--- title: "Detection gains of DSB-FC, DSB-SC and SSB" subject: "Communication Systems" chapter: 9 key: "semester_6/communication_systems:detection-gain-dsbfc-dsbsc-ssb" source: https://www.bctengineer.com/q/semester_6/communication_systems/detection-gain-dsbfc-dsbsc-ssb site: BCT Engineering --- # Detection gains of DSB-FC, DSB-SC and SSB Communication Systems · Chapter 9: Noise in Communication Systems - Asked in: 2073 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-dsbfc-dsbsc-ssb - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9 ## Question Calculate detection gains for DSB-FC, DSB-SC and SSB and also compare them. ## Answer **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 $n(t)$: $\overline{n_c^2}=\overline{n_s^2}=\overline{n^2}=N_0B_T$. **Definitions.** Message power $P=\overline{m^2(t)}$. - Input SNR $\text{SNR}_i=\dfrac{\text{received signal power}}{\text{noise power in } B_T}$ - Output SNR $\text{SNR}_o=\dfrac{\text{message power at output}}{\text{noise power at output}}$ - **Detection gain** $\gamma_d=\dfrac{\text{SNR}_o}{\text{SNR}_i}$; **figure of merit** $=\dfrac{\text{SNR}_o}{\text{SNR}_{ref}}$, where $\text{SNR}_{ref}$ is the channel SNR measured in the message band $W$ (baseband reference). Both show how much the demodulator improves or worsens SNR. **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. **Single tone, 100% modulation.** $m(t)=A_m\cos\omega_mt$, $\mu=k_aA_m=1$, so $k_a^2P=\frac{\mu^2}{2}=\frac12$: $$ \begin{aligned} \gamma_{DSB\text{-}FC} &= \frac{2(\mu^2/2)}{1+\mu^2/2}=\frac{2\mu^2}{2+\mu^2}=\frac{2}{3}\approx0.667 \\ \text{FOM} &= \frac{\mu^2}{2+\mu^2}=\frac13 \end{aligned} $$ **DSB-SC (coherent detection).** $s(t)=A_cm(t)\cos\omega_ct$, $B_T=2W$. - Signal power $=\frac{A_c^2P}{2}$; noise in $2W$ $=2N_0W$, so $\text{SNR}_i=\dfrac{A_c^2P}{4N_0W}$. - Multiply $x(t)=s(t)+n(t)$ by $\cos\omega_ct$ and low-pass filter: $$ y(t)=\frac{A_c}{2}m(t)+\frac12n_c(t) $$ - $\text{SNR}_o=\dfrac{A_c^2P/4}{2N_0W/4}=\dfrac{A_c^2P}{2N_0W}$ $$ \gamma_{DSB\text{-}SC}=\frac{\text{SNR}_o}{\text{SNR}_i}=2 $$ The quadrature noise $n_s$ is rejected by coherent detection, giving a 3 dB gain. With baseband reference $\text{SNR}_{ref}=\frac{A_c^2P}{2N_0W}$, figure of merit $=1$. **SSB-SC (coherent detection).** $s(t)=\frac{A_c}{2}[m(t)\cos\omega_ct\mp\hat m(t)\sin\omega_ct]$, $B_T=W$. - Signal power $=\frac{A_c^2}{4}\cdot\frac{P+P}{2}=\frac{A_c^2P}{4}$ (since $\overline{\hat m^2}=P$); noise in $W$ $=N_0W$, so $\text{SNR}_i=\dfrac{A_c^2P}{4N_0W}$. - Multiply by $\cos\omega_ct$ and low-pass filter: $$ y(t)=\frac{A_c}{4}m(t)+\frac12n_c(t) $$ - Signal power $\frac{A_c^2P}{16}$, noise power $\frac{N_0W}{4}$, so $\text{SNR}_o=\dfrac{A_c^2P}{4N_0W}$ $$ \gamma_{SSB}=\frac{\text{SNR}_o}{\text{SNR}_i}=1 $$ With baseband reference ($\text{SNR}_{ref}=\frac{A_c^2P}{4N_0W}$, same transmitted power), figure of merit $=1$. **Comparison:** | Parameter | DSB-FC (AM) | DSB-SC | SSB-SC | |---|---|---|---| | Transmission bandwidth $B_T$ | $2W$ | $2W$ | $W$ | | Detector | Envelope or coherent | Coherent | Coherent | | Detection gain $\text{SNR}_o/\text{SNR}_i$ | $\frac{2k_a^2P}{1+k_a^2P}$ ($=\frac23$ at $\mu=1$) | 2 | 1 | | Figure of merit (baseband ref.) | $\frac{k_a^2P}{1+k_a^2P}$ ($=\frac13$ at $\mu=1$) | 1 | 1 | | Threshold effect | Yes (envelope detector) | No | No | | Power wasted in carrier | Yes (at least 1/2; 2/3 for a 100% tone) | No | No | **Conclusions:** for the same transmitted power and noise PSD, DSB-SC and SSB give the **same output SNR** (equal to baseband); DSB-SC's detection gain of 2 is offset by its double noise bandwidth. SSB achieves this in **half the bandwidth**. DSB-FC is at least **3 dB worse** (since $k_a^2P\le1$ gives FOM $\le\frac12$, with equality only for a 100% square wave) and **4.77 dB worse** (a factor of 3) for 100% single-tone modulation, because the carrier carries no information; its advantage is that it allows a cheap envelope detector. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the detection gain of DSB-SC with coherent detection? A. 1 B. 2 C. 2/3 D. 1/2 Answer: B. SNR_o = A_c^2 P/(2N0W) is twice SNR_i = A_c^2 P/(4N0W), because the quadrature noise n_s is rejected. (Key point: DSB-SC detection gain) 2. For SSB-SC, B_T = W and coherent detection gives SNR_o equal to SNR_i. The detection gain is: A. 2 B. 1/2 C. 1 D. 2/3 Answer: C. Both SNR_i and SNR_o equal A_c^2 P/(4N0W), so the gain is 1. (Key point: SSB detection gain) 3. Why is the DSB-FC detection gain lower than DSB-SC? A. The envelope has more bandwidth B. The carrier consumes transmitted power but carries no message, so it counts in SNR_i but not SNR_o C. The noise power is higher in DSB-FC D. Coherent detection cannot be used for DSB-FC Answer: B. At μ = 1 for a tone, two-thirds of the power is in the carrier, so the gain is 2/3. (Key point: Why DSB-FC is worse) 4. For the same transmitted power and noise PSD, which is correct? A. SSB has half the output SNR of DSB-SC B. DSB-SC and SSB give the same output SNR, but SSB uses half the bandwidth C. DSB-SC has twice the output SNR of SSB D. DSB-FC has the best output SNR Answer: B. DSB-SC's gain of 2 is offset by its doubled noise bandwidth; both match baseband, with SSB needing only W. 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