Something doesn't look right to me here. You show the green wave, a sine wave input going into the bridge rectifier. What comes out of that should be both sides of the sinewave in a positive direction - sort of a double-hump waveform that looks kinda like "mmmm". Each of those double positive pulses should turn on the opto's diode, which in turn should turn on the phototransistor inside the optocoupler at that same doubled rate.
I expect to see the output - what you called signal2 - to have double the frequency of signal1. The red waveform appears to be just a clipped version of the sinewave at the same frequency. I would expect to see that somewhere on the other side of the circuit if a diode was shorted and burned out.
However, if the frequency was double the input, then that red waveform would appear more normal, in that the phototransistor is only able to generate a voltage and conduct in one direction and not the other. Still, I don't see any charge held over from the 100nF cap. Are you sure that signal2 is really not from the other side of the circuit? It just doesn't look right to me.