Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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

Like all my questions, I fear this will be very naive, because my physics background is very limited. Please bear with me.

I think of the electromagnetic field as a section of a vector bundle over spacetime, but I think nothing will be lost if we just treat it as a function $f:{\mathbb R}^4\rightarrow{\mathbb R}^6$.

Now suppose I want to send you a wireless message (by radio, cellphone, TV, whatever). I encode my message by jigglinig some electrons around so as to change the electromagnetic field from $f$ to a new function $f+g$. You are able to observe some values of this function $f+g$, or at least to infer something about some values of that function from the way the electrons in your receiver are jiggling around.

So at the very most, you are able to observe the function $f+g$. How on earth, then, can you infer anything about the function $g$?

I understand that you can decompose the field $f+g$ into Fourier components, but given that the ambient field $f$ can be anything at all, I don't see how that helps. The ambient field $f$ is affected by all sorts of things, from other people's communications to cosmic rays, to the fact that I happen to have just carried a 9-volt battery across my living room, much of which you have no knowledge of. Now I add a function $g$ to this completely unknown function $f$, and you're supposed to recover $g$ by observing the sum.

Clearly, this can't work. Clearly it goes ahead and works anyway. What am I missing?

share|cite|improve this question
I noticed there's no accepted answer here. Are you still looking for something additional beyond what's already written below? – DanielSank Feb 29 at 5:33

$f$ is not uniform. There are peaks caused by atomic nuclei, natural processes, and other transmitters, but they do not fill $f$'s domain completely. One can choose a suitable $g$ such that it occupies one of the quieter sections, and then filter on that.

share|cite|improve this answer

Also, if you know something about the signal that you're looking for, you can do a lot better than listening into the dark. For instance, if you know that you're looking for a 1054 Hz amplitude modulated signal, you can parse the field so that you're only looking at a narrow spectrum around there, and interpret the field that way. What is happening with the X-Ray spectrum, for instance, at that moment is not important, as far as your signal is concerned. Nor is the static electric and magnetic field at that moment.

share|cite|improve this answer

There are two components to making communication possible. As you correctly identified, the clue lies in making a Fourier transformation, i.e. looking at things in the frequency domain. The two components are:

(1) Spectrum Management. The authorities in charge of the radio spectrum make sure that there is no overlap (in frequency) of transmissions unless the transmitters are far enough away from each other so their target audiences don't overlap.

(2) The filter in your radio receiver. If you take the transmission (say at Jerry's 1054MHz), the center frequency is 1054MHz and the information modulating this carrier gives a small spread in frequency around this value, so you get a little hump instead of the delta function you'd have if there was only an unmodulated carrier. The filter in your receiver is tunable to select just this small spread and reject everything else. So there can even be large powers from other transmitters giving a large variability in the composite electric field at a point, but your filter will reject all the power except the power lying in the band of interest. (Well in reality it won't do this perfectly but that is the design target).

Minor nitpick - I think you meant "curvature of connection on a vector bundle", rather than "section of a vector bundle"?

share|cite|improve this answer
This is where , modulation comes into play. You should read about modulation, need for modulation also. – BigSack Sep 2 '12 at 8:53

Here is one way to look at the problem of signal detection.

Suppose we had two quantities. $X$ represents the signal that was transmitted. It does not change over time and is a scalar. $w$ represents random noise, which persists across all frequencies. Since $w$ is random, we can only represent it as a random variable (function which maps a real number to a real number). Typically $w \sim \mathcal{N}(0, \sigma^2)$. Typically, $\sigma^2$ increases as the temperature at the receiver increases.

There are two cases, either the signal was transmitted or it was not transmitted. This problem can be expressed as a hypothesis testing problem:

$$ \mathcal{H}_0 : Y = X + w$$ $$ \mathcal{H}_1 : Y = w$$

where under the hypothesis 0, the signal was transmitted and under hypothesis 1, the signal was not transmitted.

A simple test would be to declare a signal was transmitted every time $Y$ was above a certain value, and declare a signal was not transmitted every time $Y$ was below a certain value.

Notice by choosing this `certain value' (threshold), we are choosing the chance of falsely declaring a signal was present, when it was not present (this happens with a low threshold). We also choose the chance of missing the signal when it is present (this happens with a high threshold). Note, then increasing the false detection probability decreases the detection probability. Then, we have a tradeoff between these two probabilities, which depends on the threshold.

Note that earlier we said we had complete control over $X$. If we chose $X$ to be very large (transmitting with a large amount of power), we are increasing the signal to noise ratio, which is defined as the amplitude of the signal $X$, squared, divided by $\sigma^2$. In general, we can avoid the tradeoff in the previous paragraph if the signal to noise ratio is high. Signal detection performance increases as the signal to noise ratio increases (for optimal Neyman-Pearson detection methods).

To relate to your question, if $X$ was very small, then the problem does become much harder. We are more likely to miss the signal, when it is present. Therefore, different statistical techniques, modulation techniques (how $X$ is chosen), coding (ways to recover from detection errors), must be developed.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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