Modulation, detection and coding principles

Define the signal alphabet and decision rule

A symbol selects one signal possibility during a defined interval. Four equally usable symbols can represent two coded bits; sixteen can represent four. Symbol rate counts these selections per second. Actual usable bit rate also depends on coding and framing. More densely packed signal alternatives can demand better discrimination under the same noise and power conditions.

As a simple baseband illustration, allow voltage levels−3, −1, +1, +3 V and choose the nearest level at the sampling instant. Midpoint thresholds lie at−2, 0, +2 V under equal-prior, equal-noise assumptions. A−3 V symbol observed at−2.4 V is classified correctly; observed at−1.8 V it is classified as−1 V. Those numbers do not specify a real interface or its permissible voltage.

Synchronize and preserve the distinctions

A practical receiver also needs timing, frequency and often phase information. Correct amplitude thresholds cannot repair sampling at the wrong instant. Filtering can suppress some unwanted components but may spread symbols in time, producing intersymbol interference. Detection should account for the channel’s memory rather than assume each sample is isolated.

Amplitude, frequency and phase modulation offer different signal sets. Quadrature descriptions use two perpendicular carrier components to describe amplitude and phase conveniently. Their coordinates are signal parameters, not two independent physical wires in the radio path. A phase rotation can move a constellation relative to the receiver’s decision regions unless synchronization or another suitable scheme compensates for it.

Compare codewords before claiming correction

Hamming distance counts bit positions that differ between equal-length words. Separating valid codewords allows some corrupted words to be associated with a unique nearby valid word. A minimum distance of at least 2 t+1 permits correction of any pattern of up to t bit errors with nearest-word decoding. Additional errors can cause a wrong decision.

Detection and correction are different capabilities. A checksum can reveal many corruptions without identifying the intended message. Retransmission requires a protocol and adds time and channel use. Codes should be evaluated for their actual noise, burst and synchronization conditions; repetition is easy to explain but wastes many channel symbols compared with more capable constructions.

Count the failure patterns before adding their probabilities

In the five-copy example, label transmitted positions 1–5. Let p be the probability that one coded bit flips, with equal probabilities and independent flips across positions. Suppose exactly positions 1, 2 and 3 flip while 4 and 5 do not. Independence makes the probability of that particular pattern p×p×p×(1−p)×(1−p)=p³(1−p)². The same three-flip count can occur at other positions. Choose the two positions that did not flip: 12, 13, 14, 15, 23, 24, 25, 34, 35 or 45. These ten distinct pairs account for all ten exactly-three-flip patterns.

Exactly four flips leave one unflipped position, which has five possible locations. All five flips form one pattern. A received word cannot have exactly three and exactly four flips at once. These failure cases are mutually exclusive, so their probabilities add: ten copies of the three-flip probability, five copies of the four-flip probability and one all-flip probability. This constructs the polynomial used below rather than treating its coefficients as unexplained constants.

Why is two-error correction a separate guarantee? A received word at distance at most two from 00000 cannot also be within two of 11111: going through that word would make the two valid words differ in at most four positions, yet they differ in five. The two neighbourhoods do not overlap. Three flips from 00000 produce a word closer to 11111, giving the wrong majority. This positional argument does not require independent noise; independence is needed to multiply the pattern probabilities.

Five copies reduce error under one explicit model

Encode one information bit as 00000 or 11111 and decode by majority. Minimum distance 5 guarantees correction of any two flipped copies. Under independent symmetric flip probability p=.1 per coded bit, failure probability is 10 p³(1−p)²+5 p⁴(1−p)+p⁵=.00856. Three or more flipped copies cause the wrong majority.

Five transmitted bits represent one information bit, so code rate is 1/5. A four-level mapping at 1000 symbols/s carries 2000 coded bits/s and only 400 information bits/s with this code before framing. The comparison fixes per-coded-bit error probability; it does not hold total message energy, latency and bandwidth all equal.

For comparison, three copies fail when exactly two or all three flip. There are three choices of the unflipped position, so P(failure)=3p²(1−p)+p³. At p=0.1 this is 0.027+0.001=0.028. Five copies give 0.0081+0.00045+0.00001=0.00856 under the same per-coded-bit model, at the cost of more transmitted copies. Neither comparison fixes total message energy or delay.

Try the reasoning

1. Can the five-copy code always correct a burst of three flipped copies?

   Show the answer and reasoning

   No. That produces a wrong majority. Independence is needed for the probability calculation, while the two-error guarantee is a separate combinatorial statement.

2. Do four signal levels give four bits per symbol?

   Show the answer and reasoning

   No. Four choices give log₂4=2 bits. Four bits require sixteen distinct choices before coding overhead.

3. If a check reports corruption, is the original data known?

   Show the answer and reasoning

   Not necessarily. Detection may only establish disagreement with the check; correction or retransmission needs additional structure.

4. A three-copy majority code is used with independent flip probability p=0.2. Derive its failure probability and explain why it cannot guarantee correction of two flips.

   Show the answer and reasoning

   Exactly two flips have three possible positions for the unflipped bit, each with probability p²(1−p); all three flip with probability p³. The sum is 3×0.2²×0.8+0.2³=0.104. Two flips leave a majority of the wrong value. The valid words differ in three positions, so radius-one neighbourhoods are disjoint, but the two-error guarantee of the five-copy code does not transfer.

5. In a different five-copy channel, all five bits flip together with probability 0.1, and otherwise none flips. Each bit still has marginal flip probability 0.1. Is failure still 0.00856?

   Show the answer and reasoning

   No. Every all-flip event produces the wrong majority, so failure probability is 0.1. The marginal probability for each bit is unchanged, but the joint patterns are completely correlated. Products such as p³(1−p)² describe the independent model, not this channel. The guarantee for any actual pattern of at most two flips still holds.

What to carry forward

Specify the alphabet, synchronization, code rate and error model together.