Aufgaben:Exercise 4.8: Numerical Analysis of the AWGN Channel Capacity

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$C$  as a function of  $E_{\rm S}/{N_0}$

There are two different equations for the channel capacity  $C$  of the AWGN channel as an upper bound for the code rate  $R$  in digital signal transmission:

$\text{Channel capacity as a function of energy per symbol}$:

$$C( E_{\rm S}/{N_0}) = {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 1 + \frac { 2 \cdot E_{\rm S}}{N_0}) .$$

Here, the following abbreviations are used:

• $E_{\rm S}$  denotes the (average) energy per symbol of the digital signal,
• $N_0$  indicates the AWGN noise power density.

$\text{Channel capacity as a function of energy per bit}$:

$$C( E_{\rm B}/{N_0}) = {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 1 + \frac { 2 \cdot R \cdot E_{\rm B}}{N_0}) .$$

• The relation  $E_{\rm S} = R \cdot E_{\rm B}$  has to be considered,  where  $R$  is the code rate of the best possible channel coding.
• Error-free transmission  (with optimal code)  is possible for the given  $E_{\rm B}/N_0$  as long as  $R \le C$    ⇒   Shannon's channel coding theorem  holds.

Given by the table is the curve of channel capacity as a function of  $E_{\rm S}/N_0$.  The focus of this exercise is the numerical evaluation of the second equation.

Hints:

• The exercise belongs to the chapter  AWGN channel capacity for discrete value input.
• Reference is made in particular to the pages 
  • The channel capacity $C$ as a function of $E_{\rm S}/{N_0}$,
  • The channel capacity $C$ as a function of $E_{\rm B}/{N_0}$.
• Since the results are to be expressed in  "bit",  "log"   ⇒   "log₂"  is used in the equations.

Questions

1

Which equations accurately describe the relationship between  $E_{\rm B}/{N_0}$  and the rate  $R$  for the AWGN channel?

	$R = 1/2 \cdot \log_2 (1 + 2 \cdot R \cdot E_{\rm B}/{N_0})$.
	$2^{2R} = 1 + 2 \cdot R \cdot E_{\rm B}/{N_0}$.
	$E_{\rm B}/{N_0} = (2^{2R} -1)/(2R) $.

2

Specify the smallest possible value for  $E_{\rm B}/{N_0}$  that can still be used to transmit over the AWGN channel without errors.

$\text{Min} \ \big[E_{\rm B}/{N_0}\big] \ = \ $

3

Which result is obtained in  $\rm dB$?

$\text{Min} \ \big[10 \cdot \lg (E_{\rm B}/{N_0})\big] \ = \ $ $ \ \rm dB$

4

Give the AWGN channel capacity  $C$  for  $10 \cdot \lg (E_{\rm B}/{N_0}) = 0$  dB.

$C \ = \ $ $ \ \rm bit/channel\:use$

5

Specify the required  $E_{\rm B}/{N_0}$  for error-free transmission with  $R = 1$ .  
Note: The solution can be found in the table on the information page.

$\text{Min} \ \big[E_{\rm B}/{N_0}\big] \ = \ $

6

How can a point of the  $C(E_{\rm B}/{N_0})$  curve be determined more easily?

	Calculation of the channel capacity  $C$  for the given  $E_{\rm B}/{N_0}$.
	Calculation of the required  $E_{\rm B}/{N_0}$  for the given  $C$.

Solution

Solution

(1)  All proposed solutions are correct:

• Based on the equation

$$C = {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 1 + { 2 \cdot E_{\rm S}}/{N_0}) $$

we obtain with  $C = R$  and  $E_{\rm S} = R · E_{\rm B}$  the equation according to solution suggestion 1:

$$R = {1}/{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 1 + { 2 \cdot R \cdot E_{\rm B}}/{N_0})\hspace{0.05cm}. $$

• Bringing the factor  $1/2$  to the left side of the equation and forming the power to the base  $2$,  we get the suggestion 2:

$$2^{2R} = 1 + 2 \cdot R \cdot E_{\rm B}/{N_0}\hspace{0.05cm}. $$

• If we solve this equation for  $E_{\rm B}/{N_0}$,  we get

$$E_{\rm B}/{N_0} = \frac{2^{2R} - 1} { 2 R} \hspace{0.05cm}. $$

(2)  Error-free transmission is possible over a channel with channel capacity  $C$  as long as the code rate is $R ≤ C$.

• The absolute limit is obtained in the limiting case  $C=R = 0$.

• Or more precisely:  For an arbitrarily small positive  $ε$,  the following must hold:   $C=R =ε$  with  $ε → 0$.

• Using the result of subtask  (1),  the governing equation is:

$${\rm Min}\hspace{0.1cm}\big[E_{\rm B}/{N_0}\big] = \lim\limits_{R \hspace{0.05cm}\rightarrow \hspace{0.05cm}0}\frac{2^{2R} - 1} { 2 R} \hspace{0.05cm}. $$

• Since here the quotient in the boundary transition  $ R → 0$  yields the result  "0 divided by 0",  the  l'Hospital's rule  is to be applied here:  
  Differentiate numerator and denominator, form the quotient and finally put  $R = 0$ . 
• With  $x = 2R$  the result is:

$${\rm Min}\hspace{0.1cm}\big[E_{\rm B}/{N_0}\big] = \lim\limits_{x \hspace{0.05cm}\rightarrow \hspace{0.05cm}0}\frac{2^{x} - 1} { x} = \frac{{\rm ln}\hspace{0.1cm} (2) \cdot 2^{x} } { 1} \hspace{0.05cm}\bigg |_{x=0} = {\rm ln}\hspace{0.1cm} (2) \hspace{0.15cm}\underline{= 0.693} \hspace{0.05cm}.$$

(3)  In logarithmic form,  we obtain::

$${\rm Min}\hspace{0.1cm}\big[10\cdot {\rm lg} \hspace{0.1cm}(E_{\rm B}/{N_0})\big] = 10\cdot {\rm lg} \hspace{0.1cm}(0.693) \hspace{0.15cm}\underline{= -1.59\,{\rm dB}} \hspace{0.05cm}. $$

(4)  Thus,  the abscissa value in non-logarithmic form is:   $E_{\rm B}/{N_0} = 1$.  It follows with  $C=R$:

$$\frac{2^{2C} - 1} { 2 C} \stackrel{!}{=} 1 \hspace{0.3cm}\Rightarrow\hspace{0.3cm}\underline{C = 0.5} \hspace{0.05cm}. $$

(5)  For  $R = 1$  ⇒  $E_{\rm B} = E_{\rm S}$.  Therefore:

$$ C(E_{\rm B}/{N_0}) = 1 \hspace{0.3cm}\Longleftrightarrow \hspace{0.3cm} C(E_{\rm S}/{N_0}) = 1 \hspace{0.05cm}.$$

• From the table on the information page we can read:

$$ C(E_{\rm S}/{N_0}) = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} E_{\rm S}/{N_0} = 1.5 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \underline{E_{\rm B}/{N_0} = 1.5}\hspace{0.05cm}.$$

• The corresponding dB value is  $10 \cdot \lg (E_{\rm B}/{N_0}) = 1.76 \ \rm dB$.

• The same result is obtained with  $R = 1$  via the equation

$$E_{\rm B}/{N_0} = \frac{2^{2R} - 1} { 2 \cdot R} = \frac{4 - 1} { 2 } = 1.5 \hspace{0.05cm}.$$

(6) Proposed solution 2 2, is correct, as will be shown by an example:

(a)  The channel capacity  $C$  for  $10 \cdot \lg (E_{\rm B}/{N_0}) = 15 \ \rm dB$   ⇒    $\E_{\rm B}/{N_0} = 31.62$ is sought.

Capacity curves as a function of  $10 \cdot \lg (E_{\rm S}/{N_0})$  and  $10 \cdot \lg (E_{\rm B}/{N_0})$

• Then, according to the proposed solution 1 with  $x = 2C$:

$$31.62 = \frac{2^{x} - 1} { x} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 31.62 \cdot x = 2^{x} - 1 \hspace{0.05cm}. $$

• The solution $x = 7.986$   ⇒   $C = 3.993 \ \rm (bit/use)$ can only be found graphically or iteratively.

(b)  The necessary abscissa value  $10 \cdot \lg (E_{\rm B}/{N_0})$  for the capacity  $C = 4 \ \rm bit/symbol$ is sought:

$$E_{\rm B}/{N_0} = \frac{2^{2C} - 1} { 2 \cdot C} = \frac{2^8 - 1} { 8 } = 31.875$$

$$\Rightarrow \hspace{0.3cm} 10\cdot {\rm lg} \hspace{0.1cm}(E_{\rm B}/{N_0}) = 15.03\,{\rm dB} \hspace{0.05cm}.$$

The graph shows the AWGN channel capacity as a function of 

• $10 \cdot \lg (E_{\rm S}/{N_0})$  ⇒  red curve and numbers;
  these indicate the channel capacity  $C$  for the given  $10 \cdot \lg (E_{\rm S}/{N_0})$ ;
• $10 \cdot \lg (E_{\rm B}/{N_0})$  ⇒  green curve and numbers;
  these indicate the required  $10 \cdot \lg (E_{\rm B}/{N_0})$  for the given channel capacity  $C$ .

The intersection of the two curves is at  $1.76\ \rm dB$.