Aufgaben:Exercise 4.9: Costas Rule Loop - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:54:46Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:54, 16 March 2026
Line 87:		Line 87:

	[[Category:Modulation Methods: Exercises\|^4.2 Linear Digital Modulation^]]		[[Category:Modulation Methods: Exercises\|^4.2 Linear Digital Modulation^]]
	[[de:Aufgaben:4.~~8_Costas–Regelschleife~~]]		[[de:Aufgaben:Aufgabe 4.9: Costas–Regelschleife]]

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:48:46Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:48, 16 March 2026
Line 74:		Line 74:
	'''(3)'''  <u>Solutions 2 and 3</u>  are correct:		'''(3)'''  <u>Solutions 2 and 3</u>  are correct:
	*By multiplying  $y_1(t)$  and  $y_2(t)$  we obtain:		*By multiplying  $y_1(t)$  and  $y_2(t)$  we obtain:
	:$$x(t) = y_1(t) \cdot y_2(t)= \frac{s_0^2}{4} \cdot \cos ( \phi - \theta) \cdot \sin ( \phi - \theta)		:$$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}.$$
	= \frac{s_0^2}{8} \cdot \sin ( 2\cdot\phi - 2\cdot\theta) \hspace{0.05cm}.$$
	*Using the small angle approximation  $\sin(α) ≈ α$  it follows:		*Using the small angle approximation  $\sin(α) ≈ α$  it follows:
	:$$x(t) \approx \frac{s_0^2}{4} \cdot ( \phi - \theta) \hspace{0.05cm}.$$		:$$x(t) \approx \frac{s_0^2}{4} \cdot ( \phi - \theta) \hspace{0.05cm}.$$

Maintenance script: Add German interlanguage link

2026-03-16T12:33:53Z

Add German interlanguage link

← Older revision		Revision as of 14:33, 16 March 2026
Line 88:		Line 88:

	[[Category:Modulation Methods: Exercises\|^4.2 Linear Digital Modulation^]]		[[Category:Modulation Methods: Exercises\|^4.2 Linear Digital Modulation^]]
			[[de:Aufgaben:4.8_Costas–Regelschleife]]

Guenter at 16:51, 15 April 2022

2022-04-15T16:51:10Z

← Older revision		Revision as of 18:51, 15 April 2022
Line 57:		Line 57:
	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The <u>second solution</u> is correct:		'''(1)'''  The  <u>second solution</u>  is correct:
	*Using the addition theorem of trigonometry, we obtain:		*Using the addition theorem of trigonometry,  we obtain:
	:$$ m_1(t) = \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi) \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \theta) = \pm \frac{s_0}{2} \cdot \left [ \cos ( \phi - \theta) + \cos (4 \pi \cdot f_{\rm T} \cdot t + \phi +\theta)\right]\hspace{0.05cm}.$$		:$$ m_1(t) = \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi) \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \theta) = \pm \frac{s_0}{2} \cdot \left [ \cos ( \phi - \theta) + \cos (4 \pi \cdot f_{\rm T} \cdot t + \phi +\theta)\right]\hspace{0.05cm}.$$
	*After the low-pass, only the DC component  $y_1(t) = ± s_0/2 · \cos (\phi - θ)$ remains.		*After the low-pass,  only the DC component  $y_1(t) = ± s_0/2 · \cos (\phi - θ)$  remains.



	'''(2)'''  Here the <u>last solution</u> is correct:		'''(2)'''  Here the  <u>last solution</u>  is correct:
	*Analogous to question  '''(1)''',  the result for the input signal of the lower low-pass filter is:		*Analogous to question  '''(1)''',  the result for the input signal of the lower low-pass filter is:
	:$$ m_2(t) = \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi) \cdot \left [-\sin (2 \pi \cdot f_{\rm T} \cdot t + \theta) \right]= \pm \frac{s_0}{2} \cdot \left [ \sin ( \phi - \theta) + \sin (4 \pi \cdot f_{\rm T} \cdot t + \phi +\theta)\right].$$		:$$ m_2(t) = \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi) \cdot \left [-\sin (2 \pi \cdot f_{\rm T} \cdot t + \theta) \right]= \pm \frac{s_0}{2} \cdot \left [ \sin ( \phi - \theta) + \sin (4 \pi \cdot f_{\rm T} \cdot t + \phi +\theta)\right].$$
Line 72:		Line 72:


	'''(3)'''  <u>Solutions 2 and 3</u> are correct:		'''(3)'''  <u>Solutions 2 and 3</u>  are correct:
	*By multiplying  $y_1(t)$  and  $y_2(t)$  we obtain:		*By multiplying  $y_1(t)$  and  $y_2(t)$  we obtain:
	:$$x(t) = y_1(t) \cdot y_2(t)= \frac{s_0^2}{4} \cdot \cos ( \phi - \theta) \cdot \sin ( \phi - \theta)		:$$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  $\sin(α) ≈ α$  it follows:		*Using the small angle approximation  $\sin(α) ≈ α$  it follows:
	:$$x(t) \approx \frac{s_0^2}{4} \cdot ( \phi - \theta) \hspace{0.05cm}.$$		:$$x(t) \approx \frac{s_0^2}{4} \cdot ( \phi - \theta) \hspace{0.05cm}.$$
	*The rule signal  $x(t)$  is thus proportional to the phase error  $\phi - θ$, which is controlled to zero by the Costas rule loop.		*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  $r(t)$.		*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.		*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  $x(t) = 0$.		*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  $x(t) = 0$.

	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 16:46, 15 April 2022

2022-04-15T16:46:47Z

← Older revision		Revision as of 18:46, 15 April 2022
Line 4:		Line 4:

	[[File:EN_Mod_A_4_8.png\|right\|frame\|Costas rule loop]]		[[File:EN_Mod_A_4_8.png\|right\|frame\|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.		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  (BPSK), the received signal can be expressed as		In binary phase modulation  $\rm (BPSK)$,  the received signal can be expressed as
	:$$ r(t) = \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi)$$		:$$ r(t) = \pm s_0 \cdot \cos (2 \pi \cdot f_{\rm T} \cdot t + \phi)$$
	The phase rotation  $ϕ$  on the transmission channel is always assumed to be unknown.  The ~~specification~~  "±"  describes the phase jumps of the BPSK signal.		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		The task of the circuit indicated by the diagram is to generate a carrier signal
	:$$z(t) = \cos (2 \pi \cdot f_{\rm T} \cdot t + \theta)$$		:$$z(t) = \cos (2 \pi \cdot f_{\rm T} \cdot t + \theta)$$
	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.		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  $θ$.		*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  $θ = \phi$.		*However,  the Costas rule loop iteratively achieves the desired result  $θ = \phi$.




			Notes:
			*The exercise belongs to the chapter  [[Modulation_Methods/Linear_Digital_Modulation\|"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:
	''Notes:''
	*The exercise belongs to the chapter  [[Modulation_Methods/Linear_Digital_Modulation\|Linear Digital Modulation]].

	*In the diagram, "TP" denotes low-passes, 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:
	:$$\cos (\omega_{\rm 0} \cdot t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\cos (\omega_{\rm 0} \cdot t + 90^\circ) = -\sin (\omega_{\rm 0} \cdot t)\hspace{0.05cm}.$$		:$$\cos (\omega_{\rm 0} \cdot t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\cos (\omega_{\rm 0} \cdot t + 90^\circ) = -\sin (\omega_{\rm 0} \cdot t)\hspace{0.05cm}.$$
	*Further, the following trigonometric relations hold:		*Further,  the following trigonometric relations hold:
	:$$\cos (\alpha) \cdot \cos (\beta) = {1} /{2} \cdot \big [ \cos (\alpha - \beta) + \cos (\alpha + \beta)\big]\hspace{0.05cm},$$		:$$\cos (\alpha) \cdot \cos (\beta) = {1} /{2} \cdot \big [ \cos (\alpha - \beta) + \cos (\alpha + \beta)\big]\hspace{0.05cm},$$
	:$$\sin (\alpha) \cdot \cos (\beta) = {1} /{2} \cdot \big [ \sin (\alpha - \beta) + \sin (\alpha + \beta)\big]\hspace{0.05cm}.$$		:$$\sin (\alpha) \cdot \cos (\beta) = {1} /{2} \cdot \big [ \sin (\alpha - \beta) + \sin (\alpha + \beta)\big]\hspace{0.05cm}.$$

Hwang at 17:19, 21 March 2022

2022-03-21T17:19:29Z

Show changes

Guenter at 14:03, 16 March 2022

2022-03-16T14:03:34Z

← Older revision		Revision as of 16:03, 16 March 2022
Line 92:		Line 92:


	[[Category:Modulation Methods: Exercises\|^4.2 ~~Lineare digitale~~ Modulation^]]		[[Category:Modulation Methods: Exercises\|^4.2 Linear Digital Modulation^]]

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 Aufgaben:Aufgabe 4.9: Costas Rule Loop to Aufgaben:Exercise 4.9: Costas Rule Loop

← Older revision		Revision as of 16:01, 16 March 2022
(No difference)

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 Aufgaben:Aufgabe 4.9: Costas–Regelschleife to Aufgaben:Aufgabe 4.9: Costas Rule Loop

← Older revision		Revision as of 16:00, 16 March 2022
(No difference)

Guenter at 16:27, 9 November 2021

2021-11-09T16:27:55Z

← Older revision		Revision as of 18:27, 9 November 2021
Line 3:		Line 3:
	}}		}}

	[[File:~~Mod_A_4_8_vers2~~.png\|right\|frame\|Costas–Regelschleife]]		[[File:EN_Mod_A_4_8.png\|right\|frame\|Costas–Regelschleife]]
	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.		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.