Electronic – What happens if SINR is < 1

communicationdigital-communications

What happens if the SINR of a signal is less than 1 in a (wireless) network? Is that practical at all to have a SINR < 1? Using the Shannon capacity formula, it seems that we must be able to have SINR < 1, but we get a rate < 1 bits/sec/Hz.

Best Answer

Yes, it is practical.

For example, if you have an ASK signal with modulation depth 60%:

>> am = [ (ones(1,100) * 0.2) (ones(1,200) * 0.8) (ones(1,200) * 0.2) (ones(1,100) * 0.8)];

ASK signal

Using a low-pass filter

>> d = fdesign.lowpass(0.01, 0.02, 0.01, 100);
>>  hd = design(d)

hd =

 FilterStructure: 'Direct-Form FIR'
      Arithmetic: 'double'         
       Numerator: [1x946 double]   
PersistentMemory: false

you can reduce the output signal bandwidth in order to be nice to the neighbors:

>> amlp = filter2(hd.Numerator, sig);

AM after lowpass

The recipient gets a noisy signal

>> amno = amlp + 2*rand(1,600) - 0.5;

AM with noise

and also uses a lowpass filter to reconstruct it:

>> amre = filter2(hd.Numerator, amno) - mean(amno) + 0.5;

Reconstructed signal

This signal is sufficiently similar to the original signal that you can decide between 0 and 1 bits, but you need a rather narrow filter here in order to remove the noise -- in my case, 1% of the sampling rate (that is the .01 above). Note that we're only interested in the signal at the points 50, 150, 250, ..., 550, i.e. the middle of each symbol.

In order to be able to reconstruct that signal, I had to use rather long symbols (100 samples). That is, with 100 Hz sampling rate, which would allow me to express frequencies up to 50 Hz, I can only transfer 1 bit per second.

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