Aufgaben:Exercise 4.9: Costas Rule Loop

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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

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  $x(t) = 0$.