How it works
ASK (OOK): s(t) = A_n·cos(2πf_c·t) where A_n is 0 or A. Bandwidth = 2R_b for basic NRZ baseband. BER for coherent ASK: P_e = Q(√(E_b/N₀)). FSK: two carriers f_c ± Δf represent bits 0 and 1. Minimum frequency separation for orthogonality is 1/(2T_b) = R_b/2 — this gives minimum-shift keying (MSK). Bandwidth of binary FSK ≈ 2(Δf + R_b). PSK: phase carries information; BPSK has phases 0° and 180°. BPSK BER = Q(√(2E_b/N₀)), which is 3 dB better than ASK at the same bit energy.
Key points to remember
BPSK constellation has 2 points on the I-axis separated by 2√E_b; QPSK has 4 points at ±45°, ±135°, transmitting 2 bits/symbol and using the same bandwidth as BPSK at double the bit rate. BER of QPSK equals BPSK at the same E_b/N₀ — this is the key advantage of QPSK over BPSK. Coherent detection requires carrier phase synchronisation; non-coherent FSK (using envelope detectors) has BER = ½e^(−E_b/2N₀), slightly worse than coherent. DPSK (differential PSK) avoids the need for absolute phase reference by encoding data as phase changes — BER = ½e^(−E_b/N₀). Bandwidth efficiency (bits/s/Hz) increases from ASK/FSK to PSK to higher-order QAM.
Exam tip
The examiner always asks you to draw the constellation diagram and write the BER expression for BPSK and QPSK, then compare bandwidth efficiency — show that QPSK doubles the spectral efficiency of BPSK at the same BER.