Coding teaser: source coding

Make sure to read the Wiki page about the basics of a channel. Several discussions and interpretations are important for this chapter 😊.

Complete Channel Model

Previously, we showed a simple channel model that consists of the source, channel, and receiver. In modern communication systems, we also add an encoder after the source and a decoder before the receiver. Figure 1 shows the complete channel model. The encoder is used to transform the source symbols into a code that suitable for the channel. The receiver decodes the received message coming out of the channel.

Figure 1: Complete channel model with the encoder added after the source and the decoder added before the receiver.

Let's bring back Bob and Alice into the scene again. This time let's use the complete channel model. Since most of our communication systems use binary representations, Bob needs to encode his data into binary digits before he sends it over the channel.

Let's look at a few examples. Suppose our source alphabet $S$ has the symbols of all combinations of flipping three fair coins. Then our symbols would be $\{TTT,TTH,THT,THH,HTT,HTH,HHT,HHH\}$. Bob needs to map each symbol into a binary representation such that $\{TTT,TTH,THT,THH,HTT,HTH,HHT,HHH\}\rightarrow \{000,001,010,011,100,101,110,111\}$. This is the encoding process. Let's add new definitions:

• The binary set $\{0,1\}$ are the code symbols.
• The encoded patterns $\{000,001,010,011,100,101,110,111\}$ are called the codewords. For example, $000$ is the codeword for the symbol $TTT$. $101$ is the codeword for $HTH$. In practice, we let some random variable $X$ whose outcomes are codewords $\{x_{1},x_{2},x_{3},...,x_{n}\}$ associated with probabilities $\{P(x_{1}),P(x_{2}),P(x_{3}),...,P(x_{n})\}$. For example, $x_{1}=000$ is the codeword for $TTT$. Moreover, from our example, the source symbols are equiprobable and we have a direct mapping from source symbol to codeword (i.e., $s_{i}\rightarrow x_{i}$ ) then it follows that the codewords are equiprobable too: $P(x_{1})=P(x_{2})=P(x_{3})=P(x_{4})=P(x_{5})=P(x_{6})=P(x_{7})=P(x_{8})={\frac {1}{8}}$. The probability distribution of $X$ can change depending on the encoding scheme.
• We usually tabulate the encoding or mapping. We call this table the codebook. It's a simple look-up-table (LUT). Let's call the table below as codebook-1.

Source Symbol	Codeword
TTT	000
TTH	001
THT	010
THH	011
HTT	100
HTH	101
HHT	110
HHH	111

The encoder looks at this table and endcodes message before it gets into the channel. For example, suppose we want to send the message $TTTHHHTHTHTH$. The encoder uses codebook-1 to translate the source message into an encoded message: $000111010101$. Receivers should also know the same codebook-1 so that they can decode the message. For example, if the encoded message is $101011001110$, then we can decode the message as $HTHTHHTTHHHT$. Choosing the encoding where the $T\rightarrow 0$ and $H\rightarrow 1$ is simple. This encoding is arbitrary and it seems convenient for our needs. However, nothing stops us from encoding the source symbols with a different representation. For example, we can choose an encoding where $T\rightarrow 1$ and $H\rightarrow 0$. We can go for more complex ones like codebook-2 as shown below:

Source Symbol	Codeword
TTT	01
TTH	001
THT	0001
THH	00001
HTT	000001
HTH	0000001
HHT	00000001
HHH	000000001

Where a source message $TTTHHHTHTHTH$ translates into $0100000000100010000001$. Codebook-2 can also decode an encoded message. For example, when we receive $00000010000100100000001$ we get $HTHTHHTTHHHT$. The choice of the encoding process is purely arbitrary. However, we are interested if the encoding efficient. One metric is the average code length defined as:

$L(X)=\sum _{i=1}^{n}P(x_{i})l_{i}$

(1)

Where $l_{i}$ is the code length of the codeword $x_{i}$. For example, the average code length of codebook-1 is:

$L(X)=8\cdot {\frac {1}{8}}\cdot 3=3$

The average code length of codebook-2 is:

$L(X)={\frac {1}{8}}\cdot 2+{\frac {1}{8}}\cdot 3+{\frac {1}{8}}\cdot 4+{\frac {1}{8}}\cdot 5+{\frac {1}{8}}\cdot 6+{\frac {1}{8}}\cdot 7+{\frac {1}{8}}\cdot 8+{\frac {1}{8}}\cdot 9=5.5$

Clearly, codebook-1 has a shorter codelength than codebook-2. Let's introduce another metric called the coding efficiency. Coding efficiency is defined as:

$\eta ={\frac {H(S)}{L(X)}}$

(2)

Where $H(S)$ is the source entropy. The coding efficiency $\eta$ has a range of $0\leq \eta \leq 1$. If you recall our "go-to" interpretation of entropy, given some entropy $H(S)$ means we can represent $m=2^{H(S)}$ outcomes. Essentially, $H(S)$ is the minimum number of bits we can use to represent all outcomes of a source regardless of its distribution. Of course, a smaller coding length is desirable if we want to compress representations. For our simple coin flip example, $H(S)=3$. The efficiency of codebook-1 is $\eta ={\frac {3}{3}}=1$. Codebook-1 is said to be maximally efficient because it matches the minimum number of bits to represent the source symbols. The efficiency of codebook-2 is $\eta ={\frac {3}{5.5}}=0.5455$.

$L(X)$ should never be lower than $H(S)$. Otherwise, that doesn't make sense. How can you represent $H(S)=3$ with $L(X)=2$? Something's wrong with your codebook if this happens.

Channel Rates and Channel Capacity

Let's bring back Bob and Alice into the scene again. When Bob sends his file to Alice, it takes a while before she gets the entire message because the channel has limited bandwidth. Bandwidth is the data rate that the channel supports. Let's define $C$ as the maximum channel capacity in units of bits per second (bits/s). We'll provide a quantitative approach later. Channels have different rates because of their physical limitations and noise. For example, an ethernet bandwidth can achieve a bandwidth of 10 Gbps (gigabits per second), while Wifi can only achieve at a maximum of 6.9 Gbps. It really is slower to communicate wirelessly. It is possible to transmit data to the channel at a much slower rate such that we are not utilizing the full capacity of the channel. We'd like to encode our source symbols into some codewords that utilizes the full capacity of a channel. Shannon's source coding theorem says:

"Let a source have entropy $H$ (bits per symbol) and a channel have a capacity $C$ (bits per second). Then it is possible to encode the output of the source in such a way as to transmit at the average rate $C/H-\epsilon$ symbols per second over the channel where $\epsilon$ is arbitrarily small. It is not possible to transmit at an average rate greater than $C/H$."

Take note that, $\epsilon$ in Shannon's statement is not the noise probability that we know. It's just a term that is used to indicate a small margin of error. In a nutshell, Shannon's source coding theorem says that we can calculate a maximum symbol rate (symbols/s) that utilizes the maximum channel rate $C$ while knowing the source entropy $H(S)$. This maximum symbol rate is given by:

$R_{\textrm {max}}={\frac {C}{H(S)}}$

(3)

Let's look at a few examples to appreciate this.

Channel Rate Example 1: Three Coin Flips

Let's re-use the three coin flips example again. Suppose we want to send the three coin flip symbols over some channel that has a channel capacity $C=3$ bits/s. We also know (from the above example) that we need a minimum of $H(S)=3$ bits per symbol to represent each outcome. This translates to:

$R_{\textrm {max}}={\frac {3\ {\textrm {bits/s}}}{3\ {\textrm {bits/symbols}}}}=1\ {\textrm {symbols/s}}$

If we are to utilize the maximum capacity of the channel, we need to encode the source symbols $\{TTT,TTH,THT,THH,HTT,HTH,HHT,HHH\}$ to its binary representation such that each binary digit in each codeword $x_{i}$ provides 1 bit of information . Luckily, codebook-1 does this because we have a coding efficiency of $\eta =1$ which means that every binary digit of the codeword provides at least 1-bit of information. We are maximally efficient with this encoding. If you think about it, if we send 1 symbol/s to the channel and our encoding has 1 symbol (e.g., TTT) is 3 bits long then essentially we are able to send 3 bits/s to the channel. Since $C=3$ bits/s, then we utilized the entire maximum channel capacity.

From the above example, can we actually go beyond $R_{\textrm {max}}={\frac {C}{H(S)}}$ symbols/s? NO. Shannon explicitly said that "It is not possible to transmit at an average rate greater than $C/H$." This is the maximum symbol rate already.

Channel Rate Example 2: A 6-sided Die

This time, suppose we want to send the symbols of a fair 6-sided die on a channel that only has $C=1$ bits/s. Such that $S$ has outcomes $\{1,2,3,4,5,6\}.Ofcourse,alloutcomesareequiprobablewith<math>P(s_{i})={\frac {1}{6}}$. From here, we can determined $H(S)=\log _{2}(6)=2.58$ bits/symbol. Therefore:

$R_{\textrm {max}}={\frac {1\ {\textrm {bits/s}}}{2.58\ {\textrm {bits/symbols}}}}=0.387\ {\textrm {symbols/s}}$

Again, Shannon says that with the given $C$ and $H(S)$ we can only transmit at a maximum symbol rate of $R_{\textrm {max}}$ symbols/s. Now, let's think of an encoding scheme that suits our needs. For convenience, to represent 6 unique outcomes, we can use 3 bits to represent each outcome. Let's say we have codebook-3:

Source Symbol	Codeword
1	000
2	001
3	010
4	011
5	100
6	101

The coding efficiency for codebook-3 and for the 6-die example is:

$\eta ={\frac {H(S)}{L(X)}}={\frac {2.58}{3}}=0.86$

This is not that efficient but it will suffice for now.

Figure 1: $I(R,S)$ of a BSC with varying $p$ and parametrizing $\epsilon$.

Previously, we comprehensively investigated the measures of $H(R)$, $H(R|S)$, and $I(R,S)$ on a BSC. Recall that:

$I(R,S)=H(R)-H(R|S)$

If you think about it, $I(R,S)$ is a measure of channel quality. It tells us how much we can estimate the input from an observed output. In English, we said that $I(R,S)$ is a measure of the received information contained in the source information. Ideally if there is no noise, then $H(R|S)=0$ resulting in $I(R,S)=H(S)=H(R)$. We want this because this means all the information from the source gets to the receiver! When noise is present $H(R|S)\neq 0$, it degrades the quality of our $I(R,S)$ and the information about the receiver contained in the source is now being shared with noise. This only means one thing: we want $I(R,S)$ to be at maximum. From the previous discussion, we presented how $I(R,S)$ varies with $p$ and parametrizing the noise probability $\epsilon$. Figure 1 shows this plot (it's the same image from the last discussion). We can visualize $I(R,S)$ like a pipe that constricts the flow of information depending on the noise parameter $\epsilon$. Figure 2 GIF shows the pipe visualization.

Figure 2: Pipe visualization for $I(R,S)$

Let's modify our definition of $I(R,S)$ a bit. In communication theory, $I(R,S)$ is a metric for transmission rate. So the units is really in bits/second. Figure 1 demonstrates how much data can be transmitted per second. Interchanging between bits or bits/second might be confusing but for now, when we talk about channels (and also $I(R,S)$), let's use bits/s. Bits/s is also a measure of bandwidth. From the discussions, it is clear that we can define the maximum channel capacity $C$ as:

$C=\max(I(S,R))$

(1)

We can re-write this as:

$C=\max(H(R)-H(R|S))$

(1)

Here, $C$ is in units of bits/second. It tells us the maximum number of data that we can transmit per unit of time. Both $H(R)$ and $H(R|S)$ are affected by noise parameter $\epsilon$. In a way, noise limits $C$. For example, a certain $\epsilon$ can only provide a certain maximum $I(R,S)$. Looking at figure 1, when $\epsilon =0$ (or $\epsilon =1$) we can get a maximum $I(R,S)=1$ bits/s. When $\epsilon =0.25$ (or $\epsilon =0.75$) we can get a maximum $I(R,S)=0.189$ bits/s. These occur when $p=0.5$. We can prove this mathematically. Let's determine $C$ of a BSC given some $p$ and $\epsilon$.

$C=\max(H(R)-H(R|S))$

Since we know $H(R)$ follows the Bernoulli entropy $H_{b}(q)$ where $q=p+\epsilon -2\epsilon p$. Then we can just write:

$H(R)=H_{b}(q)=-q\log _{2}(q)-(1-q)\log _{2}(1-q)$

$\max(H(R))=1$ bits when $q=0.5$. Since $\epsilon$ is fixed depending on the channel, we can manipulate the input probability $p$ such that we can force $q=0.5$ and achieve the maximum $H(R)$. Hold on to this very important thought. We also know that $H(R|S)=H_{b}(\epsilon )$. Unfortunately, since $\epsilon$ is fixed then we can't immediately assume $H_{b}(\epsilon )=1$ not unless $\epsilon =0.5$. Therefore we can write $C$ as:

$C=\max(H_{b}(q)-H_{b}(\epsilon ))$

Leading to:

$C=1-H_{b}(\epsilon )$

(2)

Equation 2 shows how the maximum capacity is limited by the noise $\epsilon$. It describes the pipe visualization in figure 1. When $\epsilon =0$, $H_{b}(\epsilon )=0$ and our BSC can transmit $C=1.0$ bits/s of data. When $\epsilon =0.5$, $H_{b}(\epsilon )=1$, the noise entropy is at maximum and our channel can transmit $C=0$ bits/s. Clearly, we can't transmit data this way. Lastly, when $\epsilon =1$, then $H_{b}(\epsilon )=0$ again and we're back to transmitting $C=1$ bits/s.

Again, we can assume $\max(H_{b}(q))=1$ because we can transform the input to have a new probability distribution $p$ that maximizes $H_{b}(q)$! If we have the In fact, this is a prelude Shannon's source coding theorem:

Note that the above statement only says that we can recode the source into something that can maximize the channel capacity. Also note that the $\epsilon$ is an arbitrary error. Not the noise probability. Figure 1 and our idea