Power saving of SSB at 100% and 50% modulation

Power saving is measured against conventional DSB-FC AM with the same carrier and modulation index.

Formulas

With carrier power $P_c = A_c^2/2$ and modulation index $\mu$:

$$P_{DSB\text{-}FC} = P_c\left(1 + \frac{\mu^2}{2}\right), \qquad P_{SSB} = P_c\frac{\mu^2}{4}$$

since one sideband has power $\left(\frac{\mu A_c}{2}\right)^2/2 = P_c\mu^2/4$.

$$\text{Saving} = \frac{P_{DSB\text{-}FC} - P_{SSB}}{P_{DSB\text{-}FC}} = \frac{1 + \frac{\mu^2}{2} - \frac{\mu^2}{4}}{1 + \frac{\mu^2}{2}} = \frac{4 + \mu^2}{4 + 2\mu^2}$$

(i) $\mu = 1$ (100%)

$$\begin{aligned} P_{DSB\text{-}FC} &= P_c(1 + 0.5) = 1.5P_c \\ P_{SSB} &= 0.25P_c \\ \text{Saving} &= \frac{1.5 - 0.25}{1.5} = \frac{1.25}{1.5} = 0.8333 = 83.33\% \end{aligned}$$

(ii) $\mu = 0.5$ (50%)

$$\begin{aligned} P_{DSB\text{-}FC} &= P_c(1 + 0.125) = 1.125P_c \\ P_{SSB} &= \frac{0.25}{4}P_c = 0.0625P_c \\ \text{Saving} &= \frac{1.125 - 0.0625}{1.125} = \frac{1.0625}{1.125} = 0.9444 = 94.44\% \end{aligned}$$

Answer: power saving = 83.33% at 100% modulation and 94.44% at 50% modulation.

The saving is larger at low modulation because most of the DSB-FC power is in the carrier, which SSB does not send.