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京都大学 情報学研究科 通信情報システム専攻 2020年8月実施 専門基礎A [A-5]

Author

SUN

Description

Answer all the following questions.

(1)

SAS_A and SBS_B are independent and stationary memoryless information sources.
SAS_A generates information symbols 00 and 11 with probabilities 0.70.7 and 0.30.3, respectively, while SBS_B generates 00 and 11 with probabilities 0.60.6 and 0.40.4, respectively. Answer the following questions. log23=1.6\log_2 3 = 1.6, log25=2.3\log_2 5 = 2.3, and log27=2.8\log_2 7 = 2.8 may be used.

(a) Find the value of the entropy of SAS_A.

(b) Consider the nnth extension of SAS_A. Find a binary Huffman code for the second extension (n=2n=2) of SAS_A and the expected codeword length per symbol.

(c) Compare the entropy in Question (a) and the expected codeword length per symbol in Question (b). Explain which should be larger and the reason.

(d) Explain whether the expected codeword length per symbol in Question (b) increases or decreases as nn in Question (b) increases and the reason.

(e) An information source SXS_X has two states and generates information symbols by following SAS_A and SBS_B when its state is sAs_A and sBs_B, respectively.
SXS_X transits from a state to the other state when it generates 11. Draw the state diagram of SXS_X.

(f) Find the stationary distribution of SXS_X in Question (e).

(g) Find the value of the entropy of SXS_X in Question (e). Round down to one decimal place.

(2)

Answer the following questions related to channel coding. Let CC be the binary cyclic code of length 1515 that has generator polynomial

G(x)=x4+x+1.G(x) = x^4 + x + 1.

(a) Determine whether x10+x7+x4+x3+x2+x+1x^{10} + x^7 + x^4 + x^3 + x^2 + x + 1 is a codeword polynomial of CC or not.

(b) Find the codeword polynomial for the message polynomial x5+x3+xx^5 + x^3 + x in a systematic form.

(c) Find the minimum distance of CC.

(d) Find the maximum number of error bits corrected by CC.

(e) Consider communications with CC through a memoryless binary symmetric channel with crossover probability pp. Evaluate the probability of decoding failure assuming that any correctable errors are corrected.

(f) Explain the channel coding theorem.

题目描述

回答全部问题。

  1. SA,SBS_A,S_B 是彼此独立的平稳无记忆信息源。SAS_A 以概率 0.7,0.30.7,0.3 产生符号 0,10,1SBS_B 以概率 0.6,0.40.6,0.4 产生符号 0,10,1。可使用 log23=1.6\log_2 3=1.6log25=2.3\log_2 5=2.3log27=2.8\log_2 7=2.8
    1. SAS_A 的熵。
    2. 考虑 SAS_Ann 次扩展。为二次扩展(n=2n=2)构造二元 Huffman 码,并求每个原始符号的平均码长。
    3. 比较第 1 小问的熵与第 2 小问的平均码长,说明哪个更大及原因。
    4. 当扩展次数 nn 增大时,每个符号的平均码长会增大还是减小?说明原因。
    5. 信息源 SXS_X 有状态 sA,sBs_A,s_B;处于相应状态时分别按 SA,SBS_A,S_B 产生符号,每当产生 11 就转移到另一个状态。画出 SXS_X 的状态图。
    6. SXS_X 的平稳分布。
    7. SXS_X 的熵,并向下取到一位小数。
  2. 回答信道编码问题。设 CC 为长度 1515、生成多项式
    G(x)=x4+x+1G(x)=x^4+x+1
    的二元循环码。
    1. 判断 x10+x7+x4+x3+x2+x+1x^{10}+x^7+x^4+x^3+x^2+x+1 是否为 CC 的码字多项式。
    2. 求消息多项式 x5+x3+xx^5+x^3+x 的系统形式码字多项式。
    3. CC 的最小距离。
    4. CC 最多能纠正的错误比特数。
    5. 通过交叉概率为 pp 的无记忆二元对称信道使用 CC 通信,假设所有可纠正错误均被纠正,求译码失败概率。
    6. 说明信道编码定理。

Kai

(1)

(a)

H(SA)=ipilog21pi=0.7log210.7+0.3log210.3=0.86 bits/symbol\begin{aligned} H(S_A) &= \sum_i p_i \log_2 \frac{1}{p_i} \\ &= 0.7 \cdot \log_2 \frac{1}{0.7} + 0.3 \cdot \log_2 \frac{1}{0.3} \\ &= 0.86 \text{ bits/symbol} \end{aligned}

(b)

The symbol probabilities are:

  • p(00)=0.7×0.7=0.49p(00) = 0.7 \times 0.7 = 0.49
  • p(01)=0.7×0.3=0.21p(01) = 0.7 \times 0.3 = 0.21
  • p(10)=0.3×0.7=0.21p(10) = 0.3 \times 0.7 = 0.21
  • p(11)=0.3×0.3=0.09p(11) = 0.3 \times 0.3 = 0.09

Construct Huffman Code:

Average Code Length (Lˉ\bar{L}):

Lˉ=1×0.49+2×0.21+3×0.21+3×0.09=0.49+0.42+0.63+0.27=1.81 bits/sequence\begin{aligned} \bar{L} &= 1 \times 0.49 + 2 \times 0.21 + 3 \times 0.21 + 3 \times 0.09 \\ &= 0.49 + 0.42 + 0.63 + 0.27 \\ &= 1.81 \text{ bits/sequence} \end{aligned}

Average Code Length per Symbol (Lˉs\bar{L}_s):

Lˉs=Lˉ2=1.812=0.905 bits/symbol\bar{L}_s = \frac{\bar{L}}{2} = \frac{1.81}{2} = 0.905 \text{ bits/symbol}

(c)

H(SA)<LˉsH(S_A) < \bar{L}_s.

According to Shannon's source coding theorem, the entropy of a source provides the ultimate lower bound on the average length of any lossless code for that source.

The source SAS_A doesn't meet with dyadic distribution (powers of 1/21/2), so the Huffman coding will never reach the lower bound exactly.

(d)

The expected code length per symbol decreases as block length nn increases. According to Shannon's source coding theorem for the nn-th extension of a discrete memoryless source (DMS), the average codeword length per source symbol satisfies:

H(X)Lˉnn<H(X)+1nH(X) \le \frac{\bar{L}^n}{n} < H(X) + \frac{1}{n}

As nn increases, the length per symbol will approach the entropy bound.

(e)

(f)

Let the stationary probabilities be πa,πb\pi_a, \pi_b.

{πa=0.7πa+0.4πbπb=0.3πa+0.6πbπa+πb=1{0.3πa=0.4πbπa+πb=1{πa=47πb=37\begin{cases} \pi_a = 0.7\pi_a + 0.4\pi_b \\ \pi_b = 0.3\pi_a + 0.6\pi_b \\ \pi_a + \pi_b = 1 \end{cases} \Rightarrow \begin{cases} 0.3\pi_a = 0.4\pi_b \\ \pi_a + \pi_b = 1 \end{cases} \Rightarrow \begin{cases} \pi_a = \frac{4}{7} \\ \pi_b = \frac{3}{7} \end{cases}

(g)

H(SX)=πaH(A)+πbH(B)H(S_X) = \pi_a H(A) + \pi_b H(B)
  • H(A)=H(SA)0.88 bits/symbolH(A) = H(S_A) \approx 0.88 \text{ bits/symbol}
  • H(B)=0.6log20.60.4log20.40.97 bits/symbolH(B) = -0.6 \log_2 0.6 - 0.4 \log_2 0.4 \approx 0.97 \text{ bits/symbol}

Substitute into H(SX)H(S_X):

H(SX)=47×0.88+37×0.970.50+0.41=0.91 bits/symbol\begin{aligned} H(S_X) &= \frac{4}{7} \times 0.88 + \frac{3}{7} \times 0.97 \\ &\approx 0.50 + 0.41 \\ &= 0.91 \text{ bits/symbol} \end{aligned}

(2)

(a)

A polynomial is a codeword if it is divisible by the generator polynomial g(x)g(x).

(x10+x7+x4+x3+x2+x+1)0(modg(x))(x^{10} + x^7 + x^4 + x^3 + x^2 + x + 1) \equiv 0 \pmod{g(x)}

So this is a codeword polynomial.

(b)

The systematic form is given by c(x)=xrm(x)+r(x)c(x) = x^r m(x) + r(x). Given message m(x)=x5+x3+xm(x) = x^5 + x^3 + x. Degree of g(x)g(x) is r=4r=4.

  1. Shift: xrm(x)=x4(x5+x3+x)=x9+x7+x5x^r m(x) = x^4 (x^5 + x^3 + x) = x^9 + x^7 + x^5.
  2. Modulo: x9+x7+x5x2+x+1(modg(x))x^9 + x^7 + x^5 \equiv x^2 + x + 1 \pmod{g(x)}.
    • So remainder r(x)=x2+x+1r(x) = x^2 + x + 1.
  3. Codeword: c(x)=x9+x7+x5+x2+x+1c(x) = x^9 + x^7 + x^5 + x^2 + x + 1.

(c)

For (15,11)(15, 11) cyclic code:

  • r=1511=4r = 15 - 11 = 4.
  • n=2r1=15n = 2^r - 1 = 15, k=2r1r=11k = 2^r - 1 - r = 11.
  • Since g(x)g(x) is a primitive polynomial, the (15,11)(15, 11) cyclic code is a Hamming code.
  • Minimum distance dmin=3d_{\min} = 3.

(d)

Nerrordmin12=312=1N_{\text{error}} \le \left\lfloor \frac{d_{\min} - 1}{2} \right\rfloor = \left\lfloor \frac{3 - 1}{2} \right\rfloor = 1

The maximum number of correctable error bits is 1.

(e)

Probability of error occurring in a codeword (assuming correction of 1 bit):

Perror=1(1p)1515p(1p)14P_{\text{error}} = 1 - (1-p)^{15} - 15p(1-p)^{14}

(This represents the probability of having 2 or more errors).

(f)

  • Achievability: For any data transmission rate R<CR < C (channel capacity), it is possible to design a coding scheme that allows for communication with an arbitrarily low probability of error.
  • Converse: If R>CR > C, it is impossible to achieve arbitrarily low probability of error.