What is the significance of the AM wave equation in amplitude modulation?

[atomic_light11]

AM amplitude modulation---

Hi everyone, I am learning AM recently, but I have some question that keep troubling me..

1. We know that AM wave is produced by the multiplication of carrier wave and modulation wave. But why does the equation of AM wave is not a direct multiplication of V_m×V_c but consist of a constant? Can anyone explain the general concept of the AM wave equation?

Why it is
V_am = [E_c + E_mcos(ω_mt)]cos(ω_ct)

but not
V_am = [E_ccos(ω_ct)] [E_mcos(ω_mt)]

2. In single sideband AM, why does the cancelling of one sideband doesn't affect the information in the wave? And what is the effect to the wave pattern after the AM wave has been transform from double sideband (DSBFC) to single sideband (SSBFC)?

Thanks!

 

[ShreyasR]

Hi!

The amplitude of the carrier is E_C. This is the base amplitude of the carrier. when it is modulated with a modulating signal e_M, It's amplitude is changing according to the instantaneous value of e_M.

Have a look at the waveforms:

V_am = [E_c + E_mcos(ω_mt)]cos(ω_ct)

In this equation,
[E_c + E_mcos(ω_mt)] refers to the instantaneous amplitude of the am wave, which is, you know, changing with time...
cos(ω_ct) is the frequency component of the wave.

Compare with v = V_m sin (ωt + $\phi$)

If it was V_am = [E_ccos(ω_ct)] [E_mcos(ω_mt)]

That means [Eccos(ωct)] is the amplitude. In this case, the amplitude goes negative in one half cycle. It is meaningless. So the main amplitude (with zero modulating signal) is E_M. But the instantaneous value of the modulating signal ADDS to this main amplitude E_M. So that's why you have that equation! Have a closer and more detailed look at that waveform, it will be clear to you!