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

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	[[Category:Modulation Methods: Exercises\|^3.1 Phase Modulation^]]		[[Category:Modulation Methods: Exercises\|^3.1 Phase Modulation^]]
	[[de:Aufgaben:3.~~3_Summe_zweier_Schwingungen~~]]		[[de:Aufgaben:Aufgabe 3.3: Summe zweier Schwingungen]]

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	:$$q_1(t) = 0.9\,{\rm V} \cdot \sin(2 \pi \cdot 1\,{\rm kHz} \cdot t)$$		:$$q_1(t) = 0.9\,{\rm V} \cdot \sin(2 \pi \cdot 1\,{\rm kHz} \cdot t)$$
	The weights of the Bessel-Dirac delta lines when  $η_1 = 0.9$  are obtained as follows:		The weights of the Bessel-Dirac delta lines when  $η_1 = 0.9$  are obtained as follows:
	:$${\rm J}_0 (0.9) = 0.808 \approx 0.8,\hspace{1cm}		:$${\rm J}_0 (0.9) = 0.808 \approx 0.8,\hspace{1cm}{\rm J}_1 (0.9) = 0.406 \approx 0.4,$$
	{\rm J}_1 (0.9) = 0.406 \approx 0.4,$$		:$${\rm J}_2 (0.9) = 0.095 \approx 0.1,\hspace{1cm}{\rm J}_3 (0.9) \approx {\rm J}_4 (0.9) \approx\ \text{ ...} \ \approx 0 \hspace{0.05cm}.$$
	:$${\rm J}_2 (0.9) = 0.095 \approx 0.1,\hspace{1cm}
	{\rm J}_3 (0.9) \approx {\rm J}_4 (0.9) \approx\ \text{ ...} \ \approx 0 \hspace{0.05cm}.$$
	Use the approximations given in the graph to simplify the calculations.		Use the approximations given in the graph to simplify the calculations.

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	'''(4)'''  Now, before multiplication and summation there needs to be a frequency shift of   $B_2(f)$  to the right – or of  $B_1(f)$  to the left– by   $1 \ \rm kHz$ .  This gives:		'''(4)'''  Now, before multiplication and summation there needs to be a frequency shift of   $B_2(f)$  to the right – or of  $B_1(f)$  to the left– by   $1 \ \rm kHz$ .  This gives:
	:$$S_{\rm TP}(f = 1\,{\rm kHz}) = B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f = 3\,{\rm kHz})		:$$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)= 0.1 \cdot {\rm j} \cdot 0.3 + 0.4 \cdot 0.9\hspace{0.15cm} = 0.36 + {\rm j} \cdot 0.03$$
	+ B_{1}(f = 1\,{\rm kHz}) \cdot B_{2}(f = 0)
	= 0.1 \cdot {\rm j} \cdot 0.3 + 0.4 \cdot 0.9\hspace{0.15cm} = 0.36 + {\rm j} \cdot 0.03$$
	:$$\Rightarrow \hspace{0.3cm} {\rm Re}[S_{\rm TP}(f = 1\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.36} \hspace{0.05cm},\hspace{0.3cm} {\rm Im}[S_{\rm TP}(f = 1\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.03} \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} {\rm Re}[S_{\rm TP}(f = 1\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.36} \hspace{0.05cm},\hspace{0.3cm} {\rm Im}[S_{\rm TP}(f = 1\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.03} \hspace{0.05cm}.$$

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	'''(5)'''  The Dirac delta line  $S_+(f = 98 \ \rm kHz)$  corresponds to the   $S_{\rm TP}(f)$–line at  $f = -2 \ \rm kHz$.  This is		'''(5)'''  The Dirac delta line  $S_+(f = 98 \ \rm kHz)$  corresponds to the   $S_{\rm TP}(f)$–line at  $f = -2 \ \rm kHz$.  This is
	:$$S_{\rm TP}(f \hspace{-0.05cm}=\hspace{-0.05cm} -2\,{\rm kHz}) \hspace{-0.03cm}=\hspace{-0.03cm} B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} 0) +		:$$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$$
	B_{1}(f \hspace{-0.05cm}=\hspace{-0.05cm} 1\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} -3\,{\rm kHz})= 0.1 \cdot 0.9 + 0.4 \cdot {\rm j} \cdot 0.3 \hspace{0.15cm}\hspace{-0.03cm}=\hspace{-0.03cm} 0.09 + {\rm j} \cdot 0.12$$
	:$$\Rightarrow \hspace{0.3cm} {\rm Re}[S_{\rm +}(f = 98\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.09} \hspace{0.05cm}, \hspace{0.3cm} {\rm Im}[S_{\rm +}(f = 98\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.12} \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} {\rm Re}[S_{\rm +}(f = 98\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.09} \hspace{0.05cm}, \hspace{0.3cm} {\rm Im}[S_{\rm +}(f = 98\,{\rm kHz})] \hspace{0.15cm}\underline {= 0.12} \hspace{0.05cm}.$$

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	[[Category:Modulation Methods: Exercises\|^3.1 Phase Modulation^]]		[[Category:Modulation Methods: Exercises\|^3.1 Phase Modulation^]]
			[[de:Aufgaben:3.3_Summe_zweier_Schwingungen]]

Hwang at 14:21, 18 January 2023

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

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Solution

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

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Solution

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

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	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Phasenmodulation~~ (PM)		{{quiz-Header\|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
	}}		}}

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Solution

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	'''(1)'''  The <u>third answer</u> is correct:		'''(1)'''  The <u>third answer</u> is correct:
	*~~Bei Winkelmodulation bewegt sich der komplexe Zeiger~~  $s_{\rm TP}(t)$  ~~stets auf einem Kreisbogen mit folgendem Öffnungswinkel~~:		*In angle modulation, the complex pointer   $s_{\rm TP}(t)$  always moves on a circular arc the following opening angle:
	:$$2 · K_{\rm PM} · q_{\rm max} = 2 \cdot {\rm 1/V} \cdot 1.45 \ \rm V = 2.9 \ \rm rad \approx 166^\circ.$$		:$$2 · K_{\rm PM} · q_{\rm max} = 2 \cdot {\rm 1/V} \cdot 1.45 \ \rm V = 2.9 \ \rm rad \approx 166^\circ.$$
	*~~Mit der~~ (~~zugegebenermaßen sehr groben~~) ~~Näherung~~  $166^\circ \approx 180^\circ$  ~~ergibt sich tatsächlich ein Halbkreis~~.		*Using the (admittedly very rough) approximation  $166^\circ \approx 180^\circ$  we indeed get a semicircle.



	'''(2)'''  ~~Es gilt allgemein~~  $S_{\rm TP}(f) = B_1(f) ∗ B_2(f)$.		'''(2)'''  In general,   $S_{\rm TP}(f) = B_1(f) ∗ B_2(f)$ holds.
	*Da  $B_1(f)$  ~~auf Frequenzen~~  $\|f\| ≤ 2 \ \rm kHz$  ~~und~~  $B_2(f)$  ~~auf den Bereich~~  $±3 \ \rm kHz$  ~~begrenzt sind~~, ~~ist das Faltungsprodukt auf~~  $\|f\| ≤ 5 \ \rm kHz$  ~~beschränkt~~.		*Since  $B_1(f)$  is limited to the frequencies   $\|f\| ≤ 2 \ \rm kHz$  and  $B_2(f)$  is limited to the range   $±3 \ \rm kHz$ , the convolution product is limited to  $\|f\| ≤ 5 \ \rm kHz$ .
	*~~Daraus folgt~~:		*It follows that:
	:$$f_{\rm min} \hspace{0.15cm}\underline {= -5 \ \rm kHz},$$		:$$f_{\rm min} \hspace{0.15cm}\underline {= -5 \ \rm kHz},$$
	:$$f_{\rm max} \hspace{0.15cm}\underline {=+5 \ \rm kHz}.$$		:$$f_{\rm max} \hspace{0.15cm}\underline {=+5 \ \rm kHz}.$$
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	'''(3)'''  ~~Das Faltungsprodukt für die Frequenz~~  $f = 0$  ~~ergibt sich durch Multiplikation von~~  $B_1(f)$  ~~mit~~  $B_2(f)$  ~~und anschließender Summation~~.		'''(3)'''  The convolution product for frequency   $f = 0$  results from multiplying  $B_1(f)$  with  $B_2(f)$  and summing.
	*~~Nur für~~  $f = 0$  ~~sind sowohl~~  $B_1(f)$  ~~als auch~~  $B_2(f)$  ~~von Null verschieden~~.		*Only for   $f = 0$  are both  $B_1(f)$  and  $B_2(f)$  non-zero.
	*~~Damit erhält man~~:		*Thus, we get:
	:$$ S_{\rm TP}(f = 0) = B_{1}(f = 0) \cdot B_{2}(f = 0)= 0.8 \cdot 0.9 \hspace{0.15cm}\underline {= 0.72}\hspace{0.2cm}{\rm (rein \hspace{0.15cm} reell)} \hspace{0.05cm}.$$		:$$ S_{\rm TP}(f = 0) = B_{1}(f = 0) \cdot B_{2}(f = 0)= 0.8 \cdot 0.9 \hspace{0.15cm}\underline {= 0.72}\hspace{0.2cm}{\rm (rein \hspace{0.15cm} reell)} \hspace{0.05cm}.$$



	'''(4)'''  ~~Nun muss vor der Multiplikation und Summation noch eine Frequenzverschiebung von~~  $B_2(f)$  ~~nach rechts~~ – ~~oder von~~  $B_1(f)$  ~~nach links – um~~  $1 \ \rm kHz$  erfolgen.  ~~Somit erhält man~~:		'''(4)'''  Now, before multiplication and summationm there needs to be a frequency shift of   $B_2(f)$  to the right – or of  $B_1(f)$  to the left– by   $1 \ \rm kHz$  erfolgen.  This gives:
	:$$S_{\rm TP}(f = 1\,{\rm kHz}) = B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f = 3\,{\rm kHz})		:$$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)		+ B_{1}(f = 1\,{\rm kHz}) \cdot B_{2}(f = 0)
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	'''(5)'''  ~~Die Diraclinie~~  $S_+(f = 98 \ \rm kHz)$  ~~entspricht der~~  $S_{\rm TP}(f)$~~–Linie bei~~  $f = -2 \ \rm kHz$.  ~~Diese ist~~		'''(5)'''  The Dirac line  $S_+(f = 98 \ \rm kHz)$  corresponds to the   $S_{\rm TP}(f)$–line at  $f = -2 \ \rm kHz$.  This is
	:$$S_{\rm TP}(f \hspace{-0.05cm}=\hspace{-0.05cm} -2\,{\rm kHz}) \hspace{-0.03cm}=\hspace{-0.03cm} B_{1}(f = -2\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} 0) +		:$$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$$		B_{1}(f \hspace{-0.05cm}=\hspace{-0.05cm} 1\,{\rm kHz}) \cdot B_{2}(f \hspace{-0.05cm}=\hspace{-0.05cm} -3\,{\rm kHz})= 0.1 \cdot 0.9 + 0.4 \cdot {\rm j} \cdot 0.3 \hspace{0.15cm}\hspace{-0.03cm}=\hspace{-0.03cm} 0.09 + {\rm j} \cdot 0.12$$

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	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Richtig ist die~~ <u>~~dritte Antwort~~</u>:		'''(1)'''  The <u>third answer</u> is correct:
	*Bei Winkelmodulation bewegt sich der komplexe Zeiger  $s_{\rm TP}(t)$  stets auf einem Kreisbogen mit folgendem Öffnungswinkel:		*Bei Winkelmodulation bewegt sich der komplexe Zeiger  $s_{\rm TP}(t)$  stets auf einem Kreisbogen mit folgendem Öffnungswinkel:
	:$$2 · K_{\rm PM} · q_{\rm max} = 2 \cdot {\rm 1/V} \cdot 1.45 \ \rm V = 2.9 \ \rm rad \approx 166^\circ.$$		:$$2 · K_{\rm PM} · q_{\rm max} = 2 \cdot {\rm 1/V} \cdot 1.45 \ \rm V = 2.9 \ \rm rad \approx 166^\circ.$$

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