Why we do that in AM demodulation?

[sophiecentaur]

I woke up in the middle of the night thinking about this thread.
old jim

OMG Jim. you're taking your work home with you again!
Aren't you supposed to be retired?

 

[jim hardy]

OMG Jim. you're taking your work home with you again!
Aren't you supposed to be retired?

Yes, old firehorse syndrome i guess ...

thouht maybe i'd learn something new !

old jim

 

[0xDEADBEEF]

why there will be noise at only the carrier frequency and it's harmonics, assuming that the noise added by the channel is a white noise?

Imagine the 1/cos function. Those spikes are terrible! And you multiply your function with it. If there is any signal at all there due to white or whatever noise, it will produce a giant signal. So you will see the same spikes again after dividing your signal. These spikes have twice the period of your carrier so they will be full of harmonics of the carrier frequency.

And finally we have the answer why this stuff isn't done. Three things will kill the scheme:

1) Noise. If your signal is f(t)cos(kt)+e(t) where e is noise, you will have divergences in the spikes of the 1/cos function, because e(t) is surely not zero at the crossings. As I said: you amplify your signal infinitely in the region with the worst signal to noise ratio.

2) Phase and frequency. If your local oscillator is not perfectly phase and frequency locked against the sending oscillator, the zero crossings will be off, the spikes will not get canceled and will dominate whatever comes out

3) Offset voltages. It is pretty much impossible to have components without any offset voltage. So even if your frequency and phase would match perfectly, your zero crossings will be off again producing spikes.

Infinite spikes are simply a bad idea. This type of problem happens also in the Wiener deconvolution, where you also divide the blurring of the signal away, but due to the same signal to noise problems you need to dampen the method at the frequencies were the noise is large compared to the signal.

 

[NascentOxygen]

You substitute the multiplication with cos(x) by a multiplication with 1/cos(x).

That is a very clear-eyed way of expressing it.

Multiplying by cos x looks much smoother than multiplying by this: