Aufgaben:Exercise 4.9: Costas Rule Loop - Revision history

Guenter at 16:51, 15 April 2022

2022-04-15T16:51:10Z

Guenter at 16:46, 15 April 2022

2022-04-15T16:46:47Z

Hwang at 17:19, 21 March 2022

2022-03-21T17:19:29Z

Guenter at 14:03, 16 March 2022

2022-03-16T14:03:34Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.9: Costas Rule Loop to Aufgaben:Exercise 4.9: Costas Rule Loop

2022-03-16T14:01:09Z

Guenter moved page Aufgabe 4.9: Costas Rule Loop to Exercise 4.9: Costas Rule Loop

Guenter: Guenter moved page Aufgaben:Aufgabe 4.9: Costas–Regelschleife to Aufgaben:Aufgabe 4.9: Costas Rule Loop

2022-03-16T14:00:01Z

Guenter moved page Aufgabe 4.9: Costas–Regelschleife to Aufgabe 4.9: Costas Rule Loop

Guenter at 16:27, 9 November 2021

2021-11-09T16:27:55Z

Javier: Text replacement - "Category:Aufgaben zu Modulationsverfahren" to "Category:Modulation Methods: Exercises"

2021-03-23T12:49:26Z

Text replacement - "Category:Aufgaben zu Modulationsverfahren" to "Category:Modulation Methods: Exercises"

Javier: Text replacement - "[[Modulationsverfahren" to "[[Modulation_Methods"

2020-07-09T12:40:44Z

Text replacement - "[[Modulationsverfahren" to "[[Modulation_Methods"

Guenter at 11:42, 20 April 2020

2020-04-20T11:42:48Z

@@ Line 57: / Line 57: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The <u>second solution</u> is correct:
+'''(1)'''&nbsp; The&nbsp; <u>second solution</u>&nbsp; is correct:
-*Using the addition theorem of trigonometry, we obtain:
+*Using the addition theorem of trigonometry,&nbsp; we obtain:
 :$$ m_1(t)  =  \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi) \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \theta) =  \pm \frac{s_0}{2} \cdot \left [ \cos ( \phi - \theta) + \cos (4 \pi \cdot f_{\rm T} \cdot t + \phi +\theta)\right]\hspace{0.05cm}.$$
-*After the low-pass, only the DC component&nbsp; $y_1(t) = ± s_0/2 · \cos (\phi - θ)$ remains.
+*After the low-pass,&nbsp; only the DC component&nbsp; $y_1(t) = ± s_0/2 · \cos (\phi - θ)$&nbsp; remains.
-'''(2)'''&nbsp; Here the <u>last solution</u> is correct:
+'''(2)'''&nbsp; Here the&nbsp; <u>last solution</u>&nbsp; is correct:
 *Analogous to question&nbsp; '''(1)''',&nbsp; the result for the input signal of the lower low-pass filter is:
 :$$ m_2(t)  =  \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi) \cdot \left [-\sin (2 \pi \cdot f_{\rm T} \cdot t + \theta) \right]= \pm \frac{s_0}{2} \cdot \left [ \sin ( \phi - \theta) + \sin (4 \pi \cdot f_{\rm T} \cdot t + \phi +\theta)\right].$$
@@ Line 72: / Line 72: @@
-'''(3)'''&nbsp; <u>Solutions 2 and 3</u> are correct:
+'''(3)'''&nbsp; <u>Solutions 2 and 3</u>&nbsp; are correct:
 *By multiplying&nbsp; $y_1(t)$&nbsp; and&nbsp; $y_2(t)$&nbsp; we obtain:
 :$$x(t)  =  y_1(t) \cdot y_2(t)= \frac{s_0^2}{4} \cdot \cos ( \phi - \theta) \cdot \sin ( \phi - \theta)
@@ Line 78: / Line 78: @@
 *Using the small angle approximation&nbsp; $\sin(α) ≈ α$&nbsp; it follows:
 :$$x(t) \approx \frac{s_0^2}{4} \cdot ( \phi - \theta) \hspace{0.05cm}.$$
-*The rule signal&nbsp; $x(t)$&nbsp; is thus proportional to the phase error&nbsp; $\phi - θ$, which is controlled to zero by the Costas rule loop.&nbsp;
+*The rule signal&nbsp; $x(t)$&nbsp; is thus proportional to the phase error&nbsp; $\phi - θ$,&nbsp; which is controlled to zero by the Costas rule loop.&nbsp;
-*Thus, in the steady state, the oscillator signal&nbsp; $z(t)$&nbsp; immediately follows the received signal&nbsp; $r(t)$.
+*Thus,&nbsp; in the steady state,&nbsp; the oscillator signal&nbsp; $z(t)$&nbsp; immediately follows the received signal&nbsp; $r(t)$.
 *To achieve the required initial condition&nbsp; $θ ≈ \phi$,&nbsp; a training sequence is usually transmitted first and the phase is initialized accordingly.
-*This is also because the phase is only controlled modulo&nbsp; $π$,&nbsp; so that, for example,&nbsp; $\phi - θ = π$&nbsp; would incorrectly lead to the rule signal&nbsp; $x(t) = 0$.&nbsp;
+*This is also because the phase is only controlled modulo&nbsp; $π$,&nbsp; so that,&nbsp; for example,&nbsp; $\phi - θ = π$&nbsp; would incorrectly lead to the rule signal&nbsp; $x(t) = 0$.&nbsp;
 {{ML-Fuß}}

@@ Line 4: / Line 4: @@
 [[File:EN_Mod_A_4_8.png|right|frame|Costas rule loop]]
-An important prerequisite for coherent demodulation is in-phase carrier recovery.&nbsp; One possibility for this is the so-called&nbsp; ''Costas rule loop'', which is shown in simplified form by the adjacent block diagram.
+An important prerequisite for coherent demodulation is&nbsp; "in-phase carrier recovery".&nbsp; One possibility for this is the so-called&nbsp; "Costas rule loop",&nbsp; which is shown in simplified form by the adjacent block diagram.
-In binary phase modulation&nbsp; (BPSK), the received signal can be expressed as
+In binary phase modulation&nbsp; $\rm (BPSK)$,&nbsp; the received signal can be expressed as
 :$$ r(t) = \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi)$$
-The phase rotation &nbsp;$ϕ$&nbsp; on the transmission channel is always assumed to be unknown.&nbsp; The specification&nbsp; "±"&nbsp; describes the phase jumps of the BPSK signal.
+The phase rotation &nbsp;$ϕ$&nbsp; on the transmission channel is always assumed to be unknown.&nbsp; The factor&nbsp; "±"&nbsp; describes the phase jumps of the BPSK signal.
 The task of the circuit indicated by the diagram is to generate a carrier signal
 :$$z(t) = \cos (2 \pi \cdot f_{\rm T} \cdot t + \theta)$$
-where the phase error &nbsp;$\phi - θ$&nbsp; between the BPSK received signal &nbsp;$r(t)$&nbsp; and the oscillation &nbsp;$z(t)$&nbsp; generated at the receiver must be compensated.
+where the phase error &nbsp; $\phi - θ$ &nbsp; between the BPSK received signal &nbsp;$r(t)$&nbsp; and the oscillation &nbsp;$z(t)$&nbsp; generated at the receiver must be compensated.
-*For this purpose, a ''Voltage Controlled Oscillator''&nbsp; $($'''VCO'''$)$&nbsp; is used to generate an oscillation of frequency &nbsp;$f_{\rm T}$,&nbsp; initially with arbitrary phase &nbsp;$θ$.
+*For this purpose, a&nbsp; "Voltage Controlled Oscillator"&nbsp; $($'''VCO'''$)$&nbsp; is used to generate an oscillation of frequency &nbsp;$f_{\rm T}$,&nbsp; initially with arbitrary phase &nbsp;$θ$.
-*However, the Costas rule loop iteratively achieves the desired result &nbsp;$θ = \phi$.&nbsp;
+*However,&nbsp; the Costas rule loop iteratively achieves the desired result &nbsp;$θ = \phi$.&nbsp;
 Notes:
+*The exercise belongs to the chapter&nbsp; [[Modulation_Methods/Linear_Digital_Modulation|"Linear Digital Modulation"]].
+*In the diagram,&nbsp; "TP"&nbsp; denotes low-passes&nbsp; (German:&nbsp; "Tiefpass" &nbsp; &rArr; &nbsp; subscript:&nbsp; "TP"),&nbsp; which are assumed to be ideal.
+*The square labeled &nbsp;$π/2$&nbsp; denotes a phase rotation by &nbsp;$π/2 \ (90^\circ)$,&nbsp; so that,&nbsp; for example,&nbsp; a cosine signal becomes a&nbsp; "minus-sine" signal:
 :$$\cos (\omega_{\rm 0} \cdot t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\cos (\omega_{\rm 0} \cdot t + 90^\circ) = -\sin (\omega_{\rm 0} \cdot t)\hspace{0.05cm}.$$
-*Further, the following trigonometric relations hold:
+*Further,&nbsp; the following trigonometric relations hold:
 :$$\cos (\alpha) \cdot \cos (\beta)  =  {1} /{2} \cdot \big [ \cos (\alpha - \beta) + \cos (\alpha + \beta)\big]\hspace{0.05cm},$$
 :$$\sin (\alpha) \cdot \cos (\beta)  =   {1} /{2} \cdot \big [ \sin (\alpha - \beta) + \sin (\alpha + \beta)\big]\hspace{0.05cm}.$$

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Modulationsverfahren/Lineare digitale Modulation
+{{quiz-Header|Buchseite=Modulation_Methods/Linear_Digital_Modulation
 }}
-[[File:EN_Mod_A_4_8.png|right|frame|Costas&ndash;Regelschleife]]
+[[File:EN_Mod_A_4_8.png|right|frame|Costas rule loop]]
-Eine wichtige Voraussetzung für kohärente Demodulation ist die phasenrichtige Trägerrückgewinnung.&nbsp; Eine Möglichkeit hierfür bietet die so genannte&nbsp; ''Costas–Regelschleife'', die vereinfacht durch das nebenstehende Blockschaltbild dargestellt ist.
+An important prerequisite for coherent demodulation is in-phase carrier recovery.&nbsp; One possibility for this is the so-called&nbsp; ''Costas rule loop'', which is shown in simplified form by the adjacent block diagram.
-Das Empfangssignal kann bei der binären Phasenmodulation&nbsp; (BPSK) als
+In binary phase modulation&nbsp; (BPSK), the received signal can be expressed as
 :$$ r(t) = \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi)$$
-geschrieben werden.&nbsp; Die Phasendrehung &nbsp;$ϕ$&nbsp; auf dem Übertragungskanal wird dabei stets als unbekannt angenommen.&nbsp; Die Angabe&nbsp; „±”&nbsp; beschreibt die Phasensprünge des BPSK–Signals.
+The phase rotation &nbsp;$ϕ$&nbsp; on the transmission channel is always assumed to be unknown.&nbsp; The specification&nbsp; "±"&nbsp; describes the phase jumps of the BPSK signal.
-Aufgabe der durch die Grafik angegebenen Schaltung ist es, ein Trägersignal
+The task of the circuit indicated by the diagram is to generate a carrier signal
 :$$z(t) = \cos (2 \pi \cdot f_{\rm T} \cdot t + \theta)$$
-zu generieren, wobei der Phasenfehler &nbsp;$\phi - θ$&nbsp; zwischen dem BPSK–Empfangssignal &nbsp;$r(t)$&nbsp; und der am Empfänger generierten Schwingung &nbsp;$z(t)$&nbsp; ausgeregelt werden muss.
+where the phase error &nbsp;$\phi - θ$&nbsp; between the BPSK received signal &nbsp;$r(t)$&nbsp; and the oscillation &nbsp;$z(t)$&nbsp; generated at the receiver must be compensated.
-*Hierzu wird mit einem regelbaren Oszillator&nbsp; $($'''VCO''',&nbsp; ''Voltage Controlled Oscillator''$)$&nbsp; eine Schwingung der Frequenz &nbsp;$f_{\rm T}$&nbsp; erzeugt, zunächst mit beliebiger Phase &nbsp;$θ$.
+*For this purpose, a ''Voltage Controlled Oscillator''&nbsp; $($'''VCO'''$)$&nbsp; is used to generate an oscillation of frequency &nbsp;$f_{\rm T}$,&nbsp; initially with arbitrary phase &nbsp;$θ$.
-*Durch die Costas–Regelschleife wird jedoch iterativ das Wunschergebnis &nbsp;$θ = \phi$&nbsp; erreicht.
+*However, the Costas rule loop iteratively achieves the desired result &nbsp;$θ = \phi$.&nbsp;
@@ Line 23: / Line 23: @@
-''Hinweise:''
+''Notes:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Modulation_Methods/Lineare_digitale_Modulation|Lineare digitale Modulation]].
+*The exercise belongs to the chapter&nbsp; [[Modulation_Methods/Linear_Digital_Modulation|Linear Digital Modulation]].
-*In der Grafik  bezeichnet „TP” Tiefpässe, die als ideal angenommen werden.
+*In the diagram, "TP" denotes low-passes, which are assumed to be ideal.
-*Das mit &nbsp;$π/2$&nbsp; beschriftete Quadrat kennzeichnet eine Phasendrehung um &nbsp;$π/2 \ (90^\circ)$, so dass beispielsweise aus einem Cosinus–Signal ein Minus–Sinus–Signal wird:
+*The square labeled &nbsp;$π/2$&nbsp; denotes a phase rotation by &nbsp;$π/2 \ (90^\circ)$, so that, for example, a cosine signal becomes a minus-sine signal:
 :$$\cos (\omega_{\rm 0} \cdot t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\cos (\omega_{\rm 0} \cdot t + 90^\circ) = -\sin (\omega_{\rm 0} \cdot t)\hspace{0.05cm}.$$
-*Weiter gelten folgende trigonometrischen Beziehungen:
+*Further, the following trigonometric relations hold:
 :$$\cos (\alpha) \cdot \cos (\beta)  =  {1} /{2} \cdot \big [ \cos (\alpha - \beta) + \cos (\alpha + \beta)\big]\hspace{0.05cm},$$
 :$$\sin (\alpha) \cdot \cos (\beta)  =   {1} /{2} \cdot \big [ \sin (\alpha - \beta) + \sin (\alpha + \beta)\big]\hspace{0.05cm}.$$
@@ Line 35: / Line 35: @@
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Berechnen Sie das Signal &nbsp;$y_1(t)$&nbsp; nach dem Tiefpass im oberen Zweig.&nbsp; Welche der folgenden Aussagen ist richtig?
+{Calculate the signal &nbsp;$y_1(t)$&nbsp; after the low-pass in the upper branch.&nbsp; Which of the following statements is correct?
 |type="()"}
 - $y_1(t) = ± s_0/2 · \big[\cos (\phi - θ) + \cos (4 π · f_{\rm T} · t +\phi + θ)\big],$
@@ Line 44: / Line 44: @@
 - $y_1(t) = ± s_0/2 · \sin (\phi - θ).$
-{Berechnen Sie das Signal &nbsp;$y_2(t)$&nbsp; nach dem Tiefpass im unteren Zweig.&nbsp; Welche der folgenden Aussagen ist richtig?
+{Calculate the signal &nbsp;$y_2(t)$&nbsp; after the low-pass in the lower branch.&nbsp; Which of the following statements is correct?
 |type="()"}
 - $y_2(t) = ± s_0/2 ·  \big[\cos (\phi - θ) + \cos (4 π · f_{\rm T} · t +\phi + θ)\big],$
@@ Line 50: / Line 50: @@
 + $y_2(t) = ± s_0/2 · \sin (\phi - θ).$
-{Berechnen Sie das Regelsignal &nbsp;$x(t)$&nbsp; und geben Sie eine Näherung für kleine Phasenabweichung &nbsp;$\phi - θ$&nbsp; an.&nbsp; Welche Gleichungen sind richtig?
+{Calculate the rule signal &nbsp;$x(t)$&nbsp; and give an approximation for small phase deviation &nbsp;$\phi - θ$.&nbsp; Which equations are correct?
 |type="[]"}
 - $x(t) = s_0^2/8 · \cos(\phi + θ)$,
@@ Line 60: / Line 60: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Richtig ist der <u>zweite Lösungsvorschlag</u>:
+'''(1)'''&nbsp; The <u>second solution</u> is correct:
-*Mit dem Additionstheorem der Trigonometrie erhält man:
+*Using the addition theorem of trigonometry, we obtain:
 :$$ m_1(t)  =  \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi) \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \theta) =  \pm \frac{s_0}{2} \cdot \left [ \cos ( \phi - \theta) + \cos (4 \pi \cdot f_{\rm T} \cdot t + \phi +\theta)\right]\hspace{0.05cm}.$$
-*Nach dem Tiefpass verbleibt nur der Gleichanteil&nbsp; $y_1(t) = ± s_0/2 · \cos (\phi - θ).$
+*After the low-pass, only the DC component&nbsp; $y_1(t) = ± s_0/2 · \cos (\phi - θ)$ remains.
-'''(2)'''&nbsp; Richtig ist hier der <u>letzte Lösungsvorschlag</u>:
+'''(2)'''&nbsp; Here the <u>last solution</u> is correct:
-*Analog zu Teilaufgabe&nbsp; '''(1)'''&nbsp; ergibt sich für das Eingangssignal des unteren Tiefpasses:
+*Analogous to question&nbsp; '''(1)''',&nbsp; the result for the input signal of the lower low-pass filter is:
 :$$ m_2(t)  =  \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi) \cdot \left [-\sin (2 \pi \cdot f_{\rm T} \cdot t + \theta) \right]= \pm \frac{s_0}{2} \cdot \left [ \sin ( \phi - \theta) + \sin (4 \pi \cdot f_{\rm T} \cdot t + \phi +\theta)\right].$$
-*Dies führt zu folgendem  Ausgangssignal:
+*This leads to the following output signal:
 :$$ y_2(t) = \pm {s_0}/{2} \cdot\sin ( \phi - \theta) \hspace{0.05cm}.$$
-'''(3)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 2 und 3</u>:
+'''(3)'''&nbsp; <u>Solutions 2 and 3</u> are correct:
-*Durch Multiplikation von&nbsp; $y_1(t)$&nbsp; und&nbsp; $y_2(t)$&nbsp; erhält man:
+*By multiplying&nbsp; $y_1(t)$&nbsp; and&nbsp; $y_2(t)$&nbsp; we obtain:
 :$$x(t)  =  y_1(t) \cdot y_2(t)= \frac{s_0^2}{4} \cdot \cos ( \phi - \theta) \cdot \sin ( \phi - \theta)
   =  \frac{s_0^2}{8} \cdot \sin ( 2\cdot\phi - 2\cdot\theta) \hspace{0.05cm}.$$
-*Mit der Kleinwinkelnäherung&nbsp; $\sin(α) ≈ α$&nbsp; folgt daraus:
+*Using the small angle approximation&nbsp; $\sin(α) ≈ α$&nbsp; it follows:
 :$$x(t) \approx \frac{s_0^2}{4} \cdot ( \phi - \theta) \hspace{0.05cm}.$$
-*Das Regelsignal&nbsp; $x(t)$&nbsp; ist also proportional zum Phasenfehler&nbsp; $\phi - θ$, der mit der Costas–Regelschleife zu Null geregelt wird.&nbsp;
+*The rule signal&nbsp; $x(t)$&nbsp; is thus proportional to the phase error&nbsp; $\phi - θ$, which is controlled to zero by the Costas rule loop.&nbsp;
-*Im eingeschwungenen Zustand folgt somit das Oszillatorsignal&nbsp; $z(t)$&nbsp; unmittelbar dem Empfangssignal&nbsp; $r(t)$.
+*Thus, in the steady state, the oscillator signal&nbsp; $z(t)$&nbsp; immediately follows the received signal&nbsp; $r(t)$.
-*Um die erforderliche Startbedingung&nbsp; $θ ≈ \phi$&nbsp; zu erreichen, wird meist zunächst eine Trainigssequenz übertragen und die Phase entsprechend initialisiert.
+*To achieve the required initial condition&nbsp; $θ ≈ \phi$,&nbsp; a training sequence is usually transmitted first and the phase is initialized accordingly.
-*Dies auch, weil die Phase nur modulo&nbsp; $π$&nbsp; ausgeregelt wird, so dass beispielsweise&nbsp; $\phi - θ = π$&nbsp; fälschlicherweise zum Regelsignal&nbsp; $x(t) = 0$&nbsp; führen würde.
+*This is also because the phase is only controlled modulo&nbsp; $π$,&nbsp; so that, for example,&nbsp; $\phi - θ = π$&nbsp; would incorrectly lead to the rule signal&nbsp; $x(t) = 0$.&nbsp;
 {{ML-Fuß}}

← Older revision		Revision as of 14:03, 16 March 2022
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−	[[Category:Modulation Methods: Exercises\|^4.2 ~~Lineare digitale~~ Modulation^]]	+	[[Category:Modulation Methods: Exercises\|^4.2 Linear Digital Modulation^]]

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 }}
-[[File:Mod_A_4_8_vers2.png|right|frame|Costas&ndash;Regelschleife]]
+[[File:EN_Mod_A_4_8.png|right|frame|Costas&ndash;Regelschleife]]
 Eine wichtige Voraussetzung für kohärente Demodulation ist die phasenrichtige Trägerrückgewinnung.&nbsp; Eine Möglichkeit hierfür bietet die so genannte&nbsp; ''Costas–Regelschleife'', die vereinfacht durch das nebenstehende Blockschaltbild dargestellt ist.

← Older revision	Revision as of 14:01, 16 March 2022
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← Older revision	Revision as of 14:00, 16 March 2022
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← Older revision		Revision as of 12:49, 23 March 2021
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−	[[Category:~~Aufgaben zu Modulationsverfahren~~\|^4.2 Lineare digitale Modulation^]]	+	[[Category:Modulation Methods: Exercises\|^4.2 Lineare digitale Modulation^]]