# Multiple channels of information in single electromagnetic wave?

I'm trying to understand how can multiple radio stations transmit information just by transmitting using different frequency. The way I understand it all those different frequency waves add up to a single wave? If so how can that not corrupt information on other frequencies?

Say if I transmit: $\sin(2\pi x)$

And separately: $\sin(2\pi x\times 2)$

Does it end up as a single wave of: $\sin(2\pi x)+\sin(2\pi x\times 2)$?

Or does it work in a different way completely?

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– dmckee Aug 15 '11 at 20:28

Say if I transmit: $\sin(2\pi x)$

And separately: $\sin(2\pi x\times 2)$

Does it end up as a single wave of: $\sin(2\pi x)+\sin(2\pi x\times 2)$?

Yes, that's exactly how it works. This is called superposition. There are electromagnetic waves at hundreds of different frequencies all filling the air simultaneously.

The way something like a radio is able to pick out the station it wants is by using a resonant circuit (also known as a band-pass filter), which damps out all frequencies of EM waves except within a small range.

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Thanks a lot! Could not find answer to this so I was not sure if that was even a smart question :) Knowing this I can continue to meaningfully study signal processing. – Rytis Aug 16 '11 at 19:36

It's possible to split this combined signal into the original components again. You can do that because the sine and cosine functions form a base of a Hilbert space, a space called $L^2(\mathbb{R})$.

Now what does this mean? The word "space" is perhaps confusing, a mathematical space is essentially just a set of mathematical objects, in this case the space contains all possible combined signals (each signal is a "point" in the space).

In a Hilbert space, it is possible to calculate the scalar product of two elements: let $\varphi,\psi$ be two possible signals, (for instance it could be $\varphi(x)=\sin(x)$, $\psi(x)=\cos(2x)$), then we have $$\langle \varphi| \psi\rangle_{\!\!\!\!_{L^2(\mathbb{R})}} = \int_\mathbb{R}\!\mathrm{d}x\ \varphi(x)\cdot\psi(x)$$ This yields a single number. In case you don't understand what this expression means, don't worry. You can think of the scalar product as measuring "how much the signal $\varphi$ is 'in' the signal $\psi$", in a sense.

Suppose now you have given a signal $$\varphi(x) = a\cdot \sin(x) + b\cdot \sin(2x)$$ where $a$ and $b$ are unknown prefactors. It is these prefactors that carry the actual information in a radio signal, so you want to know them. And that's where we can use our scalar product: we calculate $$\langle \sin(x) | \varphi\rangle$$ and $$\langle \sin(2x) | \varphi\rangle.$$ Actually, we're running into a bit of a problem right here: the sine functions oscillate literally forever, infinite. That makes the result of the scalar product also infinite. But in reality, we don't have infinite signals, they are actually limited to some finite time duration. Suppose the signals start at $x=0$ and end at the time $x=4\pi$, (before and after, both are just zero all the time). Then we have \begin{align} \langle \sin(x) | \varphi\rangle &= \int\limits_0^{4\pi}\!\mathrm{d}x\ \sin(x)\cdot\psi(x) \\ &= \int\limits_0^{4\pi}\!\mathrm{d}x\ \sin(x)\cdot\sin(x) + \int\limits_0^{4\pi}\!\mathrm{d}x\ \sin(x)\cdot\sin(2x) \end{align} The integrals can be calculated with help of some known trigonometric equations. It turns out the result is $\langle \sin(x) | \varphi\rangle = 2\pi\cdot a$, whereas $\langle \sin(2x) | \varphi\rangle = 2\pi\cdot b$. We calculated the prefactor of each of the sine functions!

I called the "time"-variable $x$ here, like you did. In physics, we would usually call it $t$ for radio signals.

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I'll need to go through this myself with examples, but this is really useful information for me! Thank you! – Rytis Aug 16 '11 at 19:34