Detection gain of DSB-AM is less than 1

Detection gain is the ratio of the output signal-to-noise ratio of a demodulator to its input (pre-detection) SNR:

$$\gamma_d=\frac{\text{SNR}_o}{\text{SNR}_i}$$

It shows whether the detector improves ($\gamma_d>1$) or degrades ($\gamma_d<1$) the SNR.

Receiver model. The received signal $s(t)$ plus white noise ($N_0/2$) passes through a band-pass (IF) filter of bandwidth $B_T$, then a demodulator and a low-pass filter of bandwidth $W$ (message bandwidth).

 s(t)  +   +-------+    +-------+   +-----+
 ---->(+)-->|  BPF  |--->| demod |-->| LPF |--> y(t)
       ^    |  B_T  |    |       |   |  W  |
      w(t)  +-------+    +-------+   +-----+

The filtered noise is narrowband and is written in in-phase/quadrature form

$$n(t)=n_c(t)\cos\omega_ct-n_s(t)\sin\omega_ct$$

where $n_c$ and $n_s$ are low-pass, each with the same power as $n(t)$: $\overline{n_c^2}=\overline{n_s^2}=\overline{n^2}=N_0B_T$.

DSB-FC (conventional AM). $s(t)=A_c[1+k_am(t)]\cos\omega_ct$, $B_T=2W$. (Coherent and envelope detection give the same result at high SNR.)

• Signal power $=\frac{A_c^2}{2}(1+k_a^2P)$; noise in $2W$ $=2N_0W$:

$$\text{SNR}_i=\frac{A_c^2(1+k_a^2P)}{4N_0W}$$

• Coherent detection (or envelope detection at high SNR) and DC blocking:

$$y(t)=\frac{A_c}{2}k_am(t)+\frac12n_c(t)\ \Rightarrow\ \text{SNR}_o=\frac{A_c^2k_a^2P/4}{2N_0W/4}=\frac{A_c^2k_a^2P}{2N_0W}$$ $$\gamma_{DSB\text{-}FC}=\frac{\text{SNR}_o}{\text{SNR}_i}=\frac{2k_a^2P}{1+k_a^2P},\qquad \text{FOM}=\frac{k_a^2P}{1+k_a^2P}$$

The carrier uses power but carries no information, so the gain is reduced.

Proof that $\gamma_d<1$. To avoid over-modulation, $\lvert k_am(t)\rvert\le1$, so $k_a^2P=\overline{k_a^2m^2(t)}\le1$, with equality only for a square-wave message at 100% modulation. Hence

$$\gamma_d=\frac{2k_a^2P}{1+k_a^2P}\le\frac{2(1)}{1+1}=1$$

For any practical message (sinusoid, speech) $k_a^2P<1$ and $\gamma_d<1$.

Single tone, 100% modulation. $m(t)=A_m\cos\omega_mt$, $\mu=k_aA_m=1$, so $k_a^2P=\frac{\mu^2}{2}=\frac12$:

$$\begin{aligned} \gamma_{DSB\text{-}FC} &= \frac{2(\mu^2/2)}{1+\mu^2/2}=\frac{2\mu^2}{2+\mu^2}=\frac{2}{3}\approx0.667 \\ \text{FOM} &= \frac{\mu^2}{2+\mu^2}=\frac13 \end{aligned}$$

So for 100% single-tone DSB-AM the detection gain is $2/3<1$: the detector output SNR is about $1.76$ dB lower than the input SNR. The reason is that two-thirds of the transmitted power is in the carrier, which is counted in the input signal power but contributes nothing to the output message. (Compare DSB-SC, $\gamma_d=2$.)