Skip to main content

Noise performance of DSB-SC and SSB-SC

Short answer

DSB-SC has detection gain 2; SSB-SC has gain 1, but both achieve figure of merit 1. For the same transmitted power, they provide identical output SNR because SSB’s halved input noise compensates for the lack of sideband coherent addition. SSB, however, uses half the bandwidth.

With necessary derivations, compare the noise performance of DSB-SC and SSB-SC modulations in analog communication system.

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.

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.

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.

Comparison:

QuantityDSB-SCSSB-SC
Bandwidth BTB_T2W2WWW
Input noise power2N0W2N_0WN0WN_0W
Detection gain21
Figure of merit11
Output SNR (same transmitted power)EqualEqual

DSB-SC has twice the detection gain because its two sidebands add coherently, but it collects twice the noise. For the same transmitted power both give identical output SNR; SSB is preferred because it needs half the bandwidth.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Communication Systems chapter 9 questions →