Difference between revisions of "Aufgaben:Exercise 4.9: Costas Rule Loop"

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{{quiz-Header|Buchseite=Modulationsverfahren/Lineare digitale Modulation
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[[File:Mod_A_4_8_vers2.png|right|frame|Costas–Regelschleife]]
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[[File:EN_Mod_A_4_8.png|right|frame|Costas rule loop]]
Eine wichtige Voraussetzung für kohärente Demodulation ist die phasenrichtige Trägerrückgewinnung.  Eine Möglichkeit hierfür bietet die so genannte  ''Costas–Regelschleife'', die vereinfacht durch das nebenstehende Blockschaltbild dargestellt ist.
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An important prerequisite for coherent demodulation is  "in-phase carrier recovery".  One possibility for this is the so-called  "Costas rule loop",  which is shown in simplified form by the adjacent block diagram.
  
Das Empfangssignal kann bei der binären Phasenmodulation  (BPSK) als
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In binary phase modulation  $\rm (BPSK)$,  the received signal can be expressed as
 
:$$ r(t) = \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi)$$
 
:$$ r(t) = \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi)$$
geschrieben werden.  Die Phasendrehung  $ϕ$  auf dem Übertragungskanal wird dabei stets als unbekannt angenommen.  Die Angabe  „±”  beschreibt die Phasensprünge des BPSK–Signals.
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The phase rotation  $ϕ$  on the transmission channel is always assumed to be unknown.  The factor  "±"  describes the phase jumps of the BPSK signal.
  
Aufgabe der durch die Grafik angegebenen Schaltung ist es, ein Trägersignal
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The task of the circuit indicated by the diagram is to generate a carrier signal
 
:$$z(t) = \cos (2 \pi \cdot f_{\rm T} \cdot t + \theta)$$
 
:$$z(t) = \cos (2 \pi \cdot f_{\rm T} \cdot t + \theta)$$
zu generieren, wobei der Phasenfehler  $\phi - θ$  zwischen dem BPSK–Empfangssignal  $r(t)$  und der am Empfänger generierten Schwingung  $z(t)$  ausgeregelt werden muss.  
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where the phase error   $\phi - θ$   between the BPSK received signal  $r(t)$  and the oscillation  $z(t)$  generated at the receiver must be compensated.
*Hierzu wird mit einem regelbaren Oszillator  $($'''VCO''',  ''Voltage Controlled Oscillator''$)$  eine Schwingung der Frequenz  $f_{\rm T}$  erzeugt, zunächst mit beliebiger Phase  $θ$.  
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*For this purpose, a  "Voltage Controlled Oscillator"  $($'''VCO'''$)$  is used to generate an oscillation of frequency  $f_{\rm T}$,  initially with arbitrary phase  $θ$.  
*Durch die Costas–Regelschleife wird jedoch iterativ das Wunschergebnis  $θ = \phi$  erreicht.
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*However,  the Costas rule loop iteratively achieves the desired result  $θ = \phi$. 
  
  
  
  
 
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Notes:  
 
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*The exercise belongs to the chapter  [[Modulation_Methods/Linear_Digital_Modulation|"Linear Digital Modulation"]].  
 
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*In the diagram,  "TP"  denotes low-passes  (German:  "Tiefpass"   ⇒   subscript:  "TP"),  which are assumed to be ideal.
 
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*The square labeled  $π/2$  denotes a phase rotation by  $π/2 \ (90^\circ)$,  so that,  for example,  a cosine signal becomes a  "minus-sine" signal:
''Hinweise:''
 
*Die Aufgabe gehört zum  Kapitel  [[Modulation_Methods/Lineare_digitale_Modulation|Lineare digitale Modulation]].
 
 
*In der Grafik  bezeichnet „TP” Tiefpässe, die als ideal angenommen werden.  
 
*Das mit  $π/2$  beschriftete Quadrat kennzeichnet eine Phasendrehung um  $π/2 \ (90^\circ)$, so dass beispielsweise aus einem Cosinus–Signal ein Minus–Sinus–Signal wird:
 
 
:$$\cos (\omega_{\rm 0} \cdot t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\cos (\omega_{\rm 0} \cdot t + 90^\circ) = -\sin (\omega_{\rm 0} \cdot t)\hspace{0.05cm}.$$
 
:$$\cos (\omega_{\rm 0} \cdot t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\cos (\omega_{\rm 0} \cdot t + 90^\circ) = -\sin (\omega_{\rm 0} \cdot t)\hspace{0.05cm}.$$
*Weiter gelten folgende trigonometrischen Beziehungen:
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*Further,  the following trigonometric relations hold:
 
:$$\cos (\alpha) \cdot \cos (\beta)  =  {1} /{2} \cdot \big [ \cos (\alpha - \beta) + \cos (\alpha + \beta)\big]\hspace{0.05cm},$$  
 
:$$\cos (\alpha) \cdot \cos (\beta)  =  {1} /{2} \cdot \big [ \cos (\alpha - \beta) + \cos (\alpha + \beta)\big]\hspace{0.05cm},$$  
 
:$$\sin (\alpha) \cdot \cos (\beta)  =  {1} /{2} \cdot \big [ \sin (\alpha - \beta) + \sin (\alpha + \beta)\big]\hspace{0.05cm}.$$
 
:$$\sin (\alpha) \cdot \cos (\beta)  =  {1} /{2} \cdot \big [ \sin (\alpha - \beta) + \sin (\alpha + \beta)\big]\hspace{0.05cm}.$$
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===Fragebogen===
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===Questions===
  
 
<quiz display=simple>
 
<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?
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{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="()"}
 
|type="()"}
 
- $y_1(t) = ± s_0/2 · \big[\cos (\phi - θ) + \cos (4 π · f_{\rm T} · t +\phi + θ)\big],$
 
- $y_1(t) = ± s_0/2 · \big[\cos (\phi - θ) + \cos (4 π · f_{\rm T} · t +\phi + θ)\big],$
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- $y_1(t) = ± s_0/2 · \sin (\phi - θ).$
 
- $y_1(t) = ± s_0/2 · \sin (\phi - θ).$
  
{Berechnen Sie das Signal &nbsp;$y_2(t)$&nbsp; nach dem Tiefpass im unteren Zweig.&nbsp; Welche der folgenden Aussagen ist richtig?
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{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="()"}
 
|type="()"}
 
- $y_2(t) = ± s_0/2 ·  \big[\cos (\phi - θ) + \cos (4 π · f_{\rm T} · t +\phi + θ)\big],$
 
- $y_2(t) = ± s_0/2 ·  \big[\cos (\phi - θ) + \cos (4 π · f_{\rm T} · t +\phi + θ)\big],$
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+ $y_2(t) = ± s_0/2 · \sin (\phi - θ).$
 
+ $y_2(t) = ± s_0/2 · \sin (\phi - θ).$
  
{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?
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{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="[]"}
 
|type="[]"}
 
- $x(t) = s_0^2/8 · \cos(\phi + θ)$,
 
- $x(t) = s_0^2/8 · \cos(\phi + θ)$,
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</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Richtig ist der <u>zweite Lösungsvorschlag</u>:
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'''(1)'''&nbsp; The&nbsp; <u>second solution</u>&nbsp; is correct:
*Mit dem Additionstheorem der Trigonometrie erhält man:
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*Using the addition theorem of trigonometry,&nbsp; we obtain:
 
:$$ m_1(t)  =  \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi) \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \theta) =  \pm \frac{s_0}{2} \cdot \left [ \cos ( \phi - \theta) + \cos (4 \pi \cdot f_{\rm T} \cdot t + \phi +\theta)\right]\hspace{0.05cm}.$$
 
:$$ m_1(t)  =  \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi) \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \theta) =  \pm \frac{s_0}{2} \cdot \left [ \cos ( \phi - \theta) + \cos (4 \pi \cdot f_{\rm T} \cdot t + \phi +\theta)\right]\hspace{0.05cm}.$$
*Nach dem Tiefpass verbleibt nur der Gleichanteil&nbsp; $y_1(t) = ± s_0/2 · \cos (\phi - θ).$
+
*After the low-pass,&nbsp; only the DC component&nbsp; $y_1(t) = ± s_0/2 · \cos (\phi - θ)$&nbsp; remains.  
  
  
  
'''(2)'''&nbsp; Richtig ist hier der <u>letzte Lösungsvorschlag</u>:
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'''(2)'''&nbsp; Here the&nbsp; <u>last solution</u>&nbsp; is correct:
*Analog zu Teilaufgabe&nbsp; '''(1)'''&nbsp; ergibt sich für das Eingangssignal des unteren Tiefpasses:
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*Analogous to question&nbsp; '''(1)''',&nbsp; the result for the input signal of the lower low-pass filter is:
 
:$$ m_2(t)  =  \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi) \cdot \left [-\sin (2 \pi \cdot f_{\rm T} \cdot t + \theta) \right]= \pm \frac{s_0}{2} \cdot \left [ \sin ( \phi - \theta) + \sin (4 \pi \cdot f_{\rm T} \cdot t + \phi +\theta)\right].$$
 
:$$ m_2(t)  =  \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi) \cdot \left [-\sin (2 \pi \cdot f_{\rm T} \cdot t + \theta) \right]= \pm \frac{s_0}{2} \cdot \left [ \sin ( \phi - \theta) + \sin (4 \pi \cdot f_{\rm T} \cdot t + \phi +\theta)\right].$$
*Dies führt zu folgendem  Ausgangssignal:  
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*This leads to the following output signal:
 
:$$ y_2(t) = \pm {s_0}/{2} \cdot\sin ( \phi - \theta) \hspace{0.05cm}.$$
 
:$$ y_2(t) = \pm {s_0}/{2} \cdot\sin ( \phi - \theta) \hspace{0.05cm}.$$
  
  
  
'''(3)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 2 und 3</u>:
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'''(3)'''&nbsp; <u>Solutions 2 and 3</u>&nbsp; are correct:
*Durch Multiplikation von&nbsp; $y_1(t)$&nbsp; und&nbsp; $y_2(t)$&nbsp; erhält man:
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*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)  
 
:$$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}.$$
 
  =  \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:
 
:$$x(t) \approx \frac{s_0^2}{4} \cdot ( \phi - \theta) \hspace{0.05cm}.$$
 
:$$x(t) \approx \frac{s_0^2}{4} \cdot ( \phi - \theta) \hspace{0.05cm}.$$
*Das Regelsignal&nbsp; $x(t)$&nbsp; ist also proportional zum Phasenfehler&nbsp; $\phi - θ$, der mit der Costas–Regelschleife zu Null geregelt wird.&nbsp;  
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*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;  
*Im eingeschwungenen Zustand folgt somit das Oszillatorsignal&nbsp; $z(t)$&nbsp; unmittelbar dem Empfangssignal&nbsp; $r(t)$.
+
*Thus,&nbsp; in the steady state,&nbsp; the oscillator signal&nbsp; $z(t)$&nbsp; immediately follows the received signal&nbsp; $r(t)$.
*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,&nbsp; for example,&nbsp; $\phi - θ = π$&nbsp; would incorrectly lead to the rule signal&nbsp; $x(t) = 0$.&nbsp;  
  
 
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[[Category:Modulation Methods: Exercises|^4.2 Lineare digitale Modulation^]]
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[[Category:Modulation Methods: Exercises|^4.2 Linear Digital Modulation^]]

Latest revision as of 17:51, 15 April 2022

Costas rule loop

An important prerequisite for coherent demodulation is  "in-phase carrier recovery".  One possibility for this is the so-called  "Costas rule loop",  which is shown in simplified form by the adjacent block diagram.

In binary phase modulation  $\rm (BPSK)$,  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  $ϕ$  on the transmission channel is always assumed to be unknown.  The factor  "±"  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   $\phi - θ$   between the BPSK received signal  $r(t)$  and the oscillation  $z(t)$  generated at the receiver must be compensated.

  • For this purpose, a  "Voltage Controlled Oscillator"  $($VCO$)$  is used to generate an oscillation of frequency  $f_{\rm T}$,  initially with arbitrary phase  $θ$.
  • However,  the Costas rule loop iteratively achieves the desired result  $θ = \phi$. 



Notes:

  • The exercise belongs to the chapter  "Linear Digital Modulation".
  • In the diagram,  "TP"  denotes low-passes  (German:  "Tiefpass"   ⇒   subscript:  "TP"),  which are assumed to be ideal.
  • The square labeled  $π/2$  denotes a phase rotation by  $π/2 \ (90^\circ)$,  so that,  for example,  a cosine signal becomes a  "minus-sine" signal:
$$\cos (\omega_{\rm 0} \cdot t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\cos (\omega_{\rm 0} \cdot t + 90^\circ) = -\sin (\omega_{\rm 0} \cdot t)\hspace{0.05cm}.$$
  • Further,  the following trigonometric relations hold:
$$\cos (\alpha) \cdot \cos (\beta) = {1} /{2} \cdot \big [ \cos (\alpha - \beta) + \cos (\alpha + \beta)\big]\hspace{0.05cm},$$
$$\sin (\alpha) \cdot \cos (\beta) = {1} /{2} \cdot \big [ \sin (\alpha - \beta) + \sin (\alpha + \beta)\big]\hspace{0.05cm}.$$


Questions

1

Calculate the signal  $y_1(t)$  after the low-pass in the upper branch.  Which of the following statements is correct?

$y_1(t) = ± s_0/2 · \big[\cos (\phi - θ) + \cos (4 π · f_{\rm T} · t +\phi + θ)\big],$
$y_1(t) = ± s_0/2 · \cos (\phi - θ),$
$y_1(t) = ± s_0/2 · \sin (\phi - θ).$

2

Calculate the signal  $y_2(t)$  after the low-pass in the lower branch.  Which of the following statements is correct?

$y_2(t) = ± s_0/2 · \big[\cos (\phi - θ) + \cos (4 π · f_{\rm T} · t +\phi + θ)\big],$
$y_2(t) = ± s_0/2 · \cos (\phi - θ),$
$y_2(t) = ± s_0/2 · \sin (\phi - θ).$

3

Calculate the rule signal  $x(t)$  and give an approximation for small phase deviation  $\phi - θ$.  Which equations are correct?

$x(t) = s_0^2/8 · \cos(\phi + θ)$,
$x(t) = s_0^2/8 · \sin(2 \phi - 2θ),$
$x(t) ≈ s_0^2/4 · (\phi - θ),$
$x(t) ≈ s_0^2/4 · (\phi - θ)^2.$


Solution

(1)  The  second solution  is correct:

  • Using the addition theorem of trigonometry,  we obtain:
$$ m_1(t) = \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi) \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \theta) = \pm \frac{s_0}{2} \cdot \left [ \cos ( \phi - \theta) + \cos (4 \pi \cdot f_{\rm T} \cdot t + \phi +\theta)\right]\hspace{0.05cm}.$$
  • After the low-pass,  only the DC component  $y_1(t) = ± s_0/2 · \cos (\phi - θ)$  remains.


(2)  Here the  last solution  is correct:

  • Analogous to question  (1),  the result for the input signal of the lower low-pass filter is:
$$ m_2(t) = \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi) \cdot \left [-\sin (2 \pi \cdot f_{\rm T} \cdot t + \theta) \right]= \pm \frac{s_0}{2} \cdot \left [ \sin ( \phi - \theta) + \sin (4 \pi \cdot f_{\rm T} \cdot t + \phi +\theta)\right].$$
  • This leads to the following output signal:
$$ y_2(t) = \pm {s_0}/{2} \cdot\sin ( \phi - \theta) \hspace{0.05cm}.$$


(3)  Solutions 2 and 3  are correct:

  • By multiplying  $y_1(t)$  and  $y_2(t)$  we obtain:
$$x(t) = y_1(t) \cdot y_2(t)= \frac{s_0^2}{4} \cdot \cos ( \phi - \theta) \cdot \sin ( \phi - \theta) = \frac{s_0^2}{8} \cdot \sin ( 2\cdot\phi - 2\cdot\theta) \hspace{0.05cm}.$$
  • Using the small angle approximation  $\sin(α) ≈ α$  it follows:
$$x(t) \approx \frac{s_0^2}{4} \cdot ( \phi - \theta) \hspace{0.05cm}.$$
  • The rule signal  $x(t)$  is thus proportional to the phase error  $\phi - θ$,  which is controlled to zero by the Costas rule loop. 
  • Thus,  in the steady state,  the oscillator signal  $z(t)$  immediately follows the received signal  $r(t)$.
  • To achieve the required initial condition  $θ ≈ \phi$,  a training sequence is usually transmitted first and the phase is initialized accordingly.
  • This is also because the phase is only controlled modulo  $π$,  so that,  for example,  $\phi - θ = π$  would incorrectly lead to the rule signal  $x(t) = 0$.