Aufgaben:Exercise 4.6: AWGN Channel Capacity

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Flowchart of the information

We start from the   AWGN channel model :

• $X$  denotes the input (transmitter).
• $N$  stands for a Gaussian distributed noise.
• $Y = X +N$  describes the output (receiver) in case of additive noise.

For the probability density function  $\rm (PDF)$  of the noise,  let hold:

$$f_N(n) = \frac{1}{\sqrt{2\pi \hspace{0.03cm}\sigma_{\hspace{-0.05cm}N}^2}} \cdot {\rm e}^{ - \hspace{0.05cm}{n^2}\hspace{-0.05cm}/{(2 \hspace{0.03cm} \sigma_{\hspace{-0.05cm}N}^2) }} \hspace{0.05cm}.$$

Since the random variable  $N$  is zero mean   ⇒   $m_{N} = 0$,  we can equate the variance $\sigma_{\hspace{-0.05cm}N}^2$  with the power  $P_N$ .  In this case, the differential entropy of the random variable  $N$  is specifiable  (with the pseudo–unit "bit")  as follows:

$$h(N) = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 2\pi {\rm e} \cdot P_N \right )\hspace{0.05cm}.$$

In this exercise,  $P_N = 1 \hspace{0.15cm} \rm mW$  is given.  It should be noted:

• The power  $P_N$  in the above equation, like the variance  $\sigma_{\hspace{-0.05cm}N}^2$ , must be dimensionless.

• To work with this equation, the physical quantity  $P_N$  must be suitably normalized, for example corresponding to  $P_N = 1 \hspace{0.15cm} \rm mW$    ⇒    $P_N\hspace{0.01cm}' = 1$.

• With other normalization, for example  $P_N = 1 \hspace{0.15cm} \rm mW$     ⇒     $P_N\hspace{0.01cm}' = 0.001$  a completely different numerical value would result fo  $h(N)$ .

Further,  you can consider for the solution of this exercise:

• The channel capacity is defined as the maximum mutual information between input  $X$  and output  $Y$  with the best possible input distribution:

$$C = \max_{\hspace{-0.15cm}f_X:\hspace{0.05cm} {\rm E}[X^2] \le P_X} \hspace{-0.2cm} I(X;Y) \hspace{0.05cm}.$$

• The channel capacity of the AWGN channel is:

$$C_{\rm AWGN} = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 1 + \frac{P_X}{P_N} \right ) = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 1 + \frac{P_{\hspace{-0.05cm}X}\hspace{0.01cm}'}{P_{\hspace{-0.05cm}N}\hspace{0.01cm}'} \right )\hspace{0.05cm}.$$

It can be seen:  The channel capacity  $C$  and also the mutual information  $I(X; Y)$  are independent of the above normalization, in contrast to the differential entropies.

• With Gaussian noise PDF  $f_N(n)$,  an Gaussian input PDF $f_X(x)$  leads to the maximum mutual information and thus to the channel capacity.

Hints:

• The exercise belongs to the chapter  AWGN channel capacity with continuous input.
• Since the results are to be given in  "bit",  "log"  ⇒  "log₂"  is used in the equations.

Questions

1

What transmission power is required for  $C = 2 \ \rm bit$?

$P_X \ = \ $ $\ \rm mW$

2

Under which conditions is  $I(X; Y) = 2 \ \rm bit$  achievable at all?

	$P_X$  is determined as in  (1)  or larger.
	The random variable  $X$  is Gaussian distributed.
	The random variable  $X$  is zero mean.
	The random variables  $X$  and  $N$  are uncorrelated.
	The random variables  $X$  and  $Y$  are uncorrelated.

3

Calculate the differential entropies of the random variables  $N$,  $X$  and  $Y$  with appropriate normalization,
for example,  $P_N = 1 \hspace{0.15cm} \rm mW$    ⇒    $P_N\hspace{0.01cm}' = 1$.

$h(N) \ = \ $ $\ \rm bit$

$h(X) \ = \ $ $\ \rm bit$

$h(Y) \ = \ $ $\ \rm bit$

4

What are the other information-theoretic descriptive quantities?

$h(Y|X) \ = \ $ $\ \rm bit$

$h(X|Y) \ = \ $ $\ \rm bit$

$h(XY) \ = \ $ $\ \rm bit$

5

What quantities would result for the same  $P_X$  in the limiting case   $P_N\hspace{0.01cm} ' \to 0$ ?

$h(X) \ = \ $ $\ \rm bit$

$h(Y|X) \ = \ $ $\ \rm bit$

$h(Y) \ = \ $ $\ \rm bit$

$I(X;Y) \ = \ $ $\ \rm bit$

$h(X|Y) \ = \ $ $\ \rm bit$

Solution

Solution

(1)  The equation for the AWGN channel capacity in  "bit"  is:

$$C_{\rm bit} = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 1 + {P_X}/{P_N} \right )\hspace{0.05cm}.$$

With  $C_{\rm bit} = 2$  this results in:

$$4 \stackrel{!}{=} {\rm log}_2\hspace{0.05cm}\left ( 1 + {P_X}/{P_N} \right ) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 1 + {P_X}/{P_N} \stackrel {!}{=} 2^4 = 16 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} P_X = 15 \cdot P_N \hspace{0.15cm}\underline{= 15\,{\rm mW}} \hspace{0.05cm}. $$

(2)  Correct are  proposed solutions 1 through 4.  Justification:

• For  $P_X < 15 \ \rm mW$  the mutual information  $I(X; Y)$  will always be less than  $2$  bit,  regardless of all other conditions.
• With  $P_X = 15 \ \rm mW$  the maximum mutual information  $I(X; Y) = 2$  bit is only achievable if the input quantity  $X$  is Gaussian distributed. 
  The output quantity  $Y$  is then also Gaussian distributed.
• If the random variable  $X$  has a constant proportion  $m_X$  then the variance  $\sigma_X^2 = P_X - m_X^2 $  for given  $P_X$  is smaller, and it holds  
  $I(X; Y) = 1/2 · \log_2 \ (1 + \sigma_X^2/P_N) < 2$  bit.
• The precondition for the given channel capacity equation is that  $X$  and  $N$  are uncorrelated.  On the other hand, if the random variables  $X$  and  $N$  were uncorrelated, then  $I(X; Y) = 0$ would result.

Informationstheoretische Größen für den AWGN-Kanal

(3)  Die angegebene Gleichung für die differentielle Entropie macht nur bei dimensionsloser Leistung Sinn.  Mit der vorgeschlagenen Normierung erhält man:

• Für  $P_N = 1 \ \rm mW$   ⇒   $P_N\hspace{0.05cm}' = 1$:

$$h(N) \ = \ {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 2\pi {\rm e} \cdot 1 \right ) = \ {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 17.08 \right ) \hspace{0.15cm}\underline{= 2.047\,{\rm bit}}\hspace{0.05cm},$$

• Für  $P_X = 15 \ \rm mW$   ⇒   $P_X\hspace{0.01cm}' = 15$:

$$h(X) \ = \ {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 2\pi {\rm e} \cdot 15 \right ) = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 2\pi {\rm e} \right ) + {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left (15 \right ) \hspace{0.15cm}\underline{= 4.000\,{\rm bit}}\hspace{0.05cm}, $$

• Für  $P_Y = P_X + P_N = 16 \ \rm mW$   ⇒  $P_Y\hspace{0.01cm}' = 16$:

$$h(Y) = 2.047\,{\rm bit} + 2.000\,{\rm bit} \hspace{0.15cm}\underline{= 4.047\,{\rm bit}}\hspace{0.05cm}.$$

(4)  Für die differentielle Irrelevanz gilt beim AWGN–Kanal:

$$h(Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X) = h(N) \hspace{0.15cm}\underline{= 2.047\,{\rm bit}}\hspace{0.05cm}.$$

• Entsprechend nebenstehender Grafik gilt aber auch:

$$h(Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X) = h(Y) - I(X;Y) = 4.047 \,{\rm bit} - 2 \,{\rm bit} \hspace{0.15cm}\underline{= 2.047\,{\rm bit}}\hspace{0.05cm}. $$

• Daraus kann die differentielle Äquivokation wie folgt berechnet werden:

$$h(X \hspace{-0.05cm}\mid \hspace{-0.05cm} Y) = h(X) - I(X;Y) = 4.000 \,{\rm bit} - 2 \,{\rm bit} \hspace{0.15cm}\underline{= 2.000\,{\rm bit}}\hspace{0.05cm}.$$

• Abschließend wird auch noch die differentielle Verbundentropie angegeben, die aus obigem Schaubild nicht direkt ablesbar ist:

$$h(XY) = h(X) + h(Y) - I(X;Y) = 4.000 \,{\rm bit} + 4.047 \,{\rm bit} - 2 \,{\rm bit} \hspace{0.15cm}\underline{= 6.047\,{\rm bit}}\hspace{0.05cm}.$$

Informationstheoretische Größen beim idealen Kanal

(5)  Beim idealen Kanal erhält man mit $h(X)\hspace{0.15cm}\underline{= 4.000 \,{\rm bit}}$:

$$h(Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X) \ = \ h(N) \hspace{0.15cm}\underline{= 0\,{\rm (bit)}}\hspace{0.05cm},$$

$$h(Y) \ = \ h(X) \hspace{0.15cm}\underline{= 4\,{\rm bit}}\hspace{0.05cm},$$

$$I(X;Y) \ = \ h(Y) - h(Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X)\hspace{0.15cm}\underline{= 4\,{\rm bit}}\hspace{0.05cm},$$ $$ h(X \hspace{-0.05cm}\mid \hspace{-0.05cm} Y) \ = \ h(X) - I(X;Y)\hspace{0.15cm}\underline{= 0\,{\rm (bit)}}\hspace{0.05cm}.$$

• Die Grafik zeigt diese Größen in einem Flussdiagramm.

• Das gleiche Diagramm ergäbe sich auch im wertdiskreten Fall mit  $M = 16$  gleichwahrscheinlichen Symbolen   ⇒   $H(X)= 4.000 \,{\rm bit}$.
• Man müsste nur jedes  $h$  durch ein  $H$  ersetzen.