- 2071 Shrawan (CS II) · 5+3 marks
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 bits. Take M-ary baseband PAM: the receiver output sample (after the matched filter) takes one of equally spaced levels
so adjacent levels are apart. Assumptions: symbols equally likely, additive white Gaussian noise of PSD , so the sampled noise is Gaussian with zero mean and variance (unit-energy pulse, matched filter). Decision thresholds are placed midway between levels:
-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 :
Outer level (one neighbour, e.g. ): an error occurs only if :
There are inner and outer levels, so the average symbol error probability is
In terms of energy. The average symbol energy is
With :
Check, M = 2: and , the polar binary result.
With Gray coding, bit error rate .
Binary vs M-ary
| Point | Binary | M-ary |
|---|---|---|
| Bits per symbol | 1 | |
| Symbol rate for bit rate | ||
| Bandwidth needed | (Nyquist) | |
| Bandwidth efficiency | 2 bit/s/Hz (ideal) | bit/s/Hz |
| Power for same | lower | higher (levels closer) |
| Noise immunity | better | poorer |
| Transmitter/receiver | simple (one threshold) | complex ( thresholds, AGC, precise level control) |
So M-ary signalling saves bandwidth by a factor 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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