Lecture 10 - 2025 / 4 / 22

Encoding

{0,1}4C=Null(H)\{0, 1\}^4 \to C = \text{Null}(H)

SSG\quad S \quad \mapsto SG

standard form G=(I4,P4×3)G = (I_4, P_{4\times 3})

Decoding

xHx{=0xx0which column of Hx \mapsto Hx \begin{cases} =0 & x \mapsto x\\ \ne 0 & \text{which column of } H \end{cases}

Extending

[7,4,3]entend[8,4,4][7, 4, 3] \xrightarrow{\rm entend} [8, 4, 4]

[m,k,d (odd)]entend[m+1,k,d+1 (even)][m, k, d \text{ (odd)}] \xrightarrow{\rm entend} [m + 1, k, d + 1 \text{ (even)}]

Hamming [7,4,3][7, 4, 3] code is equivalent to the code of the following generator matrix (Cyclic code)

G=[1101000011010000110100001101]G = \begin{bmatrix} 1101000\\0110100\\0011010\\0001101 \end{bmatrix}, "shift 1 bit, shift 1 bit, shift 1 bit"

Channel Capacity

Raw message Source coding\xrightarrow{\rm Source\ coding} Source code Channel coding\xrightarrow{\rm Channel\ coding} Channel Code Noisy channel\xrightarrow{\rm Noisy \ channel} ...(Decoding)...

  1. How to formulate a noisy channel: XP(YX)YX \xrightarrow{P(Y | X)} Y

  2. Given a noisy channel P(YX)P(Y | X), how much information the receiver obtain about XX?

I(X;Y)=D(PXYPXPY)I(X; Y) = D(P_{XY} \| P_X P_Y)

Definition (Channel Capacity): The channel capacity for a noisy channel P(YX)P(Y | X) is C:=maxPXI(X;Y)C := \max_{P_X} I(X; Y).

It is possible that H(Y)>H(X)H(Y) > H(X).

Channel Coding Theorem (Informal)

R:R : rate, number of bits transmitted each time. e.g. 000000,1111110 \to 00000, 1\to 11111

if R<CR < C, \exists error correcting code, error0\textit{error} \to 0
if R>CR > C, \nexists error correcting code, error0\textit{error} \to 0