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written 7.3 years ago by |

$l_{avg} = \sum\limits_{i = 1}^n f_i l_i$

Where $l_i$ = actual codeword length of symbol.

$L_{avg} = (0.4)1 + (0.19) \times 3 + (0.16) \times 3 + (0.15)3 + (0.1) \times 3\\ L_{avg} = 2.2 bit/symbol\\ H = \sum\limits_{i = 1}^{n = 5} P_i log = \bigg( \dfrac{1}{P_i} \bigg)\\ H = 2.149 \hspace{2cm} \text{bit/symbol}\\ \eta = \dfrac{H}{L} \times 100\\ \eta = \dfrac{2.149}{2.2} \times 100 \\ \eta = 97.6 \%$

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