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M-ary error probability; binary vs M-ary

Short answer

For M-ary PAM, symbol error probability is P_e = 2(1-1/M) Q(√(6E_s/((M^2-1)N_0))) with E_s = E_b log2 M. M-ary signaling improves bandwidth efficiency by a factor of log2 M but requires higher power and more complex hardware (M-1 thresholds) for the same error rate.

Derive the expression of error probability in case of M-ary system. Compare binary and M-ary scheme in terms of bandwidth efficiency and system complexity.

Error probability of an M-ary (PAM) system

In an M-ary system each symbol carries n=log⁡2Mn = \log_2 M bits. Take M-ary baseband PAM: the receiver output sample (after the matched filter) takes one of MM equally spaced levels

ak∈{±A, ±3A, …, ±(M−1)A}a_k \in \{\pm A,\ \pm 3A,\ \dots,\ \pm(M-1)A\}

so adjacent levels are 2A2A apart. Assumptions: symbols equally likely, additive white Gaussian noise of PSD N0/2N_0/2, so the sampled noise nn is Gaussian with zero mean and variance σ2=N0/2\sigma^2 = N_0/2 (unit-energy pulse, matched filter). Decision thresholds are placed midway between levels: 0,±2A,±4A,…0, \pm 2A, \pm 4A, \dots

  -3A     -A      +A     +3A      (M = 4)
   x  |    x   |   x   |   x
     -2A       0      +2A        <- thresholds
 outer         inner        outer

Inner level (two neighbours): an error occurs if ∣n∣>A|n| > A:

Pe,inner=P(n>A)+P(n<−A)=2Q ⁣(Aσ)P_{e,\text{inner}} = P(n>A) + P(n<-A) = 2Q\!\left(\frac{A}{\sigma}\right)

Outer level (one neighbour, e.g. +(M−1)A+(M-1)A): an error occurs only if n<−An < -A:

Pe,outer=Q ⁣(Aσ)P_{e,\text{outer}} = Q\!\left(\frac{A}{\sigma}\right)

There are M−2M-2 inner and 22 outer levels, so the average symbol error probability is

Pe=1M[(M−2) 2Q ⁣(Aσ)+2 Q ⁣(Aσ)]=2(1−1M)Q ⁣(Aσ)\begin{aligned} P_e &= \frac{1}{M}\left[(M-2)\,2Q\!\left(\frac{A}{\sigma}\right) + 2\,Q\!\left(\frac{A}{\sigma}\right)\right] \\ &= 2\left(1-\frac{1}{M}\right)Q\!\left(\frac{A}{\sigma}\right) \end{aligned}

In terms of energy. The average symbol energy is

Es=2M∑i=1M/2(2i−1)2A2=(M2−1)A23  ⇒  A2=3EsM2−1E_s = \frac{2}{M}\sum_{i=1}^{M/2}(2i-1)^2A^2 = \frac{(M^2-1)A^2}{3} \;\Rightarrow\; A^2 = \frac{3E_s}{M^2-1}

With σ2=N0/2\sigma^2 = N_0/2:

Pe=2(1−1M)Q ⁣(6Es(M2−1)N0),Es=Eblog⁡2MP_e = 2\left(1-\frac{1}{M}\right)Q\!\left(\sqrt{\frac{6E_s}{(M^2-1)N_0}}\right), \qquad E_s = E_b\log_2 M

Check, M = 2: Es=EbE_s = E_b and Pe=Q(2Eb/N0)P_e = Q\left(\sqrt{2E_b/N_0}\right), the polar binary result.

With Gray coding, bit error rate ≈Pe/log⁡2M\approx P_e/\log_2 M.

Binary vs M-ary

PointBinaryM-ary
Bits per symbol1log⁡2M\log_2 M
Symbol rate for bit rate RbR_bRbR_bRb/log⁡2MR_b/\log_2 M
Bandwidth needed≈Rb/2\approx R_b/2 (Nyquist)≈Rb/(2log⁡2M)\approx R_b/(2\log_2 M)
Bandwidth efficiency2 bit/s/Hz (ideal)2log⁡2M2\log_2 M bit/s/Hz
Power for same PeP_elowerhigher (levels closer)
Noise immunitybetterpoorer
Transmitter/receiversimple (one threshold)complex (M−1M-1 thresholds, AGC, precise level control)

So M-ary signalling saves bandwidth by a factor log⁡2M\log_2 M but needs more signal power and a more complex system for the same error rate. It is used where bandwidth is scarce (telephone-line modems, digital radio).

✓ Answer checked by Claude Opus 5.5 (high effort)

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