written 8.4 years ago by
teamques10
★ 68k
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•
modified 8.4 years ago
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$P ( a_4‘)= P (a_4 ) + P (a_5 ) = 0.1 + 0.1 = 0.2$
$C (a_4 ) = α_1 * 0$
$C (a_5 ) = α_1 * 1$
$P (a_3’ ) = P (a_3 ) + P (a_4’ ) = 0.2 + 0.2 =0.4$
$C ( a_3 ) =α_2 * 0$
$C ( a_4 ‘ ) = α_2 * 1 = α_1$
$C ( a_4 ) = α_1 * 0 = α_2 * 1 * 0 = α_2 * 10$
$C (a_5) = α_1 * 1 = α_2 * 1 * 1 = α_2 * 1$
$P (a_3” ) = P (a_3’ ) + P (a_1 ) = 0.4 + 0.2 = 0.6$
$C (a_3’ ) = α_2 = α_3 * 0$
$C ( a_1 ) = α_3 * 1 = 0 * 1 = 01$
$C ( a_3 ) = α_3 * 00 = 0 * 00 =000$
$C (a_4 ) = α_3 * 010 = 0 * 010 = 0010$
$C (a_5 ) = α_3 *011 = 0 * 011 =0011$
Symbols |
Codeword |
$a_1$ |
01 |
$a_2$ |
1 |
$a_3$ |
000 |
$a_4$ |
0010 |
$a_5$ |
0011 |
Verification
Symbols |
Codeword |
$a_1$ |
01 |
$a_2$ |
1 |
$a_3$ |
000 |
$a_4$ |
0010 |
$a_5$ |
0011 |
$Entropy =_I log_2 (P_i)$
$= - (0.2 log_2 0.2 + 0.4 log_2 0.4 + 0.2 log_2 0.2 + 0.1 log_2 0.1 + 0.1 log_2 0.1 )$
Entropy = 2.122 bits/ symbol