1
$\begingroup$

Im sudying this concepts, but i doesn't find anything clear about it, because I don't have a clear interpretation of the bandidth concept, and the number of bands in a beating wave.

$\bullet$ Bandwidth I think the definition of the bandwidth is the lower frequency, i.e. the modulator freqency, we can call it $\Delta \omega$

$\bullet$ Bandwith to transmit in a certain range For this I assumed that the bandwith $\Delta \omega=\omega_{\max}$ where $\omega_{\max}$ is the higher frequency of the range we want to transmit.

$\bullet$ Number of bands with fixed bandwidth For a certain other frequency $\omega_0$ I'm not sure if this is well assumed, but for this i calculated 'how many frequencies of the bandwidth fit in one $\omega_0$' i.e. $$N=\frac{\omega_0}{\Delta \omega}$$

I dont know if this assumptions are well, because I didn't find lot of information about this in internet, and i need someone to confirm it, thanks.

$\endgroup$
2
  • 1
    $\begingroup$ Can you give a more explicit example? For example either the formula, the spectrum, or the waveform of the "beating wave" you're asking about? $\endgroup$
    – The Photon
    Jun 27 at 17:59
  • 1
    $\begingroup$ You might also want to search for existing questions about the term bandwidth on Electrical Engineering Stackexchange. We've had several that might answer your question (although I don't understand your question well enough to be sure). $\endgroup$
    – The Photon
    Jun 27 at 18:01
5
$\begingroup$

A quick review of beats: Adding 2 pure frequencies gives:

$$f(t)=e^{i\omega_1 t}+ e^{i\omega_2 t}$$

with the average frequency:

$$ \omega_0\equiv\frac 1 2(\omega_1+\omega_2)$$

and half-difference:

$$ \Delta\omega\equiv\frac 1 2(\omega_2-\omega_1)\equiv \frac{BW} 2$$

you get:

$$f(t)=e^{i(\omega_0-\Delta\omega )t}+e^{i(\omega_0+\Delta\omega )t}$$

$$f(t)=e^{i\omega_0 t}e^{-i\Delta\omega t}+e^{i\omega_0 t}e^{i\Delta\omega t}$$

$$f(t)=e^{i\omega_0 t}[e^{i\Delta\omega t}+e^{-i\Delta\omega t}]$$ $$f(t)=e^{i\omega_0 t}[\cos(-\Delta\omega t)+i\sin(-\Delta\omega t)+\cos(\Delta\omega t)-i\sin(\Delta\omega t)]$$ $$f(t)= [2\cos(\Delta\omega t)e^{i\omega_0 t}]$$

Which is a "pure" tone at $\omega_0$ modulated at $\Delta\omega$, as shown in wikipedia picture (https://en.wikipedia.org/wiki/Beat_(acoustics) ):

enter image description here

If we write it as:

$$ f(t)= A(t)\times e^{i\omega_0 t}$$

then

$$ A(t) = 2\cos(\Delta\omega t) $$

There is not a lot of information in that signal.

Suppose you can modulate $A(t)$ at all frequencies from $-\Delta\omega$ to $+\Delta\omega$? Fourier analysis tells that:

$$ A(t)=\int_{-\Delta\omega}^{+\Delta\omega}\tilde{A} (\omega')e^{i\omega't}d\omega'$$

Then:

$$f(t)=A(t)e^{i\omega_0 t} $$

and the frequency content is the convolution of the Fourier transforms of each multiplicand:

$$\tilde{f}(\omega)=(\tilde{A}(\omega)\circledast \delta(\omega-\omega_0))$$

The convolution just shifts the frequency content in the bandwidth from being zero-centered over to $\omega_0$.

In practice, that means we can encode information with bandwidth $BW=2\Delta\omega$ into $A(t)$. That signal is then used to modulate a carrier wave $\exp{i\omega_0 t}$ (with $\omega_0 \gg \Delta\omega$).

The ratio:

$$ \frac{\omega_0}{\Delta\omega}$$

doesn't really mean much, as long as it's bigger than one. That's usually an implementation consideration. For example, if you're operating an L-band radar at 1,500 MHz, you would like all of your signal to be 'radar', so you may go $+/-100$ MHz.

$\endgroup$
4
  • $\begingroup$ What I mean with the before relation, is for example, given a bandwidth $BW$ fixed, I dont know how to calculate the number of bands that will 'fit' in this bandwidth for the visible spectrum. So what I have assumed is that the number of bands will be about $N=\omega' / BW$, that in this case as we are treating with visible, we can pick $\omega'=2\pi 500$nm. If is not like this, how we could get the number of bands that fit in a certain bandwith? $\endgroup$
    – euler33
    Jun 27 at 18:42
  • $\begingroup$ @euler33 why? what is the physical significance of number-of-bands? $\endgroup$
    – JEB
    Jun 27 at 20:11
  • $\begingroup$ Is more or less the thing that i dont have clear, this comes from a problem, where we have a $BW=2 \omega'$ a bandwidth of a beating. They ask me for the bandwidth for transmit in audible spectrum $[20Hz-20kHz]$, so in this case $BW=2 \cdot 2 \pi 20 kHz$. But later they ask for how many bands will fit in the visible spectrum, and i dont know at all also the physical significance of this number of bands $\endgroup$
    – euler33
    Jun 27 at 21:23
  • 1
    $\begingroup$ the point of that is to show how much bandwidth is available in optical communication. You need 40 kHz to capture all of human hearing, so on an optical system (320 THz), you can hold almost 1 billion channels. $\endgroup$
    – JEB
    Jun 28 at 0:39
1
$\begingroup$

To a radio engineer, bandwidth is the frequency spectrum occupied by the information content of a modulated signal. A simple example is a single voice channel, where the bandwidth is 3.1kHz from the lowest note to the highest note. What you’re discussing is the sum and difference components of two pure sine waves. Strictly speaking, the bandwidth of these is zero because they don’t contain a spread of information.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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