For a project of mine(uhf radios, pic32mx450f), I went with bpsk and 8b/10b encoding, with a 10 sine + 8b/10b k_28_1 preamble. This allows syncing to the correct phase, and then I run the moving bitstream through the detector until it hits the special code.

For phase detection, I looked at quite a few different ways, but ended up doing simple peak tracking and phase adjustment on each peak detect, which could be construed as a form of locked loop. I ended up locking clock adjustment after the preamble, but it would work to keep it open as well. Incoming audio is filtered with low and high pass fir filters to clean the signal and remove sub audibles. Everything I read seemed to point to the filter being an important part of the detector.

I think locking onto a raised root sine filtered drifting qpsk is going to be a challenge, bpsk less so, but probably more doable. The best options will depend on channel quality, I'd say.

I suppose that since this is a university project, perhaps you have chosen some of these complex items for the challenge? One alternative that comes to mind in place of directly sharing voice with data, would be to use a sub audible, or low frequency carrier to differentiate between repeated voice and data, similar to sub-audible tones used by uhf radio systems.

This isn't really a direct answer for all your questions, but this is too lengthy for a comment.

Here are some links I studied from dsp.stackexchange.com, which incidentally may be better suited to this question, or have more folks familiar with your questions nuances.

https://dsp.stackexchange.com/questions/8456/how-to-perform-carrier-phase-recovery-in-software

https://dsp.stackexchange.com/questions/3158/how-to-demodulate-an-afsk-signal-in-software?rq=1

1.414 is the square root of 2. If the same signal is sent over the 2 independent channels, then your received Noise PSD is the same, but your signal is now twice as powered. You add the signals, but the noise is not additive because it is IID / uncorrelated. Twice as much power is a 3dB Gain. I'm a little hazy on BER curves, but if that Q(sqrt(g)) is the likelyhood function, then instead of g you have 2*g, and Sqrt(2*g) = Sqrt(2)*sqrt(g)

Yeah. The argument under the square root is the received Signal to noise ratio in Eb/No (energy-per-bit/Noise-Power-in-1Hz-bandwidth). One receiver and one channel is R = S + N. In this example R = S1 + S2 + N. If S1 and S2 are equal then you have 2S + N, thus twice the SNR.

## Best Answer

The AWGN channel with noise power spectral density $$\frac{N}{2}$$ And the signals: $$s_1(t)=\sqrt{2\frac{E}{T}}\sin(t)\qquad t\in[0,T]$$ $$s_2(t)=-\sqrt{2\frac{E}{T}}\sin(t)=-s_1(t)$$

In a normal coherent oscillator, you would receive something like the following in the constellation diagram (noise not represented):

(the picture may be wrong because of the usage of sine or cosine, but you get the idea).

The red dashed line is the decision boundary, and the distance from the signal to this boundary represents the actual energy that the receiver can use to make the decision. So, in this case, it is obvious that the BER will be $$Q\bigg(\sqrt{\frac{2E}{N}}\bigg)$$

If the local oscillator in a coherent receiver is ahead in phase by 45°, it means that the receiver has is decision boundary rotated in the constellation diagram:

So the distance from the signal to the boundary is actually less than before, therefore the BER increases. How much? It depends on the distance to the boundary, so the new BER is: $$Q\bigg(\sqrt{\frac{E}{N}}\bigg)$$