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

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{{quiz-Header|Buchseite=Modulationsverfahren/Lineare digitale Modulationsverfahren
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{{quiz-Header|Buchseite=Modulation_Methods/Linear_Digital_Modulation
 
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[[File:|right|]]
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[[File:EN_Mod_A_4_8.png|right|frame|Costas rule loop]]
 +
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; $\rm (BPSK)$,&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 factor&nbsp; "±"&nbsp; describes the phase jumps of the BPSK signal.
  
===Fragebogen===
+
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.
 +
*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,&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,&nbsp; the following trigonometric relations hold:
 +
:$$\cos (\alpha) \cdot \cos (\beta)  =  {1} /{2} \cdot \big [ \cos (\alpha - \beta) + \cos (\alpha + \beta)\big]\hspace{0.05cm},$$
 +
:$$\sin (\alpha) \cdot \cos (\beta)  =  {1} /{2} \cdot \big [ \sin (\alpha - \beta) + \sin (\alpha + \beta)\big]\hspace{0.05cm}.$$
 +
 
 +
 
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Multiple-Choice Frage
+
{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="()"}
 +
- $y_1(t) = ± s_0/2 · \big[\cos (\phi - θ) + \cos (4 π · f_{\rm T} · t +\phi + θ)\big],$
 +
+ $y_1(t) = ± s_0/2 · \cos (\phi - θ),$
 +
- $y_1(t) = ± s_0/2 · \sin (\phi - θ).$
 +
 
 +
{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="()"}
 +
- $y_2(t) = ± s_0/2 ·  \big[\cos (\phi - θ) + \cos (4 π · f_{\rm T} · t +\phi + θ)\big],$
 +
- $y_2(t) = ± s_0/2 · \cos (\phi - θ),$
 +
+ $y_2(t) = ± s_0/2 · \sin (\phi - θ).$
 +
 
 +
{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="[]"}
- Falsch
+
- $x(t) = s_0^2/8 · \cos(\phi + θ)$,
+ Richtig
+
+ $x(t) = s_0^2/8 · \sin(2 \phi - 2θ),$
 +
+ $x(t) ≈ s_0^2/4 · (\phi - θ),$
 +
- $x(t) ≈ s_0^2/4 · (\phi - θ)^2.$
  
  
{Input-Box Frage
+
</quiz>
|type="{}"}
+
 
$\alpha$ = { 0.3 }
+
===Solution===
 +
{{ML-Kopf}}
 +
'''(1)'''&nbsp; The&nbsp; <u>second solution</u>&nbsp; is correct:
 +
*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,&nbsp; only the DC component&nbsp; $y_1(t) = ± s_0/2 · \cos (\phi - θ)$&nbsp; remains.
 +
 
 +
 
 +
 
 +
'''(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:
 +
:$$ m_2(t)  =  \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi) \cdot \left [-\sin (2 \pi \cdot f_{\rm T} \cdot t + \theta) \right]= \pm \frac{s_0}{2} \cdot \left [ \sin ( \phi - \theta) + \sin (4 \pi \cdot f_{\rm T} \cdot t + \phi +\theta)\right].$$
 +
*This leads to the following output signal:
 +
:$$ y_2(t) = \pm {s_0}/{2} \cdot\sin ( \phi - \theta) \hspace{0.05cm}.$$
  
  
  
</quiz>
+
'''(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:
 +
:$$x(t)  =  y_1(t) \cdot y_2(t)= \frac{s_0^2}{4} \cdot \cos ( \phi - \theta) \cdot \sin ( \phi - \theta)
 +
=  \frac{s_0^2}{8} \cdot \sin ( 2\cdot\phi - 2\cdot\theta) \hspace{0.05cm}.$$
 +
*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; $\phi - θ$,&nbsp; which is controlled to zero by the Costas rule loop.&nbsp;
 +
*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,&nbsp; for example,&nbsp; $\phi - θ = π$&nbsp; would incorrectly lead to the rule signal&nbsp; $x(t) = 0$.&nbsp;
  
===Musterlösung===
 
{{ML-Kopf}}
 
'''1.'''
 
'''2.'''
 
'''3.'''
 
'''4.'''
 
'''5.'''
 
'''6.'''
 
'''7.'''
 
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Modulationsverfahren|^4.2 Lineare digitale Modulationsverfahren^]]
+
[[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  $x(t) = 0$.