Aufgaben:Exercise 2.5Z: Nyquist Equalization - Revision history

Guenter at 08:36, 6 October 2021

2021-10-06T08:36:50Z

Guenter at 08:29, 6 October 2021

2021-10-06T08:29:01Z

Guenter: Guenter moved page Aufgaben:Exercise 2.5Z: Nyquist Equalisation to Aufgaben:Exercise 2.5Z: Nyquist Equalization

2021-10-05T12:23:05Z

Guenter moved page Exercise 2.5Z: Nyquist Equalisation to Exercise 2.5Z: Nyquist Equalization

Javier: Text replacement - "equalise" to "equalize"

2021-09-21T12:51:27Z

Text replacement - "equalise" to "equalize"

Guenter: Guenter moved page Aufgaben:Aufgabe 2.5Z: Nyquistentzerrung to Aufgaben:Exercise 2.5Z: Nyquist Equalisation

2021-09-20T13:14:58Z

Guenter moved page Aufgabe 2.5Z: Nyquistentzerrung to Exercise 2.5Z: Nyquist Equalisation

Guenter at 10:12, 20 September 2021

2021-09-20T10:12:05Z

Oezer at 01:28, 17 September 2021

2021-09-17T01:28:08Z

Oezer at 01:26, 17 September 2021

2021-09-17T01:26:01Z

Oezer at 01:09, 17 September 2021

2021-09-17T01:09:01Z

Oezer at 00:55, 17 September 2021

2021-09-17T00:55:06Z

@@ Line 80: / Line 80: @@
 {{ML-Kopf}}
 [[File:P_ID922__LZI_Z_2_5_a.png|right|frame|Cosine-square spectrum]]
-'''(1)'''&nbsp; Using the constant spectrum&nbsp; $X(f) = T$&nbsp; the following is obtained for the spectral function of the receiver output signal&nbsp; $y(t)$:
+'''(1)'''&nbsp; Using the constant spectrum&nbsp; $X(f) = T$&nbsp; the following is obtained for the spectral function of the output signal&nbsp; $y(t)$:
 :$$Y(f)=  T \cdot {H(f)}.$$
 *The signal value at&nbsp; $t = 0$&nbsp; is equal to the area under $Y(f)$.
-*As can be seen from the adjacent sketch, this is equal to&nbsp; $1$. From this it follows that:
+*As can be seen from the adjacent sketch, this is equal to&nbsp; $1$.
 :$$y(t = 0)\; \underline{= 1}.$$
@@ Line 95: / Line 96: @@
 *Due to&nbsp; $\cos(0) = 1$&nbsp; and&nbsp; ${\rm si}(0) = 1$,&nbsp; &nbsp; $H_{\rm E}(f = 0)\;\underline{=1}$ also holds.
-*Considering the given trigonometric transformation it further holds that:
+*Considering the given trigonometric transformation it further holds:
 :$$H_{\rm  E}(f) = {\pi f T}/{2} \cdot {\rm cot}\left( {\pi f
   T}/{2}\right),$$
@@ Line 111: / Line 112: @@
 :$$ H_{\rm  E}(f \cdot T = 1.00)= { \pi }/{2} \cdot {\rm cot}\left( 90^{\circ}\right) ={ \pi }/{2} \cdot 0 \hspace{0.15cm}\underline{ = 0}.$$
 '''(3)'''&nbsp; Considering the Gaussian channel the following holds:
-$$H_{\rm  E}(f)=  \frac{H(f)}{H_{\rm  S}(f) \cdot H_{\rm  K}(f)} =  H_{\rm
+:$$H_{\rm  E}(f)=  \frac{H(f)}{H_{\rm  S}(f) \cdot H_{\rm  K}(f)} =  H_{\rm
   E}^{(2)}(f)\cdot {\rm e}^{\pi (f\hspace{0.05cm}\cdot \hspace{0.05cm} T)^2}.$$
-Here,&nbsp; $H_{\rm E}^{(2)}(f)$&nbsp; denotes the equalizer frequency response computed in the subtask&nbsp; '''(2)'''&nbsp; assuming an ideal channel. The following numerical results are obtained:
+Here,&nbsp; $H_{\rm E}^{(2)}(f)$&nbsp; denotes the equalizer frequency response computed in the subtask&nbsp; '''(2)'''&nbsp; assuming an ideal channel.
 :$$H_{\rm  E}(f\cdot T = 0) =  1 \cdot {\rm e}^{0} \hspace{0.15cm}\underline{= 1},$$
 :$$H_{\rm  E}(f \cdot T = 0.25) =  0.948 \cdot 1.217 \hspace{0.15cm}\underline{=  1.154},$$

@@ Line 6: / Line 6: @@
 A digital baseband transmission system is modelled by the depicted block diagram.
-*The "transmitter", "channel" and "receiver" components are described in the frequency domain by&nbsp; $H_{\rm S}(f)$,&nbsp; $H_{\rm K}(f)$&nbsp; and&nbsp; $H_{\rm E}(f)$&nbsp;.
+*The transmitter, channel and receiver components are described in the frequency domain by&nbsp; $H_{\rm S}(f)$,&nbsp; $H_{\rm K}(f)$&nbsp; and&nbsp; $H_{\rm E}(f)$&nbsp;.
 *The overall frequency response&nbsp; $H(f) = H_{\rm S}(f) \cdot H_{\rm K}(f) \cdot H_{\rm E}(f)$&nbsp; has a&nbsp; $\cos^2$&ndash;shaped curve:
@@ Line 17: / Line 17: @@
 \end{array}$$
 *The signal&nbsp; $y(t)$&nbsp; before the decision circuit thus exhibits equidistant zero crossings at intervals of&nbsp; $T$&nbsp;.
-*It is assumed here that the source emits a&nbsp; [[Signal_Representation/Special_Cases_of_Impulse_Signals#Dirac_delta_or_impulse|Dirac-delta]]&nbsp; $x(t)$&nbsp; with weight&nbsp; $T$&nbsp; (see graph).
+*It is assumed here that the source transmits a&nbsp; [[Signal_Representation/Special_Cases_of_Impulse_Signals#Dirac_delta_or_impulse|Dirac-delta]]&nbsp; $x(t)$&nbsp; with weight&nbsp; $T$&nbsp; (see graph).
-It is pointed out that this is a so-called "Nyquist system".
+It is pointed out that this is a so-called&nbsp; "Nyquist system".
-As will be discussed in detail in the book&nbsp; [[Digital_Signal_Transmission]],&nbsp; these Nyquist systems represent an important class of digital transmission systems since the sequentially transmitted symbols do not influence each other in such systems.
+As will be discussed in detail in the book&nbsp; [[Digital_Signal_Transmission/Linear_Nyquist_Equalization|Digital Signal Transmission]],&nbsp; these Nyquist systems represent an important class of digital transmission systems since the sequentially transmitted symbols do not influence each other in such systems.
 However, these far-reaching aspects are not needed for the solution of this task.
@@ Line 29: / Line 29: @@
 *the transmission pulse&nbsp; $s(t)$&nbsp; be rectangular with pulse duration&nbsp; $T$:
-:$$H_{\rm S}(f) = {\rm si}(\pi f T),$$
+:$$H_{\rm S}(f) = {\rm si}(\pi f T)= {\rm sinc}(f T)\hspace{1.0cm} \Rightarrow \hspace{0.5cm}
 *the channel is assumed to be ideal up to and including subtask&nbsp; '''(2)'''&nbsp; while for the last subtask&nbsp; '''(3)'''&nbsp; the following shall hold:
 :$$H_{\rm K}(f) = H_{\rm G}(f) = {\rm e}^{-\pi(f \cdot T)^2} .$$
-For both channels, the receiver and simultaneously the equalizer frequency response&nbsp; $H_{\rm E}(f)$ are searched-for so that the overall frequency response has the desired Nyquist shape.
+For both channels, the receiver and the equalizer frequency response&nbsp; $H_{\rm E}(f)$ are searched-for so that the overall frequency response has the desired Nyquist shape.
@@ Line 55: / Line 56: @@
-{First, let&nbsp; $H_{\rm K}(f) = 1$ hold &nbsp; &rArr; &nbsp; <u>ideal channel</u>. Compute the frequency response&nbsp; $H_{\rm E}(f)$ for this case. <br>What values are obtained at the frequencies given below?
+{First,&nbsp; let&nbsp; $H_{\rm K}(f) = 1$&nbsp; hold &nbsp; &rArr; &nbsp; <u>ideal channel</u>.&nbsp; Compute the frequency response&nbsp; $H_{\rm E}(f)$ for this case. <br>What values are obtained at the frequencies given below?
 |type="{}"}
 $|H_{\rm E}(f \cdot T = 0)| \ = \ $  { 1 3% }

@@ Line 33: / Line 33: @@
 :$$H_{\rm K}(f) = H_{\rm G}(f) = {\rm e}^{-\pi(f \cdot T)^2} .$$
-For both channels, the receiver and simultaneously the equaliser frequency response&nbsp; $H_{\rm E}(f)$ are searched-for so that the overall frequency response has the desired Nyquist shape.
+For both channels, the receiver and simultaneously the equalizer frequency response&nbsp; $H_{\rm E}(f)$ are searched-for so that the overall frequency response has the desired Nyquist shape.
@@ Line 88: / Line 88: @@
-[[File:EN_LZI_Z_2_5c.png|right|frame|Frequency response of the Nyquist equaliser]]
+[[File:EN_LZI_Z_2_5c.png|right|frame|Frequency response of the Nyquist equalizer]]
 '''(2)'''&nbsp; From the condition&nbsp; $H_{\rm S}(f) \cdot H_{\rm E}(f) = H(f)$&nbsp; it follows in the considered range:
 :$$H_{\rm  E}(f)=  \frac{H(f)}{H_{\rm  S}(f)} =   \frac{\cos^2(\pi f T/2)}{\sin(\pi f T)/(\pi f T)}.$$
@@ Line 116: / Line 116: @@
   E}^{(2)}(f)\cdot {\rm e}^{\pi (f\hspace{0.05cm}\cdot \hspace{0.05cm} T)^2}.$$
-Here,&nbsp; $H_{\rm E}^{(2)}(f)$&nbsp; denotes the equaliser frequency response computed in the subtask&nbsp; '''(2)'''&nbsp; assuming an ideal channel. The following numerical results are obtained:
+Here,&nbsp; $H_{\rm E}^{(2)}(f)$&nbsp; denotes the equalizer frequency response computed in the subtask&nbsp; '''(2)'''&nbsp; assuming an ideal channel. The following numerical results are obtained:
 :$$H_{\rm  E}(f\cdot T = 0) =  1 \cdot {\rm e}^{0} \hspace{0.15cm}\underline{= 1},$$
 :$$H_{\rm  E}(f \cdot T = 0.25) =  0.948 \cdot 1.217 \hspace{0.15cm}\underline{=  1.154},$$

← Older revision		Revision as of 10:12, 20 September 2021
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−	[[Category:Linear and Time-Invariant Systems: Exercises\|^2.3 ~~Lineare Verzerrungen~~^]]	+	[[Category:Linear and Time-Invariant Systems: Exercises\|^2.3 Linear Distortions^]]

← Older revision		Revision as of 01:28, 17 September 2021
Line 33:		Line 33:
	:$$H_{\rm K}(f) = H_{\rm G}(f) = {\rm e}^{-\pi(f \cdot T)^2} .$$		:$$H_{\rm K}(f) = H_{\rm G}(f) = {\rm e}^{-\pi(f \cdot T)^2} .$$

−	For both channels, the receiver~~–~~ and simultaneously the equaliser frequency response  $H_{\rm E}(f)$ are searched-for so that the overall frequency response has the desired Nyquist shape.	+	For both channels, the receiver and simultaneously the equaliser frequency response  $H_{\rm E}(f)$ are searched-for so that the overall frequency response has the desired Nyquist shape.

← Older revision	Revision as of 12:23, 5 October 2021
(No difference)

← Older revision	Revision as of 13:14, 20 September 2021
(No difference)

@@ Line 55: / Line 55: @@
-{First, let&nbsp; $H_{\rm K}(f) = 1$ hold &nbsp; &rArr; &nbsp; <u>ideal channel</u>. Berechnen Sie für diesen Fall den Frequenzgang&nbsp; $H_{\rm E}(f)$. <br>Welche Werte ergeben sich bei den nachfolgend genannten Frequenzen?
+{First, let&nbsp; $H_{\rm K}(f) = 1$ hold &nbsp; &rArr; &nbsp; <u>ideal channel</u>. Compute the frequency response&nbsp; $H_{\rm E}(f)$ for this case. <br>What values are obtained at the frequencies given below?
 |type="{}"}
 $|H_{\rm E}(f \cdot T = 0)| \ = \ $  { 1 3% }
@@ Line 64: / Line 64: @@
-{Berechnen Sie&nbsp; $H_{\rm E}(f)$&nbsp;  für den gaußförmigen Kanal&nbsp; $H_{\rm K}(f) = H_{\rm G}(f)$&nbsp; entsprechend der Angabe.
+{Compute&nbsp; $H_{\rm E}(f)$&nbsp; for the Gaussian-shaped channel&nbsp; $H_{\rm K}(f) = H_{\rm G}(f)$&nbsp; according to the description.
 |type="{}"}
 $|H_{\rm E}(f \cdot T = 0)|\ = \ $ { 1 3% }
@@ Line 78: / Line 78: @@
 ===Solution===
 {{ML-Kopf}}
-[[File:P_ID922__LZI_Z_2_5_a.png|right|frame|Cosinus–Quadrat–Spektrum]]
+[[File:P_ID922__LZI_Z_2_5_a.png|right|frame|Cosine-square spectrum]]
 '''(1)'''&nbsp; Mit dem konstanten Spektrum&nbsp; $X(f) = T$&nbsp; erhält man für die Spektralfunktion des Empfängerausgangssignals&nbsp; $y(t)$:
 :$$Y(f)=  T \cdot {H(f)}.$$
@@ Line 88: / Line 88: @@
-[[File:EN_LZI_Z_2_5c.png|right|frame|Frequenzgang des Nyquistentzerrers]]
+[[File:EN_LZI_Z_2_5c.png|right|frame|Frequency Response of the Nyquist equaliser]]
 '''(2)'''&nbsp; Aus der Bedingung&nbsp; $H_{\rm S}(f) \cdot H_{\rm E}(f) = H(f)$&nbsp; folgt im betrachteten Bereich:
 :$$H_{\rm  E}(f)=  \frac{H(f)}{H_{\rm  S}(f)} =   \frac{\cos^2(\pi f T/2)}{\sin(\pi f T)/(\pi f T)}.$$

@@ Line 30: / Line 30: @@
 *the transmission pulse&nbsp; $s(t)$&nbsp; be rectangular with pulse duration&nbsp; $T$:
 :$$H_{\rm S}(f) = {\rm si}(\pi f T),$$
-* der Kanal bis einschließlich Teilaufgabe&nbsp; '''(2)'''&nbsp; als ideal vorausgesetzt wird, während für die last subtask&nbsp; '''(3)'''&nbsp; gelten soll:
+*the channel is assumed to be ideal up to and including subtask&nbsp; '''(2)'''&nbsp; while for the last subtask&nbsp; '''(3)'''&nbsp; the following shall hold:
 :$$H_{\rm K}(f) = H_{\rm G}(f) = {\rm e}^{-\pi(f \cdot T)^2} .$$
-Gesucht ist für beide Kanäle der Empfänger&ndash; und gleichzeitig Entzerrerfrequenzgang&nbsp; $H_{\rm E}(f)$, damit der Gesamtfrequenzgang die gewünschte Nyquistform aufweist.
+For both channels, the receiver&ndash; and simultaneously the equaliser frequency response&nbsp; $H_{\rm E}(f)$ are searched-for so that the overall frequency response has the desired Nyquist shape.
@@ Line 43: / Line 43: @@
 *The task belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/Linear_Distortions|Linear Distortions]].
-*Als bekannt vorausgesetzt wird die folgende trigonometrische Beziehung:
+*The following trigonometric relation is assumed to be known:
 :$$\frac{\cos^2(\alpha /2)}{\sin(\alpha )} = {1}/{2} \cdot {\rm cot}(\alpha /2) .$$
@@ Line 50: / Line 50: @@
 <quiz display=simple>
-{Berechnen Sie den Ausgangssignalwert zum Zeitpunkt&nbsp; $t = 0$.
+{Compute the output signal value at time&nbsp; $t = 0$.
 |type="{}"}
 $y(t = 0) \ = \ $  { 1 3% }
-{Zunächst sei&nbsp; $H_{\rm K}(f) = 1$ &nbsp; &rArr; &nbsp; <u>idealer Kanal</u>. Berechnen Sie für diesen Fall den Frequenzgang&nbsp; $H_{\rm E}(f)$. <br>Welche Werte ergeben sich bei den nachfolgend genannten Frequenzen?
+{First, let&nbsp; $H_{\rm K}(f) = 1$ hold &nbsp; &rArr; &nbsp; <u>ideal channel</u>. Berechnen Sie für diesen Fall den Frequenzgang&nbsp; $H_{\rm E}(f)$. <br>Welche Werte ergeben sich bei den nachfolgend genannten Frequenzen?
 |type="{}"}
 $|H_{\rm E}(f \cdot T = 0)| \ = \ $  { 1 3% }