Aufgaben:Exercise 5.5Z: About the Rake Receiver - Revision history

Guenter at 15:03, 7 February 2023

2023-02-07T15:03:42Z

Hwang at 08:25, 29 December 2021

2021-12-29T08:25:53Z

Guenter at 16:39, 20 December 2021

2021-12-20T16:39:34Z

Guenter at 15:09, 17 December 2021

2021-12-17T15:09:16Z

Guenter at 14:49, 17 December 2021

2021-12-17T14:49:42Z

Guenter at 14:48, 17 December 2021

2021-12-17T14:48:55Z

Guenter at 14:46, 17 December 2021

2021-12-17T14:46:41Z

Hwang at 12:16, 13 December 2021

2021-12-13T12:16:40Z

Hwang at 10:57, 13 December 2021

2021-12-13T10:57:34Z

Javier: Text replacement - "RAKE" to "Rake"

2021-12-02T10:32:20Z

Text replacement - "RAKE" to "Rake"

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Mod_Z_5_5.png|right|frame|2-way channel <br>& rake receiver]]
+[[File:EN_Mod_Z_5_5.png|right|frame|Two-way channel <br>& rake receiver]]
 The diagram shows a two-way channel&nbsp; (yellow background).&nbsp; The corresponding descriptive equation is:
 :$$ r(t) =0.6 \cdot s(t) + 0.4 \cdot s (t - \tau) \hspace{0.05cm}.$$

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Mod_Z_5_5.png|right|frame|2-way channel <br>& rake receiver '''KORREKTUR''': two-way channel, rake receiver]]
+[[File:EN_Mod_Z_5_5.png|right|frame|2-way channel <br>& rake receiver]]
 The diagram shows a two-way channel&nbsp; (yellow background).&nbsp; The corresponding descriptive equation is:
 :$$ r(t) =0.6 \cdot s(t) + 0.4 \cdot s (t - \tau) \hspace{0.05cm}.$$

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1888__Mod_Z_5_5.png|right|frame|2-way channel <br>& rake receiver '''KORREKTUR''': two-way channel, rake receiver]]
+[[File:EN_Mod_Z_5_5.png|right|frame|2-way channel <br>& rake receiver '''KORREKTUR''': two-way channel, rake receiver]]
 The diagram shows a two-way channel&nbsp; (yellow background).&nbsp; The corresponding descriptive equation is:
 :$$ r(t) =0.6 \cdot s(t) + 0.4 \cdot s (t - \tau) \hspace{0.05cm}.$$

@@ Line 68: / Line 68: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; <u>Solution 1</u> is correct:
-*The impulse response&nbsp; $h_{\rm K}(t)$&nbsp; is obtained as the received signal&nbsp; $r(t)$ when there is a dirac pulse at the input &nbsp; ⇒ &nbsp;  $s(t) = δ(t)$.&nbsp; It follows that:
+*The impulse response&nbsp; $h_{\rm K}(t)$&nbsp; is obtained as the received signal&nbsp; $r(t)$&nbsp; when there is a Dirac delta pulse at the input &nbsp; ⇒ &nbsp;  $s(t) = δ(t)$.&nbsp; It follows that:
-:$$ h_(t) = 0.6 \cdot \delta (t ) + 0.4 \cdot \delta (t - \tau) \hspace{0.05cm}.$$
+:$$ h_{\rm K}(t) = 0.6 \cdot \delta (t ) + 0.4 \cdot \delta (t - \tau) \hspace{0.05cm}.$$
-'''(2)'''&nbsp; <u>Solutions 2 and 3</u> are correct:
+'''(2)'''&nbsp; <u>Solutions 2 and 3</u>&nbsp; are correct:
-*By definition, the channel frequency response&nbsp; $H_{\rm K}(f)$&nbsp; is the Fourier transform of the impulse response&nbsp; $h_{\rm K}(t)$.&nbsp; With the shift theorem this results in:
+*By definition,&nbsp; the channel frequency response&nbsp; $H_{\rm K}(f)$&nbsp; is the Fourier transform of the impulse response&nbsp; $h_{\rm K}(t)$.&nbsp; With the shift theorem this results in:
 :$$H_{\rm K}(f) = 0.6 + 0.4 \cdot {\rm e}^{ \hspace{0.03cm}{\rm j} \hspace{0.03cm} \cdot \hspace{0.03cm}2 \pi f \tau}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H_{\rm K}(f= 0) = 0.6 + 0.4 = 1 \hspace{0.05cm}.$$
-*Accordingly, the first proposed solution is incorrect in contrast to the other two: &nbsp; $H_{\rm K}(f)$ is complex-valued and the magnitude is periodic with&nbsp; $1/τ$, as the following calculation shows:
+*Accordingly,&nbsp; the first proposed solution is incorrect in contrast to the other two:
-:$$|H_{\rm K}(f)|^2  =  \left [0.6 + 0.4 \cdot \cos(2 \pi f \tau) \right ]^2 + \left [ 0.4 \cdot \sin(2 \pi f \tau) \right ]^2 = \left [0.6^2 + 0.4^2 \cdot \left ( \cos^2(2 \pi f \tau) + \sin^2(2 \pi f \tau)\right ) \right ] +  2 \cdot 0.6 \cdot 0.4 \cdot \cos(2 \pi f \tau)$$
+#&nbsp;  $H_{\rm K}(f)$ is complex-valued and
 *For&nbsp; $f = 0$,&nbsp;&nbsp; $|H_{\rm K}(f)| = 1$.&nbsp; This value is repeated in the respective frequency spacing&nbsp; $1/τ$.&nbsp;
@@ Line 85: / Line 88: @@
 *To satisfy the given&nbsp; $h_{\rm KR}(t)$ equation, either&nbsp; $τ_0 = 0$&nbsp; must hold or&nbsp; $τ_1 = 0$.&nbsp; With&nbsp; $τ_0 = 0$&nbsp; we obtain for the impulse response:
 :$$h_{\rm KR}(t)  =  0.6 \cdot h_0 \cdot \delta (t ) + 0.4 \cdot h_0 \cdot \delta (t - \tau) +  0.6 \cdot h_1 \cdot \delta (t -\tau_1) + 0.4 \cdot h_1 \cdot \delta (t - \tau-\tau_1) \hspace{0.05cm}.$$
-*To be able to focus the "main energy" on a time point,&nbsp; $τ_1 = τ$&nbsp; would then have to be chosen.&nbsp; With&nbsp; $h_0 = 0.6$&nbsp; and&nbsp; $h_1 = 0.4$,&nbsp; we then obtain&nbsp; $A_0 ≠ A_2$:
+*To be able to focus the&nbsp; "main energy"&nbsp; at a certain time point,&nbsp; $τ_1 = τ$&nbsp; would have to be chosen.&nbsp;
 :$$h_{\rm KR}(t) = 0.36 \cdot \delta (t ) +0.48 \cdot \delta (t - \tau) + 0.16 \cdot \delta (t - 2\tau)\hspace{0.05cm}.$$
 *In contrast, with&nbsp; $h_0 = 0.6$,&nbsp; $h_1 = 0.4$,&nbsp; $τ_0 = τ$&nbsp; and&nbsp; $τ_1 = 0$:
 :$$h_{\rm KR}(t)  =  0.6 \cdot h_0 \cdot \delta (t - \tau ) + 0.4 \cdot h_0 \cdot \delta (t - 2\tau) +  0.6 \cdot h_1 \cdot \delta (t) + 0.4 \cdot h_1 \cdot \delta (t - \tau)=  0.24 \cdot \delta (t ) +0.52 \cdot \delta (t - \tau) + 0.24 \cdot \delta (t - 2\tau) \hspace{0.05cm}.$$
-*Here, the additional condition&nbsp; $A_0 = A_2$&nbsp; is satisfied.&nbsp; Thus, the result we are looking for is:
+*Here,&nbsp; the additional condition&nbsp; $A_0 = A_2$&nbsp; is satisfied.&nbsp; Thus,&nbsp; the result we are looking for is:
 :$$ \underline{\tau_0 = \tau = 1\,{\rm &micro; s} \hspace{0.05cm},\hspace{0.2cm}\tau_1 =0} \hspace{0.05cm}.$$
@@ Line 96: / Line 100: @@
 '''(4)'''&nbsp; The following must apply to the normalization factor:
 :$$K= \frac{1}{h_0^2 + h_1^2} = \frac{1}{0.6^2 + 0.4^2} = \frac{1}{0.52} \hspace{0.15cm}\underline {\approx 1.923} \hspace{0.05cm}.$$
-*This gives for the common impulse response&nbsp; $($it holds $0.24/0.52 = 6/13)$:
+*This gives for the common impulse response&nbsp; $($it holds&nbsp; $0.24/0.52 = 6/13)$:
 :$$ h_{\rm KR}(t) = \frac{6}{13} \cdot \delta (t ) + 1.00 \cdot \delta (t - \tau) + \frac{6}{13} \cdot \delta (t - 2\tau)\hspace{0.05cm}.$$
@@ Line 103: / Line 107: @@
 [[File:P_ID1902__Mod_Z_5_5e.png|right|frame|Signals to illustrate the rake receiver]]
 '''(5)'''&nbsp; <u>Statements 1 and 4</u> are correct, as shown in the diagram:
-*For the received signal&nbsp; $r(t)$&nbsp; and for the rake output signal&nbsp; $b(t)$&nbsp; holds:
+*For the received signal&nbsp; $r(t)$&nbsp;  holds:
 :$$r(t)  =  0.6 \cdot s(t) + 0.4 \cdot s (t - 1\,{\rm &micro; s})\hspace{0.05cm},$$
 :$$b(t)  = \frac{6}{13} \cdot s(t) + 1 \cdot s (t - 1\,{\rm &micro; s}) + \frac{6}{13} \cdot s (t - 2\,{\rm &micro; s}) \hspace{0.05cm}.$$
 *The overshoot of the output signal  &nbsp; ⇒  &nbsp; $b(t) > 1$&nbsp; is due to the normalization factor&nbsp; $K = 25/13$.&nbsp;

@@ Line 23: / Line 23: @@
 Notes:
 *The exercise belongs to the chapter&nbsp; [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation|Error Probability of Direct-Sequence Spread Spectrum Modulation]].
 *Reference is made in particular to the section&nbsp; [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation#Principle_of_the_rake_receiver |Principle of the rake receiver]].

@@ Line 6: / Line 6: @@
 The diagram shows a two-way channel&nbsp; (yellow background).&nbsp; The corresponding descriptive equation is:
 :$$ r(t) =0.6 \cdot s(t) + 0.4 \cdot s (t - \tau) \hspace{0.05cm}.$$
-Let the delay on the secondary path be &nbsp;$τ = 1 \ \rm &micro; s$.&nbsp; Drawn below is the structure of a rake receiver&nbsp; (green background)&nbsp; with general coefficients &nbsp;$K$, &nbsp;$h_0$, &nbsp;$h_1$, &nbsp;$τ_0$&nbsp; and &nbsp;$τ_1$.
+Let the delay on the secondary path be &nbsp;$τ = 1 \ \rm &micro; s$.&nbsp;
-The purpose of the rake receiver is to combine the energy of the two signal paths,&nbsp; making the decision more reliable.&nbsp;
+Drawn below is the structure of a rake receiver&nbsp; (green background)&nbsp; with general coefficients &nbsp;$K$, &nbsp;$h_0$, &nbsp;$h_1$, &nbsp;$τ_0$&nbsp; and &nbsp;$τ_1$.
-The combined impulse response of the channel&nbsp; (German:&nbsp; "Kanal" &nbsp; &rArr; &nbsp; subscript "K")&nbsp; and the rake receiver can be expressed in the form
+*The purpose of the rake receiver is to combine the energy of the two signal paths,&nbsp; making the decision more reliable.&nbsp;
 :$$h_{\rm KR}(t) = A_0 \cdot \delta (t ) + A_1 \cdot \delta (t - \tau) + A_2 \cdot \delta (t - 2\tau)$$
-but only if the rake coefficients &nbsp;$h_0$, &nbsp;$h_1$, &nbsp;$τ_0$&nbsp; and &nbsp;$τ_1$&nbsp; are appropriately chosen.&nbsp; The main part of &nbsp;$h_{\rm KR}(t)$&nbsp; is supposed to be at &nbsp;$t = τ$.&nbsp;
+:but only if the rake coefficients &nbsp;$h_0$, &nbsp;$h_1$, &nbsp;$τ_0$&nbsp; and &nbsp;$τ_1$&nbsp; are appropriately chosen.&nbsp;
+*The main part of &nbsp;$h_{\rm KR}(t)$&nbsp; is supposed to be at &nbsp;$t = τ$.&nbsp;
-The constant &nbsp;$K$&nbsp; is to be chosen so that the amplitude of the main path &nbsp;$A_1 = 1$&nbsp;:
+*The constant &nbsp;$K$&nbsp; is to be chosen so that the amplitude of the main path &nbsp;$A_1 = 1$&nbsp;:
 :$$K= \frac{1}{h_0^2 + h_1^2}.$$
 Apart from the rake parameters,&nbsp; the signals &nbsp;$r(t)$&nbsp; and &nbsp;$b(t)$ are sought when&nbsp;$s(t)$&nbsp; is a rectangle of height &nbsp;$s_0 = 1$&nbsp; and width &nbsp;$T = \ \rm 5 &micro; s$.&nbsp;

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1888__Mod_Z_5_5.png|right|frame|Two-way channel <br>& rake '''KORREKTUR''': two-way channel, rake receiver]]
+[[File:P_ID1888__Mod_Z_5_5.png|right|frame|2-way channel <br>& rake receiver '''KORREKTUR''': two-way channel, rake receiver]]
 The diagram shows a two-way channel&nbsp; (yellow background).&nbsp; The corresponding descriptive equation is:
 :$$ r(t) =0.6 \cdot s(t) + 0.4 \cdot s (t - \tau) \hspace{0.05cm}.$$
 Let the delay on the secondary path be &nbsp;$τ = 1 \ \rm &micro; s$.&nbsp; Drawn below is the structure of a rake receiver&nbsp; (green background)&nbsp; with general coefficients &nbsp;$K$, &nbsp;$h_0$, &nbsp;$h_1$, &nbsp;$τ_0$&nbsp; and &nbsp;$τ_1$.
-The purpose of the rake receiver is to combine the energy of the two signal paths, making the decision more reliable.&nbsp;
+The purpose of the rake receiver is to combine the energy of the two signal paths,&nbsp; making the decision more reliable.&nbsp;
-The combined impulse response of the channel and the rake receiver can be expressed in the form
+The combined impulse response of the channel&nbsp; (German:&nbsp; "Kanal" &nbsp; &rArr; &nbsp; subscript "K")&nbsp; and the rake receiver can be expressed in the form
 :$$h_{\rm KR}(t) = A_0 \cdot \delta (t ) + A_1 \cdot \delta (t - \tau) + A_2 \cdot \delta (t - 2\tau)$$
 but only if the rake coefficients &nbsp;$h_0$, &nbsp;$h_1$, &nbsp;$τ_0$&nbsp; and &nbsp;$τ_1$&nbsp; are appropriately chosen.&nbsp; The main part of &nbsp;$h_{\rm KR}(t)$&nbsp; is supposed to be at &nbsp;$t = τ$.&nbsp;
@@ Line 16: / Line 16: @@
 The constant &nbsp;$K$&nbsp; is to be chosen so that the amplitude of the main path &nbsp;$A_1 = 1$&nbsp;:
 :$$K= \frac{1}{h_0^2 + h_1^2}.$$
-Apart from the rake parameters, the signals &nbsp;$r(t)$&nbsp; and &nbsp;$b(t)$ are sought when&nbsp;$s(t)$&nbsp; is a rectangle of height &nbsp;$s_0 = 1$&nbsp; and width &nbsp;$T = \ \rm 5 &micro; s$.&nbsp;
+Apart from the rake parameters,&nbsp; the signals &nbsp;$r(t)$&nbsp; and &nbsp;$b(t)$ are sought when&nbsp;$s(t)$&nbsp; is a rectangle of height &nbsp;$s_0 = 1$&nbsp; and width &nbsp;$T = \ \rm 5 &micro; s$.&nbsp;
@@ Line 36: / Line 36: @@
 {Which statements are valid for the channel impulse response &nbsp;$h_{\rm K}(t)$?
 |type="[]"}
-+ $h_{\rm K}(t)$&nbsp; consists of two Dirac functions.
++ $h_{\rm K}(t)$&nbsp; consists of two Dirac delta functions.
 - $h_{\rm K}(t)$&nbsp; is complex-valued.
 - $h_{\rm K}(t)$&nbsp; is a function periodic with delay time &nbsp;$\tau$.&nbsp;
@@ Line 42: / Line 42: @@
 {Which statements are true for the channel frequency response &nbsp;$H_{\rm K}(f)$?
 |type="[]"}
-- &nbsp;$H_{\rm K}(f = 0) = 2$ is true.
+- $H_{\rm K}(f = 0) = 2$&nbsp; is true.
 + $H_{\rm K}(f)$&nbsp; is complex-valued.
 + $|H_{\rm K}(f)|$&nbsp; is a function periodic with frequency &nbsp;$1/τ$.&nbsp;

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Modulationsverfahren/BER der PN–Modulation
+{{quiz-Header|Buchseite=Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation
 }}
-[[File:P_ID1888__Mod_Z_5_5.png|right|frame|Zweiwegekanal <br>& Rake]]
+[[File:P_ID1888__Mod_Z_5_5.png|right|frame|Two-way channel <br>& rake '''KORREKTUR''': two-way channel, rake receiver]]
-Die Grafik zeigt einen Zweiwegekanal&nbsp; (gelbe Hinterlegung).&nbsp; Die entsprechende Beschreibungsgleichung lautet:
+The diagram shows a two-way channel&nbsp; (yellow background).&nbsp; The corresponding descriptive equation is:
 :$$ r(t) =0.6 \cdot s(t) + 0.4 \cdot s (t - \tau) \hspace{0.05cm}.$$
-Die Verzögerung auf dem Nebenpfad sei &nbsp;$τ = 1 \ \rm &micro; s$.&nbsp; Darunter gezeichnet ist die Struktur eines Rake–Empfängers&nbsp; (grüne Hinterlegung)&nbsp; mit den allgemeinen Koeffizienten &nbsp;$K$, &nbsp;$h_0$, &nbsp;$h_1$, &nbsp;$τ_0$&nbsp; und &nbsp;$τ_1$.
+Let the delay on the secondary path be &nbsp;$τ = 1 \ \rm &micro; s$.&nbsp; Drawn below is the structure of a rake receiver&nbsp; (green background)&nbsp; with general coefficients &nbsp;$K$, &nbsp;$h_0$, &nbsp;$h_1$, &nbsp;$τ_0$&nbsp; and &nbsp;$τ_1$.
-Der Rake–Empfänger hat die Aufgabe, die Energie der beiden Signalpfade zu bündeln und dadurch die Entscheidung sicherer zu machen.&nbsp;
+The purpose of the rake receiver is to combine the energy of the two signal paths, making the decision more reliable.&nbsp;
-Die gemeinsame Impulsantwort von Kanal und Rake–Empfänger kann in der Form
+The combined impulse response of the channel and the rake receiver can be expressed in the form
 :$$h_{\rm KR}(t) = A_0 \cdot \delta (t ) + A_1 \cdot \delta (t - \tau) + A_2 \cdot \delta (t - 2\tau)$$
-angegeben werden, allerdings nur dann, wenn die Rake–Koeffizienten &nbsp;$h_0$, &nbsp;$h_1$, &nbsp;$τ_0$&nbsp; und &nbsp;$τ_1$&nbsp; geeignet gewählt werden.&nbsp; Der Hauptanteil von &nbsp;$h_{\rm KR}(t)$&nbsp; soll bei &nbsp;$t = τ$&nbsp; liegen.
+but only if the rake coefficients &nbsp;$h_0$, &nbsp;$h_1$, &nbsp;$τ_0$&nbsp; and &nbsp;$τ_1$&nbsp; are appropriately chosen.&nbsp; The main part of &nbsp;$h_{\rm KR}(t)$&nbsp; is supposed to be at &nbsp;$t = τ$.&nbsp;
-Die Konstante &nbsp;$K$&nbsp; ist so zu wählen, dass die Amplitude des Hauptpfads &nbsp;$A_1 = 1$&nbsp; ist:
+The constant &nbsp;$K$&nbsp; is to be chosen so that the amplitude of the main path &nbsp;$A_1 = 1$&nbsp;:
 :$$K= \frac{1}{h_0^2 + h_1^2}.$$
-Gesucht sind außer den Rake–Parametern die Signale &nbsp;$r(t)$&nbsp; und &nbsp;$b(t)$, wenn &nbsp;$s(t)$&nbsp; ein Rechteck der Höhe &nbsp;$s_0 = 1$&nbsp; und der Breite &nbsp;$T = \ \rm 5 &micro; s$&nbsp; ist.
+Apart from the rake parameters, the signals &nbsp;$r(t)$&nbsp; and &nbsp;$b(t)$ are sought when&nbsp;$s(t)$&nbsp; is a rectangle of height &nbsp;$s_0 = 1$&nbsp; and width &nbsp;$T = \ \rm 5 &micro; s$.&nbsp;
@@ Line 25: / Line 25: @@
-''Hinweise:''
+''Notes:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Modulation_Methods/Fehlerwahrscheinlichkeit_der_PN%E2%80%93Modulation|Fehlerwahrscheinlichkeit der PN-Modulation]].
+*The exercise belongs to the chapter&nbsp; [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation|Error Probability of Direct-Sequence Spread Spectrum Modulation]].
-*Bezug genommen wird insbesondere auf den Abschnitt&nbsp; [[Modulation_Methods/Fehlerwahrscheinlichkeit_der_PN–Modulation#Prinzip_des_Rake.E2.80.93Empf.C3.A4ngers |Prinzip des Rake-Empfängers]].
+*Reference is made in particular to the section&nbsp; [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation#Principle_of_the_rake_receiver |Principle of the rake receiver]].
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Welche Aussagen gelten für die Kanalimpulsantwort &nbsp;$h_{\rm K}(t)$?
+{Which statements are valid for the channel impulse response &nbsp;$h_{\rm K}(t)$?
 |type="[]"}
-+ $h_{\rm K}(t)$&nbsp; besteht aus zwei Diracfunktionen.
++ $h_{\rm K}(t)$&nbsp; consists of two Dirac functions.
-- $h_{\rm K}(t)$&nbsp; ist komplexwertig.
+- $h_{\rm K}(t)$&nbsp; is complex-valued.
-- $h_{\rm K}(t)$&nbsp; ist eine mit der Verzögerungszeit &nbsp;$\tau$&nbsp; periodische Funktion.
+- $h_{\rm K}(t)$&nbsp; is a function periodic with delay time &nbsp;$\tau$.&nbsp;
-{Welche Aussagen gelten für den Kanalfrequenzgang &nbsp;$H_{\rm K}(f)$?
+{Which statements are true for the channel frequency response &nbsp;$H_{\rm K}(f)$?
 |type="[]"}
-- Es gilt &nbsp;$H_{\rm K}(f = 0) = 2$.
+- &nbsp;$H_{\rm K}(f = 0) = 2$ is true.
-+ $H_{\rm K}(f)$&nbsp; ist komplexwertig.
++ $H_{\rm K}(f)$&nbsp; is complex-valued.
-+ $|H_{\rm K}(f)|$&nbsp; ist eine mit der Frequenz &nbsp;$1/τ$&nbsp; periodische Funktion.
++ $|H_{\rm K}(f)|$&nbsp; is a function periodic with frequency &nbsp;$1/τ$.&nbsp;
-{Setzen Sie &nbsp;$K = 1$, &nbsp;$h_0 = 0.6$&nbsp; und &nbsp;$h_1 = 0.4$.&nbsp; Bestimmen Sie die Verzögerungen &nbsp;$τ_0$&nbsp; und &nbsp;$τ_1$, damit die &nbsp;$h_{\rm KR}(t)$–Gleichung mit &nbsp;$A_0 = A_2$&nbsp; erfüllt wird.
+{Set &nbsp;$K = 1$, &nbsp;$h_0 = 0.6$&nbsp; and &nbsp;$h_1 = 0.4$.&nbsp; Determine the delays &nbsp;$τ_0$&nbsp; and &nbsp;$τ_1$ so that the &nbsp;$h_{\rm KR}(t)$ equation is satisfied with &nbsp;$A_0 = A_2$.&nbsp;
 |type="{}"}
 $τ_0 \ = \ $  { 1 3% } $\ \rm &micro; s$
 $τ_1 \ = \ $ { 0. } $\ \rm &micro; s$
-{Welcher Wert ist für die Konstante &nbsp;$K$&nbsp; zu wählen?
+{What value should be chosen for the constant &nbsp;$K$?&nbsp;
 |type="{}"}
 $K  \ = \ $ { 1.923 3% }
-{Welche Aussagen gelten für die Signale &nbsp;$r(t)$&nbsp; und &nbsp;$b(t)$?
+{Which statements are valid for the signals &nbsp;$r(t)$&nbsp; and &nbsp;$b(t)$?
 |type="[]"}
-+ Der Maximalwert von &nbsp;$r(t)$&nbsp; ist &nbsp;$1$.
++ The maximum value of &nbsp;$r(t)$&nbsp; is &nbsp;$1$.
-- Die Breite von &nbsp;$r(t)$&nbsp; ist &nbsp;$7 \ &micro; s$.
+- The width of &nbsp;$r(t)$&nbsp; is &nbsp;$7 \ &micro; s$.
-- Der Maximalwert von &nbsp;$b(t)$&nbsp; ist &nbsp;$1$.
+- The maximum value of &nbsp;$b(t)$&nbsp; is &nbsp;$1$.
-+ Die Breite von &nbsp;$b(t)$&nbsp; ist &nbsp;$7 \ &micro; s$.
++ The width of &nbsp;$b(t)$&nbsp; is &nbsp;$7 \ &micro; s$.
@@ Line 67: / Line 67: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 1</u>:
+'''(1)'''&nbsp; <u>Solution 1</u> is correct:
-*Die Impulsantwort&nbsp; $h_{\rm K}(t)$&nbsp; ergibt sich als das Empfangssignal&nbsp; $r(t)$, wenn am Eingang ein Diracimpuls anliegt &nbsp; ⇒ &nbsp;  $s(t) = δ(t)$.&nbsp; Daraus folgt:
+*The impulse response&nbsp; $h_{\rm K}(t)$&nbsp; is obtained as the received signal&nbsp; $r(t)$ when there is a dirac pulse at the input &nbsp; ⇒ &nbsp;  $s(t) = δ(t)$.&nbsp; It follows that:
 :$$ h_(t) = 0.6 \cdot \delta (t ) + 0.4 \cdot \delta (t - \tau) \hspace{0.05cm}.$$
-'''(2)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 2 und 3</u>:
+'''(2)'''&nbsp; <u>Solutions 2 and 3</u> are correct:
-*Der Kanalfrequenzgang&nbsp; $H_{\rm K}(f)$&nbsp; ist definitionsgemäß die Fouriertransformierte der Impulsantwort&nbsp; $h_{\rm K}(t)$.&nbsp; Mit dem Verschiebungssatz ergibt sich hierfür:
+*By definition, the channel frequency response&nbsp; $H_{\rm K}(f)$&nbsp; is the Fourier transform of the impulse response&nbsp; $h_{\rm K}(t)$.&nbsp; With the shift theorem this results in:
 :$$H_{\rm K}(f) = 0.6 + 0.4 \cdot {\rm e}^{ \hspace{0.03cm}{\rm j} \hspace{0.03cm} \cdot \hspace{0.03cm}2 \pi f \tau}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H_{\rm K}(f= 0) = 0.6 + 0.4 = 1 \hspace{0.05cm}.$$
-*Der erste Lösungsvorschlag ist dementsprechend falsch im Gegensatz zu den beiden anderen: &nbsp; $H_{\rm K}(f)$ ist komplexwertig und der Betrag ist periodisch mit&nbsp; $1/τ$, wie die nachfolgende Rechnung zeigt:
+*Accordingly, the first proposed solution is incorrect in contrast to the other two: &nbsp; $H_{\rm K}(f)$ is complex-valued and the magnitude is periodic with&nbsp; $1/τ$, as the following calculation shows:
 :$$|H_{\rm K}(f)|^2  =  \left [0.6 + 0.4 \cdot \cos(2 \pi f \tau) \right ]^2 + \left [ 0.4 \cdot \sin(2 \pi f \tau) \right ]^2 = \left [0.6^2 + 0.4^2 \cdot \left ( \cos^2(2 \pi f \tau) + \sin^2(2 \pi f \tau)\right ) \right ] +  2 \cdot 0.6 \cdot 0.4 \cdot \cos(2 \pi f \tau)$$
-*Für&nbsp; $f = 0$&nbsp; ist&nbsp; $|H_{\rm K}(f)| = 1$.&nbsp; Im jeweiligen Frequenzabstand&nbsp; $1/τ$&nbsp; wiederholt sich dieser Wert.
+*For&nbsp; $f = 0$,&nbsp;&nbsp; $|H_{\rm K}(f)| = 1$.&nbsp; This value is repeated in the respective frequency spacing&nbsp; $1/τ$.&nbsp;