In amplitude modulation, the amplitude of the carrier wave is changed according to the variations in the message signal, and that's what we draw inside the two envelops of the message signal, that's totally fine, but what is confusing to me, that we never changed the frequency of the carrier wave, then why do those two sidebands appear in the frequency response of the AM wave, I know that the mathematics works out to be like that, but how on the earth they appear in first place, when we never tried to mess with the frequency of the carrier?
Electronic – Why do sidebands appear after Amplitude Modulation
Modulation
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Best Answer
AM means that the amplitude of the carrier is multiplied by the payload signal plus some offset.
The multiplication theorem for sine says
\$\sin(a) \cdot \sin(b) = \frac{1}{2} ( \cos(a-b) - \cos(a+b))\$
Let \$a=2\pi f_c t\$ be the carrier phase and \$b=2\pi f_s t\$ the phase of a sinusoidal payload signal then as a result of AM (=multiplication of sines) you get a sum of two sinusoids. The frequency of one of them is a little bit below (\$f_c-f_s\$) and one of them a little bit above (\$f_c+f_s\$) the carrier frequency.
This works not only for a payload signal that is a pure sine wave but also for a whole band (any signal can be seen as a sum of sinusoids).
So as a result you get one band above and one (mirrored) below the carrier frequency.
(Note: The offset mentioned in the beginning results in the fact that also the unchanged carrier frequency will be present after AM; but I've neglected that for simplicity).
BTW:
AM does mess with the frequency. It's a missconception that multiplication of two signals wouldn't mean messing with the frequency; that's true only for multiplication by a constant; that's a linear operation; multiplcation of two time varying signals is not. Any non-linear operation introduces new frequencies.
Simple example: frequency doubling (=messing with frequency) of a sinusoid can be done by simply squaring the signal (=multiplying by itself).