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I want to design an optimal (minimizing error probability) receiver for system of two waveforms (with $0<\alpha<\frac{\pi}{2}$):

\begin{align} s_1 &=A\cos^{2}(\pi f_c t + \alpha)\\ s_2 &=A\cos^{2}(\pi f_c t - \alpha) \end{align}

It seems like a modified BPSK, as:

\begin{align} s_1&=A\cdot\ \frac{1 + \cos(2\pi f_c t + 2\alpha)}{2}\\ s_2&=A\cdot\ \frac{1 + \cos(2\pi f_c t - 2\alpha)}{2} \end{align}

But I don't know what to do next.

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  • $\begingroup$ Please edit your question to add the homework or self-study tag. $\endgroup$ Commented Nov 5, 2020 at 19:13

2 Answers 2

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Awesome, your transformation into a sum of a constant and a cosine is wonderful!

Since both signals are cosines shifted from the origin by the same $\frac A2$, that doesn't chang their euclidean distance.

Thus, subtracting $\frac A2$ doesn't change your decider's error rate. Also, it's technically super easy – just a high pass filter that removes the constant, no matter what it was.

The you're left with

\begin{align} \tilde s_1 &= \frac A2 \cos(2\pi f_c t + 2\alpha)\\ \tilde s_2 &= \frac A2 \cos(2\pi f_c t - 2\alpha)& \beta := 2\alpha\\[1.5em] \tilde s_1 &= \frac A2 \cos(2\pi f_c t + \beta)\\ \tilde s_2 &= \frac A2 \cos(2\pi f_c t - \beta),&0<\beta<\pi \end{align}

Now, IQ-mixing this down with $f_c$ will indeed give us some phase-modulation-type thing, with the two constellation points with $\pm \beta$ as phase – but that's not a BPSK, since that is not always a phase difference of 180°!

However, it's always symmetrical to the I-axis in your constellation diagram. So, you can just remove the real part of your baseband signal and end up with something where the sign contains all the information.

Note that this, geometrically, is a Maximum Likelihood decision only if your noise distribution (which haven't told us) is is symmetrical after the transformation – which is the case for the usual complex Gaussian noise we assume, but I'd be careful whether that's the right assumption here - such non-symmetric PSK constellations are (surprisingly) actually used in a couple of cases, most of which involve interesting residual carrier systems with nonlinearities that shape the noise in a non-symmetric way.

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  • $\begingroup$ The noise is AWGN with power $\frac{N_0}{2}$ Watt/Hz. So it is also a ML receiver? In addition, how can I derive the probability of error from this receiver? $\endgroup$
    – thnghh
    Commented Nov 5, 2020 at 16:33
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Contrary to what Marcus Muller says, the signal set is a BPSK signal set with a DC offset. The BPSK signal set is not an antipodal BPSK signal set unless $\alpha = \pi/4$; in which case the two signals $\cos(2\pi f_c t\pm 2\alpha)$ differ in phase by $4\alpha = \pi$ making it a BPSK signal set even in Marcus Muller's narrow view of what BPSK means. But more generally, $$\cos(2\pi f_c t\pm 2\alpha) = \cos(2\alpha)\cos(2\pi f_c t)\,\mp\, \sin(2\alpha)\sin(2\pi f_c t)$$ showing that the given signals can be expressed as

$\big($a DC component plus a $\cos(2\pi f_c t)$ term$\big)\mp$ a $\sin(2\pi f_c t)$ term

and this last part (the stuff outside the big parentheses; the stuff inside the big parentheses is the same for both possible transmitted signals) is a classical antipodal BPSK signal that can be detected in the usual way using matched filtering or correlation receivers and the error probability calculated as described in the link just cited. If anyone is dead set on finding the complex baseband $(I,Q)$ equivalent of the signals and making the decision therefrom because that is the only way of approaching the problem that the person knows, then the decision is made based entirely on the $Q$ branch output with the $I$ branch output being ignored completely.

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  • $\begingroup$ right, if one can project it to a single axis in the end, and decide between two points, then it's BPSK, you're right – I just didn't see the fact that my shift and my projection could've been absorbed in one shift. A well-deserved criticism for me, and a well-deserved upvote to you. $\endgroup$ Commented Nov 6, 2020 at 13:29

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