Difference between revisions of "Aufgaben:Exercise 3.6Z:Optimum Nyquist Equalizer for Exponential Pulse"

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{{quiz-Header|Buchseite=Digitalsignalübertragung/Lineare Nyquistentzerrung}}
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{{quiz-Header|Buchseite=Digital_Signal_Transmission/Linear_Nyquist_Equalization}}
  
[[File:P_ID1434__Dig_Z_3_6.png|right|frame|Beidseitiger Exponentialimpuls]]
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[[File:P_ID1434__Dig_Z_3_6.png|right|frame|Two-sided exponential input pulse]]
Wie in der  [[Aufgaben:Aufgabe_3.6:_Transversalfilter_des_Optimalen_Nyquistentzerrers|Aufgabe 3.6]]  betrachten wir wieder den optimalen Nyquistentzerrer, wobei nun als Eingangsimpuls  $g_x(t)$  eine beidseitig abfallende Exponentialfunktion anliegt:
+
As in  [[Aufgaben:Exercise_3.6:_Transversal_Filter_of_the_Optimal_Nyquist_Equalizer|Exercise 3.6]]  we consider again the optimal Nyquist equalizer.
 +
*The input pulse  $g_x(t)$  is a two-sided exponential function:
 
:$$g_x(t) = {\rm e }^{ -  |t|/T}\hspace{0.05cm}.$$
 
:$$g_x(t) = {\rm e }^{ -  |t|/T}\hspace{0.05cm}.$$
 
+
*Through a transversal filter of  $N$–th order with the impulse response
*Durch ein Transversalfilter  $N$–ter Ordnung mit der Impulsantwort
 
 
:$$h_{\rm TF}(t) = \sum_{\lambda = -N}^{+N} k_\lambda \cdot \delta(t - \lambda \cdot T)$$
 
:$$h_{\rm TF}(t) = \sum_{\lambda = -N}^{+N} k_\lambda \cdot \delta(t - \lambda \cdot T)$$
  
:ist es immer möglich, dass der Ausgangsimpuls  $g_y(t)$  Nulldurchgänge bei  $t/T = ±1, \ \text{...} \ , \ t/T = ±N$  aufweist und  $g_y(t = 0) = 1$  ist.
+
:it is possible that the output pulse &nbsp;$g_y(t)$&nbsp; has zero crossings at&nbsp; $t/T = &plusmn;1, \ \text{...} \ , \ t/T = &plusmn;N$,&nbsp; <br>while &nbsp;$g_y(t = 0) = 1$.&nbsp;
*Im allgemeinen Fall führen dann allerdings die Vorläufer und Nachläufer mit &nbsp;$| \nu | > N$&nbsp; zu Impulsinterferenzen.
+
 
+
*However,&nbsp; in the general case,&nbsp; the precursors and trailers with &nbsp;$| \nu | > N$&nbsp; lead to intersymbol interference.
 
 
  
  
  
  
''Hinweis:''
+
Note:&nbsp; The exercise belongs to the chapter &nbsp;[[Digital_Signal_Transmission/Linear_Nyquist_Equalization|"Linear Nyquist Equalization"]].
*Die Aufgabe gehört zum  Kapitel &nbsp;[[Digitalsignal%C3%BCbertragung/Lineare_Nyquistentzerrung|Linare Nyquistentzerrung]].
 
 
   
 
   
  
  
  
===Fragebogen===
+
===Questions===
 
<quiz display=simple>
 
<quiz display=simple>
{Geben Sie die Signalwerte &nbsp;$g_x(\nu) = g_x(t = \nu T)$&nbsp; bei Vielfachen von &nbsp;$T$&nbsp; an.
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{Give the signal values &nbsp;$g_x(\nu) = g_x(t = \nu T)$&nbsp; at multiples of &nbsp;$T$.&nbsp;  
 
|type="{}"}
 
|type="{}"}
 
$g_x(0)\ = \ $ { 1 3% }
 
$g_x(0)\ = \ $ { 1 3% }
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$g_x(2)\ = \ $ { 0.135 3% }
 
$g_x(2)\ = \ $ { 0.135 3% }
  
{Berechnen Sie die optimalen Filterkoeffizienten für &nbsp;$N = 1$.
+
{Calculate the optimal filter coefficients for &nbsp;$N = 1$.
 
|type="{}"}
 
|type="{}"}
 
$k_0 \ = \ $ { 1.313 3% }
 
$k_0 \ = \ $ { 1.313 3% }
 
$k_1 \ = \ $ { -0.43775--0.41225 }
 
$k_1 \ = \ $ { -0.43775--0.41225 }
  
{Berechnen Sie die Ausgangswerte &nbsp;$g_2 = g_{\rm \nu}(t = 2T)$&nbsp; und&nbsp; $g_3 = g_{\rm \nu}(t = 3T)$.
+
{Calculate the output values &nbsp;$g_2 = g_{y}(t = 2T)$&nbsp; and&nbsp; $g_3 = g_{y}(t = 3T)$.
 
|type="{}"}
 
|type="{}"}
 
$g_2 \ = \ $ { 0. }
 
$g_2 \ = \ $ { 0. }
 
$g_3\ = \ $ { 0. }
 
$g_3\ = \ $ { 0. }
  
{Welche der folgenden Aussagen sind zutreffend?
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{Which of the following statements is true?
|type="[]"}
+
|type="()"}
+ Beim gegebenen Eingangsimpuls &nbsp;$g_x(t)$&nbsp; ist mit einem Transversalfilter zweiter Ordnung keine Verbesserung möglich.
+
+ For the given input pulse &nbsp;$g_x(t)$,&nbsp; no improvement is possible with a second-order transversal filter.
- Die erste Aussage ist unabhängig vom Eingangsimpuls &nbsp;$g_x(t)$.
+
- The first statement is independent of the input pulse &nbsp;$g_x(t)$.
- Beim gegebenen Eingangsimpuls ergibt sich mit einem "unendlichen" Transversalfilter eine weitere Verbesserung.
+
- For the given input pulse,&nbsp; a further improvement is obtained with an&nbsp; "infinite"&nbsp; transversal filter.
 
</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Die fünf ersten Abtastwerte des Eingangsimpulses im Abstand $T$ lauten:
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'''(1)'''&nbsp; The five first samples of the input pulse at distance&nbsp; $T$&nbsp; are:
 
:$$g_x(0)\hspace{0.25cm}\underline{ = 1},\hspace{0.2cm}g_x(1) \hspace{0.25cm}\underline{= 0.368},\hspace{0.25cm}g_x(2) \hspace{0.25cm}\underline{=
 
:$$g_x(0)\hspace{0.25cm}\underline{ = 1},\hspace{0.2cm}g_x(1) \hspace{0.25cm}\underline{= 0.368},\hspace{0.25cm}g_x(2) \hspace{0.25cm}\underline{=
 
0.135},\hspace{0.2cm}g_x(3) = 0.050,\hspace{0.2cm}g_x(4)  {= 0.018}
 
0.135},\hspace{0.2cm}g_x(3) = 0.050,\hspace{0.2cm}g_x(4)  {= 0.018}
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'''(2)'''&nbsp; Entsprechend der [[Aufgaben:3.6_ONE-Transversalfilter|Musterlösung zur Aufgabe 3.6]] kommt man auf folgendes Gleichungssystem:
+
'''(2)'''&nbsp; According to&nbsp; [[Aufgaben:Exercise_3.6:_Transversal_Filter_of_the_Optimal_Nyquist_Equalizer|"solution to Exercise 3.6"]],&nbsp; we arrive at the following system of equations:
 
:$$2t = T \hspace{-0.1cm}:\hspace{0.2cm}g_1 \hspace{-0.2cm} \ = \ \hspace{-0.2cm} k_0 \cdot g_x(1) +k_1 \cdot
 
:$$2t = T \hspace{-0.1cm}:\hspace{0.2cm}g_1 \hspace{-0.2cm} \ = \ \hspace{-0.2cm} k_0 \cdot g_x(1) +k_1 \cdot
 
[g_x(0)+g_x(2)]= 0\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
 
[g_x(0)+g_x(2)]= 0\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
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\frac{1-k_0}{0.736} \hspace{0.05cm}.$$
 
\frac{1-k_0}{0.736} \hspace{0.05cm}.$$
  
*Dies führt zum Ergebnis:
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*This leads to the result:
 
:$$k_0 - 0.324 \cdot 0.736 \cdot k_0 = 1 \hspace{0.3cm}\Rightarrow
 
:$$k_0 - 0.324 \cdot 0.736 \cdot k_0 = 1 \hspace{0.3cm}\Rightarrow
 
\hspace{0.3cm} k_0 \hspace{0.15cm}\underline {= 1.313}, \hspace{0.2cm}k_1\hspace{0.15cm}\underline { = -0.425}
 
\hspace{0.3cm} k_0 \hspace{0.15cm}\underline {= 1.313}, \hspace{0.2cm}k_1\hspace{0.15cm}\underline { = -0.425}
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'''(3)'''&nbsp; Für den Zeitpunkt $t = 2T$ gilt:
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[[File:P_ID1440__Dig_Z_3_6_c.png|right|frame|Input pulse&nbsp; (top),&nbsp; output pulse for&nbsp; $N = 1$&nbsp; (bottom)]]
:$$g_2 \ = \ k_0 \cdot g_x(2) +k_1 \cdot [g_x(1)+g_x(3)]= \ 1.313 \cdot 0.050 -0.425 \cdot [0.135+0.018]\hspace{0.15cm}\underline {\approx 0}
+
'''(3)'''&nbsp; For time&nbsp; $t = 2T$&nbsp; holds:
 +
:$$g_2 \ = \ k_0 \cdot g_x(2) +k_1 \cdot [g_x(1)+g_x(3)]$$
 +
:$$\Rightarrow\hspace{0.3cm} g_2 \ = \ 1.313 \cdot 0.050 -0.425 \cdot [0.135+0.018]\hspace{0.15cm}\underline {\approx 0}
 
\hspace{0.05cm}.$$
 
\hspace{0.05cm}.$$
  
*Ebenso ist auch der Ausgangsimpuls zum Zeitpunkt $t = 3T$ gleich Null:
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*Similarly,&nbsp; the output pulse at time&nbsp; $t = 3T$&nbsp; is also zero:
:$$g_3 \ = \ k_0 \cdot g_x(3) +k_1 \cdot [g_x(2)+g_x(4)]= 1.313 \cdot 0.135 -0.425 \cdot [0.368+0.050]\hspace{0.15cm}\underline {\approx 0}
+
:$$g_3 \ = \ k_0 \cdot g_x(3) +k_1 \cdot [g_x(2)+g_x(4)$$
 +
:$$\Rightarrow\hspace{0.3cm}g_3 \ = 1.313 \cdot 0.135 -0.425 \cdot [0.368+0.050]\hspace{0.15cm}\underline {\approx 0}
 
\hspace{0.05cm}.$$
 
\hspace{0.05cm}.$$
 +
*The figure shows that&nbsp; '''for this exponentially decaying pulse, the first-order transversal filter provides complete equalization'''.
 +
 +
*Outside the interval &nbsp;$-T < t < T$,&nbsp; &nbsp;$g_y(t)$&nbsp; is identically zero.
 +
 +
*Inside it results in a triangular shape.
  
[[File:P_ID1440__Dig_Z_3_6_c.png|right|frame|Eingangsimpuls (oben), Ausgangsimpuls für <i>N</i> = 1 (unten)]]
 
  
*Die Abbildung zeigt, dass bei diesem exponentiell abfallenden Impuls das Transversalfilter erster Ordnung eine vollständige Entzerrung bewirkt.
+
'''(4)'''&nbsp; Only the&nbsp; <u>first statement</u> is correct:
*Außerhalb des Intervalls &nbsp;$-T < t < T$&nbsp; ist &nbsp;$g_y(t)$&nbsp; identisch Null, innerhalb ergibt sich eine Dreieckform.
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*Since already with a first-order delay filter all precursors and trailers are compensated,&nbsp; also with a second-order filter and also for $N &#8594; &#8734;$ no further improvements result.
  
 +
*However,&nbsp; '''this result applies exclusively to the (bilaterally) exponentially decaying input pulse'''.
  
'''(4)'''&nbsp; Richtig ist nur der <u>erste Lösungsvorschlag</u>:
+
*For almost any other pulse shape,&nbsp; the larger&nbsp; $N$&nbsp; is,&nbsp; the better the result.
*Nachdem bereits mit einem Laufzeitfilter erster Ordnung alle Vor&ndash; und Nachläufer kompensiert werden, ergeben sich auch mit einem Filter zweiter Ordnung und auch für $N &#8594; &#8734;$ keine weiteren Verbesserungen.
 
*Dieses Ergebnis gilt jedoch ausschließlich für den (beidseitig) exponentiell abfallenden Eingansgimpuls.
 
*Bei fast jeder anderen Impulsform ist das Ergebnis um so besser, je größer $N$ ist.  
 
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  

Latest revision as of 17:01, 22 June 2022

Two-sided exponential input pulse

As in  Exercise 3.6  we consider again the optimal Nyquist equalizer.

  • The input pulse  $g_x(t)$  is a two-sided exponential function:
$$g_x(t) = {\rm e }^{ - |t|/T}\hspace{0.05cm}.$$
  • Through a transversal filter of  $N$–th order with the impulse response
$$h_{\rm TF}(t) = \sum_{\lambda = -N}^{+N} k_\lambda \cdot \delta(t - \lambda \cdot T)$$
it is possible that the output pulse  $g_y(t)$  has zero crossings at  $t/T = ±1, \ \text{...} \ , \ t/T = ±N$, 
while  $g_y(t = 0) = 1$. 
  • However,  in the general case,  the precursors and trailers with  $| \nu | > N$  lead to intersymbol interference.



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



Questions

1

Give the signal values  $g_x(\nu) = g_x(t = \nu T)$  at multiples of  $T$. 

$g_x(0)\ = \ $

$g_x(1)\ = \ $

$g_x(2)\ = \ $

2

Calculate the optimal filter coefficients for  $N = 1$.

$k_0 \ = \ $

$k_1 \ = \ $

3

Calculate the output values  $g_2 = g_{y}(t = 2T)$  and  $g_3 = g_{y}(t = 3T)$.

$g_2 \ = \ $

$g_3\ = \ $

4

Which of the following statements is true?

For the given input pulse  $g_x(t)$,  no improvement is possible with a second-order transversal filter.
The first statement is independent of the input pulse  $g_x(t)$.
For the given input pulse,  a further improvement is obtained with an  "infinite"  transversal filter.


Solution

(1)  The five first samples of the input pulse at distance  $T$  are:

$$g_x(0)\hspace{0.25cm}\underline{ = 1},\hspace{0.2cm}g_x(1) \hspace{0.25cm}\underline{= 0.368},\hspace{0.25cm}g_x(2) \hspace{0.25cm}\underline{= 0.135},\hspace{0.2cm}g_x(3) = 0.050,\hspace{0.2cm}g_x(4) {= 0.018} \hspace{0.05cm}.$$


(2)  According to  "solution to Exercise 3.6",  we arrive at the following system of equations:

$$2t = T \hspace{-0.1cm}:\hspace{0.2cm}g_1 \hspace{-0.2cm} \ = \ \hspace{-0.2cm} k_0 \cdot g_x(1) +k_1 \cdot [g_x(0)+g_x(2)]= 0\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \frac{k_1}{k_0} = - \frac{g_x(1)}{g_x(0)+g_x(2)} \hspace{0.05cm},$$
$$t = 0 \hspace{-0.1cm}:\hspace{0.2cm}g_0 \hspace{-0.2cm} \ = \ \hspace{-0.2cm} k_0 \cdot g_x(0) + k_1 \cdot 2 \cdot g_x(1) = 1\hspace{0.3cm}\Rightarrow \hspace{0.3cm} k_1 = \frac{1-k_0}{0.736} \hspace{0.05cm}.$$
  • This leads to the result:
$$k_0 - 0.324 \cdot 0.736 \cdot k_0 = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} k_0 \hspace{0.15cm}\underline {= 1.313}, \hspace{0.2cm}k_1\hspace{0.15cm}\underline { = -0.425} \hspace{0.05cm}.$$


Input pulse  (top),  output pulse for  $N = 1$  (bottom)

(3)  For time  $t = 2T$  holds:

$$g_2 \ = \ k_0 \cdot g_x(2) +k_1 \cdot [g_x(1)+g_x(3)]$$
$$\Rightarrow\hspace{0.3cm} g_2 \ = \ 1.313 \cdot 0.050 -0.425 \cdot [0.135+0.018]\hspace{0.15cm}\underline {\approx 0} \hspace{0.05cm}.$$
  • Similarly,  the output pulse at time  $t = 3T$  is also zero:
$$g_3 \ = \ k_0 \cdot g_x(3) +k_1 \cdot [g_x(2)+g_x(4)$$
$$\Rightarrow\hspace{0.3cm}g_3 \ = \ 1.313 \cdot 0.135 -0.425 \cdot [0.368+0.050]\hspace{0.15cm}\underline {\approx 0} \hspace{0.05cm}.$$
  • The figure shows that  for this exponentially decaying pulse, the first-order transversal filter provides complete equalization.
  • Outside the interval  $-T < t < T$,   $g_y(t)$  is identically zero.
  • Inside it results in a triangular shape.


(4)  Only the  first statement is correct:

  • Since already with a first-order delay filter all precursors and trailers are compensated,  also with a second-order filter and also for $N → ∞$ no further improvements result.
  • However,  this result applies exclusively to the (bilaterally) exponentially decaying input pulse.
  • For almost any other pulse shape,  the larger  $N$  is,  the better the result.