SSB figure of merit and 32-QAM constellation

Figure of merit of SSB receiver

Figure of merit $=\dfrac{\text{SNR}_o}{\text{SNR}_c}$, where $\text{SNR}_c$ is the channel SNR: average received signal power divided by noise power in the message bandwidth $W$.

Coherent SSB receiver: BPF of width $W$ → product modulator with $\cos\omega_ct$ → LPF ($W$). Noise PSD $N_0/2$.

Signal: $s(t)=\frac{A_c}{2}[m(t)\cos\omega_ct-\hat m(t)\sin\omega_ct]$ (USB). Since $m$ and $\hat m$ have equal power $P$ and are orthogonal:

$$P_s=\frac{A_c^2}{4}\cdot\frac{P}{2}\cdot2=\frac{A_c^2P}{4},\qquad \text{SNR}_c=\frac{A_c^2P}{4N_0W}$$

Noise after BPF (width $W$): $n(t)=n_c\cos\omega_ct-n_s\sin\omega_ct$ with $\overline{n_c^2}=\overline{n_s^2}=N_0W$.

Detector output (multiply by $\cos\omega_ct$, LPF):

$$y(t)=\frac{A_c}{4}m(t)+\frac12n_c(t)$$ $$\begin{aligned} S_o &= \frac{A_c^2P}{16},\qquad N_o=\frac{N_0W}{4} \\ \text{SNR}_o &= \frac{A_c^2P/16}{N_0W/4}=\frac{A_c^2P}{4N_0W} \end{aligned}$$ $$\text{FOM}_{SSB}=\frac{\text{SNR}_o}{\text{SNR}_c}=1$$

SSB has the same noise performance as DSB-SC and baseband, while using only half the bandwidth.

Efficient constellation of 32-QAM

In 32-QAM each symbol carries $\log_232=5$ bits. 32 is not a perfect square, so a square grid cannot be used. The efficient arrangement is the cross constellation: a $6\times6$ square grid of points (36) with the 4 corner points removed, giving 32 points. Removing the corners (which have the highest energy) lowers the average and peak power for the same minimum distance $d$.

              Q
              |
        .  *  *  *  *  .
        *  *  *  *  *  *
        *  *  *  *  *  *
   -----------+------------ I
        *  *  *  *  *  *
        *  *  *  *  *  *
        .  *  *  *  *  .
              |
  * = point (32), . = removed corner

Points lie at $I,Q\in\{\pm1,\pm3,\pm5\}\cdot\frac d2$ except the four corners $(\pm5,\pm5)$.