View source for Aufgaben:Exercise 3.2: Eye Pattern according to Gaussian Low-Pass

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

[[File:P_ID1381__Dig_A_3_2.png|right|frame|Basic transmission pulse &nbsp;$g_s(t)$&nbsp; &rArr; &nbsp; blue curve,<br>basic detection pulse &nbsp;$g_d(t)$ &nbsp; &rArr; &nbsp; red curve ]]
Let a binary bipolar redundancy-free baseband system with bit rate &nbsp;$R_{\rm B} = 100\,{\rm Mbit/s}$&nbsp;  be given.&nbsp; Further:
* Rectangular transmission pulses,&nbsp; possible amplitude values &nbsp;$&plusmn; 1\,{\rm V}$.

* The AWGN noise power density &nbsp;$($on the resistor &nbsp;$1 \, \Omega)$:&nbsp;  $N_0=10^{\rm -9} \, {\rm V}^2/{\rm Hz}$.

* Receiver filter is a Gaussian low-pass with cutoff frequency &nbsp;$f_{\rm G} = 50 \, {\rm MHz}$.&nbsp; The frequency response is:
:$$H_{\rm G}(f) = {\rm e}^{- \pi  \hspace{0.05cm}\cdot \hspace{0.05cm}{f}^2/({2f_{\rm G}})^2}
  \hspace{0.05cm}.$$
* The basic detection pulse &nbsp;$g_d(t) = g_s(t) * h_{\rm G}(t)$&nbsp; is shown in the graph (red curve). <br>Some prominent pulse sample values are indicated.

* The detection noise power can be calculated using the following equation:
:$$\sigma_d^2 = {N_0}/{2} \cdot \int_{-\infty}^{+\infty}
|H_{\rm G}(f)|^2 \,{\rm d} f \hspace{0.05cm}.$$

The&nbsp; "eye diagram"&nbsp; can be used to determine the bit error probability&nbsp; $($in general:&nbsp; symbol error probability$)$.

# The mean symbol error probability &nbsp;$p_{\rm S}$&nbsp; is obtained from this after averaging over all possible noiseless detection samples in consideration of the noise rms value&nbsp; $\sigma_d$.
# The worst-case error probability serves as an upper bound for &nbsp;$p_{\rm S}$:&nbsp;
::$$p_{\rm U} = {\rm Q} \left( \frac{\ddot{o}(T_{\rm D})/2}{ \sigma_d}
  \right) \hspace{0.3cm}{\rm with}\hspace{0.3cm}\frac{\ddot{o}(T_{\rm D})}{ 2}= g_d(t=0) - |g_d(t=T)|- |g_d(t=-T)|-\hspace{0.01cm}\text{ ...}$$

::Here, &nbsp;$\ddot{o}(T_{\rm D})$&nbsp; denotes the vertical eye opening.&nbsp; Let the detection time &nbsp;$T_{\rm D} = 0$&nbsp; be optimally chosen.




Notes: 
*The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference|"Error Probability with Intersymbol Interference"]].
  
* Use our HTML5/JavaScript applet &nbsp;[[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|"Complementary Gaussian Error Functions"]] for the numerical evaluation of the Q&ndash;function.
 



===Questions===

<quiz display=simple>
{What is the symbol duration?
|type="{}"}
$T \ = \ $ { 10 3% } $\ {\rm ns}$

{What is the rms value of the detection noise signal?
|type="{}"}
$\sigma_d\ = \ $ { 0.188 3% } $\ {\rm V}$

{What are the basic detection pulse samples &nbsp;$g_{\rm \nu} = g_d(\nu \cdot T)$,&nbsp; in particular
|type="{}"}
$g_0\ = \ $ { 0.79 3% } $\ {\rm V}$
$g_1\ = \ $ { 0.105 3% } $\ {\rm V}$
$g_2\ = \ $ { 0 3% } $\ {\rm V}$

{Calculate the eye opening and the worst-case error probability.
|type="{}"}
$\ddot{o}(T_{\rm D})\ = \ $ { 1.16 3% } $\ {\rm V}$
$p_{\rm U}\ = \ $ { 1 3% } $\ \cdot 10^{\rm -3}$

{Calculate the average error probability &nbsp;$p_{\rm S}$&nbsp; by averaging over the possible noiseless samples.
|type="{}"}
$p_{\rm S}\ = \ $ { 0.256 3% } $\ \cdot 10^{\rm -3}$

{What would be the minimum increase in transmitted pulse amplitude &nbsp;$s_0$&nbsp; required to satisfy the condition &nbsp;$p_{\rm S} \ &#8804; 10^{\rm -10}$?&nbsp;
|type="{}"}
$s_0\ = \ ${ 1.993 3% } $\ {\rm V}$
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp; The symbol duration is the reciprocal of the bit rate:
:$$T = \frac{1}{10^8\,{\rm bit/s}} = 10^{-8}\,{\rm s}\hspace{0.15cm}\underline { = 10\,{\rm ns}}
\hspace{0.05cm}.$$

'''(2)'''&nbsp; Integration according to the given equation leads to:
:$$\sigma_d^2 = \frac{N_0}{2} \cdot \int_{-\infty}^{+\infty}
|H_{\rm G}(f)|^2 \,{\rm d} f = \frac{N_0 \cdot f_{\rm
G}}{\sqrt{2}}= \frac{10^{-9}\,{\rm V/Hz} \cdot 5 \cdot
10^{7}\,{\rm Hz} }{\sqrt{2}}\approx 0.035\,{\rm V^2}\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}\sigma_d \hspace{0.15cm}\underline { = 0.188\,{\rm
V}}\hspace{0.05cm}.$$

'''(3)'''&nbsp; These values can be obtained from the graph:
:$$g_0 = g_d(0)\hspace{0.15cm}\underline { = 0.790\,{\rm V}}, \hspace{0.2cm}g_1 = g_d(10\,{\rm
ns}) \hspace{0.15cm}\underline {= 0.105\,{\rm V}}= g_{-1}, \hspace{0.2cm}g_2  = g_{-2} \hspace{0.15cm}\underline { \approx
 0} \hspace{0.05cm}.$$

'''(4)'''&nbsp; Using the basic detection pulse values calculated in&nbsp; '''(3)''',&nbsp; we obtain for the vertical eye opening:
:$$\ddot{o}(T_{\rm D}) = 2 \cdot (g_0 - g_1 - g_{-1}) = 2 \cdot
(0.790\,{\rm V} - 2\cdot 0.105\,{\rm V})  \hspace{0.15cm}\underline {= 1.16\,{\rm
V}}\hspace{0.05cm}.$$
[[File:EN_Dig_A_3_2_neu.png|right|frame|Eye diagram with and without noise]]
*The right graph shows the noisless eye diagram.&nbsp; One can see from this the vertical eye opening in the center of the symbol:
:$$\ddot{o}(T_{\rm D} = 0) = 2 \cdot 0.58 \cdot s_0.$$

*Together with the noise rms value,&nbsp; we obtain for the worst-case error probability:
:$$p_{\rm U} = {\rm Q} \left( \frac{1.16\,{\rm
V}/2}{ 0.188\,{\rm V}}
  \right) \approx {\rm Q}(3.08)\hspace{0.15cm}\underline {\approx 10^{-3}}
  \hspace{0.05cm}.$$

'''(5)'''&nbsp; From the noisless eye diagram,&nbsp; one can see that the signal component at the detection time&nbsp; $T_{\rm D} = 0$&nbsp; can assume six different values. 

*In the upper half of the eye,&nbsp; these are:
:$$1.)\hspace{0.2cm} g_0 + g_1 + g_{-1} = 0.790\,{\rm V} + 2\cdot 0.105\,{\rm
 V}= 1\,{\rm V} = s_0$$
:$$\Rightarrow \hspace{0.3cm}
 p_{\rm 1} = {\rm Q} \left( \frac{1\,{\rm
V}}{ 0.188\,{\rm V}}
  \right) \approx 5 \cdot 10^{-8}
  \hspace{0.05cm},$$
:$$2.)\hspace{0.2cm} g_0 = 0.790\,{\rm V} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
 p_{\rm 2} = {\rm Q} \left( \frac{0.790\,{\rm
V}}{ 0.188\,{\rm V}}
  \right) \approx 1.3 \cdot 10^{-5}
  \hspace{0.05cm},$$
:$$3.)\hspace{0.2cm} g_0 - g_1 - g_{-1} = 0.580\,{\rm V} = \ddot{o}(T_{\rm
 D})/2\hspace{0.3cm}\Rightarrow \hspace{0.3cm}p_{\rm 3} = p_{\rm U} \approx  10^{-3}
  \hspace{0.05cm}.$$

*Averaging over these values with appropriate weighting &nbsp;$(p_2$&nbsp; occurs twice as often as&nbsp; $p_1$&nbsp; and&nbsp; $p_3)$&nbsp; gives:
:$$p_{\rm S} \ = \ {1}/{4} \cdot (p_{\rm 1} + 2 \cdot p_{\rm 2} + p_{\rm 3})
  = {1}/{4} \cdot (5 \cdot 10^{-8} + 2 \cdot 1.3 \cdot 10^{-5} + 10^{-3})
 \hspace{0.15cm}\underline {  \approx 0.256 \cdot 10^{-3}}
  \hspace{0.05cm}.$$
*Since&nbsp; $p_1$&nbsp; and&nbsp; $p_2$&nbsp; are much smaller than&nbsp; $p_3 = p_{\rm U}$,&nbsp; the average error probability is (almost) a factor of &nbsp;$4$&nbsp; smaller than &nbsp;$p_{\rm U}$.



'''(6)'''&nbsp; To reduce the error probability,&nbsp; $s_0$&nbsp; must be increased.&nbsp; Thus,&nbsp; the approximation &nbsp;$p_{\rm S} &asymp; p_{\rm U}/4$&nbsp; is even more accurate:
:$$p_{\rm S} \le 10^{-10}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}p_{\rm U} = {\rm Q} \left( \frac{0.58 \cdot s_0}{ 0.188\,{\rm V}}
  \right)\le 4 \cdot 10^{-10}\hspace{0.3cm}
\Rightarrow \hspace{0.3cm} \frac{0.58 \cdot s_0}{ 0.188\,{\rm V}}
  \ge 6.15 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}s_0 \ge 1.993\,{\rm V} \hspace{0.15cm}\underline { \approx 2\,{\rm V}}
  \hspace{0.05cm}.$$
{{ML-Fuß}}



[[Category:Digital Signal Transmission: Exercises|^3.2 BER with Intersymbol Interference^]]