Aufgaben:Exercise 4.15: Optimal Signal Space Allocation - Revision history

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	[[Category:Digital Signal Transmission: Exercises\|^4.4 Coherent Demodulation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.4 Coherent Demodulation^]]
	[[de:Aufgaben:4.~~15_Optimale_Signalraumbelegung~~]]		[[de:Aufgaben:Aufgabe 4.15: Optimale Signalraumbelegung]]

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	* Four further points lie offset by  $45^\circ$  on a second circle with radius  $R$,  where the following shall hold:		* Four further points lie offset by  $45^\circ$  on a second circle with radius  $R$,  where the following shall hold:
	:$$R_{\rm min} \le R \le R_{\rm max}\hspace{0.05cm},\hspace{0.2cm} R_{\rm min}= \frac{ \sqrt{3}-1}{ \sqrt{2}} \approx 0.518		:$$R_{\rm min} \le R \le R_{\rm max}\hspace{0.05cm},\hspace{0.2cm} R_{\rm min}= \frac{ \sqrt{3}-1}{ \sqrt{2}} \approx 0.518\hspace{0.05cm},\hspace{0.2cm}R_{\rm max}= \frac{ \sqrt{3}+1}{ \sqrt{2}} \approx 1.932\hspace{0.05cm}.$$
	\hspace{0.05cm},\hspace{0.2cm}
	R_{\rm max}= \frac{ \sqrt{3}+1}{ \sqrt{2}} \approx 1.932\hspace{0.05cm}.$$

	Let the two axes  ("basis functions")  be normalized respectively and denoted  $I$  and  $Q$  for simplicity.  For further simplification,  $E = 1$  can be set.		Let the two axes  ("basis functions")  be normalized respectively and denoted  $I$  and  $Q$  for simplicity.  For further simplification,  $E = 1$  can be set.
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	'''(1)'''  Because of  $M = 8$   ⇒   $b = 3$,  the average signal energy per bit is  $E_{\rm B} = E_{\rm S}/3$,  where the average signal energy per symbol  $(E_{\rm S})$  is to be calculated as the mean square distance of the signal space points from the origin.  With  $r = 1$  one obtains:		'''(1)'''  Because of  $M = 8$   ⇒   $b = 3$,  the average signal energy per bit is  $E_{\rm B} = E_{\rm S}/3$,  where the average signal energy per symbol  $(E_{\rm S})$  is to be calculated as the mean square distance of the signal space points from the origin.  With  $r = 1$  one obtains:
	:$$E_{\rm S} = {1}/{8 } \cdot ( 4 \cdot r^2 + 4 \cdot R^2) = ({1 + R^2})/{2 }$$		:$$E_{\rm S} = {1}/{8 } \cdot ( 4 \cdot r^2 + 4 \cdot R^2) = ({1 + R^2})/{2 }$$
	:$$\Rightarrow \hspace{0.3cm} E_{\rm B} = {E_{\rm S}}/{3} = ({1 + R^2})/{6}		:$$\Rightarrow \hspace{0.3cm} E_{\rm B} = {E_{\rm S}}/{3} = ({1 + R^2})/{6}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	In particular:		In particular:
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	'''(3)'''  The graphic illustrates the geometric calculation.  With  "Pythagoras"  one obtains:		'''(3)'''  The graphic illustrates the geometric calculation.  With  "Pythagoras"  one obtains:
	:$$d_{\rm min}^2 =(R/\sqrt{2})^2 + (R/\sqrt{2}-1)^2 = 1 - \sqrt{2} \cdot R + R^2$$		:$$d_{\rm min}^2 =(R/\sqrt{2})^2 + (R/\sqrt{2}-1)^2 = 1 - \sqrt{2} \cdot R + R^2$$
	:$$ \Rightarrow \hspace{0.3cm}d_{\rm min} = \sqrt{ 1 - \sqrt{2} \cdot R + R^2}		:$$ \Rightarrow \hspace{0.3cm}d_{\rm min} = \sqrt{ 1 - \sqrt{2} \cdot R + R^2}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*In particular,  for $R = 1$  ("8–PSK"):		*In particular,  for $R = 1$  ("8–PSK"):
	:$$d_{\rm min} = \sqrt{ 2 - \sqrt{2} } \hspace{0.1cm} \underline{= 0.765} \hspace{0.1cm} (= 2 \cdot \sin (22.5^{\circ}) )		:$$d_{\rm min} = \sqrt{ 2 - \sqrt{2} } \hspace{0.1cm} \underline{= 0.765} \hspace{0.1cm} (= 2 \cdot \sin (22.5^{\circ}) )\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*In contrast,  for  $\underline {R = \sqrt{2}}$  corresponding to the right graph for subtask  '''(1)''',  the minimum distance is  $d_{\rm min} \ \underline {= 1}$.		*In contrast,  for  $\underline {R = \sqrt{2}}$  corresponding to the right graph for subtask  '''(1)''',  the minimum distance is  $d_{\rm min} \ \underline {= 1}$.
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	'''(4)'''  Using the results of  '''(1)'''  and  '''(3)''',  we obtain in general or  for $R = 1$ ("8–PSK"):		'''(4)'''  Using the results of  '''(1)'''  and  '''(3)''',  we obtain in general or  for $R = 1$ ("8–PSK"):
	:$$\eta = \frac{ d_{\rm min}^2}{ 4 \cdot E_{\rm B}} = \frac{ 1 - \sqrt{2} \cdot R + R^2}{ 4 \cdot (1 + R^2)/6}		:$$\eta = \frac{ d_{\rm min}^2}{ 4 \cdot E_{\rm B}} = \frac{ 1 - \sqrt{2} \cdot R + R^2}{ 4 \cdot (1 + R^2)/6}= \frac{ 3/2 \cdot(1 - \sqrt{2} \cdot R + R^2)}{ 1 + R^2}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} R = 1: \hspace{0.2cm}\eta =\frac{ 3/2 \cdot(2 - \sqrt{2}) }{ 2} = 3/4 \cdot(2 - \sqrt{2})\hspace{0.1cm} \underline{\approx 0.439}\hspace{0.05cm}.$$
	= \frac{ 3/2 \cdot(1 - \sqrt{2} \cdot R + R^2)}{ 1 + R^2}\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} R = 1: \hspace{0.2cm}\eta =
	\frac{ 3/2 \cdot(2 - \sqrt{2}) }{ 2} = 3/4 \cdot(2 - \sqrt{2})\hspace{0.1cm} \underline{\approx 0.439}\hspace{0.05cm}.$$


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	'''(5)'''  For  $R = R_{\rm min} = (\sqrt{3}-1)/\sqrt{2}$,  the following value is obtained:		'''(5)'''  For  $R = R_{\rm min} = (\sqrt{3}-1)/\sqrt{2}$,  the following value is obtained:
	:$$\eta = \frac{ 3/2 \cdot(1 - \sqrt{2} \cdot R + R^2)}{ 1 + R^2} = 3/2 \cdot \left [ 1 - \frac{ \sqrt{2} \cdot R }{ 1 + R^2}\right ]\hspace{0.05cm},$$		:$$\eta = \frac{ 3/2 \cdot(1 - \sqrt{2} \cdot R + R^2)}{ 1 + R^2} = 3/2 \cdot \left [ 1 - \frac{ \sqrt{2} \cdot R }{ 1 + R^2}\right ]\hspace{0.05cm},$$
	:$$\sqrt{2} \cdot R = \sqrt{3}- 1\hspace{0.05cm},\hspace{0.2cm} 1 + R^2 = 3 - \sqrt{3} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$$\sqrt{2} \cdot R = \sqrt{3}- 1\hspace{0.05cm},\hspace{0.2cm} 1 + R^2 = 3 - \sqrt{3} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\eta = 3/2 \cdot \left [ 1 - \frac{ \sqrt{3}- 1 }{ 3 - \sqrt{3}}\right ]\hspace{0.1cm} \underline{\approx 0.634}\hspace{0.05cm}.$$
	\eta = 3/2 \cdot \left [ 1 - \frac{ \sqrt{3}- 1 }{ 3 - \sqrt{3}}\right ]\hspace{0.1cm} \underline{\approx 0.634}\hspace{0.05cm}.$$

	*For  $R = R_{\rm max}= (\sqrt{3}+1)/\sqrt{2}$  exactly the same value results.		*For  $R = R_{\rm max}= (\sqrt{3}+1)/\sqrt{2}$  exactly the same value results.

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	[[Category:Digital Signal Transmission: Exercises\|^4.4 Coherent Demodulation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.4 Coherent Demodulation^]]
			[[de:Aufgaben:4.15_Optimale_Signalraumbelegung]]

Guenter at 15:06, 1 October 2022

2022-10-01T15:06:29Z

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	<br clear=all>		<br clear=all>
	'''(2)'''  <u>All statements are true</u>:		'''(2)'''  <u>All statements are true</u>:
	[[File:P_ID2074__Dig_A_4_15c.png\|right\|frame\|To calculate ~~the~~ minimum distance]]		[[File:P_ID2074__Dig_A_4_15c.png\|right\|frame\|To calculate minimum distance]]

	*In the drawn example on the front page with  $R = R_{\rm max}$,  the distance between two neighboring blue points is exactly the same as the distance between a red (outer) and a blue (inner) point.		*In the drawn example on the front page with  $R = R_{\rm max}$,  the distance between two neighboring blue points is exactly the same as the distance between a red (outer) and a blue (inner) point.

Hwang at 19:45, 1 September 2022

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	{{ML-Kopf}}		{{ML-Kopf}}
	[[File:P_ID2073__Dig_A_4_15a.png\|right\|frame\|Special cases of  "8–QAM"]]		[[File:P_ID2073__Dig_A_4_15a.png\|right\|frame\|Special cases of  "8–QAM"]]
	'''(1)'''  Because of  $M = 8$   ⇒   $b = 3$,  the ~~averae~~ signal energy per bit is  $E_{\rm B} = E_{\rm S}/3$,  where the average signal energy per symbol  $(E_{\rm S})$  is to be calculated as the mean square distance of the signal space points from the origin.  With  $r = 1$  one obtains:		'''(1)'''  Because of  $M = 8$   ⇒   $b = 3$,  the average signal energy per bit is  $E_{\rm B} = E_{\rm S}/3$,  where the average signal energy per symbol  $(E_{\rm S})$  is to be calculated as the mean square distance of the signal space points from the origin.  With  $r = 1$  one obtains:
	:$$E_{\rm S} = {1}/{8 } \cdot ( 4 \cdot r^2 + 4 \cdot R^2) = ({1 + R^2})/{2 }$$		:$$E_{\rm S} = {1}/{8 } \cdot ( 4 \cdot r^2 + 4 \cdot R^2) = ({1 + R^2})/{2 }$$
	:$$\Rightarrow \hspace{0.3cm} E_{\rm B} = {E_{\rm S}}/{3} = ({1 + R^2})/{6}		:$$\Rightarrow \hspace{0.3cm} E_{\rm B} = {E_{\rm S}}/{3} = ({1 + R^2})/{6}

Guenter at 16:03, 23 August 2022

2022-08-23T16:03:46Z

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	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation}}

	[[File:P_ID2069__Dig_A_4_15.png\|right\|frame\|Considered 8–QAM]]		[[File:P_ID2069__Dig_A_4_15.png\|right\|frame\|Considered  "8–QAM"]]
	A signal space constellation with  $M = 8$  signal space points is considered here:		A signal space constellation with  $M = 8$  signal space points is considered here:
	* Four points lie on a circle with radius  $r = 1$.		* Four points lie on a circle with radius  $r = 1$.
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Because of $M = 8$  ⇒  $b = 3$, the ~~mean~~ signal energy per bit is $E_{\rm B} = E_{\rm S}/3$, where the ~~mean~~ signal energy per symbol ($E_{\rm S}$) is to be calculated as the mean square distance of the signal space points from the origin. With $r = 1$ one obtains:		[[File:P_ID2073__Dig_A_4_15a.png\|right\|frame\|Special cases of  "8–QAM"]]
	~~[[File:P_ID2073__Dig_A_4_15a.png\|right\|frame\|Special cases of 8–QAM]]~~		'''(1)'''  Because of  $M = 8$   ⇒   $b = 3$,  the averae signal energy per bit is  $E_{\rm B} = E_{\rm S}/3$,  where the average signal energy per symbol  $(E_{\rm S})$  is to be calculated as the mean square distance of the signal space points from the origin.  With  $r = 1$  one obtains:
	:$$E_{\rm S} = {1}/{8 } \cdot ( 4 \cdot r^2 + 4 \cdot R^2) = ({1 + R^2})/{2 }		:$$E_{\rm S} = {1}/{8 } \cdot ( 4 \cdot r^2 + 4 \cdot R^2) = ({1 + R^2})/{2 }$$
	~~\hspace{0.3cm}~~\Rightarrow \hspace{0.3cm} E_{\rm B} = {E_{\rm S}}/{3} = ({1 + R^2})/{6}		:$$\Rightarrow \hspace{0.3cm} E_{\rm B} = {E_{\rm S}}/{3} = ({1 + R^2})/{6}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	In particular:		In particular:
	* For $R = 1$, there is an 8–PSK   ⇒   $E_{\rm S} = 1$ and $E_{\rm B} \ \underline {= 0.333}$ (see left graph).		* For  $R = 1$,  there is an  "8–PSK"   ⇒   $E_{\rm S} = 1$  and  $E_{\rm B} \ \underline {= 0.333}$  (see left graph).
	* The right graph is valid for $R = \sqrt{2}$. In this case, $E_{\rm B} \ \underline {= 0.5}$.
			* The right graph is valid for  $R = \sqrt{2}$.  In this case,  $E_{\rm B} \ \underline {= 0.5}$.


	Note that these energies actually still have to be multiplied by the normalization energy $E$.		Note that these energies actually still have to be multiplied by the normalization energy $E$.
			<br clear=all>
			'''(2)'''  <u>All statements are true</u>:
			[[File:P_ID2074__Dig_A_4_15c.png\|right\|frame\|To calculate the minimum distance]]

			*In the drawn example on the front page with  $R = R_{\rm max}$,  the distance between two neighboring blue points is exactly the same as the distance between a red (outer) and a blue (inner) point.

			*For $R > R_{\rm max}$,  the distance between two blue points is the smallest.

			*For $R < R_{\rm min}$,  the minimum distance occurs between two red points.

	~~'''(2)'''  <u>All statements are true</u>:~~
	*In the drawn example on the information section with $R = R_{\rm max}$, the distance between two neighboring blue points is exactly the same as the distance between a red (outer) and a blue (inner) point.
	~~[[File:P_ID2074__Dig_A_4_15c.png\|right\|frame\|To calculate the minimum distance]]~~
	*For $R > R_{\rm max}$, the distance between two blue points is the smallest.
	*For $R < R_{\rm min}$, the minimum distance occurs between two red points.


	'''(3)'''  The graphic illustrates the geometric calculation. With "Pythagoras" one obtains:		'''(3)'''  The graphic illustrates the geometric calculation.  With  "Pythagoras"  one obtains:
	:$$d_{\rm min}^2 =(R/\sqrt{2})^2 + (R/\sqrt{2}-1)^2 = 1 - \sqrt{2} \cdot R + R^2 ~~\hspace{0.3cm}~~		:$$d_{\rm min}^2 =(R/\sqrt{2})^2 + (R/\sqrt{2}-1)^2 = 1 - \sqrt{2} \cdot R + R^2$$
	\Rightarrow \hspace{0.3cm}d_{\rm min} = \sqrt{ 1 - \sqrt{2} \cdot R + R^2}		:$$ \Rightarrow \hspace{0.3cm}d_{\rm min} = \sqrt{ 1 - \sqrt{2} \cdot R + R^2}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	In particular, for $R = 1$ (8–PSK):		*In particular,  for $R = 1$  ("8–PSK"):
	:$$d_{\rm min} = \sqrt{ 2 - \sqrt{2} } \hspace{0.1cm} \underline{= 0.765} \hspace{0.1cm} (= 2 \cdot \sin (22.5^{\circ}) )		:$$d_{\rm min} = \sqrt{ 2 - \sqrt{2} } \hspace{0.1cm} \underline{= 0.765} \hspace{0.1cm} (= 2 \cdot \sin (22.5^{\circ}) )
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	In contrast, for $\underline {R = \sqrt{2}}$ corresponding to the right graph for subtask '''(1)''', the minimum distance is $d_{\rm min} \ \underline {= 1}$.		*In contrast,  for  $\underline {R = \sqrt{2}}$  corresponding to the right graph for subtask  '''(1)''',  the minimum distance is  $d_{\rm min} \ \underline {= 1}$.


	'''(4)'''  Using the results of '''(1)''' and '''(3)''', we obtain in general or for $R = 1$ (8–PSK):
			'''(4)'''  Using the results of  '''(1)'''  and  '''(3)''',  we obtain in general or  for $R = 1$ ("8–PSK"):
	:$$\eta = \frac{ d_{\rm min}^2}{ 4 \cdot E_{\rm B}} = \frac{ 1 - \sqrt{2} \cdot R + R^2}{ 4 \cdot (1 + R^2)/6}		:$$\eta = \frac{ d_{\rm min}^2}{ 4 \cdot E_{\rm B}} = \frac{ 1 - \sqrt{2} \cdot R + R^2}{ 4 \cdot (1 + R^2)/6}
	= \frac{ 3/2 \cdot(1 - \sqrt{2} \cdot R + R^2)}{ 1 + R^2}\hspace{0.3cm}		= \frac{ 3/2 \cdot(1 - \sqrt{2} \cdot R + R^2)}{ 1 + R^2}\hspace{0.3cm}
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	'''(5)'''  For $R = R_{\rm min} = (\sqrt{3}-1)/\sqrt{2}$, the following value is obtained:
			'''(5)'''  For  $R = R_{\rm min} = (\sqrt{3}-1)/\sqrt{2}$,  the following value is obtained:
	:$$\eta = \frac{ 3/2 \cdot(1 - \sqrt{2} \cdot R + R^2)}{ 1 + R^2} = 3/2 \cdot \left [ 1 - \frac{ \sqrt{2} \cdot R }{ 1 + R^2}\right ]\hspace{0.05cm},$$		:$$\eta = \frac{ 3/2 \cdot(1 - \sqrt{2} \cdot R + R^2)}{ 1 + R^2} = 3/2 \cdot \left [ 1 - \frac{ \sqrt{2} \cdot R }{ 1 + R^2}\right ]\hspace{0.05cm},$$
	:$$\sqrt{2} \cdot R = \sqrt{3}- 1\hspace{0.05cm},\hspace{0.2cm} 1 + R^2 = 3 - \sqrt{3} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$$\sqrt{2} \cdot R = \sqrt{3}- 1\hspace{0.05cm},\hspace{0.2cm} 1 + R^2 = 3 - \sqrt{3} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	\eta = 3/2 \cdot \left [ 1 - \frac{ \sqrt{3}- 1 }{ 3 - \sqrt{3}}\right ]\hspace{0.1cm} \underline{\approx 0.634}\hspace{0.05cm}.$$		\eta = 3/2 \cdot \left [ 1 - \frac{ \sqrt{3}- 1 }{ 3 - \sqrt{3}}\right ]\hspace{0.1cm} \underline{\approx 0.634}\hspace{0.05cm}.$$

	For $R = R_{\rm max}= (\sqrt{3}+1)/\sqrt{2}$ exactly the same value results.		*For  $R = R_{\rm max}= (\sqrt{3}+1)/\sqrt{2}$  exactly the same value results.

	*The (always desired) maximum of the power efficiency $\eta$ results, for example, for $R = R_{\rm max}$ – i.e. for the signal space constellation in the information section.
	*In this case all triangles of two neighboring blue points and the red point in between are equilateral.
	*Also for $R = R_{\rm min}$ there are equilateral triangles, but now each formed by two red and one blue point.
	*In this case the edge length $d_{\rm min}$ is clearly smaller, but at the same time a smaller $E_{\rm B}$ results, so that the power efficiency $\eta$ has the same value.


	The previously considered special cases $R = 1$ (8–PSK, left graph in the first subtask) and $R = \sqrt{2}$ (right graph) have a noticeably smaller $\eta$ with $\eta = 0.439$ and $\eta = 0.5$, ~~respectively~~ (compared to $\eta = 0.634$).		#The  (always desired)  maximum of the power efficiency  $\eta$  results e.g. for  $R = R_{\rm max}$ – i.e. for the signal space constellation in the information section.
			#In this case all triangles of two neighboring blue points and the red point in between are equilateral.
			#Also for  $R = R_{\rm min}$  there are equilateral triangles,  but now each formed by two red and one blue point.
			#In this case the edge length  $d_{\rm min}$  is clearly smaller,  but at the same time a smaller  $E_{\rm B}$  results,  so that the power efficiency  $\eta$  has the same value.
			#The previously considered special cases  $R = 1$  ("8–PSK",  left graph in the first subtask)  and  $R = \sqrt{2}$  (right graph)  have a noticeably smaller  $\eta$  with  $\eta = 0.439$  and  $\eta = 0.5$,  resp.  $($compared to  $\eta = 0.634)$.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 15:39, 23 August 2022

2022-08-23T15:39:57Z

← Older revision		Revision as of 17:39, 23 August 2022
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	A signal space constellation with  $M = 8$  signal space points is considered here:		A signal space constellation with  $M = 8$  signal space points is considered here:
	* Four points lie on a circle with radius  $r = 1$.		* Four points lie on a circle with radius  $r = 1$.
	* Four further points lie offset by  $45^\circ$  on a second circle with radius  $R$, where the following shall hold:
			* Four further points lie offset by  $45^\circ$  on a second circle with radius  $R$,  where the following shall hold:
	:$$R_{\rm min} \le R \le R_{\rm max}\hspace{0.05cm},\hspace{0.2cm} R_{\rm min}= \frac{ \sqrt{3}-1}{ \sqrt{2}} \approx 0.518		:$$R_{\rm min} \le R \le R_{\rm max}\hspace{0.05cm},\hspace{0.2cm} R_{\rm min}= \frac{ \sqrt{3}-1}{ \sqrt{2}} \approx 0.518
	\hspace{0.05cm},\hspace{0.2cm}		\hspace{0.05cm},\hspace{0.2cm}
	R_{\rm max}= \frac{ \sqrt{3}+1}{ \sqrt{2}} \approx 1.932\hspace{0.05cm}.$$		R_{\rm max}= \frac{ \sqrt{3}+1}{ \sqrt{2}} \approx 1.932\hspace{0.05cm}.$$

	Let the two axes (basis functions) be normalized respectively and denoted  $I$  and  $Q$  for simplicity. For further simplification,  $E = 1$  can be set.		Let the two axes  ("basis functions")  be normalized respectively and denoted  $I$  and  $Q$  for simplicity.  For further simplification,  $E = 1$  can be set.

	In the question section, we speak of blue and red points. According to the diagram, the blue points lie on the circle with radius  $r = 1$, the red points on the circle with radius  $R$. The case  $R = R_{\rm max}$ is drawn.		In the question section,  we speak of  "blue"  and  "red"  points.  According to the diagram,  the blue points lie on the circle with radius  $r = 1$,  the red points on the circle with radius  $R$.  The case  $R = R_{\rm max}$ is drawn.

	The system parameter  $R$  is to be determined in this exercise in such a way that the quotient		The system parameter  $R$  is to be determined in this exercise in such a way that the quotient
	:$$\eta = \frac{ (d_{\rm min}/2)^2}{ E_{\rm B}} $$		:$$\eta = \frac{ (d_{\rm min}/2)^2}{ E_{\rm B}} $$

	becomes maximum. $\eta$  is a measure for the quality of a modulation alphabet at given transmission energy per bit ("~~Power Efficiency~~"). It is calculated from		becomes maximum.  $\eta$  is a measure for the quality of a modulation alphabet at given transmission energy per bit  ("power efficiency").  It is calculated from
	* the minimum distance  $d_{\rm min}$, ~~and~~		* the minimum distance  $d_{\rm min}$,  and
	* the bit energy  ~~$E_{\rm B}$.~~

			* the average bit energy  $E_{\rm B}$.

	~~It must be ensured that  $d_{\rm min}^2$  and  $E_{\rm B}$  are normalized in the same way, but this is already implicit in the exercise.~~

			It must be ensured that  $d_{\rm min}^2$  and  $E_{\rm B}$  are normalized in the same way,  but this is already implicit in the exercise.




	''Notes:''
	* The exercise belongs to the chapter  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation\|"Carrier Frequency Systems with Coherent Demodulation"]].		Notes:
	* Reference is made in particular to the sections  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Quadrature_amplitude_modulation_.28M-QAM.29\|"Quadrature amplitude modulation"]]  and  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#M.E2.80.93level_amplitude_shift_keying_.28M.E2.80.93ASK.29\|"~~Multilevel~~ phase modulation"]].		* The exercise belongs to the chapter  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation\|"Carrier Frequency Systems with Coherent Demodulation"]].

			* Reference is made in particular to the sections  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Quadrature_amplitude_modulation_.28M-QAM.29\|"Quadrature amplitude modulation"]]  and  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#M.E2.80.93level_amplitude_shift_keying_.28M.E2.80.93ASK.29\|"Multi-level phase modulation"]].


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	===Questions===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{Calculate the average energy  $E_{\rm B}$  per bit depending on  $R$, in particular for  $R = 1$  and  $R = \sqrt{2}$.		{Calculate the average energy  $E_{\rm B}$  per bit depending on  $R$,  in particular for  $R = 1$  and  $R = \sqrt{2}$.
	\|type="{}"}		\|type="{}"}
	$R = 1 \text{:} \hspace{0.55cm} E_{\rm B}\ = \ $ { 0.333 3% }		$R = 1 \text{:} \hspace{0.55cm} E_{\rm B}\ = \ $ { 0.333 3% }
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	+ For  $R < R_{\rm min}$,  the minimum distance occurs between two red points.		+ For  $R < R_{\rm min}$,  the minimum distance occurs between two red points.
	+ For  $R > R_{\rm max}$,  the minimum distance occurs between two blue points.		+ For  $R > R_{\rm max}$,  the minimum distance occurs between two blue points.
	+ For  $R_{\rm min} ≤ R ≤ R_{\rm max}$,  the minimum distance occurs between "red" and "blue".		+ For  $R_{\rm min} ≤ R ≤ R_{\rm max}$,  the minimum distance occurs between  "red"  and  "blue".

	{Calculate the minimum distance depending on  $R$, in particular for		{Calculate the minimum distance depending on  $R$,  in particular for
	\|type="{}"}		\|type="{}"}
	$R = 1 \text{:} \hspace{0.55cm} d_{\rm min}\ = \ $ { 0.765 3% }		$R = 1 \text{:} \hspace{0.55cm} d_{\rm min}\ = \ $ { 0.765 3% }
	$R = \sqrt{2} \text{:} \hspace{0.2cm} d_{\rm min}\ = \ $ { 1 3% }		$R = \sqrt{2} \text{:} \hspace{0.2cm} d_{\rm min}\ = \ $ { 1 3% }

	{Give the power efficiency  $\eta$  in general terms. What  $\eta$  results for  $R = 1$?		{Give the power efficiency  $\eta$  in general terms.  What  $\eta$  results for  $R = 1$?
	\|type="{}"}		\|type="{}"}
	$\eta\ = \ $ { 0.439 3% }		$\eta\ = \ $ { 0.439 3% }

	{What power efficiency values result for  $R = R_{\rm min}$  and  $R = R_{\rm max}$? Interpretation.		{What power efficiency values result for  $R = R_{\rm min}$  and  $R = R_{\rm max}$?  Interpretation.
	\|type="{}"}		\|type="{}"}
	$R = R_{\rm min} \text{:} \hspace{0.35cm} \eta\ = \ ${ 0.634 3% }		$R = R_{\rm min} \text{:} \hspace{0.35cm} \eta\ = \ ${ 0.634 3% }

Hwang at 20:48, 15 August 2022

2022-08-15T20:48:39Z

Show changes

Hwang at 20:03, 15 August 2022

2022-08-15T20:03:45Z

← Older revision		Revision as of 22:03, 15 August 2022
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	{{quiz-Header\|Buchseite=~~Digitalsignalübertragung~~/~~Trägerfrequenzsysteme mit kohärenter Demodulation~~}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation}}

	[[File:P_ID2069__Dig_A_4_15.png\|right\|frame\|~~Betrachtete~~ 8–QAM]]		[[File:P_ID2069__Dig_A_4_15.png\|right\|frame\|Considered 8–QAM]]
	~~Betrachtet wird hier eine Signalraumkonstellation mit~~  $M = 8$  ~~Signalraumpunkten~~:		A signal space constellation with  $M = 8$  signal space points is considered here:
	* ~~Vier Punkte liegen auf einem Kreis mit Radius~~  $r = 1$.		* Four points lie on a circle with radius  $r = 1$.
	* ~~Vier weitere Punkte liegen um~~  $45^\circ$  ~~versetzt auf einem zweiten Kreis mit Radius~~  $R$, ~~wobei gelten soll~~:		* Four further points lie offset by  $45^\circ$  on a second circle with radius  $R$, where the following shall hold:
	:$$R_{\rm min} \le R \le R_{\rm max}\hspace{0.05cm},\hspace{0.2cm} R_{\rm min}= \frac{ \sqrt{3}-1}{ \sqrt{2}} \approx 0.518		:$$R_{\rm min} \le R \le R_{\rm max}\hspace{0.05cm},\hspace{0.2cm} R_{\rm min}= \frac{ \sqrt{3}-1}{ \sqrt{2}} \approx 0.518
	\hspace{0.05cm},\hspace{0.2cm}		\hspace{0.05cm},\hspace{0.2cm}
	R_{\rm max}= \frac{ \sqrt{3}+1}{ \sqrt{2}} \approx 1.932\hspace{0.05cm}.$$		R_{\rm max}= \frac{ \sqrt{3}+1}{ \sqrt{2}} \approx 1.932\hspace{0.05cm}.$$

	~~Die beiden Achsen~~ (~~Basisfunktionen~~) ~~seien jeweils normiert und werden vereinfachend mit~~  $I$  ~~und~~  $Q$  ~~bezeichnet~~. ~~Zur weiteren Vereinfachung kann~~  $E = 1$  ~~gesetzt werden~~.		Let the two axes (basis functions) be normalized respectively and denoted  $I$  and  $Q$  for simplicity. For further simplification,  $E = 1$  can be set.

	~~Im Fragebogen wird von blauen und roten Punkten gesprochen~~. ~~Entsprechend der Grafik liegen die blauen Punkte auf dem Kreis mit Radius~~  $r = 1$, ~~die roten auf dem Kreis mit Radius~~  $R$. ~~Gezeichnet ist der Fall~~  $R = R_{\rm max}$.		In the question section, we speak of blue and red points. According to the diagram, the blue points lie on the circle with radius  $r = 1$, the red points on the circle with radius  $R$. The case  $R = R_{\rm max}$ is drawn.

	~~Der Systemparameter~~  $R$  ~~soll~~ in ~~dieser Aufgabe so bestimmt werden, dass der Quotient~~		The system parameter  $R$  is to be determined in this exercise in such a way that the quotient
	:$$\eta = \frac{ (d_{\rm min}/2)^2}{ E_{\rm B}} $$		:$$\eta = \frac{ (d_{\rm min}/2)^2}{ E_{\rm B}} $$

	~~maximal wird~~. $\eta$  ~~ist ein Maß für die Güte eines Modulationsalphabets bei gegebener Sendeenergie pro Bit~~ ("Power Efficiency"). ~~Es berechnet sich aus~~		becomes maximum. $\eta$  is a measure for the quality of a modulation alphabet at given transmission energy per bit ("Power Efficiency"). It is calculated from
	* ~~der minimalen Distanz~~  $d_{\rm min}$, ~~und~~		* the minimum distance  $d_{\rm min}$, and
	* ~~der Bitenergie~~  $E_{\rm B}$.		* the bit energy  $E_{\rm B}$.


	~~Es ist darauf zu achten, dass~~  $d_{\rm min}^2$  ~~und~~  $E_{\rm B}$  in ~~gleicher Weise normiert sind~~, ~~was aber bereits durch die Aufgabenstellung implizit vorgegeben ist~~.		It must be ensured that  $d_{\rm min}^2$  and  $E_{\rm B}$  are normalized in the same way, but this is already implicit in the exercise.


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	''~~Hinweise~~:''		''Notes:''
	* ~~Die Aufgabe gehört zum Kapitel~~  [[~~Digitalsignal%C3%BCbertragung~~/~~Tr%C3%A4gerfrequenzsysteme_mit_koh%C3%A4renter_Demodulation~~\| ~~Trägerfrequenzsysteme mit kohärenter~~ Demodulation]].		* The exercise belongs to the chapter  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation\|"Carrier Frequency Systems with Coherent Demodulation"]].
	* ~~Bezug genommen wird insbesondere auf die Seiten~~  [[~~Digitalsignal%C3%BCbertragung~~/~~Tr%C3%A4gerfrequenzsysteme_mit_koh%C3%A4renter_Demodulation~~#~~Quadraturamplitudenmodulation_~~.28M~~.E2.80.93QAM~~.29\| ~~Quadraturamplitudenmodulation~~]]  ~~und~~  [[~~Digitalsignal%C3%BCbertragung~~/~~Tr%C3%A4gerfrequenzsysteme_mit_koh%C3%A4renter_Demodulation~~#~~Mehrstufiges_Phase~~.E2.80.~~93Shift_Keying_~~.28M.E2.80.~~93PSK~~.29\|~~Mehrstufige Phasenmodulation~~]].		* Reference is made in particular to the sections  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Quadrature_amplitude_modulation_.28M-QAM.29\|"Quadrature amplitude modulation"]]  and  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#M.E2.80.93level_amplitude_shift_keying_.28M.E2.80.93ASK.29\|"Multilevel phase modulation"]].




	===~~Fragebogen~~===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{Berechnen Sie die mittlere Energie  $E_{\rm B}$  pro Bit abhängig von  $R$, insbesondere für  $R = 1$  und  $R = \sqrt{2}$.		{Berechnen Sie die mittlere Energie  $E_{\rm B}$  pro Bit abhängig von  $R$, insbesondere für  $R = 1$  und  $R = \sqrt{2}$.

Guenter at 11:59, 13 July 2022

2022-07-13T11:59:38Z

← Older revision		Revision as of 13:59, 13 July 2022
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	[[Category:Digital Signal Transmission: Exercises\|^4.4 ~~Kohärente~~ Demodulation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.4 Coherent Demodulation^]]