Which statement distinguishes channel coding from source coding?

Channel coding adds controlled redundancy for reliability

Channel coding removes redundancy

Channel coding reduces bit rate

Channel coding is limited by entropy

In the Huffman code for 0.55, 0.15, 0.15, 0.1, 0.05, S0 gets a 1-bit code because:

It is most probable and is not merged until the last stage

Its probability is a power of 2

2074 Asoj (CS II) · 3+2+2 marks

Source vs channel coding; Huffman for five symbols

Short answer

Source coding removes redundancy; channel coding adds it for reliability. For five symbols with probabilities 0.55,0.15,0.15,0.1,0.05, Huffman gives S0=0, others 3-bit, average length 1.90 bits/symbol, efficiency 97.04%.

Distinguish between the Source coding and Channel coding. A discrete memoryless source has an alphabet of five symbols S0, S1, S2, S3, S4 with probabilities of 0.55, 0.15, 0.15, 0.1 and 0.05 respectively. Determine the Huffman code for each symbol and calculate the coding efficiency.

Source coding vs channel coding

Source coding	Channel coding
Removes redundancy from the source output	Adds controlled redundancy
Aim: efficiency (fewer bits per symbol)	Aim: reliability (error detection/correction)
Reduces bit rate and bandwidth	Increases bit rate and bandwidth
Uses symbol probabilities	Uses algebraic structure of codes
Limited by entropy: $\bar L \ge H$	Limited by channel capacity: $R \le C$
Examples: Huffman, Shannon–Fano, LZW	Hamming, cyclic (CRC), convolutional
Placed right after the source	Placed before the modulator

Huffman code

Probabilities: S0 = 0.55, S1 = 0.15, S2 = 0.15, S3 = 0.1, S4 = 0.05.

Reduction (sum of two lowest moved up as high as possible):

Stage	Probabilities (descending)	Combined
1	0.55, 0.15, 0.15, 0.1, 0.05	0.1 + 0.05 = 0.15
2	0.55, 0.15, 0.15, 0.15	0.15 + 0.15 = 0.3
3	0.55, 0.3, 0.15	0.3 + 0.15 = 0.45
4	0.55, 0.45	0.55 + 0.45 = 1

Symbol	$p_i$	Huffman code	$l_i$
S0	0.55	0	1
S1	0.15	100	3
S2	0.15	101	3
S3	0.1	110	3
S4	0.05	111	3

Coding efficiency

Symbol	$p_i$	$\log_2(1/p_i)$	$p_i\log_2(1/p_i)$
S0	0.55	0.8625	0.4744
S1	0.15	2.7370	0.4105
S2	0.15	2.7370	0.4105
S3	0.1	3.3219	0.3322
S4	0.05	4.3219	0.2161
Total	1		1.8438

\begin{aligned} H &= 1.8438\ \text{bits/symbol} \\ \bar L &= 0.55(1)+0.15(3)+0.15(3)+0.1(3)+0.05(3) = 0.55 + 1.35 = 1.90\ \text{bits/symbol} \\ \eta &= \frac{H}{\bar L} = \frac{1.8438}{1.90} = 97.04\% \end{aligned}

Answer: S0 = 0 and the other four symbols get 3-bit codes (table); $\bar L = 1.9$ bits/symbol, efficiency 97.04 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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