Post History

Take a look at the circuit and try to see what function it will perform on the input signals to yield the output signal.  For simplicity, let's consider the diode ideal, R1 and R2 the same, and R3 much larger.

R1 and R2 will average the two input signals onto the anode of the diode.  The diode will clip this signal to the positive parts.  In other words, it performs a MAX function with 0.  The output is therefore Max(Ave(S1, S2), 0).

Now consider what this results is if you put in two sine waves.  As an example, let's use 100 Hz for S1 and 101 Hz for S2.  The average of the two is:

<img src="https://electrical.codidact.com/uploads/Av8XJBi9zqNkyAKW6oknbUMj">

When the two input signals are in phase, they add, so the result is a full-amplitude sine.  When they are out of phase, they cancel, so the result is 0.  Since the difference in frequencies is 1 Hz, this adding and canceling repeats at a 1 Hz rate.

At this point, we still only have the two original input frequencies.  Although the amplitude of the averaged signal is changing at 1 Hz, there is actually no 1 Hz component to the signal.  It is simply the sum of two sines.

Now consider what happens when this signal is clipped to only the positive values:

<img src="https://electrical.codidact.com/uploads/WN9snpfaPhCBSJHzB53vSgrM">

Now we do have a strong 1 Hz component.  Hopefully you can see that if this low pass filtered to get rid of the 100 Hz and 101 Hz components, you'd be left with a 1 Hz signal.

Another way of looking at this circuit is that by adding two frequencies together, you get a result that varies in amplitude by the difference between those frequencies.  D1 and R3 are essentially a quick and dirty AM demodulator.  The result is a signal that is the amplitude envelope of the sum of the two input signals.