Aufgaben:Exercise 2.8: Asymmetrical Channel - Revision history

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Reed: /* Solution */

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Solution

Reed at 19:40, 20 December 2021

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← Older revision		Revision as of 15:30, 31 March 2022
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−	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Hüllkurvendemodulation~~	+	{{quiz-Header\|Buchseite=Modulation_Methods/Envelope_Demodulation
	}}		}}

@@ Line 75: / Line 75: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  For a cosine-shaped source signal and attenuation of the upper sideband, it holds that:
+'''(1)'''&nbsp;  For a cosine-shaped source signal and attenuation of the upper sideband,&nbsp; it holds that:
 :$$ r_{\rm TP}(t) = A_{\rm T} + \frac{A_{\rm N}}{2} \cdot \alpha_{\rm O} \cdot{\rm e}^{{\rm j} \cdot \hspace{0.03cm}\omega_{\rm N}\cdot t} + \frac{A_{\rm N}}{2} \cdot{\rm e}^{-{\rm j} \cdot \hspace{0.03cm}\omega_{\rm N}\cdot t}\hspace{0.05cm}.$$
 *At time &nbsp; $t = 0$&nbsp; all vectors point in the direction of the real axis.
@@ Line 85: / Line 85: @@
 '''(2)'''&nbsp;  The carrier amplitude is defined by the center of the ellipse:
 &nbsp; $A_{\rm T}\hspace{0.15cm}\underline { = 10 \ \rm V}$.
-*From the equation given in the first subtask, the amplitude&nbsp; $A_{\rm N}$&nbsp; can thus also be calculated:
+*From the equation given in the first subtask,&nbsp; the amplitude&nbsp; $A_{\rm N}$&nbsp; can thus also be calculated:
 :$$ \frac{A_{\rm N}}{2} \cdot ( 1+ \alpha_0) = r_{\rm TP}(t= 0) - A_{\rm T} = 5 \,{\rm V}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}A_{\rm N} \hspace{0.15cm}\underline {= 8 \,{\rm V}} \hspace{0.05cm}.$$
 *The point marked &nbsp; '''(2)'''&nbsp; can be used as a check:
@@ Line 92: / Line 92: @@
-'''(3)'''&nbsp;  The necessary time for one cycle t&nbsp; $t_1$&nbsp; is equal to the period of the source signal, i.e.,
+'''(3)'''&nbsp;  The necessary time for one cycle &nbsp; $t_1$&nbsp; is equal to the time period of the source signal:
 :$$t_1= 1/f_{\rm N} \hspace{0.15cm}\underline {=0.5 \ \rm ms}.$$
-'''(4)'''&nbsp;  Since the lower sideband is larger than the upper sideband, the peak of the pointer composite moves clockwise around the ellipse.
+'''(4)'''&nbsp;  Since the lower sideband is larger than the upper sideband,&nbsp; the peak of the pointer composite moves clockwise around the ellipse.
-*Point&nbsp; '''(2)'''&nbsp; is first reached at time&nbsp; $t_2 = 3/4 · t_1\hspace{0.15cm}\underline { = 0.375 \ \rm ms}$&nbsp; .
+*Point&nbsp; '''(2)'''&nbsp; is first reached at time&nbsp; $t_2 = 3/4 · t_1\hspace{0.15cm}\underline { = 0.375 \ \rm ms}$.
@@ Line 107: / Line 107: @@
 *The phase function is:
 :$$\phi(t = t_2) = {\rm arctan} \frac{3 \,{\rm V}}{10 \,{\rm V}} \hspace{0.15cm}\underline {= 16.7^{\circ}}\hspace{0.05cm}.$$
-*The maximum phase&nbsp; $ϕ_{\rm max}$&nbsp; is slightly larger.&nbsp; It occurs (with a positive sign) at time &nbsp; $t_3 < t_2$&nbsp; when a straight line from the origin is tangent to the ellipse.
+*The maximum phase&nbsp; $ϕ_{\rm max}$&nbsp; is slightly larger.&nbsp;
 *By setting up the ellipse equation, this point &nbsp; $(x_3$,&nbsp; $y_3)$&nbsp; can be accurately calculated analytically.
 *From this, the following would hold for the maximum phase:
-:&nbsp; $\phi_{\rm max} = {\rm arctan} \ {y_3}/{x_3} \hspace{0.05cm}.$
+:$$\phi_{\rm max} = {\rm arctan} \ {y_3}/{x_3} \hspace{0.05cm}.$$
-'''(6)'''&nbsp;  The distortion factors of second and third order can be obtained from the equation given for $v(t)$&nbsp; $($valid for $f_{\rm N} = 2 \ \rm kHz)$&nbsp;, and are:
+'''(6)'''&nbsp;  The distortion factors of second and third order can be obtained from the equation given for&nbsp; $v(t)$&nbsp; $($valid for $f_{\rm N} = 2 \ \rm kHz)$:
 :$$ K_2 = \frac{0.148 \,{\rm V}}{2.424 \,{\rm V}} = 0.061, \hspace{0.3cm} K_3 = \frac{0.056 \,{\rm V}}{2.424 \,{\rm V}} = 0.023 \hspace{0.05cm}.$$
-*Thus, for the total distortion factor we get:
+*Thus,&nbsp; for the total distortion factor we get:
 :$$K = \sqrt{K_2^2 + K_3^2 }\hspace{0.15cm}\underline { \approx 6.6 \%}.$$
-'''(7)'''&nbsp;  From the power of the useful signal and the interference signal, we obtain:
+'''(7)'''&nbsp;  From the power of the useful signal and the interference signal,&nbsp; we obtain:
 :$$ P_{v 1} = \frac{(2.424 \,{\rm V})^2}{2} = 2.94 \,{\rm V}^2\hspace{0.05cm},\hspace{0.3cm} P_{\varepsilon} = \frac{(-0.148 \,{\rm V})^2}{2} + \frac{(0.056 \,{\rm V})^2}{2}= 0.0125 \,{\rm V}^2\hspace{0.05cm}$$
 *This gives the signal-to-noise power ratio &nbsp; $\rm (SNR)$:
 :$$\rho_{v} = \frac{P_{v 1}}{P_{\varepsilon }}= \frac{(2.94 \,{\rm V})^2}{0.0125 \,{\rm V}^2} \hspace{0.15cm}\underline {\approx 230} = \frac{1}{K^2} \hspace{0.05cm}.$$
-*If, on the other hand, the amplitude distortion were also assigned to the error signal, we would arrive at a much smaller&nbsp; $\rm SNR$.
+*If,&nbsp; on the other hand,&nbsp; the amplitude distortion were also assigned to the error signal,&nbsp; we would arrive at a much smaller&nbsp; $\rm SNR$.
-*When $P_q = A_{\rm N}^2/2 = 8 \ \rm V^2$&nbsp; and&nbsp; $P_{\varepsilon}\hspace{0.02cm}' = \overline{(v(t)-q(t))^2} = {1}/{2}\cdot ( 4 \,{\rm V} - 2.424 \,{\rm V})^2 + P_{\varepsilon}= 1.254 \,{\rm V}^2$&nbsp; one would get:
+*When&nbsp; $P_q = A_{\rm N}^2/2 = 8 \ \rm V^2$&nbsp; and&nbsp; $P_{\varepsilon}\hspace{0.02cm}' = \overline{(v(t)-q(t))^2} = {1}/{2}\cdot ( 4 \,{\rm V} - 2.424 \,{\rm V})^2 + P_{\varepsilon}= 1.254 \,{\rm V}^2$&nbsp; one would get:
 :$$\rho_{v }\hspace{0.02cm}' = \frac{8 \,{\rm V}^2}{1.254 \,{\rm V}^2} \approx 6.4\hspace{0.05cm}.$$
 '''(8)'''&nbsp;  All calculations are valid regardless of the message frequency &nbsp; $f_{\rm N}$ if the attenuation factor of the upper sideband remains at &nbsp; $α_{\rm O} = 0.25$&nbsp;.
-*Thus, the same distortion factor &nbsp; $K\hspace{0.15cm}\underline { \approx 6.6 \%}$ is obtained even for &nbsp; $f_{\rm N} = 4 \ \rm kHz$&nbsp;.
+*Thus,&nbsp; the same distortion factor &nbsp; $K\hspace{0.15cm}\underline { \approx 6.6 \%}$&nbsp; is obtained even for &nbsp; $f_{\rm N} = 4 \ \rm kHz$&nbsp;.
 {{ML-Fuß}}

@@ Line 4: / Line 4: @@
 [[File:P_ID1038__Mod_A_2_8.png|right|frame|Equivalent low-pass signal  <br>in the complex plane]]
-A cosine-shaped source signal &nbsp;$q(t)$&nbsp; with amplitude &nbsp;$A_{\rm N}$&nbsp; and frequency &nbsp;$f_{\rm N}$&nbsp; is DSB amplitude modulated, such that the modulated signal is given by:
+A cosine-shaped source signal &nbsp;$q(t)$&nbsp; with amplitude &nbsp;$A_{\rm N}$&nbsp; and frequency &nbsp;$f_{\rm N}$&nbsp; is double-sideband amplitude modulated.&nbsp; The modulated signal is given by:
 :$$ s(t) = \big[ q(t) + A_{\rm T}\big] \cdot \cos(2 \pi \cdot f_{\rm T} \cdot t ) \hspace{0.05cm}.$$
 The transmission channel exhibits linear distortions:
 *While the lower sideband $($LSB frequency: &nbsp; &nbsp;$f_{\rm T} - f_{\rm N})$&nbsp; and the carrier are transmitted undistorted,
-*the upper sideband $($USB frequency: &nbsp; &nbsp;$f_{\rm T} + f_{\rm N})$&nbsp; is weighted with the attenuation factor  &nbsp;$α_{\rm O} = 0.25$&nbsp;.
+*the upper sideband $($USB frequency: &nbsp; &nbsp;$f_{\rm T} + f_{\rm N})$&nbsp; is weighted with the attenuation factor  &nbsp;$α_{\rm O} = 0.25$.
-The graph shows the locus curve, i.e., the representation of the equivalent low-pass signal &nbsp;$r_{\rm TP}(t)$&nbsp; in the complex plane.
+The graph shows the locus curve, i.e.,&nbsp; the representation of the equivalent low-pass signal &nbsp;$r_{\rm TP}(t)$&nbsp; in the complex plane.
-Evaluating the signal &nbsp;$r(t)$&nbsp; with an ideal envelope demodulator, we obtain a sink signal&nbsp;$v(t)$, which can be approximated as follows:
+Evaluating the signal &nbsp;$r(t)$&nbsp; with an ideal envelope demodulator,&nbsp; we obtain a sink signal&nbsp;$v(t)$,&nbsp; which can be approximated as follows:
 :$$v(t) = 2.424 \,{\rm V} \cdot \cos(\omega_{\rm N} \cdot t ) -0.148 \,{\rm V} \cdot \cos(2\omega_{\rm N} \cdot t )+ 0.056 \,{\rm V} \cdot \cos(3\omega_{\rm N} \cdot t )-\text{ ...}$$
-For this measurement, the message frequency &nbsp;$f_{\rm N} = 2 \ \rm kHz$&nbsp; was used.
+For this measurement,&nbsp; the message frequency &nbsp;$f_{\rm N} = 2 \ \rm kHz$&nbsp; was used.
 In subtask&nbsp; '''(7)'''&nbsp; the signal-to-noise power ratio &nbsp; $\rm (SNR)$&nbsp; should be calculated as follows:
-:$$ \rho_{v } = \frac{P_{v 1}}{P_{\varepsilon }} \hspace{0.05cm}.$$
+:$$ \rho_{v } = P_{v 1}/P_{\varepsilon } \hspace{0.05cm}.$$
-Here, &nbsp;$P_{v1} = α^2 · P_q$&nbsp; and &nbsp;$P_ε$&nbsp; denote the "powers" of both signals:
+Here, &nbsp;$P_{v1} = α^2 · P_q$&nbsp; and &nbsp;$P_ε$&nbsp; denote the&nbsp; "powers"&nbsp; of both signals:
 :$$ v_1(t)  =  2.424 \,{\rm V} \cdot \cos(\omega_{\rm N} \cdot t )\hspace{0.05cm},$$
 :$$ \varepsilon(t)  =  v(t) - v_1(t) \approx -0.148 \,{\rm V} \cdot \cos(2\omega_{\rm N} \cdot t )+ 0.056 \,{\rm V} \cdot \cos(3\omega_{\rm N} \cdot t ) \hspace{0.05cm}.$$
@@ Line 27: / Line 27: @@
 Hints:
 *This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Envelope_Demodulation|Envelope Demodulation]].
-*Particular reference is made to the page&nbsp;   [[Modulation_Methods/Envelope_Demodulation#Description_using_the_equivalent_low-pass_signal|Description using the equivalent low-pass signal]].
+*Particular reference is made to the section&nbsp;   [[Modulation_Methods/Envelope_Demodulation#Description_using_the_equivalent_low-pass_signal|Description using the equivalent low-pass signal]].
@@ Line 41: / Line 38: @@
 <quiz display=simple>
-{Give the low-pass signal &nbsp;$r_{\rm TP}(t)$&nbsp; in its analytical form. What value results for time &nbsp;$t = 0$?
+{Give the low-pass signal &nbsp;$r_{\rm TP}(t)$&nbsp; in its analytical form.&nbsp; What value results for time &nbsp;$t = 0$?
 |type="{}"}
 $r_{\rm TP}(t=0) \ = \ $ { 15 3% } $\ \rm V$
@@ Line 50: / Line 47: @@
 $A_{\rm N} \ = \ $ { 8 3% } $\ \rm V$
-{Let &nbsp;$f_{\rm N} \hspace{0.15cm}\underline{= 2 \ \rm kHz}$.&nbsp; At which time &nbsp;$t_1$&nbsp; is the starting point &nbsp; '''(1)'''&nbsp; first reached again after &nbsp;$t = 0$&nbsp;?
+{Let &nbsp;$f_{\rm N} \hspace{0.15cm}\underline{= 2 \ \rm kHz}$.&nbsp; At which time &nbsp;$t_1$&nbsp; is the starting point &nbsp; '''(1)'''&nbsp; first reached again after &nbsp;$t = 0$?
 |type="{}"}
 $t_1 \ = \ $ { 0.5 3% } $\ \rm ms$
-{At which point in time &nbsp;$t_2$&nbsp; is the elliptical point &nbsp; '''(2)'''&nbsp; with value &nbsp;$\rm j · 3\ V$&nbsp; first reached?
+{At which time  &nbsp;$t_2$&nbsp; is the elliptical point &nbsp; '''(2)'''&nbsp; with value &nbsp;$\rm j · 3\ V$&nbsp;  reached first?
 |type="{}"}
 $t_2 \ = \ $ { 0.375 3% } $\ \rm ms$
-{Calculate the  magnitude function (envelope) &nbsp;$a(t)$&nbsp; and the phase function &nbsp;$ϕ(t)$&nbsp; for this time point &nbsp;$t_2$.
+{Calculate the  magnitude function&nbsp; (envelope) &nbsp;$a(t)$&nbsp; and the phase function &nbsp;$ϕ(t)$&nbsp; for this time point &nbsp;$t_2$.
 |type="{}"}
 $a(t = t_2) \ = \ $ { 10.44 3% } $\ \rm V$

@@ Line 119: / Line 119: @@
 '''(6)'''&nbsp;  The distortion factors of second and third order can be obtained from the equation given for $v(t)$&nbsp; $($valid for $f_{\rm N} = 2 \ \rm kHz)$&nbsp;, and are:
 :$$ K_2 = \frac{0.148 \,{\rm V}}{2.424 \,{\rm V}} = 0.061, \hspace{0.3cm} K_3 = \frac{0.056 \,{\rm V}}{2.424 \,{\rm V}} = 0.023 \hspace{0.05cm}.$$
-*Thus for the total distortion factor we get:
+*Thus, for the total distortion factor we get:
 :$$K = \sqrt{K_2^2 + K_3^2 }\hspace{0.15cm}\underline { \approx 6.6 \%}.$$
-'''(7)'''&nbsp;  Frot he power of the useful signal and the interference signal, we obtain:
+'''(7)'''&nbsp;  From the power of the useful signal and the interference signal, we obtain:
 :$$ P_{v 1} = \frac{(2.424 \,{\rm V})^2}{2} = 2.94 \,{\rm V}^2\hspace{0.05cm},\hspace{0.3cm} P_{\varepsilon} = \frac{(-0.148 \,{\rm V})^2}{2} + \frac{(0.056 \,{\rm V})^2}{2}= 0.0125 \,{\rm V}^2\hspace{0.05cm}$$
 *This gives the signal-to-noise power ratio &nbsp; $\rm (SNR)$:

@@ Line 8: / Line 8: @@
 The transmission channel exhibits linear distortions:
 *While the lower sideband $($LSB frequency: &nbsp; &nbsp;$f_{\rm T} - f_{\rm N})$&nbsp; and the carrier are transmitted undistorted,
-*the upper sideband $($USB-Frequenz: &nbsp; &nbsp;$f_{\rm T} + f_{\rm N})$&nbsp; is weighted with the attenuation factor  &nbsp;$α_{\rm O} = 0.25$&nbsp;.
+*the upper sideband $($USB frequency: &nbsp; &nbsp;$f_{\rm T} + f_{\rm N})$&nbsp; is weighted with the attenuation factor  &nbsp;$α_{\rm O} = 0.25$&nbsp;.

@@ Line 117: / Line 117: @@
-'''(6)'''&nbsp;  The distortion factors of second and third ordern can be obtained from the given equation for nbsp; $v(t)$&nbsp; $($valid for $f_{\rm N} = 2 \ \rm kHz)$&nbsp;, and are:
+'''(6)'''&nbsp;  The distortion factors of second and third order can be obtained from the given equation for nbsp; $v(t)$&nbsp; $($valid for $f_{\rm N} = 2 \ \rm kHz)$&nbsp;, and are:
 :$$ K_2 = \frac{0.148 \,{\rm V}}{2.424 \,{\rm V}} = 0.061, \hspace{0.3cm} K_3 = \frac{0.056 \,{\rm V}}{2.424 \,{\rm V}} = 0.023 \hspace{0.05cm}.$$
 *Thus for the total distortion factor we get:

@@ Line 78: / Line 78: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  Bei cosinusförmigem Quellensignal und Dämpfung des oberen Seitenbandes gilt:
+'''(1)'''&nbsp;  For a cosine-shaped source signal and attenuation of the upper sideband, it holds that:
 :$$ r_{\rm TP}(t) = A_{\rm T} + \frac{A_{\rm N}}{2} \cdot \alpha_{\rm O} \cdot{\rm e}^{{\rm j} \cdot \hspace{0.03cm}\omega_{\rm N}\cdot t} + \frac{A_{\rm N}}{2} \cdot{\rm e}^{-{\rm j} \cdot \hspace{0.03cm}\omega_{\rm N}\cdot t}\hspace{0.05cm}.$$
-*Zum Zeitpunkt&nbsp; $t = 0$&nbsp; zeigen alle Vektoren in Richtung der reellen Achse.
+*At time &nbsp; $t = 0$&nbsp; all vectors point in the direction of the real axis.
-*Somit kann aus der Grafik auf der Angabenseite&nbsp; $r_{\rm TP}(t = 0)\hspace{0.15cm}\underline { = 15 \ \rm V}$&nbsp; abgelesen werden.
+*Thus &nbsp; $r_{\rm TP}(t = 0)\hspace{0.15cm}\underline { = 15 \ \rm V}$&nbsp; can be read from the graph on the exercise page.
-'''(2)'''&nbsp;  Die Trägeramplitude ist durch den Ellipsenmittelpunkt festgelegt:&nbsp; $A_{\rm T}\hspace{0.15cm}\underline { = 10 \ \rm V}$.
+'''(2)'''&nbsp;  The carrier amplitude is defined by the center of the ellipse:
-*Aus der in der ersten Teilaufgabe angegebenen Gleichung kann somit auch die Amplitude&nbsp; $A_{\rm N}$&nbsp; berechnet werden:
+&nbsp; $A_{\rm T}\hspace{0.15cm}\underline { = 10 \ \rm V}$.
 :$$ \frac{A_{\rm N}}{2} \cdot ( 1+ \alpha_0) = r_{\rm TP}(t= 0) - A_{\rm T} = 5 \,{\rm V}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}A_{\rm N} \hspace{0.15cm}\underline {= 8 \,{\rm V}} \hspace{0.05cm}.$$
-*Zur Kontrolle kann der in der Grafik markierte Punkt&nbsp; '''(2)'''&nbsp; herangezogen werden:
+*The point marked &nbsp; '''(2)'''&nbsp; can be used as a check:
 :$$\frac{A_{\rm N}}{2} \cdot ( 1- \alpha_0) = 3 \,{\rm V}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}A_{\rm N} = 8 \,{\rm V} \hspace{0.05cm}.$$
-'''(3)'''&nbsp;  Die für einen Umlauf benötigte Zeit&nbsp; $t_1$&nbsp; ist gleich der Periodendauer des Quellensignals, also
+'''(3)'''&nbsp;  The necessary time for one cycle t&nbsp; $t_1$&nbsp; is equal to the period of the source signal, i.e.,
 :$$t_1= 1/f_{\rm N} \hspace{0.15cm}\underline {=0.5 \ \rm ms}.$$
-'''(4)'''&nbsp;  Da das USB größer ist als das OSB, bewegt sich die Spitze des Zeigerverbundes auf der Ellipse im Uhrzeigersinn.
+'''(4)'''&nbsp;  Since the lower sideband is larger than the upper sideband, the peak of the pointer composite moves clockwise around the ellipse.
-*Der Punkt&nbsp; '''(2)'''&nbsp; wird zum Zeitpunkt&nbsp; $t_2 = 3/4 · t_1\hspace{0.15cm}\underline { = 0.375 \ \rm ms}$&nbsp; zum ersten Mal erreicht.
+*Point&nbsp; '''(2)'''&nbsp; is first reached at time&nbsp; $t_2 = 3/4 · t_1\hspace{0.15cm}\underline { = 0.375 \ \rm ms}$&nbsp; .
-[[File:P_ID1039__Mod_A_2_8_e.png|right|frame|Zur Berechnung von&nbsp; $t_2$&nbsp; und&nbsp; $t_3$]]
+[[File:P_ID1039__Mod_A_2_8_e.png|right|frame|Calculation of &nbsp; $t_2$&nbsp; and&nbsp; $t_3$]]
-'''(5)'''&nbsp;  Die Zeigerlänge zur Zeit&nbsp; $t_2$&nbsp; kann mit dem&nbsp; [https://de.wikipedia.org/wiki/Satz_des_Pythagoras Satz von Pythagoras]&nbsp; bestimmt werden:
+'''(5)'''&nbsp;  The pointer length at time &nbsp; $t_2$&nbsp; can be determined with the &nbsp; [https://en.wikipedia.org/wiki/Pythagorean_theorem Pythagorean Theorem]&nbsp;:
 :$$ a(t = t_2) = \sqrt{(10 \,{\rm V})^2 + (3 \,{\rm V})^2}\hspace{0.15cm}\underline { = 10.44 \,{\rm V}}\hspace{0.05cm}.$$
-*Für die Phasenfunktion gilt:
+*The phase function is:
 :$$\phi(t = t_2) = {\rm arctan} \frac{3 \,{\rm V}}{10 \,{\rm V}} \hspace{0.15cm}\underline {= 16.7^{\circ}}\hspace{0.05cm}.$$
-*Die maximale Phase&nbsp; $ϕ_{\rm max}$&nbsp; ist geringfügig größer.&nbsp; Sie tritt (mit positivem Vorzeichen) zum Zeitpunkt&nbsp; $t_3 < t_2$&nbsp; dann auf, wenn eine Gerade vom Koordinatenursprung die Ellipse tangiert.
+*The maximum phase&nbsp; $ϕ_{\rm max}$&nbsp; is slightly larger.&nbsp; It occurs (with a positive sign) at time &nbsp; $t_3 < t_2$&nbsp; when a straight line from the origin is tangent to the ellipse.
-*Durch Aufstellen der Ellipsengleichung kann dieser Punkt&nbsp; $(x_3$,&nbsp; $y_3)$&nbsp; analytisch exakt berechnet werden.
+*By setting up the ellipse equation, this point &nbsp; $(x_3$,&nbsp; $y_3)$&nbsp; can be accurately calculated analytically.
-*Daraus würde für die maximale Phase gelten:&nbsp; $\phi_{\rm max} = {\rm arctan} \ {y_3}/{x_3} \hspace{0.05cm}.$
+*From this, the following would hold for the maximum phase:
-'''(6)'''&nbsp;  Die Klirrfaktoren zweiter und dritter Ordnung können aus der angegebenen Gleichung für&nbsp; $v(t)$&nbsp; $($gültig für $f_{\rm N} = 2 \ \rm kHz)$&nbsp; ermittelt werden und  lauten:
+'''(6)'''&nbsp;  The distortion factors of second and third ordern can be obtained from the given equation for nbsp; $v(t)$&nbsp; $($valid for $f_{\rm N} = 2 \ \rm kHz)$&nbsp;, and are:
 :$$ K_2 = \frac{0.148 \,{\rm V}}{2.424 \,{\rm V}} = 0.061, \hspace{0.3cm} K_3 = \frac{0.056 \,{\rm V}}{2.424 \,{\rm V}} = 0.023 \hspace{0.05cm}.$$
-*Damit erhält man für den Gesamtklirrfaktor:
+*Thus for the total distortion factor we get:
 :$$K = \sqrt{K_2^2 + K_3^2 }\hspace{0.15cm}\underline { \approx 6.6 \%}.$$
-'''(7)'''&nbsp;  Für die Leistungen von Nutz– und Störsignal erhält man:
+'''(7)'''&nbsp;  Frot he power of the useful signal and the interference signal, we obtain:
 :$$ P_{v 1} = \frac{(2.424 \,{\rm V})^2}{2} = 2.94 \,{\rm V}^2\hspace{0.05cm},\hspace{0.3cm} P_{\varepsilon} = \frac{(-0.148 \,{\rm V})^2}{2} + \frac{(0.056 \,{\rm V})^2}{2}= 0.0125 \,{\rm V}^2\hspace{0.05cm}$$
-*Damit ergibt sich für das Signal–zu–Stör–Leistungsverhältnis&nbsp; $\rm (SNR)$:
+*This gives the signal-to-noise power ratio &nbsp; $\rm (SNR)$:
 :$$\rho_{v} = \frac{P_{v 1}}{P_{\varepsilon }}= \frac{(2.94 \,{\rm V})^2}{0.0125 \,{\rm V}^2} \hspace{0.15cm}\underline {\approx 230} = \frac{1}{K^2} \hspace{0.05cm}.$$
-*Würde man dagegen die Amplitudenverfälschung ebenfalls dem Fehlersignal zuweisen, so käme man zu einem deutlich kleineren&nbsp; $\rm SNR$.&nbsp; &nbsp;
+*If, on the other hand, the amplitude distortion were also assigned to the error signal, we would arrive at a much smaller&nbsp; $\rm SNR$.
-*Mit $P_q = A_{\rm N}^2/2 = 8 \ \rm V^2$&nbsp; und&nbsp; $P_{\varepsilon}\hspace{0.02cm}' = \overline{(v(t)-q(t))^2} = {1}/{2}\cdot ( 4 \,{\rm V} - 2.424 \,{\rm V})^2 + P_{\varepsilon}= 1.254 \,{\rm V}^2$&nbsp; würde man dann erhalten:
+*When $P_q = A_{\rm N}^2/2 = 8 \ \rm V^2$&nbsp; and&nbsp; $P_{\varepsilon}\hspace{0.02cm}' = \overline{(v(t)-q(t))^2} = {1}/{2}\cdot ( 4 \,{\rm V} - 2.424 \,{\rm V})^2 + P_{\varepsilon}= 1.254 \,{\rm V}^2$&nbsp; one would get:
 :$$\rho_{v }\hspace{0.02cm}' = \frac{8 \,{\rm V}^2}{1.254 \,{\rm V}^2} \approx 6.4\hspace{0.05cm}.$$
-'''(8)'''&nbsp;  Alle Berechnungen gelten unabhängig von der Nachrichtenfrequenz&nbsp; $f_{\rm N}$, wenn der Dämpfungsfaktor des OSB weiterhin&nbsp; $α_{\rm O} = 0.25$&nbsp; beträgt.
+'''(8)'''&nbsp;  All calculations are valid regardless of the message frequency &nbsp; $f_{\rm N}$ if the attenuation factor of the upper sideband remains at &nbsp; $α_{\rm O} = 0.25$&nbsp;.
-*Damit erhält man auch für&nbsp; $f_{\rm N} = 4 \ \rm kHz$&nbsp; den gleichen Klirrfaktor&nbsp; $K\hspace{0.15cm}\underline { \approx 6.6 \%}$.
+*Thus, the same distortion factor &nbsp; $K\hspace{0.15cm}\underline { \approx 6.6 \%}$ is obtained for &nbsp; $f_{\rm N} = 4 \ \rm kHz$&nbsp;.
 {{ML-Fuß}}

@@ Line 58: / Line 58: @@
 $t_2 \ = \ $ { 0.375 3% } $\ \rm ms$
-{Calculate the  Sie die Betragsfunktion (Hüllkurve)  &nbsp;$a(t)$&nbsp; und die Phasenfunktion &nbsp;$ϕ(t)$&nbsp; für diesen Zeitpunkt &nbsp;$t_2$.
+{Calculate the  magnitude function (envelope) &nbsp;$a(t)$&nbsp; and the phase function &nbsp;$ϕ(t)$&nbsp; for this time point &nbsp;$t_2$.
 |type="{}"}
 $a(t = t_2) \ = \ $ { 10.44 3% } $\ \rm V$
-$ϕ(t = t_2)\ = \ $ { 16.7 3% } $\ \rm Grad$
+$ϕ(t = t_2)\ = \ $ { 16.7 3% } $\ \rm degrees$
-{Berechnen Sie den Klirrfaktor &nbsp;$K$&nbsp; für &nbsp;$f_{\rm N} \hspace{0.15cm}\underline{= 2 \ \rm kHz}$.
+{Calculate the distortion factor &nbsp;$K$&nbsp; for &nbsp;$f_{\rm N} \hspace{0.15cm}\underline{= 2 \ \rm kHz}$.
 |type="{}"}
 $K \ = \ $ { 6.6 3% } $\ \text{%}$
-{Berechnen Sie für &nbsp;$f_{\rm N}\hspace{0.15cm}\underline{ = 2 \ \rm kHz}$&nbsp; das Signal–zu–Stör–Leistungsverhältnis &nbsp;$\rm (SNR)$&nbsp;  gemäß der angegebenen Definition.
+{Calculate the signal-to-noise power ratio &nbsp;$\rm (SNR)$&nbsp;for &nbsp;$f_{\rm N}\hspace{0.15cm}\underline{ = 2 \ \rm kHz}$&nbsp; according to the given definition.
 |type="{}"}
 $ρ_v \ = \ $ { 230 3% }
-{ Welcher Klirrfaktor &nbsp;$K$&nbsp; ergibt sich bei ansonsten gleichen Bedingungen mit der Nachrichtenfrequenz &nbsp;$f_{\rm N} \hspace{0.15cm}\underline{= 4 \ \rm kHz}$?
+{What distortion factor results from otherwise equal conditions for the message frequency &nbsp;$f_{\rm N} \hspace{0.15cm}\underline{= 4 \ \rm kHz}$?
 |type="{}"}
 $K \ = \ $ { 6.6 3% } $\ \text{%}$
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp;  Bei cosinusförmigem Quellensignal und Dämpfung des oberen Seitenbandes gilt: