Aufgaben:Exercise 2.4: Frequency and Phase Offset - Revision history

Reed at 15:25, 25 March 2022

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Reed at 15:25, 25 March 2022

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Reed: /* Solution */

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Solution

← Older revision		Revision as of 15:25, 25 March 2022
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	*Thus,  the modulated signal  $s(t)$  has spectral components at  $±45$ kHz,  $±48$ kHz,  $±52$ kHz  and  $±55$ kHz.		*Thus,  the modulated signal  $s(t)$  has spectral components at  $±45$ kHz,  $±48$ kHz,  $±52$ kHz  and  $±55$ kHz.
−	*It is also known,  that the transmitter-side carrier  $z(t)$  is sinusoidal  $(ϕ_{\rm T} = -90^\circ)$.	+	*It is also known  that the transmitter-side carrier  $z(t)$  is sinusoidal  $(ϕ_{\rm T} = -90^\circ)$.

@@ Line 7: / Line 7: @@
 :$$ A_1  =  2\,{\rm V}, \hspace{0.15cm}f_1 = 2\,{\rm kHz} \hspace{0.05cm},$$
 :$$A_2  =  1\,{\rm V}, \hspace{0.15cm}f_2 = 5\,{\rm kHz}\hspace{0.05cm}.$$
-This signal is DSB amplitude modulated.
+This signal is DSB amplitude-modulated.
 *Thus,&nbsp; the modulated signal &nbsp;$s(t)$&nbsp; has spectral components at &nbsp;$±45$ kHz, &nbsp;$±48$ kHz, &nbsp;$±52$ kHz&nbsp; and &nbsp;$±55$ kHz.&nbsp;

@@ Line 92: / Line 92: @@
 :$$v(t) = \cos(\Delta \phi_{\rm T}) \cdot q(t) \hspace{0.05cm}.$$
 *Even insufficient phase synchronisation does not lead to distortions,&nbsp; only to a frequency-independent attenuation.
-*With&nbsp; $ϕ_{\rm T} =-90^\circ$&nbsp; and&nbsp; $ϕ_{\rm T} = -120^\circ$&nbsp; &rArr; &nbsp; $Δϕ_{\rm T} = -30^\circ$:
+*With&nbsp; $ϕ_{\rm T} =-90^\circ$&nbsp; and&nbsp; $ϕ_{\rm E} = -120^\circ$&nbsp; &rArr; &nbsp; $Δϕ_{\rm T} = -30^\circ$:
 :$$ v(t) = \cos(30^{\circ}) \cdot q(t)= 0.866 \cdot q(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} v(t= 0) = 0.866 \cdot A_1 \hspace{0.15cm}\underline {= 1.732\,{\rm V}}\hspace{0.05cm}.$$
@@ Line 110: / Line 110: @@
 :as follows:
 :$$ b(t) = q(t) \cdot \cos((\omega_{\rm T} - \omega_{\rm E}) \cdot t ) + q(t) \cdot \cos((\omega_{\rm T} + \omega_{\rm E}) \cdot t ) \hspace{0.05cm}.$$
-*The second term lies in the vicinity of &nbsp; $2f_{\rm T}$&nbsp; for &nbsp; $f_{\rm E} ≈ f_{\rm T}$&nbsp; and is removed by the low-pass.
+*The second term lies in the vicinity of &nbsp; $2f_{\rm T}$&nbsp; for &nbsp; $f_{\rm E} = f_{\rm T}$&nbsp; and is removed by the low-pass.
 *With the frequency difference &nbsp; $Δ\hspace{-0.05cm}f_{\rm T} = f_{\rm E} - f_{\rm T}= 1$ kHz,&nbsp; this leaves:
 :$$ v(t) = q(t) \cdot \cos(2 \pi \cdot \Delta \hspace{-0.05cm}f_{\rm T} \cdot t) \hspace{0.05cm}.$$

@@ Line 50: / Line 50: @@
 $A_{\rm E} \ = \ $ { 2 3% }
 $f_{\rm E} \ \hspace{0.05cm} = \ $ { 50 3% } $\ \text{kHz}$
-$\phi_{\rm E} \ = \ $ { -94--86 } $\ \text{Grad}$
+$\phi_{\rm E} \ = \ $ { -94--86 } $\ \text{deg}$
 {Let &nbsp;$f_{\rm E} = f_{\rm T}$&nbsp; (no frequency offset).&nbsp; Which sink signal &nbsp;$v(t)$&nbsp; results with &nbsp;$ϕ_{\rm E} = - 120^\circ$? <br>Give its signal value at &nbsp;$t = 0$&nbsp;.

@@ Line 73: / Line 73: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  <u>Answers 2, 3 and 5</u> are correct:
+'''(1)'''&nbsp;  <u>Answers 2, 3 and 5</u>&nbsp; are correct:
-*Envelope demodulation is not applicable for ZSB-AM without carrier and a modulation depth of &nbsp; $m > 1$&nbsp;.
+*Envelope demodulation is not applicable for DSB-AM without carrier and a modulation depth of &nbsp; $m > 1$.
-*The performance of the synchronous demodulator is not increased by the additional carrier component, but only leads to an unnecessary increase in the transmit power to be applied.
+*The performance of the synchronous demodulator is not increased by the additional carrier component,&nbsp; but only leads to an unnecessary increase in the transmit power to be applied.
-*The third statement is also correct. The solution to [[Aufgaben:Exercise_2.4Z:__Low-pass_Influence_with_Synchronous_Demodulation|Exercise 2.4Z]]&nbsp; shows the effects of omitting or incorrectly dimensioning&nbsp; $H_{\rm E} (f)$&nbsp;.
+*The third statement is also correct.&nbsp; The solution to&nbsp; [[Aufgaben:Exercise_2.4Z:__Low-pass_Influence_with_Synchronous_Demodulation|Exercise 2.4Z]]&nbsp; shows the effects of omitting or incorrectly dimensioning&nbsp; $H_{\rm E} (f)$.
-'''(2)'''&nbsp; As the name "Synchronous demodulator” already implies, the signals &nbsp; $z(t)$&nbsp; and&nbsp; $z_{\rm E} (t)$&nbsp;
+'''(2)'''&nbsp; As the name&nbsp; "synchronous demodulator"&nbsp; already implies,&nbsp; the signals &nbsp; $z(t)$&nbsp; and&nbsp; $z_{\rm E} (t)$&nbsp;
 must be synchronous in frequency and phase:
 :$$f_{\rm E} = f_{\rm T} \hspace{0.15cm}\underline {= 50\,{\rm kHz}}, \hspace{0.15cm}\phi_{\rm E} = \phi_{\rm T} \hspace{0.15cm}\underline {= - 90^{\circ}} \hspace{0.05cm}.$$
-*The carrier frequency&nbsp; $f_{\rm T} $&nbsp; at the transmitter can be determined from the transmission spectrum data&nbsp; $S(f)$&nbsp;.&nbsp; In the case of perfect synchronisation:
+*The carrier frequency&nbsp; $f_{\rm T} $&nbsp; at the transmitter can be determined from the transmission spectrum&nbsp; $S(f)$.&nbsp; In the case of perfect synchronisation:
 :$$v(t) = {A_{\rm E}}/{2} \cdot q(t) + {A_{\rm E}}/{2} \cdot q(t)\cdot \cos(2 \cdot \omega_{\rm T} \cdot t ) \hspace{0.05cm}.$$
-*The second term is removed by the low-pass filter.&nbsp; Thus, with&nbsp; $A_{\rm E}\hspace{0.15cm}\underline{ = 2}$&nbsp;,&nbsp; $v(t) = q(t)$ holds.
+*The second term is removed by the low-pass filter.&nbsp; Thus,&nbsp; with&nbsp; $A_{\rm E}\hspace{0.15cm}\underline{ = 2}$,&nbsp; $v(t) = q(t)$ holds.
-'''(3)'''&nbsp;  In the theory section, it was shown that in general, for DSB-AM and synchronous demodulation:
+'''(3)'''&nbsp;  In the theory section,&nbsp; it was shown that in general for DSB-AM and synchronous demodulation:
 :$$v(t) = \cos(\Delta \phi_{\rm T}) \cdot q(t) \hspace{0.05cm}.$$
-*Even insufficient phase synchronisation does not lead to distortions, only frequency-independent attenuation.
+*Even insufficient phase synchronisation does not lead to distortions,&nbsp; only to a frequency-independent attenuation.
-*Mit&nbsp; $ϕ_{\rm T} =-90^\circ$&nbsp; und&nbsp; $ϕ_{\rm T} = -120^\circ$&nbsp; ist&nbsp; $Δϕ_{\rm T} = -30^\circ$&nbsp; und man erhält:
+*With&nbsp; $ϕ_{\rm T} =-90^\circ$&nbsp; and&nbsp; $ϕ_{\rm T} = -120^\circ$&nbsp; &rArr; &nbsp; $Δϕ_{\rm T} = -30^\circ$:
 :$$ v(t) = \cos(30^{\circ}) \cdot q(t)= 0.866 \cdot q(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} v(t= 0) = 0.866 \cdot A_1 \hspace{0.15cm}\underline {= 1.732\,{\rm V}}\hspace{0.05cm}.$$
@@ Line 100: / Line 100: @@
 *It is pointless to discuss whether this is still a distortion-free system.
 *The result&nbsp; $v(t) \equiv 0$&nbsp; is due to the fact that cosine and sine are orthogonal functions.
-*This principle is made use of, for example, in what is known as &nbsp; [[Modulation_Methods/Quadrature_Amplitude_Modulation|quadrature amplitude modulation.]]&nbsp;.
+*This principle is made use of,&nbsp; for example,&nbsp; in what is known as &nbsp; [[Modulation_Methods/Quadrature_Amplitude_Modulation|quadrature amplitude modulation.]].
@@ Line 111: / Line 111: @@
 :$$ b(t) = q(t) \cdot \cos((\omega_{\rm T} - \omega_{\rm E}) \cdot t ) + q(t) \cdot \cos((\omega_{\rm T} + \omega_{\rm E}) \cdot t ) \hspace{0.05cm}.$$
 *The second term lies in the vicinity of &nbsp; $2f_{\rm T}$&nbsp; for &nbsp; $f_{\rm E} ≈ f_{\rm T}$&nbsp; and is removed by the low-pass.
+*With the frequency difference &nbsp; $Δ\hspace{-0.05cm}f_{\rm T} = f_{\rm E} - f_{\rm T}= 1$ kHz,&nbsp; this leaves:
 :$$ v(t) = q(t) \cdot \cos(2 \pi \cdot \Delta \hspace{-0.05cm}f_{\rm T} \cdot t) \hspace{0.05cm}.$$
-*The first statement is correct.&nbsp; This states that now the message signal &nbsp; $v(t)$&nbsp; becomes quieter and louder again after demodulation according to a cosine function (a "beat").
+*The first statement is correct.&nbsp; This states that now the signal &nbsp; $v(t)$&nbsp; becomes quieter and louder again after demodulation according to a cosine function&nbsp; (a&nbsp; "beat").
-Diese besagt, dass nun das Nachrichtensignal nach der Demodulation gemäß einer Cosinusfunktion leiser und wieder lauter wird („Schwebung”).
+*The cosine component of &nbsp; $q(t)$&nbsp; with frequency &nbsp; $f_1 = 2\text{ kHz}$&nbsp; now becomes two components&nbsp; (each of half the amplitude)&nbsp; at &nbsp; $1\text{ kHz}$ and $3\text{ kHz}$.
-*The cosine component of &nbsp; $q(t)$&nbsp; with frequency &nbsp; $f_1 = 2\text{ kHz}$&nbsp; now becomes two components (each of half the amplitude) at &nbsp; $1\text{ kHz}$ and $3\text{ kHz}$.
+*Similarly,&nbsp; the sink signal does not include a component at&nbsp; $f_2 = 5\text{ kHz}$,&nbsp; only components at &nbsp; $4\text{ kHz}$&nbsp; and at&nbsp; $6\text{ kHz}$:
 :$$1\,{\rm V} \cdot \sin(2 \pi \cdot 5\,{\rm kHz} \cdot t)\cdot \cos(2 \pi \cdot 1\,{\rm kHz} \cdot t)  =
 .5\,{\rm V} \cdot \sin(2 \pi \cdot 4\,{\rm kHz} \cdot t)
 +  0.5\,{\rm V} \cdot \sin(2 \pi \cdot 6\,{\rm kHz} \cdot t)\hspace{0.05cm}.$$
- <u>Answers 1, 3 and 4</u> are correct.
+<u>Answers 1, 3 and 4</u>&nbsp; are correct.
 {{ML-Fuß}}

@@ Line 25: / Line 25: @@
 Hints:
 *This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Synchronous_Demodulation|Synchronous Demodulation]].
-*Particular reference is made to the pages  &nbsp;[[Modulation_Methods/Synchronous_Demodulation##Influence_of_a_frequency_offset|Influence of a frequency offset]]&nbsp; and &nbsp;[[Modulation_Methods/Synchronous_Demodulation##Influence_of_a_phase_offset|Influence of a phase offset]].
+*Particular reference is made to the pages  &nbsp;[[Modulation_Methods/Synchronous_Demodulation#Influence_of_a_frequency_offset|Influence of a frequency offset]]&nbsp; and &nbsp;[[Modulation_Methods/Synchronous_Demodulation#Influence_of_a_phase_offset|Influence of a phase offset]].
 *Take the following trigonometric transformations into account:

@@ Line 13: / Line 13: @@
-The demodulation  to be performed with the circuit sketched here,&nbsp; which is defined by the following parameters:
+The demodulation  to be performed with the circuit sketched here,&nbsp; which is defined by the following parameters <br>("E" &nbsp; &rArr;  &nbsp; "empfägerseitig" &nbsp; &rArr;  &nbsp; "receiver-side"):
 #&nbsp; Amplitude &nbsp;$A_{\rm E}$&nbsp; (no unit),
 #&nbsp; frequency &nbsp;$f_{\rm E}$,
@@ Line 23: / Line 23: @@
 Hints:
 *This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Synchronous_Demodulation|Synchronous Demodulation]].
 *Particular reference is made to the pages  &nbsp;[[Modulation_Methods/Synchronous_Demodulation##Influence_of_a_frequency_offset|Influence of a frequency offset]]&nbsp; and &nbsp;[[Modulation_Methods/Synchronous_Demodulation##Influence_of_a_phase_offset|Influence of a phase offset]].
@@ Line 51: / Line 46: @@
-{How should the signal parameters of the receiver-side carrier signal &nbsp;$z_{\rm E}(t)$&nbsp; be chosen, so that &nbsp;$v(t) = q(t)$&nbsp; holds?
+{How should the signal parameters of the receiver-side carrier signal &nbsp;$z_{\rm E}(t)$&nbsp; be chosen,&nbsp; so that &nbsp;$v(t) = q(t)$&nbsp; holds?
 |type="{}"}
 $A_{\rm E} \ = \ $ { 2 3% }
@@ Line 62: / Line 57: @@
-{Now also let &nbsp;$f_{\rm E} = f_{\rm T}$.&nbsp; Which sink signal &nbsp;$v(t)$&nbsp; results with &nbsp;$ϕ_{\rm E} = 0^\circ$? <br>Give the signal value at &nbsp;$t = 0$&nbsp;.
+{Let &nbsp;$f_{\rm E} = f_{\rm T}$&nbsp; again.&nbsp; Which sink signal &nbsp;$v(t)$&nbsp; results with &nbsp;$ϕ_{\rm E} = 0^\circ$? <br>Give the signal value at &nbsp;$t = 0$.
 |type="{}"}
 $v(t = 0)\ = \ $ { 0. }  $\ \text{V}$

@@ Line 9: / Line 9: @@
 This signal is DSB amplitude modulated.
-Thus, the modulated signal &nbsp;$s(t)$&nbsp; has spectral components at &nbsp;$±45$ kHz, &nbsp;$±48$ kHz, &nbsp;$±52$ kHz&nbsp; and &nbsp;$±55$ kHz.&nbsp; It is also known, that the transmitter-side carrier Trägersignal is sinusoidal &nbsp;$(ϕ_{\rm T} = -90^\circ)$.
+*Thus,&nbsp; the modulated signal &nbsp;$s(t)$&nbsp; has spectral components at &nbsp;$±45$ kHz, &nbsp;$±48$ kHz, &nbsp;$±52$ kHz&nbsp; and &nbsp;$±55$ kHz.&nbsp;
 The &nbsp;$H_{\rm E}(f)$&nbsp; block represents an ideal, rectangular low-pass filter, which is suitably dimensioned.