Communication Systems
Communication Systems tests the chain from source to receiver: modulation, noise and link budgets, sampling and quantisation, and digital modulation performance. ESECE problems are arithmetic-heavy, so keep dB, dBm and Hz units explicit throughout.
Analog modulation
- AM-DSB-LC: s(t) = A_c[1 + m·cos(2πf_m t)]cos(2πf_c t); bandwidth is 2f_m and power efficiency cannot exceed 1/3 at m = 1.
- FM: Carson's rule gives B ≈ 2(Δf + f_m), where Δf is the peak deviation. FM trades bandwidth for output SNR.
- Noise: available noise power is N = kTB. Noise figures add; they do not multiply. Friis for a cascade: F = F₁ + (F₂ − 1)/G₁ + …, so the first stage dominates.
Digital links
- Shannon: C = B·log₂(1 + S/N), with S/N = 10^(SNR_dB/10).
- PCM: bit rate = n·f_s; uniform quantiser SNR_q ≈ 6.02n + 1.76 dB for a full-scale sinusoid. Companding exploits the non-uniform distribution of speech amplitudes.
- Link budget: P_r = P_t + gains − losses; Eb/N0 = P_r/(R_b·N0). Coherent detection beats non-coherent by about 3 dB, and QPSK matches BPSK in Eb/N0 while doubling bandwidth efficiency.
Worked example
A 1 MHz channel with 20 dB SNR. Then S/N = 100 and C = 10⁶ × log₂(101) = 10⁶ × 6.66 ≈ 6.66 Mbit/s. Now quantise a 4 kHz voice channel with 8-bit PCM: the bit rate is 64 kbit/s and SNR_q = 6.02 × 8 + 1.76 ≈ 49.9 dB, comfortably above the 30 dB toll-quality target.
Traps
Adding noise figures in dB in cascade instead of converting to linear gain and noise factor first; treating Shannon capacity as an achievable rate rather than a bound; using 10log where 20log applies to voltage ratios; equating Δf with bandwidth in FM; forgetting that a practical Nyquist filter needs slightly more than the symbol rate.