Aufgaben:Exercise 3.3: Sum of two Oscillations - Revision history

Hwang at 14:21, 18 January 2023

2023-01-18T14:21:30Z

Reed at 15:01, 9 April 2022

2022-04-09T15:01:48Z

Reed: /* Solution */

2022-04-09T15:01:24Z

Solution

Reed: /* Solution */

2022-04-09T15:00:54Z

Solution

Reed at 14:51, 9 April 2022

2022-04-09T14:51:38Z

Reed: /* Solution */

2022-03-14T17:05:30Z

Solution

Reed at 16:18, 14 March 2022

2022-03-14T16:18:44Z

Reed: /* Questions */

2022-03-14T16:17:48Z

Questions

Reed at 15:42, 14 March 2022

2022-03-14T15:42:50Z

Reed at 15:42, 14 March 2022

2022-03-14T15:42:16Z

@@ Line 11: / Line 11: @@
 The upper graph shows the corresponding spectral function &nbsp;$B_1(f)$, when the source signal is
 :$$q_1(t) = 0.9\,{\rm V} \cdot \sin(2 \pi \cdot 1\,{\rm kHz} \cdot t)$$
-The weights of the Bessel-Dirac lines when &nbsp;$η_1 = 0.9$&nbsp; are obtained as follows:
+The weights of the Bessel-Dirac delta lines when &nbsp;$η_1 = 0.9$&nbsp; are obtained as follows:
 :$${\rm J}_0 (0.9)  = 0.808 \approx 0.8,\hspace{1cm}
 {\rm J}_1 (0.9) = 0.406 \approx 0.4,$$
@@ Line 20: / Line 20: @@
 The Bessel function &nbsp;$B_2(f)$&nbsp; is obtained for the source signal
 :$$q_2(t) = 0.65\,{\rm V} \cdot \cos(2 \pi \cdot 3\,{\rm kHz} \cdot t)$$
-The numerical values of the Dirac lines are obtained from the following:
+The numerical values of the Dirac delta lines are obtained from the following:
 :$${\rm J}_0 (0.65) = 0.897 \approx 0.9,\hspace{0.3cm}{\rm J}_1 (0.65) = 0.308 \approx 0.3, \hspace{0.3cm}{\rm J}_2 (0.65) = 0.051 \approx 0\hspace{0.05cm}.$$
 From the graph, it can be seen that due to the cosine source signal &nbsp;$q_2(t)$&nbsp; and the cosine carrier signal&nbsp;$z(t)$&nbsp;, the spectral lines at&nbsp;$±3 \ \rm kHz$&nbsp; are both positive and imaginary.
@@ Line 117: / Line 117: @@
-'''(5)'''&nbsp; The Dirac line&nbsp; $S_+(f = 98 \ \rm kHz)$&nbsp; corresponds to the &nbsp; $S_{\rm TP}(f)$–line at&nbsp; $f = -2 \ \rm kHz$.&nbsp; This is
+'''(5)'''&nbsp; The Dirac delta line&nbsp; $S_+(f = 98 \ \rm kHz)$&nbsp; corresponds to the &nbsp; $S_{\rm TP}(f)$–line at&nbsp; $f = -2 \ \rm kHz$.&nbsp; This is
 :$$S_{\rm TP}(f \hspace{-0.05cm}=\hspace{-0.05cm} -2\,{\rm kHz}) \hspace{-0.03cm}=\hspace{-0.03cm}  B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} 0) +
   B_{1}(f \hspace{-0.05cm}=\hspace{-0.05cm} 1\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} -3\,{\rm kHz})= 0.1 \cdot 0.9 + 0.4 \cdot {\rm j} \cdot 0.3 \hspace{0.15cm}\hspace{-0.03cm}=\hspace{-0.03cm} 0.09 + {\rm j} \cdot 0.12$$

@@ Line 109: / Line 109: @@
-'''(4)'''&nbsp; Now, before multiplication and summation there needs to be a frequency shift of &nbsp; $B_2(f)$&nbsp; to the right – or of&nbsp; $B_1(f)$&nbsp; to the left– by &nbsp; $1 \ \rm kHz$&nbsp; erfolgen.&nbsp; This gives:
+'''(4)'''&nbsp; Now, before multiplication and summation there needs to be a frequency shift of &nbsp; $B_2(f)$&nbsp; to the right – or of&nbsp; $B_1(f)$&nbsp; to the left– by &nbsp; $1 \ \rm kHz$&nbsp;.&nbsp; This gives:
 :$$S_{\rm TP}(f = 1\,{\rm kHz}) =  B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f = 3\,{\rm kHz})
   +  B_{1}(f = 1\,{\rm kHz}) \cdot B_{2}(f = 0)

@@ Line 109: / Line 109: @@
-'''(4)'''&nbsp; Now, before multiplication and summationm there needs to be a frequency shift of &nbsp; $B_2(f)$&nbsp; to the right – or of&nbsp; $B_1(f)$&nbsp; to the left– by &nbsp; $1 \ \rm kHz$&nbsp; erfolgen.&nbsp; This gives:
+'''(4)'''&nbsp; Now, before multiplication and summation there needs to be a frequency shift of &nbsp; $B_2(f)$&nbsp; to the right – or of&nbsp; $B_1(f)$&nbsp; to the left– by &nbsp; $1 \ \rm kHz$&nbsp; erfolgen.&nbsp; This gives:
 :$$S_{\rm TP}(f = 1\,{\rm kHz}) =  B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f = 3\,{\rm kHz})
   +  B_{1}(f = 1\,{\rm kHz}) \cdot B_{2}(f = 0)

@@ Line 88: / Line 88: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; The <u>third answer</u> is correct:
-*In angle modulation, the complex pointer &nbsp; $s_{\rm TP}(t)$&nbsp; always moves on a circular arc the following opening angle:
+*In angle modulation, the complex pointer &nbsp; $s_{\rm TP}(t)$&nbsp; always moves on a circular arc with the following opening angle:
 :$$2 · K_{\rm PM} · q_{\rm max} = 2 \cdot  {\rm 1/V} \cdot 1.45 \ \rm V = 2.9 \ \rm rad \approx 166^\circ.$$
 *Using the (admittedly very rough) approximation&nbsp; $166^\circ \approx 180^\circ$&nbsp; we indeed get a semicircle.

← Older revision		Revision as of 14:51, 9 April 2022
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−	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Phasenmodulation~~ (PM)	+	{{quiz-Header\|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
	}}		}}

@@ Line 88: / Line 88: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; The <u>third answer</u> is correct:
-*Bei Winkelmodulation bewegt sich der komplexe Zeiger&nbsp; $s_{\rm TP}(t)$&nbsp; stets auf einem Kreisbogen mit folgendem Öffnungswinkel:
+*In angle modulation, the complex pointer &nbsp; $s_{\rm TP}(t)$&nbsp; always moves on a circular arc the following opening angle:
 :$$2 · K_{\rm PM} · q_{\rm max} = 2 \cdot  {\rm 1/V} \cdot 1.45 \ \rm V = 2.9 \ \rm rad \approx 166^\circ.$$
-*Mit der (zugegebenermaßen sehr groben) Näherung&nbsp; $166^\circ \approx 180^\circ$&nbsp; ergibt sich tatsächlich ein Halbkreis.
+*Using the (admittedly very rough) approximation&nbsp; $166^\circ \approx 180^\circ$&nbsp; we indeed get a semicircle.
-'''(2)'''&nbsp; Es gilt allgemein&nbsp; $S_{\rm TP}(f) = B_1(f) ∗ B_2(f)$.
+'''(2)'''&nbsp; In general, &nbsp; $S_{\rm TP}(f) = B_1(f) ∗ B_2(f)$ holds.
-*Da&nbsp; $B_1(f)$&nbsp; auf Frequenzen&nbsp; $|f| ≤ 2 \ \rm kHz$&nbsp; und&nbsp; $B_2(f)$&nbsp; auf den Bereich&nbsp; $±3 \ \rm  kHz$&nbsp; begrenzt sind, ist das Faltungsprodukt auf&nbsp; $|f| ≤ 5 \ \rm kHz$&nbsp; beschränkt.
+*Since&nbsp; $B_1(f)$&nbsp; is limited to the frequencies &nbsp; $|f| ≤ 2 \ \rm kHz$&nbsp; and&nbsp; $B_2(f)$&nbsp; is limited to the range &nbsp; $±3 \ \rm  kHz$&nbsp;, the convolution product is limited to&nbsp; $|f| ≤ 5 \ \rm kHz$&nbsp;.
-*Daraus folgt:
+*It follows that:
 :$$f_{\rm min} \hspace{0.15cm}\underline {= -5 \ \rm kHz},$$
 :$$f_{\rm max} \hspace{0.15cm}\underline {=+5 \ \rm kHz}.$$
@@ Line 102: / Line 102: @@
-'''(3)'''&nbsp; Das Faltungsprodukt für die Frequenz&nbsp; $f = 0$&nbsp; ergibt sich durch Multiplikation von&nbsp; $B_1(f)$&nbsp; mit&nbsp; $B_2(f)$&nbsp; und anschließender Summation.
+'''(3)'''&nbsp; The convolution product for frequency &nbsp; $f = 0$&nbsp; results from multiplying&nbsp; $B_1(f)$&nbsp; with&nbsp; $B_2(f)$&nbsp; and  summing.
-*Nur für&nbsp; $f = 0$&nbsp; sind sowohl&nbsp; $B_1(f)$&nbsp; als auch&nbsp; $B_2(f)$&nbsp; von Null verschieden.
+*Only for &nbsp; $f = 0$&nbsp; are both&nbsp; $B_1(f)$&nbsp; and&nbsp; $B_2(f)$&nbsp; non-zero.
-*Damit erhält man:
+*Thus, we get:
 :$$ S_{\rm TP}(f = 0) = B_{1}(f = 0) \cdot B_{2}(f = 0)= 0.8 \cdot 0.9 \hspace{0.15cm}\underline {= 0.72}\hspace{0.2cm}{\rm (rein \hspace{0.15cm} reell)} \hspace{0.05cm}.$$
-'''(4)'''&nbsp; Nun muss vor der Multiplikation und Summation noch eine Frequenzverschiebung von&nbsp; $B_2(f)$&nbsp; nach rechts – oder von&nbsp; $B_1(f)$&nbsp; nach links – um&nbsp; $1 \ \rm kHz$&nbsp; erfolgen.&nbsp; Somit erhält man:
+'''(4)'''&nbsp; Now, before multiplication and summationm there needs to be a frequency shift of &nbsp; $B_2(f)$&nbsp; to the right – or of&nbsp; $B_1(f)$&nbsp; to the left– by &nbsp; $1 \ \rm kHz$&nbsp; erfolgen.&nbsp; This gives:
 :$$S_{\rm TP}(f = 1\,{\rm kHz}) =  B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f = 3\,{\rm kHz})
   +  B_{1}(f = 1\,{\rm kHz}) \cdot B_{2}(f = 0)
@@ Line 117: / Line 117: @@
-'''(5)'''&nbsp; Die Diraclinie&nbsp; $S_+(f = 98 \ \rm kHz)$&nbsp; entspricht der&nbsp; $S_{\rm TP}(f)$–Linie bei&nbsp; $f = -2 \ \rm kHz$.&nbsp; Diese ist
+'''(5)'''&nbsp; The Dirac line&nbsp; $S_+(f = 98 \ \rm kHz)$&nbsp; corresponds to the &nbsp; $S_{\rm TP}(f)$–line at&nbsp; $f = -2 \ \rm kHz$.&nbsp; This is
 :$$S_{\rm TP}(f \hspace{-0.05cm}=\hspace{-0.05cm} -2\,{\rm kHz}) \hspace{-0.03cm}=\hspace{-0.03cm}  B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} 0) +
   B_{1}(f \hspace{-0.05cm}=\hspace{-0.05cm} 1\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} -3\,{\rm kHz})= 0.1 \cdot 0.9 + 0.4 \cdot {\rm j} \cdot 0.3 \hspace{0.15cm}\hspace{-0.03cm}=\hspace{-0.03cm} 0.09 + {\rm j} \cdot 0.12$$

@@ Line 57: / Line 57: @@
 <quiz display=simple>
-{Es gelte &nbsp;$q(t) = q_1(t)+q_2(t)$.&nbsp; Welche geometrische Figur beschreibt die angegebene Ortskurve &nbsp;$s_{\rm TP}(t)$?
+{Let &nbsp;$q(t) = q_1(t)+q_2(t)$.&nbsp; Which geometric figure describes the given locus curve&nbsp;$s_{\rm TP}(t)$?
 |type="()"}
-- Die Ortskurve ist eine Ellipse.
+- The locus curve is an ellipse.
-- Die Ortskurve ist ein Kreis.
+- The locus curve is a circle.
-+ Die Ortskurve ist näherungsweise ein Halbkreis.
++ The locus curve is approximately a semi-circle.
-- Die Ortskurve ist ein Kreisbogen, etwa mit Öffnungswinkel &nbsp;$90^\circ$.
+- The locus curve is an arc, with an approximate opening angle of  &nbsp;$90^\circ$.
-{Berechnen Sie die Spektralfunktion &nbsp;$S_{\rm TP}(f)$.&nbsp; Zwischen welchen Frequenzen &nbsp;$f_{\rm min}$&nbsp; und &nbsp;$f_{\rm max}$&nbsp; liegen Spektrallinien?
+{Calculate the spectral function&nbsp;$S_{\rm TP}(f)$.&nbsp; Between what frequencies &nbsp;$f_{\rm min}$&nbsp; and &nbsp;$f_{\rm max}$&nbsp; do the spectral lines lie?
 |type="{}"}
 $f_{\rm min} \ = \ $ { -5.15--4.85 } $\ \rm kHz$
 $f_{\rm max} \ = \ ${ 5 3% } $\ \rm kHz$
-{Berechnen Sie das Gewicht der Diracfunktion bei &nbsp;$f = 0$.
+{Calculate the weight of the Dirac function at &nbsp;$f = 0$.
 |type="{}"}
 ${\rm Re}\big[S_{\rm TP}(f = 0)\big] \ = \ $ { 0.72 3% }
 ${\rm Im}\big[S_{\rm TP}(f = 0)\big] \ = \ $ { 0. }
-{Berechnen Sie das Gewicht der Diracfunktion bei &nbsp;$f = 1\ \rm  kHz$.
+{Calculate the weight of the Dirac function at &nbsp;$f = 1\ \rm  kHz$.
 |type="{}"}
 ${\rm Re}\big[S_{\rm TP}(f = 1 \ \rm  kHz)\big] \ = \ $ { 0.36 3% }
 ${\rm Im}\big[S_{\rm TP}(f = 1 \ \rm  kHz)\big] \ = \ $ { 0.03 3% }
-{Berechnen Sie das Gewicht der &nbsp;$S_+(f)$–Diracfunktion bei &nbsp;$f = 98 \ \rm kHz$.
+{Calculate the weight of the &nbsp;$S_+(f)$–Dirac function at &nbsp;$f = 98 \ \rm kHz$.
 |type="{}"}
 ${\rm Re}\big[S_{\rm +}(f = 98 \ \rm  kHz)\big] \ = \ $ { 0.09 3% }

@@ Line 49: / Line 49: @@
 *Particular reference is made to the page&nbsp;  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation|Equivalent low-pass signal in phase modulation]].
 *The values of the Bessel functions can be found in formula collections in table form.
-*Sie können zur Lösung dieser Aufgabe auch das interaktive Applet&nbsp; [[Applets:Besselfunktionen_erster_Art| Besselfunktion erster Art]]&nbsp; benutzen.
+*You can also use the interactive applet &nbsp; [[Applets:Bessel_functions_of_the_first_kind| Bessel functions of the first kind]]&nbsp; to solve this task.
-===Fragebogen===
+===Questions===
 <quiz display=simple>