View source for Aufgaben:Exercise 3.3Z: Optimization of a Coaxial Cable System

 
{{quiz-Header|Buchseite=Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization
}}

[[File:P_ID1409__Dig_Z_3_3.png|right|frame|Normalized system sizes for different cut-off frequencies]]
We consider a redundancy-free binary transmission system with the following specifications:
* The transmission pulses are NRZ rectangular and have energy &nbsp;$E_{\rm B} = s_0^2 \cdot T$. 
* The channel is a coaxial cable with characteristic cable attenuation &nbsp;$a_* = 40 \, {\rm dB}$.
* AWGN noise with noise power density &nbsp;$N_0 = 0.0001 \cdot E_{\rm B}$&nbsp; is present.
* The receiver frequency response &nbsp;$H_{\rm E}(f)$&nbsp; includes an ideal channel equalizer &nbsp;$H_{\rm K}^{\rm -1}(f)$&nbsp; and a Gaussian low-pass filter &nbsp;$H_{\rm G}(f)$&nbsp; with cutoff frequency &nbsp;$f_{\rm G}$&nbsp; for noise power limitation.


The table shows the eye opening &nbsp;$\ddot{o}(T_{\rm D})$&nbsp; as well as the detection noise rms value &nbsp;$\sigma_{\rm d}$&nbsp; &ndash; each normalized to the transmitted amplitude &nbsp;$s_0$&nbsp; &ndash; for different cutoff frequencies &nbsp;$f_{\rm G}$. The cutoff frequency is to be chosen such that the worst-case error probability is as small as possible, with the following definition:
:$$p_{\rm U} = {\rm Q} \left( \frac{\ddot{o}(T_{\rm D})/2}{ \sigma_d}
  \right) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q} \left( \sqrt{\rho_{\rm U}}\right)$$

*This quantity represents an upper bound for the mean error probability &nbsp;$p_{\rm S}$:&nbsp; &nbsp;  $p_{\rm S} \le p_{\rm U}$. 
*For &nbsp;$f_{\rm G} \cdot T &#8805; 0.4$,&nbsp; a lower bound can also be given: &nbsp;  $p_{\rm S} \ge p_{\rm U}/4$.





''Notes:'' 
*The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization|"Consideration of Channel Distortion and Equalization"]].
* Use the interaction module &nbsp;[[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|"Complementary Gaussian Error Functions"]] for numerical evaluation of the Q-function..
 



===Questions===
<quiz display=simple>
{Within the given grid, determine the optimal cutoff frequency with respect to the "worst-case error probability" criterion.
|type="{}"}
$f_\text{G, opt} \cdot T \  = \ $  { 0.4 3% }

{What values does this give for the worst-case signal-to-noise ratio and the worst-case error probability?
|type="{}"}
$f_\text{G} = \text{G, opt:}\hspace{0.4cm} 10 \cdot {\rm lg} \, \rho_{\rm U}  \  = \ $ { 5.41 3% } ${\ \rm dB}$
$\hspace{4.07cm}p_{\rm U}  \  = \ $ { 3.1 3% } $\ \rm \%$

{To what value would we need to reduce the noise power density &nbsp;$N_0$&nbsp; (with respect to signal energy) so that &nbsp;$p_{\rm U}$&nbsp; is no greater than &nbsp;$10^{\rm -6}$?
|type="{}"}
$N_0/E_{\rm B} \  = \ $ { 1.53 3% } $\ \cdot 10^{\rm -5}$

{For the assumptions made in '''(3)''', give a lower bound and an upper bound for the average error probability &nbsp;$p_{\rm S}$.&nbsp; 
|type="{}"}
$p_\text{ S, min}\hspace{0.02cm} \  = \ $ { 0.25 3% } $\ \cdot 10^{\rm -6}$
$p_\text{ S, max} \  = \ $ { 1 3% } $\ \cdot 10^{\rm -6}$
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp; For the optimization it is sufficient to maximize the quotient $\ddot{o}(T_{\rm D})/\sigma_d$: 
*This is maximized from the values given in the table for the cutoff frequency &nbsp;$f_{\rm G, opt} \cdot T \underline {= 0.4}$&nbsp; with $0.735/0.197 \approx 3.73$. 
*As a comparison: &nbsp; For &nbsp;$f_{\rm G} \cdot T = 0.3$&nbsp; the result is $0.192/0.094 \approx 2.04$ due to the smaller eye opening and for &nbsp;$f_{\rm G} \cdot T = 0.5$&nbsp; the quotient is also smaller than for the optimum: $1.159/0.379 \approx 3.05$.
*An even larger cutoff frequency leads to a very large noise rms value without simultaneously increasing the vertical eye opening in the same way.



'''(2)'''&nbsp; Using the result from '''(1)''', we further obtain:
:$$\rho_{\rm U} = \left ( {3.73}/{2} \right )^2 \approx 3.48 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
 10 \cdot {\rm
lg}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline { = 5.41\,{\rm dB}}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q}\left (
{3.73}/{2} \right) \hspace{0.15cm}\underline {\approx 0.031} \hspace{0.05cm}.$$


'''(3)'''&nbsp; 
*With the given $10 \cdot {\rm lg} \, E_{\rm B}/N_0 = 40 \ \rm dB$, i.e. $E_{\rm B}/N_0 = 10^4$, the worst-case signal-to-noise ratio has been found to be $10 \cdot {\rm lg} \, \rho_{\rm U} \approx 5.41 \, {\rm dB}$. 
*However, for the worst-case error probability $p_{\rm U} = 10^{\rm -6}$, $10 \cdot {\rm lg} \, \rho_{\rm U} > 13.55 \, {\rm dB}$ must be obtained.
*This is achieved by increasing the quotient $E_{\rm B}/N_0$ accordingly:
:$$10 \cdot {\rm
lg}\hspace{0.1cm}{E_{\rm B}}/{N_0} = 40\,{\rm dB}
\hspace{0.1cm}+\hspace{0.1cm}13.55\,{\rm dB}
\hspace{0.1cm}-\hspace{0.1cm}5.41\,{\rm dB}= 48.14\,{\rm dB}\hspace{0.3cm} 
\Rightarrow
\hspace{0.3cm} {E_{\rm B}}/{N_0} = 10^{4.814}\approx 65163
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}  {N_0}/{E_{\rm B}}\hspace{0.15cm}\underline {  =
1.53 \cdot 10^{-5}} \hspace{0.05cm}.$$


'''(4)'''&nbsp; 
*The upper bound for $p_{\rm S}$ is equal to the worst-case error probability $p_{\rm U} = \underline {10^{\rm -6}}$. 
*The lower bound is $\underline {0.25 \cdot 10^{\rm -6}}$, which is smaller by a factor of 4.
{{ML-Fuß}}


[[Category:Digital Signal Transmission: Exercises|^3.3 Channel Distortion and Equalization^]]