# How does multiplying the modulated signal "AGAIN" by the carrier frequency transfer the spectrum back to its original position?

I have some trouble understanding the demodulation of a signal. SI understand how we multiply the message signal by the carrier signal then chuck it through a channel, but on the receiver side I don't understand the above statement because it does not make sense to multiply the same modulated signal by the offset Ac * cos(2pifc + △f) when demodulating to get it back to where we first started.

AM modulation is basically an extension to ring modulation. Ring modulation is basically the multiplication of the message and the carrier. And AM adds the carrier to this.

Let $$\m(t)\$$ is the message (sinusoid), $$\c(t)\$$ is the carrier (sinusoid), r(t) is the ring-modulated signal, and a(t) is the amplitude-modulated signal:

$$r(t)=m(t) \ * \ c(t)\\ a(t) = r(t) + c(t)$$

So,

$$r(t)= A_m \ \sin(2\pi f_m \ t) * A_c \ \sin(2\pi f_c \ t)$$

Remember this:

$$\sin a * \sin b = \frac{1}{2} \Big(\cos (a-b)-\cos (a+b) \Big)$$

So this is where $$\+f_c\$$ and $$\-f_c\$$ offsets come from.

Now multiplication the amplitude-modulated signal, $$\a(t)\$$, with the carrier again with the same phase (phase detectors required) will produce the message signal (with some DC offset) plus another AM signal having a carrier of twice the original carrier frequency.

It's easy to get rid of the DC offset and the other component (as it requires a LPF) to get back the original message signal.

• This is perhaps counterintuitive because we tend to think of multiplying as making something bigger, but in this case we multiply by the carrier amplitude which is in the range -1 to +1 so we’re not amplifying the carrier but cancelling it out.
– Frog
Commented Aug 16, 2022 at 9:57

It is somewhat difficult "to see" equations ... and what there are "saying".

Here is a Maple sheet ... NB: dc is added to bf modulated information (signal).
We make AM modulation and then make demodulation as OP question. Real signals.
We can do this also with "complex" signals. Not used here.

The time point of view :

The frequency point of view, NB: Relative Power spectrum. EE&O.

We can see the spectrum of the demodulated wave that must be filtered.
All "dc" and "bf" information are there.