0
$\begingroup$

I started working with HF frequency. Right now, I am working with 7 MHz frequency. I have been provided a range of 7-7.5 MHz. I have to divide that range into 132 distinct frequencies to assign each frequency to a different node. Please let me know what is the minimum frequency resolution for 2 frequencies which do not interfere with each other. For example, 7.001 and 7.002 MHz freq will interfere with each other or not??

$\endgroup$
5
$\begingroup$

Very basic wave propagation theory says:

No two different frequencies ever interfere in a linear system. So, any separation, no matter how small, of the occupied bands is enough in a perfect world!

However, we don't live in a perfect world. So you need to model your system more detailledly, think about bandwidths, filters, frequency uncertainty of oscillators.

I'll be a bit blunt, in hopes it helps you find a better question, sorry: you graduated in EE with specializations in RF design. I'd have expected you to ask a more detailed question here with way more background on your overall system! I think you can do better – interference happens only under circumstances where energy of one thing ends up in the other, so maybe you'd want to analyze your proposed system more for the places where that might happen!

| improve this answer | |
$\endgroup$
  • $\begingroup$ Wasn't sure if "any separation, no matter how small" was physically possible, but it appears to be: physics.stackexchange.com/questions/73959/… (by way of physics.stackexchange.com/questions/110463/…). $\endgroup$ – natevw - AF7TB Jun 11 at 18:13
  • $\begingroup$ @natevw-AF7TB it is, but you can only observe that arbitrarily small separation if you observe infinitely long (for all other things: there's the bandwidth–duration product limit, which means your ability to determine frequency is proportional to the duration for which you observe that frequency. Manifests itself in filter design, but also in physical things like the Heisenberg uncertainty principle, which says you can't know impulse and location of a particle exactly at the same time. Impulse space and location space are linked through the Fourier transform.) $\endgroup$ – Marcus Müller Jun 11 at 19:04
1
$\begingroup$

That would depend on the bandwidth of the transmission.

enter image description here

The RSGB 'Amateur Radio Band Plans' is a very good reference.

http://www.hflink.com/bandplans/UK_bandplan.pdf

| improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ OM Nandu, would you please copy some information from the link into your answer, in case the linked document goes away? $\endgroup$ – rclocher3 Jun 10 at 17:56
  • 2
    $\begingroup$ Thank you, OM Rob. Done! 73, Nandu. $\endgroup$ – vu2nan Jun 11 at 2:47
0
$\begingroup$

If you want 2 unmodulated finite length sinusoids to be mathematically orthogonal, then the frequency separation is 1/T, where T is the length in time of a DFT used to detect them, and the period of the carriers are an integer submultiple of T. If you want a 3 dB gap between signals or the carriers are not integer periodic in T, then double that separation or a little more. If you use a non-rectangular window, then the separation roughly doubles again. If the channel has an impulse response, then you may need to add a cyclic prefix to cover the multi-path skew, which typically increases T by 25% for a given carrier frequency separation to be orthogonal.

| improve this answer | |
$\endgroup$

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.