Capacity of Memoryless Digital Channels

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Applet Description

In this applet, binary $(M=2)$ and ternary $(M=3)$ channel models without memory are considered with $M$ inputs $X$ and $M$ outputs $Y$. Such a message system is completely determined by the probability function $P_X(X)$ and the matrix $P_{\hspace{0.01cm}Y\hspace{0.03cm} \vert \hspace{0.01cm}X}(Y\hspace{0.03cm} \vert \hspace{0.03cm} X)$ of the transition probabilities.

For these binary and ternary systems, the following information-theoretic descriptive quantities are derived and clarified:

the source entropy $H(X)$ and the sink entropy $H(Y)$,
the equivocation $H(X|Y)$ and the irrelevance $H(Y|X)$,
the joint entropy $H(XY)$ as well as the mutual information $I(X; Y)$,
the channel capacity as the decisive parameter of digital channel models without memory:

$$C = \max_{P_X(X)} \hspace{0.15cm} I(X;Y) \hspace{0.05cm}.$$

These information-theoretical quantities can be calculated both in analytic–closed form or determined simulatively by evaluation of source and sink symbol sequence.

Theoretical Background

Underlying model of digital signal transmission

The set of possible source symbols is characterized by the discrete random variable $X$.

In the binary case ⇒ $M_X= |X| = 2$ holds $X = \{\hspace{0.05cm}{\rm A}, \hspace{0.15cm} {\rm B} \hspace{0.05cm}\}$ with the probability mass function $($ $\rm PMF)$ $P_X(X)= \big (p_{\rm A},\hspace{0.15cm}p_{\rm B}\big)$ and the source symbol probabilities $p_{\rm A}$ and $p_{\rm B}=1- p_{\rm A}$.
Accordingly, for a ternary source ⇒ $M_X= |X| = 3$: $X = \{\hspace{0.05cm}{\rm A}, \hspace{0.15cm} {\rm B}, \hspace{0.15cm} {\rm C} \hspace{0.05cm}\}$, $P_X(X)= \big (p_{\rm A},\hspace{0.15cm}p_{\rm B},\hspace{0.15cm}p_{\rm C}\big)$, $p_{\rm C}=1- p_{\rm A}-p_{\rm B}$.

The set of possible sink symbols is characterized by the discrete random variable $Y$. These come from the same set of symbols as the source symbols ⇒ $M_Y=M_X = M$. To simplify the following description, we denote them with lowercase letters, for example, for $M=3$: $Y = \{\hspace{0.05cm}{\rm a}, \hspace{0.15cm} {\rm b}, \hspace{0.15cm} {\rm c} \hspace{0.05cm}\}$.

The relationship between the random variables $X$ and $Y$ is given by a discrete memoryless channel model $($ $\rm DMC)$. The left graph shows this for $M=2$ and the right graph for $M=3$.

$M=2$ (left) and for $M=3$ (right).
Please note: In the right graph not all transitions are labeled

The following description applies to the simpler case $M=2$. For the calculation of all information theoretic quantities in the next section we need, besides $P_X(X)$ and $P_Y(Y)$ the two-dimensional probability functions $($each a $2\times2$–matrix$)$ of all

$\text{conditional probabilities}$ ⇒ $P_{\hspace{0.01cm}Y\hspace{0.03cm} \vert \hspace{0.01cm}X}(Y\hspace{0.03cm} \vert \hspace{0.03cm} X)$ ⇒ given by the DMC model;
$\text{joint probabilities}$ ⇒ $P_{XY}(X,\hspace{0.1cm}Y)$;
$\text{inference probabilities}$ ⇒ $P_{\hspace{0.01cm}X\hspace{0.03cm} \vert \hspace{0.03cm}Y}(X\hspace{0.03cm} \vert \hspace{0.03cm} Y)$.

Betrachtetes Modell des Binärkanals

$\text{Beispiel 1}$: We consider the sketched binary channel.

Let the falsification probabilities be:

$$\begin{align*}p_{\rm a\hspace{0.03cm}\vert \hspace{0.03cm}A} & = {\rm Pr}(Y\hspace{-0.1cm} = {\rm a}\hspace{0.05cm}\vert X \hspace{-0.1cm}= {\rm A}) = 0.95\hspace{0.05cm},\hspace{0.8cm}p_{\rm b\hspace{0.03cm}\vert \hspace{0.03cm}A} = {\rm Pr}(Y\hspace{-0.1cm} = {\rm b}\hspace{0.05cm}\vert X \hspace{-0.1cm}= {\rm A}) = 0.05\hspace{0.05cm},\\ p_{\rm a\hspace{0.03cm}\vert \hspace{0.03cm}B} & = {\rm Pr}(Y\hspace{-0.1cm} = {\rm a}\hspace{0.05cm}\vert X \hspace{-0.1cm}= {\rm B}) = 0.40\hspace{0.05cm},\hspace{0.8cm}p_{\rm b\hspace{0.03cm}\vert \hspace{0.03cm}B} = {\rm Pr}(Y\hspace{-0.1cm} = {\rm b}\hspace{0.05cm}\vert X \hspace{-0.1cm}= {\rm B}) = 0.60\end{align*}$$

$$\Rightarrow \hspace{0.3cm} P_{\hspace{0.01cm}Y\hspace{0.05cm} \vert \hspace{0.05cm}X}(Y\hspace{0.05cm} \vert \hspace{0.05cm} X) = \begin{pmatrix} 0.95 & 0.05\\ 0.4 & 0.6 \end{pmatrix} \hspace{0.05cm}.$$

Furthermore, we assume source symbols that are not equally probable:

$$P_X(X) = \big ( p_{\rm A},\ p_{\rm B} \big )= \big ( 0.1,\ 0.9 \big ) \hspace{0.05cm}.$$

Für die Wahrscheinlichkeitsfunktion der Sinke ergibt sich somit:

$$P_Y(Y) = \big [ {\rm Pr}( Y\hspace{-0.1cm} = {\rm a})\hspace{0.05cm}, \ {\rm Pr}( Y \hspace{-0.1cm}= {\rm b}) \big ] = \big ( 0.1\hspace{0.05cm},\ 0.9 \big ) \cdot \begin{pmatrix} 0.95 & 0.05\\ 0.4 & 0.6 \end{pmatrix} $$

$$\Rightarrow \hspace{0.3cm} {\rm Pr}( Y \hspace{-0.1cm}= {\rm a}) = 0.1 \cdot 0.95 + 0.9 \cdot 0.4 = 0.455\hspace{0.05cm},\hspace{1.0cm} {\rm Pr}( Y \hspace{-0.1cm}= {\rm b}) = 1 - {\rm Pr}( Y \hspace{-0.1cm}= {\rm a}) = 0.545.$$

Die Verbundwahrscheinlichkeiten $p_{\mu \kappa} = \text{Pr}\big[(X = μ) ∩ (Y = κ)\big]$ zwischen Quelle und Sinke sind:

$$\begin{align*}p_{\rm Aa} & = p_{\rm a} \cdot p_{\rm a\hspace{0.03cm}\vert \hspace{0.03cm}A} = 0.095\hspace{0.05cm},\hspace{0.5cm}p_{\rm Ab} = p_{\rm b} \cdot p_{\rm b\hspace{0.03cm}\vert \hspace{0.03cm}A} = 0.005\hspace{0.05cm},\\ p_{\rm Ba} & = p_{\rm a} \cdot p_{\rm a\hspace{0.03cm}\vert \hspace{0.03cm}B} = 0.360\hspace{0.05cm}, \hspace{0.5cm}p_{\rm Bb} = p_{\rm b} \cdot p_{\rm b\hspace{0.03cm}\vert \hspace{0.03cm}B} = 0.540\hspace{0.05cm}. \end{align*}$$

$$\Rightarrow \hspace{0.3cm} P_{XY}(X,\hspace{0.1cm}Y) = \begin{pmatrix} 0.095 & 0.005\\ 0.36 & 0.54 \end{pmatrix} \hspace{0.05cm}.$$

Für die Rückschlusswahrscheinlichkeiten erhält man:

$$\begin{align*}p_{\rm A\hspace{0.03cm}\vert \hspace{0.03cm}a} & = p_{\rm Aa}/p_{\rm a} = 0.095/0.455 = 0.2088\hspace{0.05cm},\hspace{0.5cm}p_{\rm A\hspace{0.03cm}\vert \hspace{0.03cm}b} = p_{\rm Ab}/p_{\rm b} = 0.005/0.545 = 0.0092\hspace{0.05cm},\\ p_{\rm B\hspace{0.03cm}\vert \hspace{0.03cm}a} & = p_{\rm Ba}/p_{\rm a} = 0.36/0.455 = 0.7912\hspace{0.05cm},\hspace{0.5cm}p_{\rm B\hspace{0.03cm}\vert \hspace{0.03cm}b} = p_{\rm Bb}/p_{\rm b} = 0.54/0.545 = 0.9908\hspace{0.05cm} \end{align*}$$

$$\Rightarrow \hspace{0.3cm} P_{\hspace{0.01cm}X\hspace{0.05cm} \vert \hspace{0.05cm}Y}(X\hspace{0.05cm} \vert \hspace{0.05cm} Y) = \begin{pmatrix} 0.2088 & 0.0092\\ 0.7912 & 0.9908 \end{pmatrix} \hspace{0.05cm}.$$

Definition und Interpretation verschiedener Entropiefunktionen

Im $\rm LNTwww$–Theorieteil werden alle für 2D–Zufallsgrößen relevanten Entropien definiert, die auch für die Digitalsignalübertragung gelten. Zudem finden Sie dort zwei Schaubilder, die den Zusammenhang zwischen den einzelnen Entropien illustrieren.

Für die Digitalsignalübertragung ist die rechte Darstellung zweckmäßig, bei der die Richtung von der Quelle $X$ zur Sinke $Y$ erkennbar ist.
Wir interpretieren nun ausgehend von dieser Grafik die einzelnen informationstheoretischen Größen.

Zwei informationstheoretische Modelle für die Digitalsignalübertragung.
Bitte beachten Sie: In der rechten Grafik ist $H_{XY}$ nicht darstellbar

The »source entropy« $H(X)$ denotes the average information content of the source symbol sequence. With the symbol set size $|X|$ applies:

$$H(X) = {\rm E} \left [ {\rm log}_2 \hspace{0.1cm} \frac{1}{P_X(X)}\right ] \hspace{0.1cm} = -{\rm E} \big [ {\rm log}_2 \hspace{0.1cm}{P_X(X)}\big ] \hspace{0.2cm} =\hspace{0.2cm} \sum_{\mu = 1}^{|X|} P_X(x_{\mu}) \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{P_X(x_{\mu})} \hspace{0.05cm}.$$

The »equivocation« $H(X|Y)$ indicates the average information content that an observer who knows exactly about the sink $Y$ gains by observing the source $X$ :

$$H(X|Y) = {\rm E} \left [ {\rm log}_2 \hspace{0.1cm} \frac{1}{P_{\hspace{0.05cm}X\hspace{-0.01cm}|\hspace{-0.01cm}Y}(X\hspace{-0.01cm} |\hspace{0.03cm} Y)}\right ] \hspace{0.2cm}=\hspace{0.2cm} \sum_{\mu = 1}^{|X|} \sum_{\kappa = 1}^{|Y|} P_{XY}(x_{\mu},\hspace{0.05cm}y_{\kappa}) \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{P_{\hspace{0.05cm}X\hspace{-0.01cm}|\hspace{0.03cm}Y} (\hspace{0.05cm}x_{\mu}\hspace{0.03cm} |\hspace{0.05cm} y_{\kappa})} \hspace{0.05cm}.$$

The equivocation is the portion of the source entropy $H(X)$ that is lost due to channel interference (for digital channel: transmission errors). The »mutual information« $I(X; Y)$ remains, which reaches the sink:

$$I(X;Y) = {\rm E} \left [ {\rm log}_2 \hspace{0.1cm} \frac{P_{XY}(X, Y)}{P_X(X) \cdot P_Y(Y)}\right ] \hspace{0.2cm}=\hspace{0.2cm} \sum_{\mu = 1}^{|X|} \sum_{\kappa = 1}^{|Y|} P_{XY}(x_{\mu},\hspace{0.05cm}y_{\kappa}) \cdot {\rm log}_2 \hspace{0.1cm} \frac{P_{XY}(x_{\mu},\hspace{0.05cm}y_{\kappa})}{P_{\hspace{0.05cm}X}(\hspace{0.05cm}x_{\mu}) \cdot P_{\hspace{0.05cm}Y}(\hspace{0.05cm}y_{\kappa})} \hspace{0.05cm} = H(X) - H(X|Y) \hspace{0.05cm}.$$

The »irrelevance« $H(Y|X)$ indicates the average information content that an observer who knows exactly about the source $X$ gains by observing the sink $Y$:

$$H(Y|X) = {\rm E} \left [ {\rm log}_2 \hspace{0.1cm} \frac{1}{P_{\hspace{0.05cm}Y\hspace{-0.01cm}|\hspace{-0.01cm}X}(Y\hspace{-0.01cm} |\hspace{0.03cm} X)}\right ] \hspace{0.2cm}=\hspace{0.2cm} \sum_{\mu = 1}^{|X|} \sum_{\kappa = 1}^{|Y|} P_{XY}(x_{\mu},\hspace{0.05cm}y_{\kappa}) \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{P_{\hspace{0.05cm}Y\hspace{-0.01cm}|\hspace{0.03cm}X} (\hspace{0.05cm}y_{\kappa}\hspace{0.03cm} |\hspace{0.05cm} x_{\mu})} \hspace{0.05cm}.$$

The »sink entropy« $H(Y)$, the mean information content of the sink. $H(Y)$ is the sum of the useful mutual information $I(X; Y)$ and the useless irrelevance $H(Y|X)$, which comes exclusively from channel errors:

$$H(Y) = {\rm E} \left [ {\rm log}_2 \hspace{0.1cm} \frac{1}{P_Y(Y)}\right ] \hspace{0.1cm} = -{\rm E} \big [ {\rm log}_2 \hspace{0.1cm}{P_Y(Y)}\big ] \hspace{0.2cm} =I(X;Y) + H(Y|X) \hspace{0.05cm}.$$

Die Verbundentropie $H(XY)$ gibt ist den mittleren Informationsgehalt der 2D–Zufallsgröße $XY$ an.&nbsp sie beschreibt zudem eine obere Schranke für die Summe aus Quellenentropie und Sinkenentropie:

$$H(XY) = {\rm E} \left [ {\rm log} \hspace{0.1cm} \frac{1}{P_{XY}(X, Y)}\right ] = \sum_{\mu = 1}^{|X|} \hspace{0.1cm} \sum_{\kappa = 1}^{|Y|} \hspace{0.1cm} P_{XY}(x_{\mu}\hspace{0.05cm}, y_{\kappa}) \cdot {\rm log} \hspace{0.1cm} \frac{1}{P_{XY}(x_{\mu}\hspace{0.05cm}, y_{\kappa})}\le H(X) + H(Y) \hspace{0.05cm}.$$

Betrachtetes Modell des Binärkanals

$\text{Beispiel 2}$: Es gelten die gleichen Voraussetzungen wie für das $\text{Beispiel 1}$:

(1) Die Quellensymbole sind nicht gleichwahrscheinlich:

$$P_X(X) = \big ( p_{\rm A},\ p_{\rm B} \big )= \big ( 0.1,\ 0.9 \big ) \hspace{0.05cm}.$$

(2) Die Verfälschungswahrscheinlichkeiten seien:

$$\begin{align*}p_{\rm a\hspace{0.03cm}\vert \hspace{0.03cm}A} & = {\rm Pr}(Y\hspace{-0.1cm} = {\rm a}\hspace{0.05cm}\vert X \hspace{-0.1cm}= {\rm A}) = 0.95\hspace{0.05cm},\hspace{0.8cm}p_{\rm b\hspace{0.03cm}\vert \hspace{0.03cm}A} = {\rm Pr}(Y\hspace{-0.1cm} = {\rm b}\hspace{0.05cm}\vert X \hspace{-0.1cm}= {\rm A}) = 0.05\hspace{0.05cm},\\ p_{\rm a\hspace{0.03cm}\vert \hspace{0.03cm}B} & = {\rm Pr}(Y\hspace{-0.1cm} = {\rm a}\hspace{0.05cm}\vert X \hspace{-0.1cm}= {\rm B}) = 0.40\hspace{0.05cm},\hspace{0.8cm}p_{\rm b\hspace{0.03cm}\vert \hspace{0.03cm}B} = {\rm Pr}(Y\hspace{-0.1cm} = {\rm b}\hspace{0.05cm}\vert X \hspace{-0.1cm}= {\rm B}) = 0.60\end{align*}$$

$$\Rightarrow \hspace{0.3cm} P_{\hspace{0.01cm}Y\hspace{0.05cm} \vert \hspace{0.05cm}X}(Y\hspace{0.05cm} \vert \hspace{0.05cm} X) = \begin{pmatrix} 0.95 & 0.05\\ 0.4 & 0.6 \end{pmatrix} \hspace{0.05cm}.$$

Binäre Entropiefunktion als Funktion von $p$

Wegen Voraussetzung (1) erhält man so für die Quellenentropie mit der binären Entropiefunktion $H_{\rm bin}(p)$:

$$H(X) = p_{\rm A} \cdot {\rm log_2}\hspace{0.1cm}\frac{1}{\hspace{0.1cm}p_{\rm A}\hspace{0.1cm} } + p_{\rm B} \cdot {\rm log_2}\hspace{0.1cm}\frac{1}{p_{\rm B} }= H_{\rm bin} (p_{\rm A}) = H_{\rm bin} (0.1)= 0.469 \ {\rm bit} \hspace{0.05cm};$$

$$H_{\rm bin} (p) = p \cdot {\rm log_2}\hspace{0.1cm}\frac{1}{\hspace{0.1cm}p\hspace{0.1cm} } + (1 - p) \cdot {\rm log_2}\hspace{0.1cm}\frac{1}{1 - p} \hspace{0.5cm}{\rm (Einheit\hspace{-0.15cm}: \hspace{0.15cm}bit\hspace{0.15cm}oder\hspace{0.15cm}bit/Symbol)} \hspace{0.05cm}.$$

Entsprechend gilt für die Sinkenentropie mit der PMF $P_Y(Y) = \big ( p_{\rm a},\ p_{\rm b} \big )= \big ( 0.455,\ 0.545 \big )$:

$$H(Y) = H_{\rm bin} (0.455)= 0.994 \ {\rm bit} \hspace{0.05cm}.$$

Als nächstes berechnen wir die Verbundentropie:

$$H(XY) = p_{\rm Aa} \cdot {\rm log_2}\hspace{0.1cm}\frac{1}{\hspace{0.1cm}p_{\rm Aa}\hspace{0.1cm} }+ p_{\rm Ab} \cdot {\rm log_2}\hspace{0.1cm}\frac{1}{\hspace{0.1cm}p_{\rm Ab}\hspace{0.1cm} }+p_{\rm Ba} \cdot {\rm log_2}\hspace{0.1cm}\frac{1}{\hspace{0.1cm}p_{\rm Ba}\hspace{0.1cm} }+ p_{\rm Bb} \cdot {\rm log_2}\hspace{0.1cm}\frac{1}{\hspace{0.1cm}p_{\rm Bb}\hspace{0.1cm} }$$

$$\Rightarrow \hspace{0.3cm}H(XY) = 0.095 \cdot {\rm log_2}\hspace{0.1cm}\frac{1}{0.095 } +0.005 \cdot {\rm log_2}\hspace{0.1cm}\frac{1}{0.005 }+0.36 \cdot {\rm log_2}\hspace{0.1cm}\frac{1}{0.36 }+0.54 \cdot {\rm log_2}\hspace{0.1cm}\frac{1}{0.54 }= 1.371 \ {\rm bit} \hspace{0.05cm}.$$

Entsprechend dem oberen linken Schaubild sind somit auch die restlichen informationstheoretischen Größen berechenbar:

Informationstheoretisches Modell für $\text{Beispiel 2}$

the »equivocation«:

$$H(X \vert Y) \hspace{-0.01cm} =\hspace{-0.01cm} H(XY) \hspace{-0.01cm} -\hspace{-0.01cm} H(Y) \hspace{-0.01cm} = \hspace{-0.01cm} 1.371\hspace{-0.01cm} -\hspace{-0.01cm} 0.994\hspace{-0.01cm} =\hspace{-0.01cm} 0.377\ {\rm bit} \hspace{0.05cm},$$

- the »irrelevance«:

$$H(Y \vert X) = H(XY) - H(X) = 1.371 - 0.994 = 0.902\ {\rm bit} \hspace{0.05cm}.$$

die Transinformation (englisch Mutual Information):

$$I(X;Y) = H(X) + H(Y) - H(XY) = 0.469 + 0.994 - 1.371 = 0.092\ {\rm bit} \hspace{0.05cm},$$

The results are summarized in the graph.

Anmerkung: Äquivokation und Irrelevanz könnte man (allerdings mit Mehraufwand) auch direkt aus den entsprechenden Wahrscheinlichkeitsfunktionen berechnen, zum Beispiel:

$$H(Y \vert X) = \hspace{-0.2cm} \sum_{(x, y) \hspace{0.05cm}\in \hspace{0.05cm}XY} \hspace{-0.2cm} P_{XY}(x,\hspace{0.05cm}y) \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{P_{\hspace{0.05cm}Y\hspace{-0.01cm}\vert \hspace{0.03cm}X} (\hspace{0.05cm}y\hspace{0.03cm} \vert \hspace{0.05cm} x)}= p_{\rm Aa} \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{p_{\rm a\hspace{0.03cm}\vert \hspace{0.03cm}A} } + p_{\rm Ab} \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{p_{\rm b\hspace{0.03cm}\vert \hspace{0.03cm}A} } + p_{\rm Ba} \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{p_{\rm a\hspace{0.03cm}\vert \hspace{0.03cm}B} } + p_{\rm Bb} \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{p_{\rm b\hspace{0.03cm}\vert \hspace{0.03cm}B} } = 0.902 \ {\rm bit} \hspace{0.05cm}.$$

Betrachtetes Modell des Ternärkanals:
Rote Übergänge stehen für $p_{\rm a\hspace{0.03cm}\vert \hspace{0.03cm}A} = p_{\rm b\hspace{0.03cm}\vert \hspace{0.03cm}B} = p_{\rm c\hspace{0.03cm}\vert \hspace{0.03cm}C} = q$ und blaue für $p_{\rm b\hspace{0.03cm}\vert \hspace{0.03cm}A} = p_{\rm c\hspace{0.03cm}\vert \hspace{0.03cm}A} =\text{...}= p_{\rm b\hspace{0.03cm}\vert \hspace{0.03cm}C}= (1-q)/2$

$\text{Beispiel 3}$: Nun betrachten wir ein Übertragungssystem mit $M_X = M_Y = M=3$.

(1) Die Quellensymbole seien gleichwahrscheinlich:

$$P_X(X) = \big ( p_{\rm A},\ p_{\rm B},\ p_{\rm C} \big )= \big ( 1/3,\ 1/3,\ 1/3 \big )\hspace{0.30cm}\Rightarrow\hspace{0.30cm}H(X)={\rm log_2}\hspace{0.1cm}3 \approx 1.585 \ {\rm bit} \hspace{0.05cm}.$$

(2) Das Kanalmodell ist symmetrisch ⇒ auch die Sinkensymbole sind gleichwahrscheinlich:

$$P_Y(Y) = \big ( p_{\rm a},\ p_{\rm b},\ p_{\rm c} \big )= \big ( 1/3,\ 1/3,\ 1/3 \big )\hspace{0.30cm}\Rightarrow\hspace{0.30cm}H(Y)={\rm log_2}\hspace{0.1cm}3 \approx 1.585 \ {\rm bit} \hspace{0.05cm}.$$

(3) Die Verbundwahrscheinlichkeiten ergeben sich wie folgt:

$$p_{\rm Aa}= p_{\rm Bb}= p_{\rm Cc}= q/M,$$

$$p_{\rm Ab}= p_{\rm Ac}= p_{\rm Ba}= p_{\rm Bc} = p_{\rm Ca}= p_{\rm Cb} = (1-q)/(2M)$$

$$\Rightarrow\hspace{0.30cm}H(XY) = 3 \cdot p_{\rm Aa} \cdot {\rm log_2}\hspace{0.1cm}\frac{1}{\hspace{0.1cm}p_{\rm Aa}\hspace{0.1cm} }+6 \cdot p_{\rm Ab} \cdot {\rm log_2}\hspace{0.1cm}\frac{1}{\hspace{0.1cm}p_{\rm Ab}\hspace{0.1cm} }= \ \text{...} \ = q \cdot {\rm log_2}\hspace{0.1cm}\frac{M}{q }+ (1-q) \cdot {\rm log_2}\hspace{0.1cm}\frac{M}{(1-q)/2 }.$$

Einige Ergebnisse zum $\text{Beispiel 3}$

(4) Für die Transinformation erhält man nach einigen Umformungen unter Berücksichtigung der Gleichung

$$I(X;Y) = H(X) + H(Y) - H(XY)\text{:}$$

$$\Rightarrow \hspace{0.3cm} I(X;Y) = {\rm log_2}\ (M) - (1-q) -H_{\rm bin}(q).$$

Bei fehlerfreier Ternärübertragung $(q=1)$ gilt $I(X;Y) = H(X) = H(Y)={\rm log_2}\hspace{0.1cm}3$.
Mit $q=0.8$ sinkt die Transinformaion schon auf $I(X;Y) = 0.663$ und mit $q=0.5$ auf $0.085$ bit.
Der ungünstigste Fall aus informationstheoretischer Sicht ist $q=1/3$ ⇒ $I(X;Y) = 0$.
Dagegen ist der aus der aus Sicht der Übertragungstheorie ungünstigste Fall $q=0$ ⇒ "kein einziges Übertragungssymbol kommt richtig an" aus informationstheoretischer Sicht gar nicht so schlecht.
Um dieses gute Ergebnis nutzen zu können, ist allerdings sendeseitig eine Kanalcodierung erforderlich.

Definition and meaning of channel capacity

If one calculates the mutual information $I(X, Y)$ as explained in $\text{Example 2}$, so hängt diese nicht nur vom diskreten gedächtnislosen Kanal (englisch: Discrete Memoryless Channel, kurz DMC) ab, sondern auch von der Quellenstatistik ⇒ $P_X(X)$ ab. Ergo: The mutual information $I(X, Y)$ is not a pure channel characteristic.

$\text{Definition:}$ The »channel capacity« introduced by $\text{Claude E. Shannon}$ according to his standard work [Sha48]^[1]:

$$C = \max_{P_X(X)} \hspace{0.15cm} I(X;Y) \hspace{0.05cm}.$$

The additional unit "bit/use" is often added. Since according to this definition the best possible source statistics are always the basis:

$C$ depends only on the channel properties ⇒ $P_{Y \vert X}(Y \vert X)$ but not on the source statistics ⇒ $P_X(X)$.

Shannon needed the quantity $C$ to formulate the "Channel Coding Theorem" – one of the highlights of the information theory he founded.

$\text{Shannon's Channel Coding Theorem:}$

For every transmission channel with channel capacity $C > 0$, there exists (at least) one $(k,\ n)$–block code, whose (block) error probability approaches zero as long as the code rate $R = k/n$ is less than or equal to the channel capacity:

$$R ≤ C.$$

The prerequisite for this, however, is that the following applies to the block length of this code: $n → ∞.$

$\text{Umkehrschluss von Shannons Kanalcodierungstheorem:}$

Ist die Rate $R$ des verwendeten $(n$, $k)$–Blockcodes größer als die Kanalkapazität $C$, so ist niemals eine beliebig kleine Blockfehlerwahrscheinlichkeit erreichbar.

Informationsheoretischer Größen für
verschiedene $p_{\rm A}$ und $p_{\rm B}= 1- p_{\rm A}$

$\text{Beispiel 4}$: Wir betrachten den gleichen diskreten gedächtnislosen Kanal wie im $\text{Beispiel 2}$. In diesem $\text{Beispiel 2}$ wurden die Symbolwahrscheinlichkeiten $p_{\rm A} = 0.1$ und $p_{\rm B}= 1- p_{\rm A}=0.9$ vorausgesetzt. Damit ergab sich die Transinformation zu $I(X;Y)= 0.092$ bit/Kanalzugriff ⇒ siehe erste Zeile, vierte Spalte in der Tabelle.

Die Kanalkapazität ist die Transinformation $I(X, Y)$ bei bestmöglichen Symbolwahrscheinlichkeiten $p_{\rm A} = 0.55$ und $p_{\rm B}= 1- p_{\rm A}=0.45$:

$$C = \max_{P_X(X)} \hspace{0.15cm} I(X;Y) = 0.284 \ \rm bit/Kanalzugriff \hspace{0.05cm}.$$

Aus der Tabelle erkennt man weiter (auf die Zusatzeinheit "bit/Kanalzugriff" verzichten wir im Folgenden):

Der Parameter $p_{\rm A} = 0.1$ war sehr ungünstig gewählt, weil beim vorliegenden Kanal das Symbol $\rm A$ mehr verfälscht wird als $\rm B$. Schon mit $p_{\rm A} = 0.9$ ergibt sich ein etwas besserer Wert: $I(X; Y)=0.130$.
Aus dem gleichen Grund liefert $p_{\rm A} = 0.55$, $p_{\rm B} = 0.45$ ein etwas besseres Ergebnis als gleichwahrscheinliche Symbole $p_{\rm A} = p_{\rm B} =0.5$.
Je unsymmetrischer der Kanal ist, um so mehr weicht die optimale Wahrscheinlichkeitsfunktion $P_X(X)$ von der Gleichverteilung ab. Im Umkehrschluss: Bei symmetrischem Kanal ergibt sich stets die Gleichverteilung.

Der Ternärkanal von $\text{Beispiel 3}$ ist symmetrisch. Deshalb ist hier $P_X(X) = \big ( 1/3,\ 1/3,\ 1/3 \big )$ für jeden $q$–Wert optimal, und die in der Ergebnistabelle angegebene Transinformation $I(X;Y)$ ist gleichzeitig die Kanalkapazität $C$.

Exercises

First, select the number $(1,\ 2, \text{...} \ )$ of the task to be processed. The number "$0$" corresponds to a "Reset": Same setting as at program start.
A task description is displayed. The parameter values are adjusted. Solution after pressing "Show Solution".
Source symbols are denoted by uppercase letters (binary: $\rm A$, $\rm B$), sink symbols by lowercase letters ($\rm a$, $\rm b$). Error-free transmission: $\rm A \rightarrow a$.
For all entropy values, the unit "bit/use" would have to be added.

(1) Let $p_{\rm A} = p_{\rm B} = 0.5$ and $p_{\rm b \vert A} = p_{\rm a \vert B} = 0.1$. What is the channel model? What are the entropies $H(X), \, H(Y)$ and the mutual information $I(X;\, Y)$?

Considered is the BSC model (Binary Symmetric Channel). Because of $p_{\rm A} = p_{\rm B} = 0.5$ it holds for the entropies: $H(X) = H(Y) = 1$.
Because of $p_{\rm b \vert A} = p_{\rm a \vert B} = 0.1$ eqivocation and irrelevance are also equal: $H(X \vert Y) = H(Y \vert X) = H_{\rm bin}(p_{\rm b \vert A}) = H_{\rm bin}(0.1) =0.469$.
The mutual information is $I(X;\, Y) = H(X) - H(X \vert Y)= 1-H_{\rm bin}(p_{\rm b \vert A}) = 0.531$ and the joint entropy is $H(XY) =1.469$.

(2) Let further $p_{\rm b \vert A} = p_{\rm a \vert B} = 0.1$, but now the symbol probability is $p_{\rm A} = 0. 9$. What is the capacity $C_{\rm BSC}$ of the BSC channel with $p_{\rm b \vert A} = p_{\rm a \vert B}$?
Which $p_{\rm b \vert A} = p_{\rm a \vert B}$ leads to the largest possible channel capacity and which $p_{\rm b \vert A} = p_{\rm a \vert B}$ leads to the channel capacity $C_{\rm BSC}=0$?

The capacity $C_{\rm BSC}$ is equal to the maximum mutual information $I(X;\, Y)$ considering the optimal symbol probabilities.
Due to the symmetry of the BSC model equally probable symbols $(p_{\rm A} = p_{\rm B} = 0.5)$ lead to the optimum ⇒ $C_{\rm BSC}=0.531$.
The best is the "ideal channel" $(p_{\rm b \vert A} = p_{\rm a \vert B} = 0)$ ⇒ $C_{\rm BSC}=1$. The worst BSC channel results with $p_{\rm b \vert A} = p_{\rm a \vert B} = 0.5$ ⇒ $C_{\rm BSC}=0$.
But also with $p_{\rm b \vert A} = p_{\rm a \vert B} = 1$ we get $C_{\rm BSC}=1$. Here all symbols are inverted, which is information theoretically the same as $\langle Y_n \rangle \equiv \langle X_n \rangle$.

(3) Let $p_{\rm A} = p_{\rm B} = 0.5$, $p_{\rm b \vert A} = 0.05$ and $ p_{\rm a \vert B} = 0.4$. Interpret the results in comparison to the experiment $(1)$ and to the $\text{example 2}$ in the theory section.

Unlike the experiment $(1)$ no BSC channel is present here. Rather, the channel considered here is asymmetric: $p_{\rm b \vert A} \ne p_{\rm a \vert B}$.
According to $\text{Example 2}$ it holds for $p_{\rm A} = 0.1,\ p_{\rm B} = 0.9$: $H(X)= 0.469$, $H(Y)= 0.994$, $H(X \vert Y)=0.377$, $H(Y \vert X)=0.902$, $I(X;\vert Y)=0.092$.
Now it holds $p_{\rm A} = p_{\rm B} = 0.5$ and we get $H(X)=1,000$, $H(Y)=0.910$, $H(X \vert Y)=0.719$, $H(Y \vert X)=0.629$, $I(X;\ Y)=0.281$.
All output values depend significantly on $p_{\rm A}$ and $p_{\rm B}=1-p_{\rm A}$ except for the conditional probabilities ${\rm Pr}(Y \vert X)\in \{\hspace{0.05cm}0.95,\ 0.05,\ 0.4,\ 0.6\hspace{0.05cm} \}$.

(4) Let further $p_{\rm A} = p_{\rm B}$, $p_{\rm b \vert A} = 0.05$, $ p_{\rm a \vert B} = 0.4$. What differences do you see in terms of analytical calculation and "simulation" $(N=10000)$.

The joint probabilities are $p_{\rm Aa} =0.475$, $p_{\rm Ab} =0.025$, $p_{\rm Ba} =0.200$, $p_{\rm Bb} =0.300$. Simulation: Approximation by relative frequencies:
For example, for $N=10000$: $h_{\rm Aa} =0.4778$, $h_{\rm Ab} =0.0264$, $h_{\rm Ba} =0.2039$, $h_{\rm Bb} =0.2919$. After pressing "New sequence" slightly different values.
For all subsequent calculations, no principal difference between theory and simulation, except $p \to h$. Examples:
$p_{\rm A} = 0.5 \to h_{\rm A}=h_{\rm Aa} + h_{\rm Ab} =0.5042$, $p_b = 0.325 \to h_{\rm b}=h_{\rm Ab} + h_{\rm Bb} =0. 318$, $p_{b|A} = 0.05 \to h_{\rm b|A}=h_{\rm Ab}/h_{\rm A} =0.0264/0.5042= 0.0524$,
$p_{\rm A|b} = 0.0769 \to h_{\rm A|b}=h_{\rm Ab}/h_{\rm b} =0.0264/0.318= 0.0830$. Thus, this simulation yields $I_{\rm Sim}(X;\ Y)=0.269$ instead of $I(X;\ Y)=0.281$.

(5) Setting according to $(4)$. How does $I_{\rm Sim}(X;\ Y)$ differ from $I(X;\ Y) = 0.281$ for $N=10^3$, $10^4$, $10^5$ ? In each case, averaging over ten realizations.

$N=10^3$: $0.232 \le I_{\rm Sim} \le 0.295$, mean: $0.263$ # $N=10^4$: $0.267 \le I_{\rm Sim} \le 0.293$, mean: $0.279$ # $N=10^5$: $0.280 \le I_{\rm Sim} \le 0.285$ mean: $0.282$.
With $N=10^6$ for this channel, the simulation result differs from the theoretical value by less than $\pm 0.001$.

(6) What is the capacity $C_6$ of this channel with $p_{\rm b \vert A} = 0.05$, $ p_{\rm a \vert B} = 0.4$? Is the error probability $0$ possible with the code rate $R=0.3$?

$C_6=0.284$ is the maximum of $I(X;\ Y)$ for $p_{\rm A} =0.55$ ⇒ $p_{\rm B} =0. 45$. Simulation over ten times $N=10^5$: $0.281 \le I_{\rm Sim}(X;\ Y) \le 0.289$.
With the code rate $R=0.3 > C_6$ an arbitrarily small block error probability is not achievable even with the best possible coding.

(7) Now let $p_{\rm A} = p_{\rm B}$, $p_{\rm b \vert A} = 0$, $ p_{\rm a \vert B} = 0.5$. What property does this asymmetric channel exhibit? What values result for $H(X)$, $H(X \vert Y)$, $I(X;\ Y)$ ?

The symbol $\rm A$ is never falsified, the symbol $\rm B$ with (information theoretically) maximum falsification probability $ p_{\rm a \vert B} = 0.5$
The total falsification probability is $ {\rm Pr} (Y_n \ne X_n)= p_{\rm A} \cdot p_{\rm b \vert A} + p_{\rm B} \cdot p_{\rm a \vert B}= 0.25$ ⇒ about $25\%$ of the output sink symbols are "purple".
Joint probabilities: $p_{\rm Aa}= 1/2,\ p_{\rm Ab}= 0,\ p_{\rm Ba}= p_{\rm Bb}= 1/4$, Inference probabilities: $p_{\rm A \vert a}= 1,\ p_{\rm B \vert a}= 0,\ p_{\rm A \vert b}= 1/3,\ p_{\rm B \vert b}= 2/3$.
From this we get for equivocation $H(X \vert Y)=0.689$; with source entropy $H(X)= 1$ ⇒ $I(X;\vert Y)=H(X)-H(X \vert Y)=0.311$.

(8) What is the capacity $C_8$ of this channel with $p_{\rm b \vert A} = 0.05$, $ p_{\rm a \vert B} = 035$? Is the error probability $0$ possible with the code rate $R=0.3$?

$C_8=0.326$ is the maximum of $I(X;\ Y)$ for $p_{\rm A} =0.55$. Thus, because of $C_8 >R=0.3 $ an arbitrarily small block error probability is achievable.
The only difference compared to $(6)$ ⇒ $C_6=0.284 < 0.3$ is the slightly smaller falsification probability $ p_{\rm a \vert B} = 0.35$ instead of $ p_{\rm a \vert B} = 0.4$.

(9) We consider the ideal ternary channel: $p_{\rm a \vert A} = p_{\rm b \vert B}=p_{\rm c \vert C}=1$. What is its capacity $C_9$? What is the maximum mutual information displayed by the program?

Due to the symmetry of the channel model, equally probable symbols $(p_{\rm A} = p_{\rm B}=p_{\rm C}=1/3)$ lead to the channel capacity: $C_9 = \log_2\ (3) = 1.585$.
Since in the program all parameter values can only be entered with a resolution of $0.05$ , for $I(X;\ Y)$ this maximum value is not reached.
Possible approximations: $p_{\rm A} = p_{\rm B}= 0.3, \ p_{\rm C}=0.4$ ⇒ $I(X;\ Y)= 1. 571$ # $p_{\rm A} = p_{\rm B}= 0.35, \ p_{\rm C}=0.3$ ⇒ $I(X;\ Y)= 1.581$.

(10) Let the source symbols be (nearly) equally probable. Interpret the other settings and the results.

The falsification probabilities are $p_{\rm b \vert A} = p_{\rm c \vert B}=p_{\rm a \vert C}=1$ ⇒ no single sink symbol is equal to the source symbol.
This cyclic mapping has no effect on the channel capacity: $C_{10} = C_9 = 1.585$. The program returns ${\rm Max}\big[I(X;\ Y)\big]= 1.581$.

(11) We consider up to and including $(13)$ the same ternary source. What results are obtained for $p_{\rm b \vert A} = p_{\rm c \vert B}=p_{\rm a \vert C}=0.2$ and $p_{\rm c \vert A} = p_{\rm a \vert B}=p_{\rm b \vert C}=0$?

Each symbol can only be falsified into one of the two possible other symbols. From $p_{\rm b \vert A} = p_{\rm c \vert B}=p_{\rm a \vert C}=0.2$ it follows $p_{\rm a \vert A} = p_{\rm b \vert B}=p_{\rm c \vert C}=0.8$.
This gives us for the maximum mutual information ${\rm Max}\big[I(X;\ Y)\big]= 0.861$ and for the channel capacity a slightly larger value: $C_{11} \gnapprox 0.861$.

(12) How do the results change if each symbol is $80\%$ transferred correctly and $10\%$ falsified each in one of the other two symbols?

Although the probability of correct transmission is with $80\%$ as large as in (11), here the channel capacity $C_{12} \gnapprox 0.661$ is smaller.
If one knows for the channel $(11)$ that $X = \rm A$ has been falsified, one also knows $Y = \rm b$. But not for channel $(12)$ ⇒ the channel is less favorable.

(13) Let the falsification probabilities now be $p_{\rm b \vert A} = p_{\rm c \vert A} = p_{\rm a \vert B} = p_{\rm c \vert B}=p_{\rm a \vert C}=p_{\rm b \vert C}=0.5$. Interpret this redundancy-free ternary channel.

No single sink symbol is equal to its associated source symbol; with respect to the other two symbols, a $50\hspace{-0.1cm}:\hspace{-0.1cm}50$ decision must be made.
Nevertheless, here the channel capacity is $C_{13} \gnapprox 0.584$ only slightly smaller than in the previous experiment: $C_{12} \gnapprox 0.661$.
The channel capacity $C=0$ results for the redundancy-free ternary channel exactly for the case where all nine falsification probabilities are equal to $1/3$ .

(14) What is the capacity $C_{14}$ of the ternary channel with $p_{\rm b \vert A} = p_{\rm a \vert B}= 0$ and $p_{\rm c \vert A} = p_{\rm c \vert B} = p_{\rm a \vert C}=p_{\rm b \vert C}=0. 1$ ⇒ $p_{\rm a \vert A} = p_{\rm b \vert B}=0.9$, $p_{\rm c \vert C} =0.8$?

With the default $p_{\rm A}=p_{\rm B}=0.2$ ⇒ $p_{\rm C}=0.6$ we get $I(X;\ Y)= 0.738$. Now we are looking for "better" symbol probabilities.
From the symmetry of the channel, it is obvious that $p_{\rm A}=p_{\rm B}$ is optimal. The channel capacity $C_{14}=0.995$ is obtained for $p_{\rm A}=p_{\rm B}=0.4$ ⇒ $p_{\rm C}=0.2$.
Example: Ternary transfer if the middle symbol $\rm C$ can be distorted in two directions, but the outer symbols can only be distorted in one direction at a time.

Applet Manual

(A) Auswahlmöglichkeit, ob "analytisch" oder "per Simulation"

(B) Einstellung des Parameters $N$ für die Simulation

(C) Auswahlmöglichkeit, ob "Binärquelle" oder "Ternärquelle"

(D) Einstellung der Symbolwahrscheinlichkeiten

(E) Einstellung der Übergangswahrscheinlichkeiten

(F) Numerikausgabe verschiedener Wahrscheinlichkeiten

(G) Zwei Schaubilder mit den informationstheoretischen Größen

(H) Ausgabe einer beispielhaften Quellensymbolfolge

(I) Zugehörige simulierte Sinkensymbolfolge

(J) Bereich für Übungen: Aufgabenauswahl, Fragen, Musterlösungen

About the Authors

This interactive calculation tool was designed and implemented at the Institute for Communications Engineering at the Technical University of Munich.

The first version was created in 2010 by Martin Völkl as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: Günter Söder and Klaus Eichin).

In 2020 the program was redesigned via HTML5/JavaScript by Veronika Hofmann (Ingenieurspraxis Mathematik, Supervisor: Benedikt Leible and Tasnád Kernetzky ).

Last revision and English version 2021 by Carolin Mirschina in the context of a working student activity. Translation using DEEPL.com (free version).

The conversion of this applet was financially supported by "Studienzuschüsse" (TUM Department of Electrical and Computer Engineering). We thank.

Once again: Open Applet in new Tab