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Importance of source coding; Huffman for eight symbols at 1000 symbols/s

Short answer

Source coding removes redundancy, bringing bit rate near entropy to save bandwidth. For the eight symbols, entropy is 2.6640 bits/symbol, Huffman average length = 2.72 bits/symbol, coding efficiency = 97.94%, and output bit rate = 1000 × 2.72 = 2720 bits/s.

Explain the importance of source coding in Digital Communication System. A discrete memory less source emits one of the eight symbols with probabilities P = {0.25, 0.20, 0.2, 0.15, 0.08, 0.07, 0.03, 0.02}. If the output symbols are encoded using Huffman code, find the Coding efficiency and output bit if the symbol rate of the source is 1000 symbols per second.

Importance of source coding

Source coding represents the source output with the minimum average number of bits by removing redundancy:

  • It brings the average code length Lˉ\bar L close to the entropy HH (Shannon's source coding theorem: Lˉ≥H\bar L \ge H).
  • It lowers the bit rate, so less bandwidth, power and storage are needed.
  • Variable-length prefix codes (Huffman, Shannon–Fano) give short words to likely symbols and remain uniquely decodable.

Huffman code

P={0.25,0.20,0.2,0.15,0.08,0.07,0.03,0.02}P = \{0.25, 0.20, 0.2, 0.15, 0.08, 0.07, 0.03, 0.02\} for x1…x8x_1 \dots x_8 (sum = 1).

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

StageProbabilities (descending)Combined
10.25, 0.2, 0.2, 0.15, 0.08, 0.07, 0.03, 0.020.03 + 0.02 = 0.05
20.25, 0.2, 0.2, 0.15, 0.08, 0.07, 0.050.07 + 0.05 = 0.12
30.25, 0.2, 0.2, 0.15, 0.12, 0.080.12 + 0.08 = 0.2
40.25, 0.2, 0.2, 0.2, 0.150.2 + 0.15 = 0.35
50.35, 0.25, 0.2, 0.20.2 + 0.2 = 0.4
60.4, 0.35, 0.250.35 + 0.25 = 0.6
70.6, 0.40.6 + 0.4 = 1
Symbolpip_iHuffman codelil_i
x10.25012
x20.2112
x30.20003
x40.150013
x50.081013
x60.0710004
x70.03100105
x80.02100115

Entropy:

Symbolpip_ilog⁡2(1/pi)\log_2(1/p_i)pilog⁡2(1/pi)p_i\log_2(1/p_i)
x10.252.00000.5000
x20.22.32190.4644
x30.22.32190.4644
x40.152.73700.4105
x50.083.64390.2915
x60.073.83650.2686
x70.035.05890.1518
x80.025.64390.1129
Total12.6640
H=2.6640 bits/symbolH = 2.6640\ \text{bits/symbol}

Average length and efficiency:

Lˉ=∑pili=2.72 bits/symbolη=HLˉ=2.66402.72=97.94%\begin{aligned} \bar L &= \sum p_i l_i = 2.72\ \text{bits/symbol} \\ \eta &= \frac{H}{\bar L} = \frac{2.6640}{2.72} = 97.94\% \end{aligned}

(A fixed 3-bit code would give 2.664/3=88.80%2.664/3 = 88.80\%.)

Output bit rate

Rb=rLˉ=1000×2.72=2720 bits/sR_b = r\bar L = 1000 \times 2.72 = 2720\ \text{bits/s}

(The information rate is rH=2664rH = 2664 bits/s.)

Answer: coding efficiency = 97.94 %; output bit rate = 2720 bits/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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