---
title: "Detection gain of DSB-AM is less than 1"
subject: "Communication Systems"
chapter: 9
key: "semester_6/communication_systems:detection-gain-dsbam-less-than-1"
source: https://www.bctengineer.com/q/semester_6/communication_systems/detection-gain-dsbam-less-than-1
site: BCT Engineering
---

# Detection gain of DSB-AM is less than 1

Communication Systems · Chapter 9: Noise in Communication Systems

- Asked in: 2074 Chaitra (CS II), 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/detection-gain-dsbam-less-than-1
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9

## Question

What is detecting gain? Prove that for 100% modulation of (DSB-AM), the detection gain is less than 1.

## Short answer

Detection gain is the ratio of output SNR to input SNR of a demodulator. For DSB-AM with 100% single-tone modulation, the detection gain is 2/3 ≈ 0.667, less than 1 because the carrier power contributes to input SNR but not to the output message.

## Answer

**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).

```text
 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, $\gamma_d=2$.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Detection gain of a demodulator is defined as:
   A. Output SNR divided by input (pre-detection) SNR
   B. Input SNR divided by output SNR
   C. Output signal power divided by input signal power
   D. Message bandwidth divided by transmission bandwidth
   Answer: A. Detection gain = SNR_o / SNR_i; greater than 1 means the detector improves SNR, less than 1 means it degrades it. (Key point: Definition of detection gain)

2. For DSB-FC with coherent or high-SNR envelope detection, the detection gain is:
   A. k_a^2 P / (1 + k_a^2 P)
   B. 2 k_a^2 P / (1 + k_a^2 P)
   C. 2 / (1 + k_a^2 P)
   D. 1 + k_a^2 P
   Answer: B. SNR_o = A_c^2 k_a^2 P / (2 N0 W) and SNR_i = A_c^2 (1 + k_a^2 P)/(4 N0 W), giving 2k_a^2P/(1 + k_a^2P). (Key point: DSB-FC detection gain formula)

3. For single-tone 100% modulation (μ = 1) of DSB-AM, the detection gain is:
   A. 2
   B. 1
   C. 2/3, which is less than 1
   D. 1/3
   Answer: C. k_a^2 P = μ^2/2 = 1/2, so gain = 2(1/2)/(1 + 1/2) = 2/3. The carrier takes two-thirds of the power but carries no message. (Key point: Gain is below 1 at 100% tone modulation)

## Related in Communication Systems

- [DSB-AM, DSB-SC and SSB spectra; synchronous detection](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adsbam-dsbsc-ssb-spectrum-synchronous) — Communication Systems, chapter 2
- [Phase error in coherent DSB-SC detection; AM demodulation by PLL](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aphase-error-dsbsc-am-pll) — Communication Systems, chapter 2
- [Detection gain of SSB-SC vs DSB-SC](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adetection-gain-ssbsc-dsbsc) — Communication Systems, chapter 9, asked 2×
- [Capture effect; gain parameter of DSB-FC envelope detection](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Acapture-effect-dsbfc-envelope-gain) — Communication Systems, chapter 9
- [Why SSB preferred; AM power at 100% modulation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Assb-preferred-am-power-1000w) — Communication Systems, chapter 2
- [AM wave diagram, modulation index; why near 100%](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-diagram-index-100-percent) — Communication Systems, chapter 2

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