# Is it possible to recover an over-modulated waveform?

Is it possible to recover an over-modulated waveform? I have searched the net up and down but I constantly come across the same answer 'It is not possible to fully recover a full AM signal using an envelope detector'. I do understand this as an envelope detector only tracks the positive parts of the signal.

So how do we track/demodulate an over-modulated waveform?

• This is essentially the third time you have asked this question today. Since you can't do it with an envelope detector, the answer is, use something else. And you know what that something else is, from your other questions. – Brian Drummond Feb 5 '18 at 18:37
• Wouldnt be asking if I knew that thing. Would I? – Meti Feb 5 '18 at 18:41
• It's a nice practice to edit the original question if it isn't the topic but only your knowledge about it what changes. – Janka Feb 5 '18 at 18:44
• Well you were so very close with your DSBSC questions, but they didn't clarify what bit you didn't understand. – Brian Drummond Feb 5 '18 at 18:55
• Do you know how to unrectify ( reverse polarity) of a rectified signal using the phase of the carrier reversal? It is possible, but SNR is very poor. There is no advantage in overmodulating so compression is done to prevent this. – Tony Stewart Sunnyskyguy EE75 Feb 5 '18 at 19:21

I'm no amateur radio person, just someone who took a bad course in telecommunications, so take this answer with a pinch of salt. I'll just explain how I'd do it if I was given this task.

Alright, let's go through how an AM signal might look like.

$x(t)=(1-\sin(\omega t))×\sin(\omega_ct)$

The $\omega_c$ is the carrier frequency.

This will result in a signal that looks something like this: And you can use an envelope detector to convert it back to $\sin(\omega t)$, nothing weird so far.

But let's over modulate the signal like this:

$x(t)=(0.5-\sin(\omega t))×\sin(\omega_ct)$

Which will result in a signal that looks like this: Now, how can we convert this one to something like the previous one so we can yet again apply an envelope detector to acquire our $\sin(\omega t)$?

Well, if you remember your trigonometry then you will remember that the only way to get a DC term when multiplying sine waves together is if they have the same frequency. In other words, on the receiver side you will generate the carrier signal and match the phase of the signal to maximize the DC value.

So let's say we got a local oscillator in phase with the carrier and oscillating at the same frequency, that would look like this:

$\small{y(t)=x(t)×\sin(\omega_ct)=(0.5-\sin(\omega t))×\sin(\omega_ct)^2=(0.5-\sin(\omega t))\frac{1-\cos(2\omega_c t)}{2}}$

$y(t)$ would look something like this: Now you can apply an envelope detector and get a clipped sine wave, you won't get the negative part. With some filters the clipped sine wave can look more like a regular sine wave.

There's probably something very obvious that I've missed. FYI I haven't visited your other 2 questions and read their answers that apparently can solve your problem.

Edit:

Why yes of course, just add a LP filter, add some DC value and there you go.

Yes, it is possible to recover the signal from a "over modulated" AM carrier, but not with something as simple as a envelope detector.

Think about it. Fully "overmodulated AM" is double sideband. Clearly it is possible to recover the base signal, since it has been done. Single sideband can also be recovered, as is/was routinely done by lots of ham radios.

Instead of asking basically the same question repeatedly. Sit down and actually learn what AM, and more generally, product modulation really is. Learn the math behind it. Then look up some circuits to demodulate each type.

AM is a very special case of product modulation such that the original signal can be trivially detected with a diode. That is because the amplitude envelope is the original signal. This is not true of other types of product modulation, so a simple envelope detector won't work with those.