Go Down

Topic: Optimal FSK bandwidth (Read 995 times) previous topic - next topic


I'm trying to find a simple correlation between the "optimal" bandwidth in a FSK system in relation to the "Mark" carrier frequency. In other words, if I have an xkHz mark frequency (Representing a digital High), what should my lower space frequency be?

I'm not sure if this is as simple as I think it is. I'm envisioning that Bandwidth = some% of mark frequency for most applications.



Basically,  all depends on baud rate. To reliable discriminate two symbols ( 0 or 1) at receiver's side, the higher baud rate would require wider spacing to prevent overlapping two spectum,  so there is direct proportion 


Thank you,

Im assuming this is a form or variation of Carsons rule?

Here's my confusion, and it may just be a definition confusion. According to this paper(http://www.dip.ee.uct.ac.za/~nicolls/lectures/eee482f/13_fsk_2up.pdf), Bandwidth BW=2dF + 2B, where dF is the difference in frequency between the mark and space frequency, and B is the data rate. Now, my assumption is that Bandwidth is literally the frequency deviation, or the difference between mark and space frequencies. Is this not correct?


After reading an example:
The bit period in the baseband signal is T = 1/200000 seconds, and the
baseband pulses are rectangular. The bandwidth of the baseband signal (to the
first null) is given by B = 1/(2T ). For the RF components, 2 d f = 150 kHz.
The bandwidth of the FSK signal is therefore
2 d f + 2B = 150 kHz + 200 kHz = 350 kHz.
I find it a little bit confusing, why author using coefficient 2 referring to  d f, when it's just "spacing" between marks "0" and "1" frequencies?
This is a manipulation, and if I need to make things more difficult to understand, I'd exploit this technics .

Briskly searching on my e-book shelf, I find this explanation more clear:
When only a short time duration dT is available, the uncertainty theorem puts a constraint on the accuracy of recognizing the difference between the two frequencies df.       
                                                         dF x dT >= 2 x PI();

For example, the telephone channel is less than 4 KHz wide, and so a rather large separation would be df x 2 KHz. This implies that telephone line FSK information transfer rates will not exceed around 300 b/s.

Digital Signal Processing
A Computer SciencePerspective
Jonathan (Y) Stein
p. 703

Never heard of "Carsons rule". I like http://en.wikipedia.org/wiki/Uncertainty_principle

Go Up