### 数学代写|编码理论作业代写Coding Theory代考|Convolutional Codes

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## 数学代写|编码理论作业代写Coding Theory代考|Encoding of Convolutional Codes

The easiest way to introduce convolutional codes is by means of a specific example. Consider the convolutional encoder depicted in Figure 3.1. Information bits are shifted into a register of length $m=2$, i.e. a register with two binary memory elements. The output sequence results from multiplexing the two sequences denoted by $\mathbf{b}^{(1)}$ and $\mathbf{b}^{(2)}$. Each output bit is generated by modulo 2 addition of the current input bit and some symbols of the register contents. For instance, the information sequence $\mathbf{u}=(1,1,0,1,0,0, \ldots)$ will be encoded to $\mathbf{b}^{(1)}=(1,0,0,0,1,1, \ldots)$ and $\mathbf{b}^{(2)}=(1,1,1,0,0,1, \ldots)$. The generated code sequence after multiplexing of $\mathbf{b}^{(1)}$ and $\mathbf{b}^{(2)}$ is $\mathbf{b}=(1,1,0,1,0,1,0,0,1,0,1,1, \ldots)$.
A convolutional encoder is a linear sequential circuit and therefore a Linear TimeInvariant (LTI) system. It is well known that an LTI system is completely characterized

by its impulse response. Let us therefore investigate the two impulse responses of this particular encoder. The information sequence $\mathbf{u}=1,0,0,0, \ldots$ results in the output $\mathbf{b}^{(1)}=$ $(1,1,1,0, \ldots)$ and $\mathbf{b}^{(2)}=(1,0,1,0, \ldots)$, i.e. we obtain the generator impulse responses $\mathbf{g}^{(1)}=(1,1,1,0, \ldots)$ and $\mathbf{g}^{(2)}=(1,0,1,0, \ldots)$ respectively. These generator impulse responses are helpful for calculating the output sequences for an arbitrary input sequence
\begin{aligned} &b_{i}^{(1)}=\sum_{l=0}^{m} u_{i-l} g_{l}^{(1)} \leftrightarrow \mathbf{b}^{(1)}=\mathbf{u} * \mathbf{g}^{(1)} \ &b_{i}^{(2)}=\sum_{l=0}^{m} u_{i-l} g_{l}^{(2)} \leftrightarrow \mathbf{b}^{(2)}=\mathbf{u} * \mathbf{g}^{(2)} \end{aligned}
The generating equations for $\mathbf{b}^{(1)}$ and $\mathbf{b}^{(2)}$ can be regarded as convolutions of the input sequence with the generator impulse responses $\mathrm{g}^{(1)}$ and $\mathrm{g}^{(2)}$. The code $\mathrm{B}$ generated by this encoder is the set of all output sequences $\mathbf{b}$ that can be produced by convolution of arbitrary input sequence $\mathbf{u}$ with the generator impulse responses. This explains the name convolutional codes.

The general encoder of a rate $R=k / n$ convolutional code is depicted in Figure 3.2. Each input corresponds to a shift register, i.e. each information sequence is shifted into its own register. In contrast to block codes, the $i$ th code block $\mathbf{b}{i}=b{i}^{1}, b_{i}^{2}, \ldots, b_{i}^{n}$ of a convolutional code sequence $\mathbf{b}=\mathbf{b}{0}, \mathbf{b}{1}, \ldots$ is a linear function of several information blocks $\mathbf{u}{j}=u{j}^{1}, u_{j}^{2}, \ldots, u_{j}^{k}$ with $j \in{i-m, \ldots, i}$ and not only of $\mathbf{b}_{i}$. The integer $m$

denotes the memory of the encoder. The $k$ encoder registers do not necessarily have the same length. We denote the number of memory elements of the $l$ th register by $v_{l}$. The $n$ output sequences may depend on any of the $k$ registers. Thus, we require $k \times n$ impulse responses to characterise the encoding. If the shift registers are feedforward registers, i.e. they have no feedback, than the corresponding generator impulse responses $\mathbf{g}{l}^{(j)}$ are limited to length $v{l}+1$. For this reason, $v_{l}$ is often called the constraint length of the lth input sequence. The memory $m$ of the encoder is
$$m=\max {l} v{l} .$$

## 数学代写|编码理论作业代写Coding Theory代考|Generator Matrix in the Time Domain

Similarly to linear block codes, the encoding procedure can be described as a multiplication of the information sequence with a generator matrix $\mathbf{b}=\mathbf{u G}$. However, the information sequence $\mathbf{u}$ and the code sequence $\mathbf{b}$ are semi-infinite. Therefore, the generator matrix of a convolutional code also has a semi-infinite structure. It is constructed from $k \times n$ submatrices $\mathbf{G}{i}$ according to Figure $3.5$, where the elements of the submatrices are the coefficients from the generator impulse responses. For instance, for the encoder in Figure $3.1$ we obtain \begin{aligned} &\mathbf{G}{0}=\left(g_{0}^{(1)}, g_{0}^{(2)}\right)=(11) \ &\mathbf{G}{1}=\left(g{1}^{(1)}, g_{1}^{(2)}\right)=\left(\begin{array}{ll} 1 & 0) \end{array}\right. \ &\mathbf{G}{2}=\left(g{2}^{(1)}, g_{2}^{(2)}\right)=\left(\begin{array}{ll} 1 & 1 \end{array}\right) \end{aligned}

Finally, the generator matrix is
$$\mathbf{G}=\left(\begin{array}{ccccccc} \mathbf{G}{0} & \mathbf{G}{1} & \ldots & \mathbf{G}{m} & \mathbf{0} & \mathbf{0} & \ldots \ \mathbf{0} & \mathbf{G}{0} & \mathbf{G}{1} & \ldots & \mathbf{G}{m} & \mathbf{0} & \ldots \ \mathbf{0} & \mathbf{0} & \mathbf{G}{0} & \mathbf{G}{1} & \ldots & \mathbf{G}_{m} & \ldots \ \mathbf{0} & \mathbf{0} & \mathbf{0} & \ddots & \ddots & & \ddots \end{array}\right)=\left(\begin{array}{cccccc} 11 & 10 & 11 & 00 & 00 & \ldots \ 00 & 11 & 10 & 11 & 00 & \ldots \ 00 & 00 & 11 & 10 & 11 & \ldots \ 00 & 00 & 00 & \ddots & & \ddots \end{array}\right) .$$
With this generator matrix we can express the encoding of an information sequence, for instance $\mathbf{u}=(1,1,0,1,0,0, \ldots)$, by a matrix multiplication
\begin{aligned} \mathbf{b} &=\mathbf{u G}=(1,1,0,1,0,0, \ldots)\left(\begin{array}{cccccc} 11 & 10 & 11 & 00 & 00 & \ldots \ 00 & 11 & 10 & 11 & 00 & \ldots \ 00 & 00 & 11 & 10 & 11 & \ldots \ 00 & 00 & 00 & \ddots & & \ddots \end{array}\right) \ &=(11,01,01,00,10,11,00, \ldots . \end{aligned}

## 数学代写|编码理论作业代写Coding Theory代考|State Diagram of a Convolutional Encoder

Up to now we have considered two methods to describe the encoding of a convolutional code, i.e. encoding with a linear sequential circuit, a method that is probably most suitable for hardware implementations, and a formal description with the generator matrix. Now we consider a graphical representation, the so-called state diagram. The state diagram will be helpful later on when we consider distance properties of convolutional codes and decoding algorithms.

The state diagram of a convolutional encoder describes the operation of the encoder. From this graphical representation we observe that the encoder is a finite-state machine. For the construction of the state diagram we consider the contents of the encoder registers as encoder state $\sigma$. The set $\mathrm{S}$ of encoder states is called the encoder state space.

Each memory element contains only 1 bit of information. Therefore, the number of encoder states is $2^{\nu}$. We will use the symbol $\sigma_{i}$ to denote the encoder state at time $i$. The state diagram is a graph that has $2^{v}$ nodes representing the encoder states. An example of the state diagram is given in Figure 3.6. The branches in the state diagram represent possible state transitions, e.g. if the encoder in Figure $3.1$ has the register contents $\sigma_{i}=(00)$ and the input bit $u_{i}$ at time $i$ is a 1 , then the state of the encoder changes from $\sigma_{i}=(00)$ to $\sigma_{i+1}=(10)$. Along with this transition, the two output bits $\mathbf{b}_{i}=(11)$ are generated. Similarly, the information sequence $\mathbf{u}=(1,1,0,1,0,0, \ldots)$ corresponds to the state sequence $\sigma=(00,10,11,01,10,01,00, \ldots)$, subject to the encoder starting in the all-zero state. The code sequence is again $\mathbf{b}=(11,01,01,00,10,11,00, \ldots)$. In general, the output bits only depend on the current input and the encoder state. Therefore, we label each transition with the $k$ input bits and the $n$ output bits (input/output).

## 数学代写|编码理论作业代写Coding Theory代考|Encoding of Convolutional Codes

b一世(1)=∑l=0米在一世−lGl(1)↔b(1)=在∗G(1) b一世(2)=∑l=0米在一世−lGl(2)↔b(2)=在∗G(2)

## 数学代写|编码理论作业代写Coding Theory代考|Generator Matrix in the Time Domain

G0=(G0(1),G0(2))=(11) G1=(G1(1),G1(2))=(10) G2=(G2(1),G2(2))=(11)

G=(G0G1…G米00… 0G0G1…G米0… 00G0G1…G米… 000⋱⋱⋱)=(1110110000… 0011101100… 0000111011… 000000⋱⋱).

b=在G=(1,1,0,1,0,0,…)(1110110000… 0011101100… 0000111011… 000000⋱⋱) =(11,01,01,00,10,11,00,….

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