Aufgaben:Exercise 2.13: Quadrature Amplitude Modulation - Revision history

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Reed at 16:26, 23 December 2021

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Reed: /* Solution */

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Solution

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← Older revision		Revision as of 14:27, 9 April 2022
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−	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Weitere AM–Varianten~~	+	{{quiz-Header\|Buchseite=Modulation Methods/Further AM Variants
	}}		}}

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 }}
-[[File:P_ID1055__Mod_A_2_11.png|right|frame|QAM model under consideration]]
+[[File:EN_Mod_A_2_11.png|right|frame|QAM model under consideration]]
 The&nbsp; "quadrature amplitude modulation"&nbsp; $\rm (QAM)$&nbsp; explained by the diagram allows the transmission of two source signals &nbsp;$q_1(t)$&nbsp; and &nbsp;$q_2(t)$&nbsp; over the same channel
 *under certain boundary conditions,

@@ Line 62: / Line 62: @@
 {Which of the following statements apply when &nbsp;$f_1 ≠ f_2$&nbsp; and &nbsp;$Δϕ_{\rm T} ≠ 0$&nbsp;  (with an arbitrary phase offset)?
-|type="[]"}
+|type="()"}
 - $v_1(t) = q_1(t)$&nbsp; and&nbsp; $v_2(t) = q_2(t)$&nbsp; both hold.
 - Linear distortions occur.

@@ Line 80: / Line 80: @@
 '''(2)'''&nbsp; With $A_1 = A_2 = 2 \ \rm V$ and $f_1 = f_2 = 5\ \rm  kHz$, the first and third cosine oscillations constructively overlap and the other two cancel completely.
-*Thus, the following simple result is obtained:
+*Thus,&nbsp; the following simple result is obtained:
 :$$ s(t) = 2\,{\rm V} \cdot \cos(2 \pi \cdot 20\,{\rm kHz} \cdot t) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} s(t = 50\,{\rm &micro; s}) \hspace{0.15cm}\underline {= 2\,{\rm V}} \hspace{0.05cm}.$$
-'''(3)'''&nbsp; The <u>first answer</u> is correct:
+'''(3)'''&nbsp; The&nbsp; <u>first answer</u>&nbsp; is correct:
-*For phase-synchronous demodulation&nbsp; $(Δϕ_T = 0)$&nbsp;, the signals before the low-pass filters according to subtask&nbsp; '''(2)''' are obtained as:
+*For phase-synchronous demodulation&nbsp; $(Δϕ_T = 0)$,&nbsp; the signals before the low-pass filters according to subtask&nbsp; '''(2)''' are obtained as:
 :$$b_1(t)  =  2\,{\rm V} \cdot \cos(\omega_{\rm 20} \cdot t)\cdot 2 \cdot \cos(\omega_{\rm 25} \cdot t) = 2\,{\rm V} \cdot \cos(\omega_{\rm 5} \cdot t) + 2\,{\rm V} \cdot \cos(\omega_{\rm 45} \cdot t),$$
 :$$ b_2(t)  =  2\,{\rm V} \cdot \cos(\omega_{\rm 20} \cdot t)\cdot 2 \cdot \sin(\omega_{\rm 25} \cdot t) = 2\,{\rm V} \cdot \sin(\omega_{\rm 5} \cdot t) + 2\,{\rm V} \cdot \sin(\omega_{\rm 45} \cdot t)\hspace{0.05cm}.$$
-*Thus, after eliminating the respective &nbsp; $45\ \rm  kHz$ components, we get &nbsp; $v_1(t) = q_1(t)$&nbsp; and&nbsp; $v_2(t) = q_2(t)$.
+*Thus,&nbsp; after eliminating the respective &nbsp; $45\ \rm  kHz$ components,&nbsp; we get &nbsp; $v_1(t) = q_1(t)$&nbsp; and&nbsp; $v_2(t) = q_2(t)$.
@@ Line 98: / Line 98: @@
 :$$b_2(t)=  2\,{\rm V} \cdot \cos(\omega_{\rm 20} \cdot t)\cdot 2 \cdot \sin(\omega_{\rm 25} \cdot t+ \Delta \phi_{\rm T})=
 \,{\rm V} \cdot \sin(\omega_{\rm 5} \cdot t + \Delta \phi_{\rm T}) + {(45 \,\rm kHz-Anteil )}\hspace{0.05cm}.$$
-*The sink signals&nbsp; $v_1(t)$&nbsp; and&nbsp; $v_2(t)$&nbsp; in this constellation exhibit delays and thus phase distortions compared with &nbsp; $q_1(t)$&nbsp; and&nbsp; $q_2(t)$&nbsp;.
+*The sink signals&nbsp; $v_1(t)$&nbsp; and&nbsp; $v_2(t)$&nbsp; in this constellation exhibit delays and thus phase distortions compared with &nbsp; $q_1(t)$&nbsp; and&nbsp; $q_2(t)$.
 *These belong to the class of linear distortions &nbsp; &rArr; &nbsp; <u>Answer 2</u>.
-'''(5)'''&nbsp; In general, it holds for the source signal that:
+'''(5)'''&nbsp; In general,&nbsp; it holds for the received signal:
 :$$r(t) = s(t) = q_1(t) \cdot \cos(\omega_{\rm T} \cdot t) + q_2(t) \cdot \sin(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$
 Multiplication by the receiver-side carrier signals&nbsp; $z_{1,\hspace{0.05cm}{\rm E}}(t)$&nbsp; and&nbsp; $z_{2,\hspace{0.05cm}{\rm E}}(t)$&nbsp; and band-limiting leads to the signals
@@ Line 109: / Line 109: @@
 :$$ v_2(t)  =  \sin(\Delta \phi_{\rm T}) \cdot q_1(t) + \cos(\Delta \phi_{\rm T}) \cdot q_2(t) \hspace{0.05cm}.$$
 From this it can be seen:
-*With a phase offset of &nbsp; $Δϕ_{\rm T}  = 30^\circ$&nbsp;, the sink signal&nbsp; $v_1(t)$&nbsp; includes not only the signal &nbsp; $q_1(t)$ attenuated by about &nbsp; $\cos(30^\circ) = 0.866$&nbsp;, but also the frequency &nbsp; $f_2$ contained in &nbsp; $q_2(t)$&nbsp;.
+*With a phase offset of &nbsp; $Δϕ_{\rm T}  = 30^\circ$,&nbsp; the sink signal&nbsp; $v_1(t)$&nbsp; includes not only the signal &nbsp; $q_1(t)$ attenuated by about &nbsp; $\cos(30^\circ) = 0.866$,&nbsp; <br>but also the frequency &nbsp; $f_2$ is contained in &nbsp; $q_2(t)$.
-*This is weighted by the factor &nbsp; $\sin(30^\circ) = 0.5$&nbsp;.
+*This is weighted by the factor &nbsp; $\sin(30^\circ) = 0.5$.
-*Thus, nonlinear distortions are present &nbsp; &rArr; &nbsp; <u>Answer 3</u>.
+*Thus,&nbsp; nonlinear distortions are present &nbsp; &rArr; &nbsp; <u>Answer 3</u>.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1055__Mod_A_2_11.png|right|frame|&nbsp; $\rm QAM$ model under consideration]]
+[[File:P_ID1055__Mod_A_2_11.png|right|frame|QAM model under consideration]]
-The ''quadrature amplitude modulation''&nbsp; $\rm (QAM)$&nbsp; explained by the diagram allows the transmission of two source signals &nbsp;$q_1(t)$&nbsp; and &nbsp;$q_2(t)$&nbsp; over the same channel, under certain boundary conditions, which are to be determined in this task.
+The&nbsp; "quadrature amplitude modulation"&nbsp; $\rm (QAM)$&nbsp; explained by the diagram allows the transmission of two source signals &nbsp;$q_1(t)$&nbsp; and &nbsp;$q_2(t)$&nbsp; over the same channel
-In this exercise, with &nbsp;$A_1 = A_2 = 2\ \rm  V$, let:
 :$$q_1(t)  = A_1 \cdot \cos(2 \pi \cdot f_{\rm 1} \cdot t),$$
 :$$q_2(t)  =  A_2 \cdot \sin(2 \pi \cdot f_{\rm 2} \cdot t)\hspace{0.05cm}.$$
-For &nbsp;$ω_{\rm T} = 2π · 25\ \rm kHz$, the four carrier signals shown in the diagram are:
+For &nbsp;$ω_{\rm T} = 2π · 25\ \rm kHz$,&nbsp; the four carrier signals shown in the diagram are:
 :$$z_1(t)  =  \cos(\omega_{\rm T} \cdot t),$$
 :$$ z_2(t)  =  \sin(\omega_{\rm T} \cdot t),$$
 :$$ z_{1,\hspace{0.05cm}{\rm E}}(t) =  2 \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}),$$
 :$$ z_{2,\hspace{0.05cm}{\rm E}}(t)  =  2 \cdot \sin(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T})\hspace{0.05cm}.$$
@@ Line 22: / Line 24: @@
-''Hints:''
+Hints:
 *This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Further_AM_Variants|Further AM Variants]].
 *Particular reference is made to the page&nbsp; [[Modulation_Methods/Further_AM_Variants#Quadrature_Amplitude_Modulation_.28QAM.29|Quadrature Amplitude Modulation (QAM)]].
 *It is worth noting that the carrier signals &nbsp;$z_2(t)$&nbsp; and &nbsp;$z_{2,\hspace{0.05cm}{\rm E}}(t)$&nbsp; are applied with positive signs here.
-*Often – as in the theory section – these carrier signals are given as "minus-sine".
+*Often – as in the theory section – these carrier signals are given as&nbsp; "minus-sine".
 *The following trigonometric transformations are given:
 :$$ \cos(\alpha) \cdot \cos(\beta)  = 1/2 \cdot \big[ \cos(\alpha - \beta)+ \cos(\alpha + \beta) \big],$$
@@ Line 44: / Line 45: @@
 - $s(t)$&nbsp; is composed of four sine oscillations.
-{What is &nbsp;$s(t)$&nbsp; when &nbsp;$f_1 = f_2 = 5 \ \rm kHz$.&nbsp; What signal value arises for &nbsp;$t = 50 \ \rm &micro; s$&nbsp;?
+{What is &nbsp;$s(t)$&nbsp; with &nbsp;$f_1 = f_2 = 5 \ \rm kHz$?&nbsp; What signal value arises for &nbsp;$t = 50 \ \rm &micro; s$&nbsp;?
 |type="{}"}
 $s(t = 50 \ \rm &micro; s) \ = \ $ { 2 3% } $\ \rm V$
@@ Line 50: / Line 51: @@
 {Calculate the sink signals &nbsp;$v_1(t)$&nbsp; and &nbsp;$v_2(t)$ for &nbsp;$f_1 = f_2$&nbsp; and &nbsp;$Δϕ_{\rm T} = 0$&nbsp; (no phase offset).&nbsp; Which statements are true?
 |type="()"}
-+ &nbsp; $v_1(t) = q_1(t)$&nbsp; and&nbsp; $v_2(t) = q_2(t)$ both hold.
++ $v_1(t) = q_1(t)$&nbsp; and&nbsp; $v_2(t) = q_2(t)$&nbsp; both hold.
 - Linear distortions occur.
 - Nonlinear distortions occur.
@@ Line 56: / Line 57: @@
 {Calculate the sink signals&nbsp; $v_1(t)$&nbsp; and&nbsp; $v_2(t)$&nbsp; for&nbsp; $f_1 = f_2$&nbsp; and a phase offset&nbsp; $Δϕ_{\rm T} = 30^\circ$.&nbsp; Which statements are true?
 |type="()"}
-- &nbsp;$v_1(t) = q_1(t)$&nbsp; and &nbsp;$v_2(t) = q_2(t)$ both hold.
+- $v_1(t) = q_1(t)$&nbsp; and&nbsp; $v_2(t) = q_2(t)$&nbsp; both hold.
 + Linear distortions occur.
 - Nonlinear distortions occur.
@@ Line 62: / Line 63: @@
 {Which of the following statements apply when &nbsp;$f_1 ≠ f_2$&nbsp; and &nbsp;$Δϕ_{\rm T} ≠ 0$&nbsp;  (with an arbitrary phase offset)?
 |type="[]"}
-- &nbsp;$v_1(t) = q_1(t)$&nbsp; and &nbsp;$v_2(t) = q_2(t)$ both hold.
+- $v_1(t) = q_1(t)$&nbsp; and&nbsp; $v_2(t) = q_2(t)$&nbsp; both hold.
 - Linear distortions occur.
 + Nonlinear distortions occur.

@@ Line 70: / Line 70: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Mit den angegebenen trigonometrischen Umformungen erhält man:
+'''(1)'''&nbsp; With the given trigonometric transformations we get:
 :$$s(t)  =  A_1 \cdot \cos(\omega_{\rm 1} \cdot t)\cdot \cos(\omega_{\rm T} \cdot t) + A_2 \cdot \sin(\omega_{\rm 2} \cdot t)\cdot \sin(\omega_{\rm T} \cdot t) $$
 :$$\Rightarrow  \hspace{0.3cm}s(t)   =  \frac{A_1}{2}\cdot \cos((\omega_{\rm T} - \omega_{\rm 1})\cdot t) + \frac{A_1}{2}\cdot \cos((\omega_{\rm T} + \omega_{\rm 1})\cdot t) +
    \frac{A_2}{2}\cdot \cos((\omega_{\rm T} - \omega_{\rm 2})\cdot t) - \frac{A_2}{2}\cdot \cos((\omega_{\rm T} + \omega_{\rm 2})\cdot t)\hspace{0.05cm}.$$
-*Richtig ist demnach der <u>zweite Lösungsvorschlag</u>.
+*The <u>second answer</u> is correct.
-'''(2)'''&nbsp; Mit $A_1 = A_2 = 2 \ \rm V$ und $f_1 = f_2 = 5\ \rm  kHz$ überlagern sich die erste und die dritte Cosinusschwingungen konstruktiv und die beiden anderen heben sich vollständig auf.
+'''(2)'''&nbsp; With $A_1 = A_2 = 2 \ \rm V$ and $f_1 = f_2 = 5\ \rm  kHz$, the first and third cosine oscillations constructively overlap and the other two cancel completely.
-*Es ergibt sich somit das folgende einfache Ergebnis:
+*Thus, the following simple result is obtained:
 :$$ s(t) = 2\,{\rm V} \cdot \cos(2 \pi \cdot 20\,{\rm kHz} \cdot t) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} s(t = 50\,{\rm &micro; s}) \hspace{0.15cm}\underline {= 2\,{\rm V}} \hspace{0.05cm}.$$
-'''(3)'''&nbsp; Richtig ist der <u>erste Lösungsvorschlag</u>:
+'''(3)'''&nbsp; The<u>first answer</u> is correct:
-*Bei phasensynchroner Demodulation&nbsp; $(Δϕ_T = 0)$&nbsp; erhält man für die Signale vor den Tiefpässen gemäß der Teilaufgabe&nbsp; '''(2)''':
+*For phase-synchronous demodulation&nbsp; $(Δϕ_T = 0)$&nbsp;, the signals before the low-pass filters according to subtask&nbsp; '''(2)''' are obtained as:
 :$$b_1(t)  =  2\,{\rm V} \cdot \cos(\omega_{\rm 20} \cdot t)\cdot 2 \cdot \cos(\omega_{\rm 25} \cdot t) = 2\,{\rm V} \cdot \cos(\omega_{\rm 5} \cdot t) + 2\,{\rm V} \cdot \cos(\omega_{\rm 45} \cdot t),$$
 :$$ b_2(t)  =  2\,{\rm V} \cdot \cos(\omega_{\rm 20} \cdot t)\cdot 2 \cdot \sin(\omega_{\rm 25} \cdot t) = 2\,{\rm V} \cdot \sin(\omega_{\rm 5} \cdot t) + 2\,{\rm V} \cdot \sin(\omega_{\rm 45} \cdot t)\hspace{0.05cm}.$$
-*Nach Eliminierung der jeweiligen&nbsp; $45\ \rm  kHz$–Anteile ergibt sich somit&nbsp; $v_1(t) = q_1(t)$&nbsp; und&nbsp; $v_2(t) = q_2(t)$.
+*Thus, after eliminating the respective &nbsp; $45\ \rm  kHz$ components, we get &nbsp; $v_1(t) = q_1(t)$&nbsp; and&nbsp; $v_2(t) = q_2(t)$.
-'''(4)'''&nbsp; Analog zur Teilaufgabe&nbsp; '''(3)'''&nbsp; gilt nun:
+'''(4)'''&nbsp; Analogously to subtask&nbsp; '''(3)'''&nbsp; it now holds that:
 :$$ b_1(t)  =  2\,{\rm V} \cdot \cos(\omega_{\rm 20} \cdot t)\cdot 2 \cdot \cos(\omega_{\rm 25} \cdot t+ \Delta \phi_{\rm T})=
 \,{\rm V} \cdot \cos(\omega_{\rm 5} \cdot t + \Delta \phi_{\rm T}) + {(45 \,\rm kHz-Anteil )},$$
 :$$b_2(t)=  2\,{\rm V} \cdot \cos(\omega_{\rm 20} \cdot t)\cdot 2 \cdot \sin(\omega_{\rm 25} \cdot t+ \Delta \phi_{\rm T})=
 \,{\rm V} \cdot \sin(\omega_{\rm 5} \cdot t + \Delta \phi_{\rm T}) + {(45 \,\rm kHz-Anteil )}\hspace{0.05cm}.$$
-*Die Sinkensignale&nbsp; $v_1(t)$&nbsp; und&nbsp; $v_2(t)$&nbsp; weisen bei dieser Konstellation gegenüber&nbsp; $q_1(t)$&nbsp; und&nbsp; $q_2(t)$&nbsp; Laufzeiten und damit Phasenverzerrungen auf.
+*The sink signals&nbsp; $v_1(t)$&nbsp; and&nbsp; $v_2(t)$&nbsp; in this constellation exhibit delays and thus phase distortions compared with &nbsp; $q_1(t)$&nbsp; and&nbsp; $q_2(t)$&nbsp;.
-*Diese gehören zur Klasse der linearen Verzerrungen &nbsp; &rArr; &nbsp; <u>Antwort 2</u>.
+*These belong to the class of linear distortions &nbsp; &rArr; &nbsp; <u>Answer 2</u>.
-'''(5)'''&nbsp; Allgemein gilt für das Empfangssignal:
+'''(5)'''&nbsp; In general, it holds for the source signal that:
 :$$r(t) = s(t) = q_1(t) \cdot \cos(\omega_{\rm T} \cdot t) + q_2(t) \cdot \sin(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$
-Die Multiplikation mit den empfängerseitigen Trägersignalen&nbsp; $z_{1,\hspace{0.05cm}{\rm E}}(t)$&nbsp; und&nbsp; $z_{2,\hspace{0.05cm}{\rm E}}(t)$&nbsp; und Bandbegrenzung führt zu den Signalen
+Multiplication by the receiver-side carrier signals&nbsp; $z_{1,\hspace{0.05cm}{\rm E}}(t)$&nbsp; and&nbsp; $z_{2,\hspace{0.05cm}{\rm E}}(t)$&nbsp; and band-limiting leads to the signals
 :$$v_1(t)  =  \cos(\Delta \phi_{\rm T}) \cdot q_1(t) - \sin(\Delta \phi_{\rm T}) \cdot q_2(t),$$
 :$$ v_2(t)  =  \sin(\Delta \phi_{\rm T}) \cdot q_1(t) + \cos(\Delta \phi_{\rm T}) \cdot q_2(t) \hspace{0.05cm}.$$
-Daraus ist zu ersehen:
+From this it can be seen:
-*Bei einem Phasenversatz von&nbsp; $Δϕ_{\rm T}  = 30^\circ$&nbsp; beinhaltet das Sinkensignal&nbsp; $v_1(t)$&nbsp; nicht nur das um&nbsp; $\cos(30^\circ) = 0.866$&nbsp; gedämpfte Signal&nbsp; $q_1(t)$, sondern auch die in&nbsp; $q_2(t)$&nbsp; enthaltene Frequenz&nbsp; $f_2$.
+*With a phase offset of &nbsp; $Δϕ_{\rm T}  = 30^\circ$&nbsp;, the sink signal&nbsp; $v_1(t)$&nbsp; includes not only the signal &nbsp; $q_1(t)$ attenuated by about &nbsp; $\cos(30^\circ) = 0.866$&nbsp;, but also the frequency &nbsp; $f_2$ contained in &nbsp; $q_2(t)$&nbsp;.
-*Diese ist mit dem Faktor&nbsp; $\sin(30^\circ) = 0.5$&nbsp; gewichtet.
+*This is weighted by the factor &nbsp; $\sin(30^\circ) = 0.5$&nbsp;.
-*Es liegen somit nichtlineare Verzerrungen vor  &nbsp; &rArr; &nbsp; <u>Antwort 3</u>.
+*Thus, nonlinear distortions are present &nbsp; &rArr; &nbsp; <u>Answer 3</u>.
 {{ML-Fuß}}

@@ Line 38: / Line 38: @@
 <quiz display=simple>
-{Calculate the transmitted signal &nbsp;$s(t)$&nbsp; in the case that &nbsp;$f_1 ≠ f_2$.&nbsp; Which of the following statements apply
+{Calculate the transmitted signal &nbsp;$s(t)$&nbsp; in the case that &nbsp;$f_1 ≠ f_2$.&nbsp; Which of the following statements apply?
 |type="()"}
 - $s(t)$&nbsp; is composed of two cosine and two sine oscillations.