Understanding the cause of sidebands in Amplitude Modulation

Question

I've read it many places that Amplitude Modulation produces sidebands in the frequency domain. But as best as I can imagine it, modulating the amplitude of a fixed-frequency carrier wave just makes that "louder" or "quieter", not higher-frequency or lower-frequency. That is, I believe I could sketch, on graph paper, a path of a wave function that touches a peak or a trough exactly every 1/f increments, regardless of the "volume". Why do the sidebands appear?

Answer 1

Let your carrier signal be $A_0 \cdot \cos(\omega_c t)$ with amplitude $A_0$ and carrier frequency $\omega_c$. Let your signal be a simple wave, $\phi(t) = A_s \cdot \cos(\omega_s t)$.

Then the modulated signal becomes $$A_0 A_s \cdot \cos(\omega_c t) \cdot \cos(\omega_s t)$$.

In addition, as pointed out by George in the comments, the carrier also gets transmitted.

Using the trigonometric identity $\cos(u) \cdot \cos(v) = \frac{1}{2} [\cos(u-v) + \cos(u+v)]$, you get the final signal: $$\frac{1}{2} A_0 A_s \cdot ( \cos((\omega_c - \omega_s) t) + \cos((\omega_c + \omega_s) t)) + A_0 \cdot \cos(\omega_c t)$$ Hence, the frequency becomes changed, you get the carrier frequency in the middle (at $\omega$) and two side-bands at $\omega_c \pm \omega_s$.

Now, in reality your signal is not a simple cosine, but you could do a Fourier decomposition of the signal and treat each frequency independently. The two frequencies then get smeared out and you get the two sidebands.

Answer 2

a path of a wave function that touches a peak or a trough exactly every 1/f increments, regardless of the "volume"

First of all, that's not correct. A peak or trough in some signal $U(t)$ is defined by $\frac{\partial U}{\partial t}|_{t_{\mathrm{peak}}}=0$. If $U$ is now a product of some carrier $U_{\mathrm{c}}(t)$ and some modulator $M(t)$, $$ U(t) = M(t)\cdot U_{\mathrm{c}}(t) $$ then $$ \frac{\partial U}{\partial t} = M\cdot \partial_t U_{\mathrm{c}} + (\partial_t M)\cdot U_{\mathrm{c}}. $$

Here, $U_{\mathrm{c}}$ itself will certainly not be zero in a peak, so for $U$ to have a peak at the same time as $U_{\mathrm{c}}$ had, the time-derivative of $M$ needs to be zero - in other words, you need a constant amplitude!

Still, your idea was quite correct: the frequency of the carrier in a purely amplitude-modulated signal can always be measured exactly, because even though the peaks and dips are moved by the modulation, the zeroes are not! So provided that your carrier signal is always positive and your carrier is a simple sine (or other simple) wave, you can just count the zeroes in $U(t)$ over a long time $T\gg\tfrac1f$, divide the number by $2\cdot T$ (two zeroes per period), and always get exactly the frequency of $U_{\mathrm{c}}$. _{(As a sidenote: you can in fact do this for arbitrary modulators, but it's in general more complicated.)}

The problem is: in real applications, you're mainly going through all this modulation/demodulation trouble so that you can have multiple signals transmitted simultaneously, without one getting in the way of the others. That means you can't actually measure any zero passings at all any more, because you don't know where $0$ is! You first have to filter out our specific signal, but this very filtering again only works perfectly if $M$ is constant. If it isn't and you try to filter out your carrier frequency too narrowly, you will filter away any information transmitted in $M$ too. In order to avoid this, you need to space your multiple carrier frequencies far enough apart so you can use a more tolerant filter, and the necessary spacing is just the width of the side bands.

Answer 3

A wave with amplitude modulation can be described with a carrier $A_0 cos(\omega_c t)$ and a modulating factor $\left [1+a_s cos(\omega_a t)\right ], |a_s| <1, \omega_s <<\omega_c$:

$A(t)=A_0 cos(\omega_c t)\cdot \left [1+a_s cos(\omega_s t) \right ]=A_0 cos(\omega_c t)+A_0a_s cos(\omega_c t)cos(\omega_s t)$

The first term is a carrier frequency and the last term is a superposition of two slightly different frequencies ($cos[(\omega_c-\omega_s)t]+cos[(\omega_c-\omega_s)t]$), as was explained by Lagerbaer. Thus you have three frequencies in the total signal - a carrier and two sidebands.

Answer 4

You are correct that the modulated signal touches a peak every 1/f and a trough every 1/f. But if you looked at the waveform of a carrier being amplitude modulated by a much lower frequency pure sinewave, it is intuitive that there is another frequency component. (That is the "feel" part.) The math in other posts shows you what those frequency relationships are.

I would add that commercial AM stations do not have signals which conform to these examples: The examples have zero crossings other than that which would be expected from the carrier (e.g., whenever the cosine term in the modulating signal = 0). Commercial AM stations do not allow their carriers to get modulated down to zero amplitude.