Digital modulation: ASK, FSK, PSK and QAM

Why it matters

Once data is in bits it still has to ride on a carrier to cross a radio, wired or optical channel. Amplitude, frequency and phase shift keying, and their combination in QAM, are the choices behind modems, RFID tags, Wi-Fi, industrial wireless sensors and optical transceivers. Choosing among them is a trade between bandwidth, power (bit-error rate at a given energy per bit) and receiver complexity.

Key ideas

Binary schemes. Over each bit interval T_b = 1/R_b the carrier is switched:

• ASK / on–off keying (OOK): s(t) = A·cos(2πf_c·t) for 1 and 0 for 0. Simple, envelope-detectable, and the basis of most intensity-modulated optical links, but sensitive to amplitude noise and fading and needs a decision threshold that tracks signal level.
• FSK: two frequencies f₁ and f₂ for 1 and 0. Constant envelope, can be detected non-coherently with two band-pass filters and envelope detectors. Bandwidth ≈ |f₁ − f₂| + 2R_b (Carson-like estimate). Minimum tone spacing for orthogonality is R_b/2 with coherent detection (MSK) and R_b non-coherently.
• BPSK: s(t) = ±A·cos(2πf_c·t), a phase of 0 or π. Antipodal signals give the best error performance of all binary schemes, but detection needs a coherent carrier (recovered with a squaring or Costas loop) and has a 180° phase ambiguity. DPSK encodes data in the phase change between bits, removing the need for an absolute phase reference at a cost of about 1 dB.

M-ary schemes. Grouping k = log₂M bits per symbol cuts the symbol rate to R_s = R_b/k, so the bandwidth falls by k.

• QPSK sends 2 bits per symbol on two quadrature BPSK carriers (I and Q). It has the same bit-error rate as BPSK for the same E_b/N₀ but half the bandwidth.
• M-PSK for M > 4 packs points on a circle; neighbours get closer, so more power is needed.
• M-QAM places points on a grid in the I–Q plane, varying both amplitude and phase. For M ≥ 16 it uses distance far better than M-PSK, which is why 16-, 64- and 256-QAM dominate high-rate links. Its non-constant envelope demands linear amplifiers.

Constellation view. Each symbol is a point (I, Q); the transmitted waveform is I·cos(2πf_c·t) − Q·sin(2πf_c·t). Its amplitude is √(I² + Q²) and its phase is atan2(Q, I). Error probability depends on the distance between neighbouring points relative to the noise, so for equal average energy, more points mean more errors.

Bandwidth. With ideal Nyquist (sinc) pulses the minimum passband bandwidth is B = R_s. With raised-cosine pulses of roll-off α, B = R_s(1 + α). With rectangular pulses the main-lobe (null-to-null) bandwidth is 2R_s. Spectral efficiency R_b/B therefore rises as log₂M.

Error performance (AWGN, coherent detection, Gray coding).

• BPSK and QPSK: P_b = Q(√(2E_b/N₀)).
• Coherent BFSK (orthogonal) and coherent OOK at equal average energy: P_b = Q(√(E_b/N₀)) — 3 dB worse.
• DPSK: P_b = ½·exp(−E_b/N₀). Non-coherent BFSK: P_b = ½·exp(−E_b/2N₀). The Q-function is the Gaussian tail probability; take its values from a table.

Formulas

T_b = 1/R_b, R_s = R_b/log₂M R_b bit rate (bit/s), R_s symbol rate (symbol/s), M number of symbols.

E_b = A²·T_b/2 E_b energy per bit (J) for a carrier of amplitude A (V) across 1 Ω.

B = R_s(1 + α) (raised cosine); B_null = 2R_s (rectangular pulses, main lobe) B passband bandwidth (Hz), α roll-off factor (0–1).

B_FSK ≈ |f₁ − f₂| + 2R_b

P_b,BPSK = P_b,QPSK = Q(√(2E_b/N₀)) N₀ one-sided noise power spectral density (W/Hz).

P_b,coh.BFSK = Q(√(E_b/N₀)), P_b,DPSK = ½·exp(−E_b/N₀), P_b,NC-BFSK = ½·exp(−E_b/(2N₀))

Q(x) = ½·erfc(x/√2)

Worked examples

Example 1 (standard). Data at 10 Mbit/s are sent with raised-cosine pulses, α = 0.25. Find the bandwidth for BPSK, QPSK and 16-QAM.

1. BPSK: k = 1, R_s = 10 Msym/s, B = 10 × 1.25 = 12.5 MHz.
2. QPSK: k = 2, R_s = 5 Msym/s, B = 5 × 1.25 = 6.25 MHz.
3. 16-QAM: k = 4, R_s = 2.5 Msym/s, B = 2.5 × 1.25 = 3.125 MHz. Answer: 12.5 MHz, 6.25 MHz and 3.125 MHz.

Example 2 (GATE level). A BPSK receiver sees a carrier of amplitude 10 mV (across 1 Ω) at 100 kbit/s in white noise with N₀ = 10⁻¹⁰ W/Hz. Find E_b/N₀ and the bit-error probability. Compare DPSK and coherent BFSK with the same E_b.

1. T_b = 1/10⁵ = 10 µs; E_b = A²T_b/2 = (0.01)² × 10⁻⁵/2 = 5 × 10⁻¹⁰ J.
2. E_b/N₀ = 5 × 10⁻¹⁰/10⁻¹⁰ = 5 (7.0 dB).
3. BPSK: P_b = Q(√(2 × 5)) = Q(3.16) ≈ 7.8 × 10⁻⁴.
4. DPSK: P_b = ½·e⁻⁵ = 3.4 × 10⁻³.
5. Coherent BFSK: P_b = Q(√5) = Q(2.24) ≈ 1.3 × 10⁻². Answer: E_b/N₀ = 5; P_b ≈ 7.8 × 10⁻⁴ (BPSK), 3.4 × 10⁻³ (DPSK), 1.3 × 10⁻² (coherent BFSK).

Example 3 (constellation). A 16-QAM symbol has I = 3 and Q = −1 (units of the grid spacing). Find its amplitude and phase.

1. Amplitude = √(3² + 1²) = √10 = 3.16.
2. Phase = atan2(−1, 3) = −18.4°. Answer: 3.16 units at −18.4°.

Common mistakes

• Saying PSK needs less bandwidth than ASK. Binary ASK and BPSK have the same spectrum width; only M-ary schemes save bandwidth.
• Thinking QPSK is worse than BPSK in bit-error rate. Per bit, with Gray coding, they are the same; QPSK just halves the bandwidth.
• Mixing E_b/N₀ and symbol SNR (E_s/N₀ = k·E_b/N₀).
• Using bit rate instead of symbol rate in bandwidth formulas for M-ary schemes.
• Expecting QAM to work through a saturated (class-C) amplifier; only constant-envelope schemes such as FSK and PSK tolerate that.

For GATE IN

• Bandwidth and symbol-rate numericals for M-ary PSK/QAM with given roll-off.
• Energy per bit from amplitude and bit rate; E_b/N₀ in dB.
• Ranking BER of BPSK, QPSK, DPSK, coherent and non-coherent FSK at equal E_b/N₀.
• Reading a constellation: amplitude, phase, minimum distance.
• Matched-filter receiver ideas: output SNR peak 2E/N₀ at the sampling instant.

Quick check

1. A 64-QAM system carries 30 Mbit/s. What is the symbol rate?
2. Which binary scheme gives the lowest BER at a given E_b/N₀?
3. Why does DPSK not need a coherent carrier?
4. What is the minimum (Nyquist) bandwidth for QPSK at 2 Mbit/s? Answers: 1. 5 Msym/s. 2. BPSK (antipodal). 3. Data are in the phase difference between successive bits, so the previous bit acts as the reference. 4. 1 MHz.