View source for Aufgaben:Exercise 4.8: Numerical Analysis of the AWGN Channel Capacity


{{quiz-Header|Buchseite=Information_Theory/AWGN_Channel_Capacity_for_Discrete_Input
}}

[[File:P_ID2936__Inf_A_4_8_Tab.png|right|frame|$C$&nbsp; as a function of&nbsp; $E_{\rm S}/{N_0}$]]
There are two different equations for the channel capacity&nbsp; $C$&nbsp; of the AWGN channel as an upper bound for the code rate&nbsp; $R$&nbsp; 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:
* $E_{\rm S}$&nbsp; denotes the (average) energy per symbol of the digital signal,
* $N_0$&nbsp; 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 &nbsp;$E_{\rm S} = R \cdot E_{\rm B}$&nbsp; has to be considered,&nbsp; where &nbsp;$R$&nbsp; is the code rate of the best possible channel coding.
*Error-free transmission&nbsp; (with optimal code)&nbsp; is possible for the given &nbsp;$E_{\rm B}/N_0$&nbsp; as long as &nbsp;$R \le C$&nbsp; &nbsp; &#8658; &nbsp;  [[Information_Theory/Anwendung_auf_die_Digitalsignalübertragung#Definition_and_meaning_of_channel_capacity|Shannon's channel coding theorem ]]&nbsp;holds. 


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







Hints:
*The exercise belongs to the chapter&nbsp; [[Information_Theory/AWGN_Channel_Capacity_for_Discrete_Input|AWGN channel capacity for discrete value input]].
*Reference is made in particular to the pages&nbsp; 
**[[Information_Theory/AWGN_Channel_Capacity_for_Discrete_Input#The_channel_capacity_.7F.27.22.60UNIQ-MathJax81-QINU.60.22.27.7F_as_a_function_of_.7F.27.22.60UNIQ-MathJax82-QINU.60.22.27.7F|The channel capacity $C$ as a function of $E_{\rm S}/{N_0}$]], 
**[[Information_Theory/AWGN–Kanalkapazität_bei_wertdiskretem_Eingang#The_channel_capacity_.7F.27.22.60UNIQ-MathJax128-QINU.60.22.27.7F_as_a_function_of_.7F.27.22.60UNIQ-MathJax129-QINU.60.22.27.7F|The channel capacity $C$ as a function of $E_{\rm B}/{N_0}$]]. 
*Since the results are to be expressed in&nbsp; "bit",&nbsp; "log" &nbsp; &#8658; &nbsp; "log<sub>2</sub>"&nbsp; is used in the equations.
 



===Questions===

<quiz display=simple>
{Which equations accurately describe the relationship between &nbsp;$E_{\rm B}/{N_0}$&nbsp; and the rate&nbsp; $R$&nbsp; for the AWGN channel?
|type="[]"}
+  &nbsp; $R = 1/2 \cdot \log_2  (1 + 2 \cdot  R \cdot E_{\rm B}/{N_0})$.
+  &nbsp; $2^{2R} = 1 + 2 \cdot  R \cdot E_{\rm B}/{N_0}$.
+  &nbsp; $E_{\rm B}/{N_0} = (2^{2R} -1)/(2R) $.


{Specify the smallest possible value for &nbsp;$E_{\rm B}/{N_0}$&nbsp; that can still be used to transmit over the AWGN channel without errors.
|type="{}"}
$\text{Min} \ \big[E_{\rm B}/{N_0}\big] \ = \ $ { 0.693 3% }

{Which result is obtained in&nbsp; $\rm dB$?
|type="{}"}
$\text{Min} \ \big[10 \cdot \lg (E_{\rm B}/{N_0})\big] \ = \ $ { -1.62--0.156 } $ \ \rm dB$


{Give the AWGN channel capacity&nbsp; $C$&nbsp; for &nbsp;$10 \cdot \lg (E_{\rm B}/{N_0})  = 0$&nbsp; dB.
|type="{}"}
$C \ = \ $ { 0.5 3% } $ \ \rm bit/channel\:use$

{Specify the required&nbsp; $E_{\rm B}/{N_0}$&nbsp; for error-free transmission with&nbsp; $R = 1$&nbsp;. &nbsp; <br>Note: The solution can be found in the table on the information page.
|type="{}"}
$\text{Min} \ \big[E_{\rm B}/{N_0}\big] \ = \ $ { 1.5 3% }


{How can a point of the &nbsp;$C(E_{\rm B}/{N_0})$&nbsp; curve be determined more easily?
|type="()"}
- Calculation of the channel capacity&nbsp; $C$&nbsp; for the given &nbsp;$E_{\rm B}/{N_0}$.
+ Calculation of the required &nbsp;$E_{\rm B}/{N_0}$&nbsp; for the given&nbsp; $C$.



</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp; <u>All proposed solutions</u> 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 &nbsp;$C = R$&nbsp; and &nbsp;$E_{\rm S} = R &middot; E_{\rm B}$&nbsp; 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&nbsp; $1/2$&nbsp; to the left side of the equation and forming the power to the base&nbsp; $2$,&nbsp; 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 &nbsp;$E_{\rm B}/{N_0}$,&nbsp; we get
:$$E_{\rm B}/{N_0} = \frac{2^{2R} -  1}   { 2 R}  \hspace{0.05cm}. $$



'''(2)'''&nbsp; Error-free transmission is possible over a channel with channel capacity&nbsp; $C$&nbsp; as long as the code rate is&nbsp;$R &#8804; C$.
*The absolute limit is obtained in the limiting case &nbsp;$C=R = 0$.

*Or more precisely:&nbsp;  For an arbitrarily small positive&nbsp; $&epsilon;$,&nbsp; the following must hold: &nbsp; $C=R =&epsilon;$&nbsp; with &nbsp;$&epsilon; &#8594; 0$.

*Using the result of subtask&nbsp; '''(1)''',&nbsp; 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 &nbsp;$ R &#8594; 0$&nbsp;  yields the result&nbsp; "0 divided by 0",&nbsp; the&nbsp; [https://https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule l'Hospital's rule]&nbsp; is to be applied here: &nbsp; <br>Differentiate numerator and denominator, form the quotient and finally put &nbsp;$R = 0$&nbsp;.&nbsp; 
*With &nbsp;$x = 2R$&nbsp; 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)'''&nbsp; In logarithmic form,&nbsp; 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)'''&nbsp; Thus,&nbsp; the abscissa value in non-logarithmic form is: &nbsp; $E_{\rm B}/{N_0} = 1$.&nbsp; It follows with &nbsp;$C=R$:
:$$\frac{2^{2C} -  1}   { 2 C} \stackrel{!}{=} 1 
\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\underline{C = 0.5}
\hspace{0.05cm}. $$



'''(5)'''&nbsp; For &nbsp;$R = 1$&nbsp; &rArr; &nbsp;$E_{\rm B} = E_{\rm S}$.&nbsp; 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 &nbsp;$10 \cdot \lg (E_{\rm B}/{N_0}) = 1.76 \ \rm dB$.

*The same result is obtained with &nbsp;$R = 1$&nbsp; 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)'''&nbsp;<u>Proposed solution 2 2</u>, is correct, as will be shown by an example:

'''(a)'''&nbsp; The channel capacity  &nbsp;$C$&nbsp; for &nbsp;$10 \cdot \lg (E_{\rm B}/{N_0}) = 15 \ \rm dB$ &nbsp; &#8658; &nbsp; &nbsp;$\E_{\rm B}/{N_0} = 31.62$ is sought.
[[File:EN_Inf_A_4_8_neu.png|right|frame|Capacity curves as a function of  &nbsp;$10 \cdot \lg (E_{\rm S}/{N_0})$&nbsp; and &nbsp;$10 \cdot \lg (E_{\rm B}/{N_0})$]] 
*Then, according to the proposed solution 1 with &nbsp;$x = 2C$:
:$$31.62 = \frac{2^{x} -  1}   { x} 
 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
31.62 \cdot x = 2^{x} -  1
\hspace{0.05cm}. $$
 
*The solution $x = 7.986$ &nbsp; &#8658; &nbsp; $C = 3.993 \ \rm (bit/use)$ can only be found graphically or iteratively.


'''(b)'''&nbsp; The necessary abscissa value &nbsp;$10 \cdot \lg (E_{\rm B}/{N_0})$&nbsp; for the capacity &nbsp;$C = 4 \ \rm bit/symbol$ 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&nbsp; 
* $10 \cdot \lg (E_{\rm S}/{N_0})$ &nbsp;&#8658;&nbsp; red curve and numbers; <br>these indicate the channel capacity &nbsp;$C$&nbsp; for the given &nbsp;$10 \cdot \lg (E_{\rm S}/{N_0})$&nbsp;;
* $10 \cdot \lg (E_{\rm B}/{N_0})$ &nbsp;&#8658;&nbsp; green curve and numbers; <br>these indicate the required &nbsp;$10 \cdot \lg (E_{\rm B}/{N_0})$&nbsp; for the given channel capacity &nbsp;$C$&nbsp;.


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


{{ML-Fuß}}



[[Category:Information Theory: Exercises|^4.3 AWGN and Value-Discrete Input^]]