Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 7 down vote accepted

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.

share|cite|improve this answer
I always thought that the carries was $A_0 cos(\omega_c t)$ and the amplitude modulation consisted in a time dependent amplitude of the same sign, say: $A_s (t)\cdot A_0 cos(\omega_c t)$, $A_s (t) \gt 0$ :-( – Vladimir Kalitvianski May 27 '11 at 13:54
We can always write $A_s (t) = 1 + a_s cos(\omega_s t)$, so we have a carrier (due to $1$ in this expression) and two side frequencies (due to $cos(\omega_s t)$, as was explained by Lagerbaer). – Vladimir Kalitvianski May 27 '11 at 14:08
@Georg You are right, I added this. – Lagerbaer May 27 '11 at 15:22
Ok, that math makes sense. But how does this conflict with the casual observation of a peak and a trough being drawn at an exact 1/f interval, regardless of amplitude, will still be of that frequency f? Is it that there is some precision lost in the sketch, some mathematically complex contortion of the wave form required by trying to force a fixed frequency while the amplitude changes? I know I'm not sounding very scientific here - I could trust the math, but I want to feel it. – Jason Kleban May 27 '11 at 15:46
Not sure I get what you mean with your sketch. Could you add such a sketch to your question? – Lagerbaer May 27 '11 at 15:54

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.

share|cite|improve this answer

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.

share|cite|improve this answer
Thank you for your contribution. – Jason Kleban May 27 '11 at 16:14
You are welcome! – Vladimir Kalitvianski May 27 '11 at 16:17
[1+ascos(ωat)] suggest a DC term, why is that? You can't transmit DC. – user9233 May 16 '12 at 1:45
@user9233: it is not a term, it is a factor. – Vladimir Kalitvianski May 16 '12 at 7:18

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.

share|cite|improve this answer
Why the downvote? – Jason Kleban May 27 '11 at 21:59
Perhaps from someone who has not taken a graduate course in RF communications? – Vintage May 31 '11 at 22:16

protected by Qmechanic Feb 2 '13 at 19:39

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.