Aufgaben:Exercise 3.6Z:Optimum Nyquist Equalizer for Exponential Pulse - Revision history

Guenter at 16:01, 22 June 2022

2022-06-22T16:01:00Z

Guenter at 15:15, 22 June 2022

2022-06-22T15:15:19Z

Guenter at 15:11, 22 June 2022

2022-06-22T15:11:45Z

Guenter at 15:06, 22 June 2022

2022-06-22T15:06:05Z

Hwang at 16:53, 25 May 2022

2022-05-25T16:53:24Z

Hwang at 16:39, 25 May 2022

2022-05-25T16:39:04Z

Guenter at 15:15, 7 May 2022

2022-05-07T15:15:08Z

Guenter: Guenter moved page Aufgaben:Aufgabe 3.6Z: Optimaler Nyquistentzerrer für Exponentialimpuls to Aufgaben:Exercise 3.6Z:Optimum Nyquist Equalizer for Exponential Pulse

2022-05-07T15:10:36Z

Guenter moved page Aufgabe 3.6Z: Optimaler Nyquistentzerrer für Exponentialimpuls to Exercise 3.6Z:Optimum Nyquist Equalizer for Exponential Pulse

Javier: Text replacement - "”" to """

2021-05-28T14:31:27Z

Text replacement - "”" to """

Javier: Text replacement - "„" to """

2021-05-28T14:30:45Z

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@@ Line 48: / Line 48: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The five first samples of the input pulse at distance $T$ are:
+'''(1)'''&nbsp; The five first samples of the input pulse at distance&nbsp; $T$&nbsp; are:
 :$$g_x(0)\hspace{0.25cm}\underline{ = 1},\hspace{0.2cm}g_x(1) \hspace{0.25cm}\underline{= 0.368},\hspace{0.25cm}g_x(2) \hspace{0.25cm}\underline{=
 .135},\hspace{0.2cm}g_x(3) = 0.050,\hspace{0.2cm}g_x(4)  {= 0.018}
@@ Line 54: / Line 54: @@
-'''(2)'''&nbsp; According to the [[Aufgaben:Exercise_3.6:_Transversal_Filter_of_the_Optimal_Nyquist_Equalizer|"solution to Exercise 3.6"]], we arrive at the following system of equations:
+'''(2)'''&nbsp; According to&nbsp; [[Aufgaben:Exercise_3.6:_Transversal_Filter_of_the_Optimal_Nyquist_Equalizer|"solution to Exercise 3.6"]],&nbsp; we arrive at the following system of equations:
 :$$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
 [g_x(0)+g_x(2)]= 0\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
@@ Line 68: / Line 68: @@
-'''(3)'''&nbsp; For time $t = 2T$ holds:
+[[File:P_ID1440__Dig_Z_3_6_c.png|right|frame|Input pulse&nbsp; (top),&nbsp; output pulse for&nbsp; $N = 1$&nbsp; (bottom)]]
-:$$g_2 \ = \ k_0 \cdot g_x(2) +k_1 \cdot [g_x(1)+g_x(3)]= \ 1.313 \cdot 0.050 -0.425 \cdot [0.135+0.018]\hspace{0.15cm}\underline {\approx 0}
+'''(3)'''&nbsp; For time&nbsp; $t = 2T$&nbsp; holds:
 \hspace{0.05cm}.$$
-*Similarly, the output pulse at time $t = 3T$ is also zero:
+*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)$$
 \hspace{0.05cm}.$$
-[[File:P_ID1440__Dig_Z_3_6_c.png|right|frame|Input pulse (top), output pulse for <i>N</i> = 1 (bottom)]]
+*Outside the interval &nbsp;$-T < t < T$,&nbsp; &nbsp;$g_y(t)$&nbsp; is identically zero.
-*The figure shows that for this exponentially decaying pulse, the first-order transversal filter provides complete equalization.
+*Inside it results in a triangular shape.
-'''(4)'''&nbsp; Only the <u>first statement</u> is correct:
+'''(4)'''&nbsp; Only the&nbsp; <u>first statement</u> 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 $N &#8594; &#8734;$ no further improvements result.
+*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, 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.
+*However,&nbsp; '''this result applies exclusively to the (bilaterally) exponentially decaying input pulse'''.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 [[File:P_ID1434__Dig_Z_3_6.png|right|frame|Two-sided exponential input pulse]]
-As in &nbsp;[[Aufgaben:Exercise_3.6:_Transversal_Filter_of_the_Optimal_Nyquist_Equalizer|Exercise 3.6]]&nbsp; we consider again the optimal Nyquist equalizer,&nbsp;$g_x(t)$ &nbsp;is a two-sided exponential function:
+As in &nbsp;[[Aufgaben:Exercise_3.6:_Transversal_Filter_of_the_Optimal_Nyquist_Equalizer|Exercise 3.6]]&nbsp; we consider again the optimal Nyquist equalizer.
 :$$g_x(t) = {\rm e }^{ -  |t|/T}\hspace{0.05cm}.$$
 *Through a transversal filter of &nbsp;$N$&ndash;th order with the impulse response
 :$$h_{\rm TF}(t) = \sum_{\lambda = -N}^{+N} k_\lambda \cdot \delta(t - \lambda \cdot T)$$

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Digital_Signal_Transmission/Linear_Nyquist_Equalization}}
-[[File:P_ID1434__Dig_Z_3_6.png|right|frame|Two-sided exponential pulse]]
+[[File:P_ID1434__Dig_Z_3_6.png|right|frame|Two-sided exponential input pulse]]
 As in &nbsp;[[Aufgaben:Exercise_3.6:_Transversal_Filter_of_the_Optimal_Nyquist_Equalizer|Exercise 3.6]]&nbsp; we consider again the optimal Nyquist equalizer,&nbsp;$g_x(t)$ &nbsp;is a two-sided exponential function:
 :$$g_x(t) = {\rm e }^{ -  |t|/T}\hspace{0.05cm}.$$

@@ Line 3: / Line 3: @@
 [[File:P_ID1434__Dig_Z_3_6.png|right|frame|Two-sided exponential pulse]]
-As in &nbsp;[[Aufgaben:Exercise_3.6:_Transversal_Filter_of_the_Optimal_Nyquist_Equalizer|"Exercise 3.6"]]&nbsp; we consider again the optimal Nyquist equalizer, but now the input pulse &nbsp;$g_x(t)$ &nbsp;is a two-sided exponential function:
+As in &nbsp;[[Aufgaben:Exercise_3.6:_Transversal_Filter_of_the_Optimal_Nyquist_Equalizer|Exercise 3.6]]&nbsp; we consider again the optimal Nyquist equalizer,&nbsp;$g_x(t)$ &nbsp;is a two-sided exponential function:
 :$$g_x(t) = {\rm e }^{ -  |t|/T}\hspace{0.05cm}.$$
@@ Line 9: / Line 9: @@
 :$$h_{\rm TF}(t) = \sum_{\lambda = -N}^{+N} k_\lambda \cdot \delta(t - \lambda \cdot T)$$
-:it is always 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; and &nbsp;$g_y(t = 0) = 1$.&nbsp;
+: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;
-*However, in the general case, the precursors and trailers with &nbsp;$| \nu | > N$&nbsp; then lead to intersymbol interference.
+*However,&nbsp; in the general case,&nbsp; the precursors and trailers with &nbsp;$| \nu | > N$&nbsp; lead to intersymbol interference.
-''Note:''
+Note:&nbsp; The exercise belongs to the chapter &nbsp;[[Digital_Signal_Transmission/Linear_Nyquist_Equalization|"Linear Nyquist Equalization"]].
@@ Line 36: / Line 34: @@
 $k_1 \ = \ $ { -0.43775--0.41225 }
-{Calculate the output values &nbsp;$g_2 = g_{\rm \nu}(t = 2T)$&nbsp; and&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="{}"}
 $g_2 \ = \ $ { 0. }
@@ Line 42: / Line 40: @@
 {Which of the following statements is true?
-|type="[]"}
+|type="()"}
 + For the given input pulse &nbsp;$g_x(t)$,&nbsp; no improvement is possible with a second-order transversal filter.
 - The first statement is independent of the input pulse &nbsp;$g_x(t)$.
-- For the given input pulse, a further improvement is obtained with an "infinite" transversal filter.
+- For the given input pulse,&nbsp; a further improvement is obtained with an&nbsp; "infinite"&nbsp; transversal filter.
 </quiz>

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Digital_Signal_Transmission/Linear_Nyquist_Equalization}}
-[[File:P_ID1434__Dig_Z_3_6.png|right|frame|Beidseitiger Exponentialimpuls]]
+[[File:P_ID1434__Dig_Z_3_6.png|right|frame|Two-sided exponential pulse]]
-Wie in der &nbsp;[[Aufgaben:Aufgabe_3.6:_Transversalfilter_des_Optimalen_Nyquistentzerrers|Aufgabe 3.6]]&nbsp; betrachten wir wieder den optimalen Nyquistentzerrer, wobei nun als Eingangsimpuls &nbsp;$g_x(t)$ &nbsp;eine beidseitig abfallende Exponentialfunktion anliegt:
+As in &nbsp;[[Aufgaben:Exercise_3.6:_Transversal_Filter_of_the_Optimal_Nyquist_Equalizer|"Exercise 3.6"]]&nbsp; we consider again the optimal Nyquist equalizer, but now the input pulse &nbsp;$g_x(t)$ &nbsp;is a two-sided exponential function:
 :$$g_x(t) = {\rm e }^{ -  |t|/T}\hspace{0.05cm}.$$
-*Durch ein Transversalfilter &nbsp;$N$&ndash;ter Ordnung mit der Impulsantwort
+*Through a transversal filter of &nbsp;$N$&ndash;th order with the impulse response
 :$$h_{\rm TF}(t) = \sum_{\lambda = -N}^{+N} k_\lambda \cdot \delta(t - \lambda \cdot T)$$
-:ist es immer möglich, dass der Ausgangsimpuls &nbsp;$g_y(t)$&nbsp; Nulldurchgänge bei&nbsp; $t/T = &plusmn;1, \ \text{...} \ , \ t/T = &plusmn;N$&nbsp; aufweist und &nbsp;$g_y(t = 0) = 1$&nbsp; ist.
+:it is always 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; and &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, in the general case, the precursors and trailers with &nbsp;$| \nu | > N$&nbsp; then lead to intersymbol interference.
@@ Line 17: / Line 17: @@
-''Hinweis:''
+''Note:''
-*Die Aufgabe gehört zum  Kapitel &nbsp;[[Digital_Signal_Transmission/Linear_Nyquist_Equalization|"Linear Nyquist Equalization"]].
+*The exercise belongs to the chapter &nbsp;[[Digital_Signal_Transmission/Linear_Nyquist_Equalization|"Linear Nyquist Equalization"]].
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Geben Sie die Signalwerte &nbsp;$g_x(\nu) = g_x(t = \nu T)$&nbsp; bei Vielfachen von &nbsp;$T$&nbsp; an.
+{Give the signal values &nbsp;$g_x(\nu) = g_x(t = \nu T)$&nbsp; at multiples of &nbsp;$T$.&nbsp;
 |type="{}"}
 $g_x(0)\ = \ $ { 1 3% }
@@ Line 31: / Line 31: @@
 $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="{}"}
 $k_0 \ = \ $ { 1.313 3% }
 $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;$g_2 = g_{\rm \nu}(t = 2T)$&nbsp; and&nbsp; $g_3 = g_{\rm \nu}(t = 3T)$.
 |type="{}"}
 $g_2 \ = \ $ { 0. }
 $g_3\ = \ $ { 0. }
-{Welche der folgenden Aussagen sind zutreffend?
+{Which of the following statements is true?
 |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, a further improvement is obtained with an "infinite" transversal filter.
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Die fünf ersten Abtastwerte des Eingangsimpulses im Abstand $T$ lauten:
+'''(1)'''&nbsp; The five first samples of the input pulse at distance $T$ are:
 :$$g_x(0)\hspace{0.25cm}\underline{ = 1},\hspace{0.2cm}g_x(1) \hspace{0.25cm}\underline{= 0.368},\hspace{0.25cm}g_x(2) \hspace{0.25cm}\underline{=
 .135},\hspace{0.2cm}g_x(3) = 0.050,\hspace{0.2cm}g_x(4)  {= 0.018}
@@ Line 56: / Line 56: @@
-'''(2)'''&nbsp; Entsprechend der [[Aufgaben:3.6_ONE-Transversalfilter|Musterlösung zur Aufgabe 3.6]] kommt man auf folgendes Gleichungssystem:
+'''(2)'''&nbsp; According to the [[Aufgaben:Exercise_3.6:_Transversal_Filter_of_the_Optimal_Nyquist_Equalizer|"solution to Exercise 3.6"]], we arrive at the following system of equations:
 :$$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}
@@ Line 64: / Line 64: @@
 \frac{1-k_0}{0.736} \hspace{0.05cm}.$$
-*Dies führt zum Ergebnis:
+*This leads to the result:
 :$$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}
@@ Line 70: / Line 70: @@
-'''(3)'''&nbsp; Für den Zeitpunkt $t = 2T$ gilt:
+'''(3)'''&nbsp; For time $t = 2T$ holds:
 :$$g_2 \ = \ k_0 \cdot g_x(2) +k_1 \cdot [g_x(1)+g_x(3)]= \ 1.313 \cdot 0.050 -0.425 \cdot [0.135+0.018]\hspace{0.15cm}\underline {\approx 0}
 \hspace{0.05cm}.$$
-*Ebenso ist auch der Ausgangsimpuls zum Zeitpunkt $t = 3T$ gleich Null:
+*Similarly, the output pulse at time $t = 3T$ is also zero:
 :$$g_3 \ = \ k_0 \cdot g_x(3) +k_1 \cdot [g_x(2)+g_x(4)]= 1.313 \cdot 0.135 -0.425 \cdot [0.368+0.050]\hspace{0.15cm}\underline {\approx 0}
 \hspace{0.05cm}.$$
-[[File:P_ID1440__Dig_Z_3_6_c.png|right|frame|Eingangsimpuls (oben), Ausgangsimpuls für <i>N</i> = 1 (unten)]]
+[[File:P_ID1440__Dig_Z_3_6_c.png|right|frame|Input pulse (top), output pulse for <i>N</i> = 1 (bottom)]]
-*Die Abbildung zeigt, dass bei diesem exponentiell abfallenden Impuls das Transversalfilter erster Ordnung eine vollständige Entzerrung bewirkt.
+*The figure shows that for this exponentially decaying pulse, the first-order transversal filter provides complete equalization.
-*Außerhalb des Intervalls &nbsp;$-T < t < T$&nbsp; ist &nbsp;$g_y(t)$&nbsp; identisch Null, innerhalb ergibt sich eine Dreieckform.
+*Outside the interval &nbsp;$-T < t < T$,&nbsp; &nbsp;$g_y(t)$&nbsp; is identically zero, inside it results in a triangular shape.
-'''(4)'''&nbsp; Richtig ist nur der <u>erste Lösungsvorschlag</u>:
+'''(4)'''&nbsp; Only the <u>first statement</u> is correct:
-*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.
+*Since already with a first-order delay filter all precursors and trailers are compensated, also with a second-order filter and also for $N &#8594; &#8734;$ no further improvements result.
-*Dieses Ergebnis gilt jedoch ausschließlich für den (beidseitig) exponentiell abfallenden Eingansgimpuls.
+*However, this result applies exclusively to the (bilaterally) exponentially decaying input pulse.
-*Bei fast jeder anderen Impulsform ist das Ergebnis um so besser, je größer $N$ ist.
+*For almost any other pulse shape, the larger $N$ is, the better the result.
 {{ML-Fuß}}

← Older revision		Revision as of 15:15, 7 May 2022
Line 92:		Line 92:


−	[[Category:Digital Signal Transmission: Exercises\|^3.5 ~~Lineare Nyquistentzerrung~~^]]	+	[[Category:Digital Signal Transmission: Exercises\|^3.5 Linear Nyquist Equalization^]]

@@ Line 45: / Line 45: @@
 + Beim gegebenen Eingangsimpuls &nbsp;$g_x(t)$&nbsp; ist mit einem Transversalfilter zweiter Ordnung keine Verbesserung möglich.
 - Die erste Aussage ist unabhängig vom Eingangsimpuls &nbsp;$g_x(t)$.
-- Beim gegebenen Eingangsimpuls ergibt sich mit einem &bdquo;unendlichen&rdquo; Transversalfilter eine weitere Verbesserung.
+- Beim gegebenen Eingangsimpuls ergibt sich mit einem "unendlichen&rdquo; Transversalfilter eine weitere Verbesserung.
 </quiz>