How does an amplifier inadvertently demodulate a radio signal?

[the_emi_guy]

My diagrams show the sum of three vectors, the carrier and the two sidebands, which, I think, matches the trig.

I guess I was mistaken when I assumed that this was direction you were heading when you brought up adding close freqs and obtaining beat freqs.

Sorry.

 

[Baluncore]

Sorry.

Emi_guy; don't be sorry for asking difficult questions. It wakes me up. Sorting out misunderstandings due to the inadequacies of the English language and it's users is a necessity. I have to agree with Fourier et al, that a real signal is the vector sum of all it's frequency component phasors.

It is interesting that the trigonometric identity sin(m) * sin(c) = ½ cos(c–m) – ½ cos(c+m) appears at first to be simple AM because on the LHS it is the product of the carrier by the modulation. But the equivalenced signal is synthesised from the linear combination of only two cosinewaves on the RHS. The truth is that this identity actually represents a double sideband signal with a suppressed carrier, DSB SC. The carrier must be reinserted by linear addition before it is possible to use a simple envelope detector to demodulate the signal.

The simple? broadcast AM that can be detected immediately with an envelope detector is represented by the_emi_guy's posted equation;
[A + M cos( ωm t )] sin( ωc t ) = A sin( ωc t ) + M/2( sin( ωc + ωm ) t) + M/2( sin( ωc – ωm ) t )
The difference here from DSB SC, is that the modulation is offset by A, sufficiently so that the modulation term [A + M cos( ωm t )] never crosses zero.

Getting back to the topic. It takes a non-linear component to make an envelope detector that might inadvertently demodulate an AM broadcast signal. That non-linear component could be the junction of an amplifier input transistor, zinc oxide on a galvanised iron roof, copper oxide between crimped wires, or germanium condensed and crystallised in the metal flue of a coal burning stove.

 

[sophiecentaur]

[A + Mcos(ωmt)]sin(ωct) = Asin(ωct) + M/2(sin(ωc+ωm)t) + M/2(sin(ωc-ωm)t)

So multiplying the stuff on the LHS produces AM, but adding the stuff on the RHS does not?

There is a hidden requirement in the two sideband phases that means you need them to be taylor made. Otherwise you could be getting phase modulation or an AM PM mixture.

 

[the_emi_guy]

There is a hidden requirement in the two sideband phases that means you need them to be taylor made. Otherwise you could be getting phase modulation or an AM PM mixture.

That is exactly what the diagrams that I attached were highlighting.

I also mentioned that this is not necessarily the practical way of generating modulation, but is a common method of *analyzing* modulation. phase noise in particular.

Unfortunately I took the thread a little off topic in the process.

 

[sophiecentaur]

Yep. The devil is in the phase detail. Fourier rules. [emoji846]