Noncoherent detection in Rayleigh fading channels

We have learned that BPSK fails in the Rayleigh fading channel. Can we design a signaling scheme that works in the fading channel?

The reason that BPSK fails is that BPSK encodes information in the phase of the transmit signal, and that the fading channel can change the phase of the signal uniformly randomly. So a natural idea is to encode the information in the magnitude of the transmit signal. One example is the orthogonal signaling, defined as

xA=(x[0]x[1])=(a0)  and  xB=(x[0]x[1])=(0a).

In this signal scheme, we use two symbols x[0] and x[1] to represent one bit. The tuples xA and xB have different magnitudes in different time slots.

The received signal is a tuple of two symbols

y=(y[0]y[1])=(h[0]x[0]+w[0]h[1]x[1]+w[1]).

If xA was sent, since x[1]=0, it is more likely that y[1] is close to zero. Therefore, we should be able to distinguish the two transmit signals by comparing the magnitudes of y[0] and y[1].

Maximum likelihood detector

We derive the maximum likelihood detector in this case:

x^={xA,if P(y|xA)P(y|xB),xB,if P(y|xA)<P(y|xB).

When xA was sent, we have y[0]CN(0,a2+N0) and y[1]CN(0,N0). When xB was sent, we have y[0]CN(0,N0) and y[1]CN(0,a2+N0). In addition, given xA or xB, y[0] and y[1] are conditionally independent because the chanenl gains h[0] and h[1] are independent and the noises w[0] and w[1] are independent. Hence, the conditional probability density function can be written explicitly as

P(y|xA)=P(y[0]|xA)P(y[1]|xA)=1π(a2+N0)e|y[0]|2a2+N01πN0e|y[1]|2N0,

and

P(y|xB)=P(y[0]|xB)P(y[1]|xB)=1πN0e|y[0]|2N01π(a2+N0)e|y[1]|2a2+N0.

Therefore, we have

P(y|xA)>P(y|xB)|y[0]|2a2+N0|y[1]|2N0>|y[0]|2N0|y[1]|2a2+N0|y[0]|2>|y[1]|2.

So the maximum likelihood detector has a simple decision rule:

x^={xA,if |y[0]|2|y[1]|2,xB,if |y[0]|2<|y[1]|2.

The maximum likelihood detector is again very intuitive: we compare the energy of the two symbols. It is also called the noncoherent detector or the energy detector.

Performance analysis

Now we analyze the error probability of such a detector. Again, due to symmetry, we can focus on the case where xA was sent but the detector output x^=xB.

pe=P(|y[0]|2<|y[1]|2|xA).

Given that xA was sent, y[0] and y[1] are circularly symmetric Gaussian random variables. Therefore, their energy z[0]=|y[0]|2 and z[1]=|y[1]|2 are exponentially distributed with means a2+N0 and N0, respectively. Therefore, the error probability is

pe=z1(1a2+N0ez0a2+N0)(1N0ez1N0)dz0dz1=12+a2/N0.

In this signal scheme, the energy per transmit symbol is a22, and the noise energy per symbol is N0. Hence, the SNR is a22N0. Therefore, the error probability can be rewritten as

pe=12(1+SNR).

In summary, in the fading channel, the orthogonal signal scheme works, but the error rate decays only linearly with the SNR. In contrast, the error rate decays exponentially with the SNR in the AWGN channel.