习题 2-1 (P 82-84): 4.5、4.6、4.10、4.16 (P 134): 6.2、6.8

4.5

1 $I (a_{1}) = lo g_{2} (0.6) = 0.737 bi t, I (a_{2}) = lo g_{2} (0.4) = 1.322 bi t$ 2 $P (b_{1}) = 0.6, P (b_{2}) = 0.4$ $I (a_{1}; b_{1}) = lo g_{2} \frac{p ( a _{1} / b _{1} )}{p ( a _{1} )} = 0.474 bi t, I (a_{1}; b_{2}) = lo g_{2} \frac{p ( a _{1} / b _{2} )}{p ( a _{1} )} = - 1.263 bi t$ $I (a_{2}; b_{1}) = lo g_{2} \frac{p ( a _{2} / b _{1} )}{p ( a _{2} )} = - 1.263 bi t, I (a_{2}; b_{2}) = lo g_{2} \frac{p ( a _{2} / b _{2} )}{p ( a _{2} )} = 0.907 bi t$ 3 $H (X) = H (0.6, 0.4) = 0.971 bi t / sy mb o l$ $H (Y) = H (0.6, 0.4) = 0.971 bi t / sy mb o l$ 4 $H (X Y) = H (0.5, 0.1, 0.1, 0.3) = 1.685 bi t / sy mb o l$ $H (X / Y) = 1.685 - 0.971 = 0.714 bi t / sy mb o l$ $H (Y / X) = 0.714 bi t / sy mb o l$ 5 $I (X; Y) = 0.971 - 0.714 = 0.257 bi t / sy mb o l$

4.6

$H (X) = H (\frac{1}{4}, \frac{3}{4}) = 0.811 bi t / sy mb o l$ $H (Y / X) = i \sum j \sum P (x_{i}) p (y_{i} / x_{i}) lo g P (y_{j} / x_{i}) = 0.92 bi t / sy mb o l$ $P (y_{1}) = 0.58, P (y_{2}) = 0.42$ $H (Y) = H (0.58, 0.42) = 0.98 bi t / sy mb o l$ $H (X ∣ Y) = H (X) - H (Y) + H (Y / X) = 0.75 bi t / sy mb o l$ $I (X; Y) = H (X) - H (X / Y) = 0.06 bi t / sy mb o l$ $C = 1 - H (1/3, 2/3) = 0.082 bi t / sy mb o l$ , 最佳输入分布为: $P_{x} = {\frac{1}{2} . \frac{1}{2}}$

4.10

这里不应该是 1500 symbol/s 才对么？ 二元对称信道的信道容量为 $C = 1 - H (p) = 1 - H (0.98, 0.02) = 0.859 bi t / sy mb o l$ 信源的信息量为 $14000 sy mb o l \times H (\frac{1}{2}) = 14000 bi t$ , 10 秒内传输的信息量： $1500 \times C \times 10 = 12880 bi t$ 小于信源的信息量，所以不能在 10 秒内无失真传输完。

4.16

$α_{i} = {α_{i 1} α_{i 2} α_{i 3}}, α_{i t} \in {0, 1}, i = 0, 1, \dots, 7, t = 1, 2, 3$

三次扩展后信源的输入符号集为： ${α_{0} = 000, α_{1} = 001, α_{2} = 010, α_{3} = 011, α_{4} = 100, α_{5} = 101, α_{6} = 110, α_{7} = 111}$ 同理，输出符号集也同样编码 $β_{i} = {β_{i 1} β_{i 2} β_{i 3}}, β_{i t} \in {0, 1}, i = 0, 1, \dots, 7, t = 1, 2, 3$ ${β_{0} = 000, β_{1} = 001, β_{2} = 010, β_{3} = 011, β_{4} = 100, β_{5} = 101, β_{6} = 110, β_{7} = 111}$ 那么 $P (β_{j} / α_{i}) = P (α_{i 1} / β_{j 1}) \cdot P (α_{i 2} / β_{j 2}) \cdot P (α_{i 3} / β_{j 3})$

方便起见，令 $q = \overset{p}{ˉ}$

q^{3} q^{2} p q^{2} p q^{2} p q^{2} p p^{2} q p^{2} q p^{3} q^{2} p q^{3} q^{2} p q^{2} p p^{2} q q^{2} p p^{3} p^{2} q q^{2} p p^{2} q q^{3} q^{2} p p^{2} q p^{3} q^{2} p p^{2} q p^{2} q q^{2} p q^{2} p q^{3} p^{3} p^{2} q p^{2} q q^{2} p q^{2} p p^{2} q q^{2} p p^{3} q^{3} q^{2} p q^{2} p p^{2} q p^{2} q p^{2} q p^{3} p^{2} q q^{2} p q^{3} p^{2} q q^{2} p p^{2} q p^{3} q^{2} p p^{2} q q^{2} p q^{2} p q^{3} q^{2} p p^{3} p^{2} q p^{2} q q^{2} p p^{2} q q^{2} p q^{2} p q^{3}

6.2

p (x) = \int_{- r^{2} - x^{2}}^{r^{2} - x^{2}} p (x y) d y = \int_{- r^{2} - x^{2}}^{r^{2} - x^{2}} \frac{1}{π r ^{2}} d y = \frac{2 r ^{2} - x ^{2}}{π r ^{2}} (- r \leq x \leq r) H_{c} (X) = - \int_{- r}^{r} p (x) lo g p (x) d x = - \int_{- r}^{r} p (x) lo g \frac{2 r ^{2} - x ^{2}}{π r ^{2}} d x = - \int_{- r}^{r} p (x) lo g \frac{2}{π r ^{2}} d x - \int_{- r}^{r} p (x) lo g r^{2} - x^{2} d x = lo g \frac{π r ^{2}}{2} - \int_{- r}^{r} p (x) lo g r^{2} - x^{2} d x = lo g_{2} π r / e bit /symbol p (y) = p (x) H_{c} (Y) = H_{c} (X) = lo g_{2} π r / e bit/symbol H_{c} (X Y) = - \iint_{K} p (x y) lo g p (x y) d x d y = - \iint_{K} p (x y) lo g \frac{1}{π r ^{2}} d x d y = lo g π r^{2} \iint_{K} p (x y) d x d y = lo g_{2} π r^{2} bit /symbol I_{c} (X; Y) = H_{c} (X) + H_{c} (Y) - H_{c} (X Y) = lo g_{2} π / e bit /symbol

6.8

(1) $F = 4 kHz$ 时，实现无差错传输则 $R \leq W lo g (1 + \frac{P}{N _{0} W})$ ，取等号，即 $R = W lo g (1 + \frac{P}{N _{0} W})$

P_{m i n} = N_{0} W (2^{R / W} - 1) = 5 \times 1 0^{- 6} \times 1 0^{- 3} \times 4 \times 1 0^{3} \times (2^{5.6 \times 1 0^{4} /4 \times 1 0^{3}} - 1) = 0.32766 W

(2) $W \to \infty$ 时，由实现无差错传输则 $R \leq lim_{W \to \infty} C = \frac{P}{N _{0} l n 2}$ ，取等号，则

P_{m i n} = R N_{0} ln 2 = 5.6 \times 104 \times 5 \times 106 \times 103 \times ln 2 = 1.941 \times 104 = 0.1941 mW

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信息论作业-2-1

4.5

4.6

4.10

4.16

6.2

6.8

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