Difference between revisions of "Aufgaben:Exercise 3.7: Optimal Nyquist Equalization once again"

Latest revision as of 11:38, 23 June 2022

Transversal filter
frequency response

We assume the following for this exercise:

binary bipolar NRZ rectangular pulses

$$|H_{\rm S}(f)|= {\rm sinc}(f T) \hspace{0.05cm},$$

coaxial cable with characteristic cable attenuation $a_* = 9.2 \ {\rm Np} \ (\approx 80 \ \rm dB)$:

$$|H_{\rm K}(f)|= {\rm e}^{ -9.2 \cdot \sqrt{2 \cdot |f| \cdot T} }\hspace{0.05cm},$$

optimal Nyquist equalizer consisting of matched filter and transversal filter:

$$H_{\rm E}(f) = H_{\rm MF}(f) \cdot H_{\rm TF}(f)$$

$$\hspace{0.8cm}{\rm where}\hspace{0.2cm}H_{\rm MF}(f) = H_{\rm S}^{\star}(f) \cdot H_{\rm K}^{\star}(f)\hspace{0.05cm},\hspace{0.2cm} H_{\rm TF}(f) = \frac{1}{\sum\limits_{\kappa = -\infty}^{+\infty} |H_{\rm SK}(f - {\kappa}/{T}) |^2}\hspace{0.05cm}.$$

Here, $H_{\rm SK}(f) = H_{\rm S}(f) \cdot H_{\rm K}(f)$ denotes the product of transmitter and channel frequency response.

Because of Nyquist equalization, the eye is maximally open. For the error probability holds:

$$p_{\rm S} \left ( = p_{\rm U} \right ) = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right ) \hspace{0.05cm}.$$

The normalized noise power at the decision is given by the following equations:

$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot \int_{-\infty}^{+\infty} |H_{\rm E}(f)|^2 \,{\rm d} f \hspace{0.5cm} = \hspace{0.5cm} \sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot \int_{-1/(2T)}^{+1/(2T)} H_{\rm TF}(f) \,{\rm d} f \hspace{0.5cm}= \hspace{0.5cm}T \cdot \int_{0}^{1/T} H_{\rm TF}(f) \,{\rm d} f \hspace{0.05cm}.$$

The validity of this equation follows from the periodicity of the transversal filter frequency response $H_{\rm TF}(f)$.

In the graph, the normalized noise power can be seen as the area highlighted in red.

As an approximation, the normalized noise power can be calculated by the triangular area shown in blue in the graph.

Notes:

The exercise belongs to the chapter "Linear Nyquist Equalization".

To determine the error probability you can use the interactive calculation module "Complementary Gaussian Error Functions".

Questions

$|H_{\rm SK} (f = 0)| \hspace{0.8cm} = \ $

$|H_{\rm SK} (f = f_{\rm Nyq})| \hspace{0.2cm} = \ $

$\ \cdot 10^{-5}$

$|H_{\rm SK} (f = 1/T)| \hspace{0.25cm} = \ $

$|H_{\rm TF} (f = f_{\rm Nyq})| \hspace{0.2cm} = \ $

$\ \cdot 10^8$

$\sigma_{d, \ \rm norm}^2 \hspace{0.2cm} = \ $

$\ \cdot 10^7$

$p_{\rm S} \hspace{0.2cm} = \ $

$\ \%$

Solution

(1) In general, for all frequencies $f \ge 0$:

$$|H_{\rm SK}(f)|= {\rm sinc}(f T) \cdot {\rm e}^{ -9.2 \cdot \sqrt{2 \cdot |f| \cdot T} }\hspace{0.05cm}.$$

This gives the special cases we are looking for:

$$f= 0 \text{:} \ \hspace{0.1cm}|H_{\rm SK}(f = 0)|= {\rm sinc}(0) \cdot {\rm e}^0 \hspace{0.15cm}\underline {= 1} \hspace{0.05cm},$$

$$ f= f_{\rm Nyq}\text{:} \ \hspace{0.1cm}|H_{\rm SK}(f = \frac{1}{2T})|= {\rm sinc}({1}/{2}) \cdot {\rm e}^{-9.2} \hspace{0.15cm}\underline { \approx 6.43 \cdot 10^{-5}} \hspace{0.05cm},$$

$$ f= {1}/{T} \text{:}\ \hspace{0.1cm}|H_{\rm SK}(f = \frac{1}{T})|= {\rm sinc}({1}) \cdot {\rm e}^{...} \hspace{0.15cm}\underline { = 0} \hspace{0.05cm}.$$

(2) The graph shows that $H_{\rm TF}(f)$ becomes maximal at $f = f_{\rm Nyq}$.

It follows with the given equation that

$${\sum\limits_{\kappa = -\infty}^{+\infty} |H_{\rm SK}(f - \frac{\kappa}{T}) |^2}$$

is minimal at the Nyquist frequency.

However, for $f = f_{\rm Nyq}$, of the infinite sum, only the terms with $\kappa = 0$ and $\kappa = 1$ contribute relevantly to the result.
From this follows further with the result from subtask (1):

$${\rm Max} \left [ H_{\rm TF}(f) \right ] \ = \ H_{\rm TF}(f = f_{\rm Nyq})= {1}/{2 \cdot |H_{\rm SK}(f = f_{\rm Nyq}) |^2} = \ \frac{1}{2 \cdot (6.43 \cdot 10^{-5})^2}= \frac{10^{10}}{82.69} \hspace{0.15cm}\underline {\approx 1.21 \cdot 10^{8}} \hspace{0.05cm}.$$

(3) Approximating the integral over $H_{\rm TF}(f)$ by the triangular area plotted in the graph, we obtain:

$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot \int_{0}^{1/T} H_{\rm TF}(f) \,{\rm d} f \approx T \cdot \frac{1}{2}\cdot 1.21 \cdot 10^{8}\cdot (0.64 -0.36)\hspace{0.15cm}\underline {\approx 1.7 \cdot 10^{7}} \hspace{0.05cm}.$$

(4) According to the given equation, we obtain for the (mean) symbol error probability:

$$p_{\rm S} = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right ) = {\rm Q}\left ( \sqrt{\frac{10^{8}}{1.7 \cdot 10^{7}}} \right ) \approx {\rm Q}(2.42)\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm S} \hspace{0.15cm}\underline {\approx 0.8 \%} \hspace{0.05cm}.$$

Since a binary Nyquist system is present, the worst–case probability $p_{\rm U}$ is just as large.

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Digitalsignalübertragung/Lineare Nyquistentzerrung}}
+{{quiz-Header|Buchseite=Digital_Signal_Transmission/Linear_Nyquist_Equalization}}
-[[File:P_ID1435__Dig_A_3_7.png|right|frame|Transversalfilterfrequenzgang]]
+[[File:P_ID1435__Dig_A_3_7.png|right|frame|Transversal filter <br>frequency response]]
-Wir gehen bei dieser Aufgabe von folgenden Voraussetzungen aus:
+We assume the following for this exercise:
-* binäre bipolare NRZ&ndash;Rechteckimpulse
+* binary bipolar NRZ rectangular pulses
-:$$|H_{\rm S}(f)|= {\rm si}(\pi f T) \hspace{0.05cm},$$
+:$$|H_{\rm S}(f)|= {\rm sinc}(f T) \hspace{0.05cm},$$
-* Koaxialkabel mit Kabeldämpfung $a_* = 9.2 \ {\rm Np} (\approx 80 \ \rm dB)$:
+* coaxial cable with characteristic cable attenuation &nbsp;$a_* = 9.2 \ {\rm Np} \ (\approx 80 \ \rm dB)$:
-:$$|H_{\rm K}(f)|= {\rm exp}\left [ -9.2
+:$$|H_{\rm K}(f)|= {\rm e}^{ -9.2
-\cdot \sqrt{2 \cdot |f| \cdot T}  \right ]\hspace{0.05cm},$$
+\cdot \sqrt{2 \cdot |f| \cdot T}  }\hspace{0.05cm},$$
-* optimaler Nyquistentzerrer, bestehend aus Matched&ndash;Filter und Transversalfilter:
+* optimal Nyquist equalizer consisting of matched filter and transversal filter:
-:$$H_{\rm E}(f) = H_{\rm MF}(f) \cdot H_{\rm TF}(f)\hspace{0.2cm}{\rm mit}$$
+:$$H_{\rm E}(f) = H_{\rm MF}(f) \cdot H_{\rm TF}(f)$$
-:$$H_{\rm MF}(f) = H_{\rm S}^{\star}(f) \cdot H_{\rm K}^{\star}(f)\hspace{0.05cm},\hspace{0.2cm}
+:$$\hspace{0.8cm}{\rm where}\hspace{0.2cm}H_{\rm MF}(f) = H_{\rm S}^{\star}(f) \cdot H_{\rm K}^{\star}(f)\hspace{0.05cm},\hspace{0.2cm}
    H_{\rm TF}(f) =
   \frac{1}{\sum\limits_{\kappa = -\infty}^{+\infty}  |H_{\rm SK}(f -
@@ Line 17: / Line 17: @@
   |^2}\hspace{0.05cm}.$$
-Hierbei bezeichnet $H_{\rm SK}(f) = H_{\rm S}(f) \cdot H_{\rm K}(f)$ das Produkt von Sender&ndash; und Kanalfrequenzgang.
+:Here, &nbsp;$H_{\rm SK}(f) = H_{\rm S}(f) \cdot H_{\rm K}(f)$&nbsp; denotes the product of transmitter and channel frequency response.
-Wegen der Nyquistentzerrung ist das Auge maximal geöffnet. Für die Fehlerwahrscheinlichkeit gilt:
+Because of Nyquist equalization,&nbsp; the eye is maximally open.&nbsp; For the error probability holds:
 :$$p_{\rm S}  \left ( = p_{\rm U} \right ) = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right )
   \hspace{0.05cm}.$$
-Die normierte Störleistung am Entscheider ist durch folgende Gleichungen gegeben:
+The normalized noise power at the decision is given by the following equations:
 :$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot
 \int_{-\infty}^{+\infty} |H_{\rm E}(f)|^2 \,{\rm d} f
-\hspace{0.05cm},$$
+\hspace{0.5cm} = \hspace{0.5cm}
-:$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot
+\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot
-\int_{-1/(2T)}^{+1/(2T)} H_{\rm TF}(f) \,{\rm d} f = T
+\int_{-1/(2T)}^{+1/(2T)} H_{\rm TF}(f) \,{\rm d} f \hspace{0.5cm}= \hspace{0.5cm}T
 \cdot \int_{0}^{1/T} H_{\rm TF}(f) \,{\rm d} f
 \hspace{0.05cm}.$$
-Die Gültigkeit dieser Gleichung ergibt sich aus der Periodizität des Transversalfilterfrequenzgangs. In der Grafik erkennt man die normierte Störleistung als die rot hinterlegte Fläche. Näherungsweise kann die normierte Störleistung durch die in der Grafik blau eingezeichnete Dreieckfläche berechnet werden.
+The validity of this equation follows from the periodicity of the transversal filter frequency response &nbsp;$H_{\rm TF}(f)$.
+*In the graph,&nbsp; the normalized noise power can be seen as the area highlighted in red.
+*As an approximation,&nbsp; the normalized noise power can be calculated by the triangular area shown in blue in the graph.
-''Hinweis:''
+Notes:
-* Die Aufgabe bezieht sich auf das Kapitel [[Digitalsignal%C3%BCbertragung/Lineare_Nyquistentzerrung|Lineare Nyquistentzerrung]].
+*The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Linear_Nyquist_Equalization|"Linear Nyquist Equalization"]].
-* Zur Bestimmung der Fehlerwahrscheinlichkeit können Sie das folgende interaktive Berechnungsmodul nutzen: [https://intern.lntwww.de/cgi-bin/extern/uni.pl?uno=hyperlink&due=block&b_id=1706&hyperlink_typ=block_verweis&hyperlink_fenstergroesse=blockverweis_gross|Komplementäre Gaußsche Fehlerfunktion]
+* To determine the error probability you can use the interactive calculation module&nbsp; [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|"Complementary Gaussian Error Functions"]].&nbsp;
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Multiple-Choice
+{Calculate the magnitude of the transmitter channel frequency response for the frequencies &nbsp;$f = 0$, &nbsp;$f  = 1/(2T)= f_{\rm Nyq}$&nbsp; and&nbsp; $f = 1/T =  2 \cdot f_{\rm Nyq}$.
-|type="[]"}
+|type="{}"}
-+ correct
+$|H_{\rm SK} (f = 0)| \hspace{0.8cm} = \ $ { 1 3% }
-- false
+$|H_{\rm SK} (f = f_{\rm Nyq})| \hspace{0.2cm} = \ $ { 6.43 3% } $\ \cdot 10^{-5}$
+$|H_{\rm SK} (f = 1/T)| \hspace{0.25cm} = \ $ { 0. }
+{Calculate the maximum value of &nbsp;$H_{\rm TF}(f)$&nbsp; at frequency &nbsp;$f = f_{\rm Nyq}$.
+|type="{}"}
+$|H_{\rm TF} (f = f_{\rm Nyq})| \hspace{0.2cm} = \ $ { 1.21 3% } $\ \cdot 10^8$
+{Calculate the normalized noise power according to the triangular approximation.
+|type="{}"}
+$\sigma_{d, \ \rm  norm}^2 \hspace{0.2cm} = \ $ { 1.7 3% } $\ \cdot 10^7$
-{Input-Box Frage
+{What is the symbol error probability with &nbsp;$s_0^2 \cdot T/N_0 = 10^8$?
 |type="{}"}
-$xyz$ = { 5.4 3% } $ab$
+$p_{\rm S} \hspace{0.2cm} = \ $ { 0.8 3% } $\ \%$
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;
+'''(1)'''&nbsp; In general,&nbsp; for all frequencies&nbsp; $f \ge  0$: &nbsp;
-'''(2)'''&nbsp;
+:$$|H_{\rm SK}(f)|= {\rm sinc}(f T) \cdot {\rm e}^{ -9.2
-'''(3)'''&nbsp;
+\cdot \sqrt{2 \cdot |f| \cdot T}  }\hspace{0.05cm}.$$
-'''(4)'''&nbsp;
+*This gives the special cases we are looking for:
-'''(5)'''&nbsp;
+:$$f= 0 \text{:} \ \hspace{0.1cm}|H_{\rm SK}(f = 0)|= {\rm sinc}(0) \cdot {\rm e}^0 \hspace{0.15cm}\underline {= 1}
+ \hspace{0.05cm},$$
+:$$ f= f_{\rm Nyq}\text{:} \ \hspace{0.1cm}|H_{\rm SK}(f = \frac{1}{2T})|= {\rm sinc}({1}/{2}) \cdot {\rm e}^{-9.2}
+\hspace{0.15cm}\underline { \approx 6.43 \cdot 10^{-5}}
+ \hspace{0.05cm},$$
+:$$ f= {1}/{T} \text{:}\ \hspace{0.1cm}|H_{\rm SK}(f = \frac{1}{T})|= {\rm sinc}({1}) \cdot {\rm e}^{...}
+\hspace{0.15cm}\underline { = 0} \hspace{0.05cm}.$$
+'''(2)'''&nbsp; The graph shows that&nbsp; $H_{\rm TF}(f)$&nbsp; becomes maximal at&nbsp; $f = f_{\rm Nyq}$.&nbsp;
+*It follows with the given equation that
+:$${\sum\limits_{\kappa = -\infty}^{+\infty}  |H_{\rm SK}(f -
+ \frac{\kappa}{T})
+ |^2}$$
+:is minimal at the Nyquist frequency.&nbsp;
+*However,&nbsp; for $f = f_{\rm Nyq}$,&nbsp; of the infinite sum, only the terms with&nbsp; $\kappa = 0$&nbsp; and&nbsp; $\kappa = 1$&nbsp; contribute relevantly to the result.
+*From this follows further with the result from subtask&nbsp; '''(1)''':
+:$${\rm Max} \left [ H_{\rm TF}(f) \right ] \ = \ H_{\rm TF}(f = f_{\rm
+ Nyq})=
+ {1}/{2 \cdot  |H_{\rm SK}(f = f_{\rm
+ Nyq}) |^2} = \ \frac{1}{2 \cdot  (6.43 \cdot 10^{-5})^2}=
+ \frac{10^{10}}{82.69} \hspace{0.15cm}\underline {\approx 1.21 \cdot 10^{8}}
+ \hspace{0.05cm}.$$
+'''(3)'''&nbsp; Approximating the integral over&nbsp; $H_{\rm TF}(f)$&nbsp; by the triangular area plotted in the graph,&nbsp; we obtain:
+:$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2  = T \cdot
+\int_{0}^{1/T} H_{\rm TF}(f) \,{\rm d} f \approx  T \cdot
+\frac{1}{2}\cdot 1.21 \cdot 10^{8}\cdot (0.64 -0.36)\hspace{0.15cm}\underline {\approx 1.7
+\cdot 10^{7}} \hspace{0.05cm}.$$
+'''(4)'''&nbsp; According to the given equation,&nbsp; we obtain for the (mean) symbol error probability:
+:$$p_{\rm S} = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right
+ ) = {\rm Q}\left ( \sqrt{\frac{10^{8}}{1.7
+\cdot 10^{7}}} \right ) \approx {\rm Q}(2.42)\hspace{0.3cm}
+\Rightarrow
+\hspace{0.3cm} p_{\rm S} \hspace{0.15cm}\underline {\approx 0.8 \%} \hspace{0.05cm}.$$
+:Since a binary Nyquist system is present,&nbsp; the worst&ndash;case probability&nbsp; $p_{\rm U}$&nbsp; is just as large.
 {{ML-Fuß}}
-[[Category:Aufgaben zu Digitalsignalübertragung|^3.5 Lineare Nyquistentzerrung^]]
+[[Category:Digital Signal Transmission: Exercises|^3.5 Linear Nyquist Equalization^]]