + \theta - \pi/2)/2) \\ &&\sin((\pi/2 t + \theta - 3\pi/2 - (\pi/2 t + \theta - \pi/2))/2)\\ &=& 2A\cos(\pi/2 t + \theta) \end{array} $$

The trick will be to think about the square wave case, and see what this derivation tells you about that.

  • $\begingroup$ I am not sure what you are asking? Does "I = 2" mean $I(t-2)$ ? $\endgroup$
    – Peter K.
    Commented Apr 9, 2013 at 17:03
  • $\begingroup$ I deleted the above comment as I was wrong (and it might be misleading). The algorithm above will not work with a square reference as the sine x square product wave does NOT phase shift the same way as the sine x sine product wave does when reference phase is adjusted. Only way to find I and Q when using square reference as far as I can see is to use a low pass filter and measure the outputs. $\endgroup$ Commented Apr 9, 2013 at 21:35
  • 1
    $\begingroup$ @PeterK. It's always good to see a problem worked out explicitly. This is not a "ping" yet, but definitely making progress on IQ - thanks! $\endgroup$
    – uhoh
    Commented Mar 5, 2017 at 14:38

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