Non coherent demodulation of pi/8 D8PSK (TETRA)

In TETRA specifications, one of the modulation technique used is $\frac{\pi}{8}$ Differential 8 Phase Shift Keying (D8PSK). We will discuss the bit error rate with non-coherent demodulation of $\frac{\pi}{8}$ D8PSK in Additive White Gaussian Noise (AWGN) channel.

pi/8 D8PSK

The incoming bit sequence is grouped into three bits and is mapped into differential phase as follows:

B(3k-2)	B(3k-1)	B(3k)	$D\phi(k)$
0	0	0	$+\frac{1\pi}{8}$
0	0	1	$+\frac{3\pi}{8}$
1	0	1	$+\frac{5\pi}{8}$
1	0	0	$+\frac{7\pi}{8}$
0	1	0	$-\frac{1\pi}{8}$
0	1	1	$-\frac{3\pi}{8}$
1	1	1	$-\frac{5\pi}{8}$
1	1	0	$-\frac{7\pi}{8}$

Table : Phase transitions for $\frac{\pi}{8}$ D8PSK modulation (Ref Table 5.2 of ETSI 301-893 V3.2.1)

The modulation symbol $S(k)$ is formed by applying a phase offset to previous symbol $S(k-1)$ and is defined as follows:

$S(k) = S(k-1) e^{jD\phi(k)}$ and

$S(0) = 1$ .

Alternately, the phase transitions can be represented as

$\phi(k) = \phi(k-1) + D\phi(k)$ .

The constellation of the $\frac{\pi}{8}$ D8PSK occupies phase values separated by $\frac{\pi}{8}$ as shown below in the blue dots. The red lines shows all possible phase transitions.

Figure: Constellation of $\frac{\pi}{8}$ D8PSK (Ref Figure 5.2 of ETSI 301-893 V3.2.1)

Channel Model

The transmitted waveform gets corrupted by noise $n$ , typically referred to as Additive White Gaussian Noise (AWGN).

Additive : As the noise gets ‘added’ (and not multiplied) to the received signal

White : The spectrum of the noise if flat for all frequencies.

Gaussian : The values of the noise $n$ follows the Gaussian probability distribution function, $p(n) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-(n-\mu)^2}{2\sigma^2}$

with mean $\mu=0$ and

variance $\sigma^2 = \frac{N_0}{2}$ .

The received symbol is,

$y(k) = S(k) + n(k)$

Non Coherent Receiver

A non-coherent receiver relies on the phase transitions between consecutive symbols to form an estimate of the transmitted bits. The sequence of operation is as follows:

a) On the received symbols estimate the phase

$\hat{\phi}(k) = \angle y(k)$

b) Find the delta of the estimated phase between consecutive symbols

$\hat{D}\phi(k) = \hat{\phi}(k)-\hat{\phi}(k-1)$

c) Quantize the estimated delta phase values lying from $[-\pi,\mbox{ } +\pi)$ as follows and convert to bits per the following encoding:

$\begin{array} <br /> -\pi & \le & \hat{D}\phi(k) & < & -\frac{3\pi}{4} & \right & -\frac{7\pi}{8}& : & 110\\ <br /> -\frac{3\pi}{4} & \le & \hat{D}\phi(k) & < & -\frac{\pi}{2} & \right & -\frac{5\pi}{8}& : & 111\\ <br /> -\frac{\pi}{2} & \le &\hat{D}\phi(k) & < & -\frac{\pi}{4} & \right & -\frac{3\pi}{8}& : & 011\\ -\frac{\pi}{4} & \le & \hat{D}\phi(k) & <& 0 & \right & -\frac{\pi}{8}& : &010\\ 0 & \le & \hat{D}\phi(k) & <& \frac{\pi}{4} & \right & +\frac{\pi}{8}& : &000\\ \frac{\pi}{4} & \le & \hat{D}\phi(k) & <& \frac{\pi}{2} & \right & +\frac{3\pi}{8}& : &001 \\ \frac{\pi}{2} & \le & \hat{D}\phi(k) & <& \frac{3\pi}{4} & \right & +\frac{5\pi}{8}& : &101 \\ \frac{3\pi}{4} & \le & \hat{D}\phi(k) & <& \pi & \right & +\frac{7\pi}{8}& : &100 \\ \end{array}$ .