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I have two sine signals with frequencies f1 and f2, and phase p1 and p2.

I modulate the phase of these signals to transmit two symbols simultaneously, chosen from the set [0,1,2,3]. For example, I represent 0 by 45 degrees phase shift, 1 by 90 degrees phase shift, etc.

The signals will pass through an optical communication channel, potentially getting distorted. This means there could be intermodulation products and harmonics.

I plan on using an FFT to determine magnitude and phase changes of the signals. I am hoping to retrieve the symbols from the phase changes I detect using the FFT. I want the magnitude to stay constant while the phase is shifting to transmit the symbols.

The sampling rate I'm working with is 7800 hz. I will potentially have to fit 16 transmission channels (frequencies) below 1500 hz. At any time, two frequencies randomly chosen from the set of 16 frequencies may be present, transmitting symbols as explained above.

Is this going to work, and what are the key issues to keep and mind and to work around? Any practical pointers or references are highly appreciated as I'm a novice in phase modulation and encoding/decoding of information in this fashion.

Thanks for helping! B

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  • $\begingroup$ Literature on OFDM (implementation, data rates, synchronization issues, etc.) might be applicable to your described method. $\endgroup$
    – hotpaw2
    Jan 11, 2014 at 20:48
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    $\begingroup$ If the channel suffers from intermodulation distortion, then a single FSK carrier may be a better way to go. $\endgroup$
    – John
    Jan 12, 2014 at 2:17
  • $\begingroup$ @John thanks for the suggestion, in the application the two frequencies are used to allow bidirectional communication. Is this something that can be done using FSK? $\endgroup$
    – b20000
    Jan 12, 2014 at 21:52
  • $\begingroup$ Is the communication half-duplex or full-duplex? $\endgroup$
    – John
    Jan 12, 2014 at 23:13
  • $\begingroup$ @john i have the luxury of designing the system how I want…. I'd prefer to make it full duplex because I can imagine doing it half duplex would make things more complicated. $\endgroup$
    – b20000
    Jan 13, 2014 at 17:13

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