Aufgaben:Exercise 4.8Z: What does the AWGN Channel Capacity Curve say?

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channel capacity as a function of  $10 \cdot \lg (E_{\rm B}/{N_0})$

We consider the channel capacity of the AWGN channel as in  Exercise 4.8 :

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

• The curve is shown on the right with logarithmic abscissa between  $-2 \ \rm dB$ und $+6 \ \rm dB$  dargestellt.
• The addition of "Gaussian" indicates that a Gaussian distribution was assumed for this curve at the AWGN input.

Three system variants are indicated by dots in the above graph:

• System $X$:    with  $10 \cdot \lg (E_{\rm B}/{N_0}) = 4 \ \rm dB$  and  $R = 1$,
• System $Y$:    with  $10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$  and  $R = 2$,
• System $Z$:    with  $10 \cdot \lg (E_{\rm B}/{N_0}) = 6 \ \rm dB$  and  $R = 1.5$.

In the questions for this exercise, we still use the following terms:

• Digital system:   symbol range  $M_X = |X|$  beliebig,
• Binary system:   symbol range  $M_X = 2$,
• Quaternary system:   symbol range  $M_X = 4$.

Hints:

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

Questions

1

What statement does the point  $X$  provide for digital signal transmission?

	For  $10 \cdot \lg (E_{\rm B}/{N_0}) = 4 \ \rm dB$  a digital system with rate  $R = 1$  and error probability zero can be imagined.
	Such a system does not require channel coding.
	Such a system uses an infinitely long code.
	A binary system can also meet the requirements.

2

What statement does the point  $Y$  provide for digital signal transmission?

	For  $10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$  a digital system with rate  $R = 2$ and error probability zero can be imagined.
	For  $10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$   $R = 0.5$  would be sufficient.
	For rate  $R = 2$  würde  $10 \cdot \lg (E_{\rm B}/{N_0}) = 5 \ \rm dB$  would be sufficient.

3

What statement does point  $Z$  provide for binary transmission?

	A binary system does not meet the requirements in any case.
	The curve  $C_\text{Gauß}(E_{\rm B}/{N_0})$  is not sufficient for this evaluation.

4

Which statement does the point  $Z$  provide for the quaternary transmission?

	A quaternary system does not meet the requirements in any case.
	The curve  $C_\text{Gauß}(E_{\rm B}/{N_0})$  is not sufficient for this evaluation.

Solution

Solution

(1)  Proposed solutions 1 and 3 are correct:

• Since the point  $X$  lies to the right of the channel capacity curve  $C_\text{Gauß}(E_{\rm B}/{N_0})$ , there is (at least) one message system of rate  $R = 1$that provides quasi–error–free transmission with  $10 \cdot \lg (E_{\rm B}/{N_0}) = 4 \ \rm dB$ .
• Despite the code rate  $R = 1$ , this system includes a channel coding with an infinitely long code, but unfortunately this code is unknown.
• However, a binary system of rate  $R = 1$  does not allow channel coding.

(2)  Only the proposed solution 2 is correct. Here the following statements are valid:

• The required  $E_{\rm B}/{N_0}$  for the rate  $R = 2$  results in

$$(E_{\rm B}/{N_0})_{\rm min} = \frac{2^{2R} - 1} { 2 \cdot R} = \frac{2^4 - 1} { 4 } = 3.75 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10\cdot {\rm lg} \hspace{0.1cm}(E_{\rm B}/{N_0})_{\rm min} = 15.74\,{\rm dB} \hspace{0.05cm}. $$

• The maximum code rate  $R_{\rm max}$  for  $10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$    ⇒   $E_{\rm B}/{N_0} = 1$  is calculated as follows:

$$C = R = \frac{1}{2} \cdot {\rm log}_2 \hspace{0.1cm} ( 1 + \frac { 2 \cdot R \cdot E_{\rm B}}{N_0}) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 2^{2R} - 1 \stackrel{!}{=} 2 R \hspace{0.3cm}\Rightarrow \hspace{0.3cm} R_{\rm max} = 0.5 \hspace{0.05cm}. $$

• Both calculations show that the point  $Y$ with characteristics  $10 \cdot \lg (E_{\rm B}/{N_0}) = 0 \ \rm dB$  and  $R = 1$  does not satisfy the channel coding theorem.

(3)  With a binary system, the rate  $R = 1.5$  can never be realized  ⇒   proposed solution 1.

(4)  Correct is the proposed solution 2:

• The point  $Z$  ies to the right of the boundary curve and for the code rate of a quaternary system  $R \le 2$ holds. 
• So the rate  $R =1.5$  would be quite realizable with  $M_X = 4$ .
• The proposed solution 1 is wrong. On the other hand, the second solution suggestion is correct:

• The given curve  $C_\text{Gauß}(E_{\rm B}/{N_0})$  always assumes a Gaussian distributed input.
• For a binary system, a different boundary curve results, namely  $C_\text{BPSK} ≤ 1 \ \rm bit/channel use$. $C_\text{Gauß}$  and  $C_\text{BPSK}$  are significantly different.
• For the quaternary system  $(M_X = 4)$  one would have to calculate and analyze the curve  $C_{M=4}$ .  Again,  $C_{M=4} ≤ C_\text{Gauß}$ .
• For small  $E_{\rm B}/{N_0}$ ,  $C_{M=4} \approx C_\text{Gauß}$, holds, after which the curve diverges significantly and ends in a horizontal at  $C_{M=4} = 2 \ \rm bit/channel use$.

The point  $Z$    ⇒   $10 \cdot \lg (E_{\rm B}/{N_0}) = 6 \ \rm dB, \ \ R = 1.5$  lies below  $C_{M=4}$.

• Such a quaternary system would thus be feasible, as will be shown in  Exercise 4.10 .
• But only from knowledge of  $C_\text{Gauß}$  the question cannot be answered (proposed solution 2).