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Detection gain of SSB-SC vs DSB-SC

Short answer

For SSB-SC, the detection gain is 1 (SNR_o/SNR_i). For DSB-SC, it is 2. However, both have a figure of merit of 1 relative to baseband, meaning for equal transmitted power and noise density, they provide the same output SNR despite different bandwidths.

Find the detection gain for SSB-SC demodulation and compare with DSB-SC.

Receiver model. The received signal s(t)s(t) plus white noise (N0/2N_0/2) passes through a band-pass (IF) filter of bandwidth BTB_T, then a demodulator and a low-pass filter of bandwidth WW (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)=nc(t)cos⁡ωct−ns(t)sin⁡ωctn(t)=n_c(t)\cos\omega_ct-n_s(t)\sin\omega_ct

where ncn_c and nsn_s are low-pass, each with the same power as n(t)n(t): nc2‾=ns2‾=n2‾=N0BT\overline{n_c^2}=\overline{n_s^2}=\overline{n^2}=N_0B_T.

Definitions. Message power P=m2(t)‾P=\overline{m^2(t)}.

  • Input SNR SNRi=received signal powernoise power in BT\text{SNR}_i=\dfrac{\text{received signal power}}{\text{noise power in } B_T}
  • Output SNR SNRo=message power at outputnoise power at output\text{SNR}_o=\dfrac{\text{message power at output}}{\text{noise power at output}}
  • Detection gain γd=SNRoSNRi\gamma_d=\dfrac{\text{SNR}_o}{\text{SNR}_i}; figure of merit =SNRoSNRref=\dfrac{\text{SNR}_o}{\text{SNR}_{ref}}, where SNRref\text{SNR}_{ref} is the channel SNR measured in the message band WW (baseband reference). Both show how much the demodulator improves or worsens SNR.

SSB-SC (coherent detection). s(t)=Ac2[m(t)cos⁡ωct∓m^(t)sin⁡ωct]s(t)=\frac{A_c}{2}[m(t)\cos\omega_ct\mp\hat m(t)\sin\omega_ct], BT=WB_T=W.

  • Signal power =Ac24⋅P+P2=Ac2P4=\frac{A_c^2}{4}\cdot\frac{P+P}{2}=\frac{A_c^2P}{4} (since m^2‾=P\overline{\hat m^2}=P); noise in WW =N0W=N_0W, so SNRi=Ac2P4N0W\text{SNR}_i=\dfrac{A_c^2P}{4N_0W}.
  • Multiply by cos⁡ωct\cos\omega_ct and low-pass filter:
y(t)=Ac4m(t)+12nc(t)y(t)=\frac{A_c}{4}m(t)+\frac12n_c(t)
  • Signal power Ac2P16\frac{A_c^2P}{16}, noise power N0W4\frac{N_0W}{4}, so SNRo=Ac2P4N0W\text{SNR}_o=\dfrac{A_c^2P}{4N_0W}
γSSB=SNRoSNRi=1\gamma_{SSB}=\frac{\text{SNR}_o}{\text{SNR}_i}=1

With baseband reference (SNRref=Ac2P4N0W\text{SNR}_{ref}=\frac{A_c^2P}{4N_0W}, same transmitted power), figure of merit =1=1.

DSB-SC (coherent detection). s(t)=Acm(t)cos⁡ωcts(t)=A_cm(t)\cos\omega_ct, BT=2WB_T=2W.

  • Signal power =Ac2P2=\frac{A_c^2P}{2}; noise in 2W2W =2N0W=2N_0W, so SNRi=Ac2P4N0W\text{SNR}_i=\dfrac{A_c^2P}{4N_0W}.
  • Multiply x(t)=s(t)+n(t)x(t)=s(t)+n(t) by cos⁡ωct\cos\omega_ct and low-pass filter:
y(t)=Ac2m(t)+12nc(t)y(t)=\frac{A_c}{2}m(t)+\frac12n_c(t)
  • SNRo=Ac2P/42N0W/4=Ac2P2N0W\text{SNR}_o=\dfrac{A_c^2P/4}{2N_0W/4}=\dfrac{A_c^2P}{2N_0W}
γDSB-SC=SNRoSNRi=2\gamma_{DSB\text{-}SC}=\frac{\text{SNR}_o}{\text{SNR}_i}=2

The quadrature noise nsn_s is rejected by coherent detection, giving a 3 dB gain. With baseband reference SNRref=Ac2P2N0W\text{SNR}_{ref}=\frac{A_c^2P}{2N_0W}, figure of merit =1=1.

Comparison:

QuantityDSB-SCSSB-SC
Transmission bandwidth2W2WWW
Noise power at input2N0W2N_0WN0WN_0W
SNRi\text{SNR}_i (same AcA_c)Ac2P4N0W\frac{A_c^2P}{4N_0W}Ac2P4N0W\frac{A_c^2P}{4N_0W}
Detection gain γ\gamma21
Figure of merit (baseband reference)11

The DSB-SC detection gain is twice that of SSB because the two sidebands add coherently (voltage) while the noise in them adds in power. But SSB lets in only half the noise bandwidth, so for the same transmitted power and same noise PSD, both give the same output SNR (figure of merit 1). SSB is preferred because it needs half the bandwidth.

✓ Answer checked by Claude Opus 5.5 (high effort)

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