Aufgaben:Exercise 4.14: 8-PSK and 16-PSK - 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.~~14_8~~-~~PSK_und_16~~-PSK]]		[[de:Aufgaben:Aufgabe 4.14: 8-PSK und 16-PSK]]

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	* ${\rm Q}(x)$  is the complementary Gaussian error function for which the following approximation holds:		* ${\rm Q}(x)$  is the complementary Gaussian error function for which the following approximation holds:
	:$${\rm Q}(x) \approx \frac{1}{\sqrt{2\pi} \cdot x} \cdot {\rm e}^{-x^2/2}		:$${\rm Q}(x) \approx \frac{1}{\sqrt{2\pi} \cdot x} \cdot {\rm e}^{-x^2/2}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	The last subtask deals with the bit error probability.  For this,  the following bound was given in the  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation\| "theory section"]]  under the assumption of a  "Gray code":		The last subtask deals with the bit error probability.  For this,  the following bound was given in the  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation\| "theory section"]]  under the assumption of a  "Gray code":
	:$$p_{\rm B} \le \frac{2}{{\rm log_2} \hspace{0.05cm}(M)} \cdot {\rm Q} \left ( \sqrt{{\rm log_2} \hspace{0.05cm}(M)} \cdot \sin ({ \pi}/{ M}) \cdot \sqrt{ { {2E_{\rm B}}}/{ N_0} }\right )		:$$p_{\rm B} \le \frac{2}{{\rm log_2} \hspace{0.05cm}(M)} \cdot {\rm Q} \left ( \sqrt{{\rm log_2} \hspace{0.05cm}(M)} \cdot \sin ({ \pi}/{ M}) \cdot \sqrt{ { {2E_{\rm B}}}/{ N_0} }\right )\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	However,  this equation is only applicable for  $M > 4$.  In contrast,  the exact solution		However,  this equation is only applicable for  $M > 4$.  In contrast,  the exact solution
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	* for  $M = 4$   from the fact that the  "4–PSK"  is identical with the  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Quadrature_amplitude_modulation_.28M-QAM.29\|"4–QAM"]]:		* for  $M = 4$   from the fact that the  "4–PSK"  is identical with the  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Quadrature_amplitude_modulation_.28M-QAM.29\|"4–QAM"]]:
	:$$p_{\rm B} ={\rm Q} \left ( \sqrt{ { {2E_{\rm B}}}/{ N_0} }\right )		:$$p_{\rm B} ={\rm Q} \left ( \sqrt{ { {2E_{\rm B}}}/{ N_0} }\right )\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$


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	* When solving the exercise, you can use the following equations:		* When solving the exercise, you can use the following equations:
	:$$\cos(\alpha + \beta) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \cos(\alpha ) \cdot \cos( \beta) - \sin(\alpha ) \cdot \sin( \beta)\hspace{0.05cm}, \hspace{0.25cm}		:$$\cos(\alpha + \beta) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \cos(\alpha ) \cdot \cos( \beta) - \sin(\alpha ) \cdot \sin( \beta)\hspace{0.05cm}, \hspace{0.25cm}1 - \cos(2\alpha ) = \sin^2(\alpha )\hspace{0.05cm},$$
	1 - \cos(2\alpha ) = \sin^2(\alpha )\hspace{0.05cm},$$
	:$$ \int_{0}^{T} \cos^2 ( 2\pi f_{\rm T}t) \,{\rm d} t \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.5\hspace{0.05cm},\hspace{0.15cm}{\rm falls}\hspace{0.15cm} f_{\rm T} \gg 1/T \hspace{0.05cm}.$$		:$$ \int_{0}^{T} \cos^2 ( 2\pi f_{\rm T}t) \,{\rm d} t \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.5\hspace{0.05cm},\hspace{0.15cm}{\rm falls}\hspace{0.15cm} f_{\rm T} \gg 1/T \hspace{0.05cm}.$$

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	'''(1)'''  <u>Solutions 2 and 4</u>  are correct:		'''(1)'''  <u>Solutions 2 and 4</u>  are correct:
	*The signal set  $\{s_i(t)\}$  can be represented in the band-pass domain with the given trigonometric transformation as follows:		*The signal set  $\{s_i(t)\}$  can be represented in the band-pass domain with the given trigonometric transformation as follows:
	:$$s_i(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} A \cdot \cos \left ( 2\pi f_{\rm T}t + { 2\pi }/{ M} \cdot i \right ) = A \cdot \cos \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \cos \left ( 2\pi f_{\rm T}t \right )-		:$$s_i(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} A \cdot \cos \left ( 2\pi f_{\rm T}t + { 2\pi }/{ M} \cdot i \right ) = A \cdot \cos \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \cos \left ( 2\pi f_{\rm T}t \right )-A \cdot \sin \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \sin \left ( 2\pi f_{\rm T}t \right )\hspace{0.05cm}.$$
	A \cdot \sin \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \sin \left ( 2\pi f_{\rm T}t \right )\hspace{0.05cm}.$$

	*The respective first terms in this difference lead to the signal space point  $\boldsymbol{s}_i$,  the respective second terms to the basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$.		*The respective first terms in this difference lead to the signal space point  $\boldsymbol{s}_i$,  the respective second terms to the basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$.

	*Here it is to be noted that these must be energy normalized in each case:		*Here it is to be noted that these must be energy normalized in each case:
	:$$\int_{0}^{T} \varphi_1 ( t)^2 \,{\rm d} t = \int_{0}^{T} \left [K \cdot \cos( 2\pi f_{\rm T}t)\right ]^2 \,{\rm d} t = 1\hspace{0.3cm}		:$$\int_{0}^{T} \varphi_1 ( t)^2 \,{\rm d} t = \int_{0}^{T} \left [K \cdot \cos( 2\pi f_{\rm T}t)\right ]^2 \,{\rm d} t = 1\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \frac{K^2 \cdot T}{2} = 2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} K = \sqrt{2/T } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\varphi_1 ( t) = \sqrt{2/T } \cdot \cos( 2\pi f_{\rm T}t)\hspace{0.05cm}.$$
	\Rightarrow \hspace{0.3cm} \frac{K^2 \cdot T}{2} = 2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} K = \sqrt{2/T } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	\varphi_1 ( t) = \sqrt{2/T } \cdot \cos( 2\pi f_{\rm T}t)\hspace{0.05cm}.$$
	*Using the same arithmetic,  we arrive at the second basis function:		*Using the same arithmetic,  we arrive at the second basis function:
	:$$\varphi_2 ( t) = -\sqrt{2/T } \cdot \sin( 2\pi f_{\rm T}t)\hspace{0.05cm}.$$		:$$\varphi_2 ( t) = -\sqrt{2/T } \cdot \sin( 2\pi f_{\rm T}t)\hspace{0.05cm}.$$
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	'''(2)'''  <u>Solutions 1 and 4</u>  are correct:		'''(2)'''  <u>Solutions 1 and 4</u>  are correct:
	*Using the basis functions just calculated,  the signal set  $s_i(t)$  can be represented as follows  $($again, limited to the range  $0 ≤ t ≤ T)$:		*Using the basis functions just calculated,  the signal set  $s_i(t)$  can be represented as follows  $($again, limited to the range  $0 ≤ t ≤ T)$:
	:$$s_i(t) = A \cdot \sqrt{T/2 } \cdot \cos ( 2\pi \cdot i /M ) \cdot \varphi_1 ( t)+		:$$s_i(t) = A \cdot \sqrt{T/2 } \cdot \cos ( 2\pi \cdot i /M ) \cdot \varphi_1 ( t)+A \cdot \sqrt{T/2 } \cdot \sin ( 2\pi \cdot i /M ) \cdot \varphi_2 ( t)\hspace{0.05cm}.$$
	A \cdot \sqrt{T/2 } \cdot \sin ( 2\pi \cdot i /M ) \cdot \varphi_2 ( t)
	\hspace{0.05cm}.$$

	*With the energy  $E = 0.5 \cdot A^2 \cdot T$,  which is the same for all  $M$  points and which is also the average signal energy per symbol  $(E_{\rm S})$,  the above equation is		*With the energy  $E = 0.5 \cdot A^2 \cdot T$,  which is the same for all  $M$  points and which is also the average signal energy per symbol  $(E_{\rm S})$,  the above equation is
	:$$s_i(t) = s_{{\rm I}i} \cdot \varphi_1 ( t)+		:$$s_i(t) = s_{{\rm I}i} \cdot \varphi_1 ( t)+s_{{\rm Q}i} \cdot \varphi_2 ( t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm} s_{{\rm I}i} = \sqrt{E_{\rm S} }\cdot \cos ( 2\pi \cdot i /M )\hspace{0.05cm},\hspace{0.2cm}s_{{\rm Q}i} = \sqrt{E_{\rm S} }\cdot \sin ( 2\pi \cdot i /M )\hspace{0.05cm}. $$
	s_{{\rm Q}i} \cdot \varphi_2 ( t)\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} s_{{\rm I}i} = \sqrt{E_{\rm S} }\cdot \cos ( 2\pi \cdot i /M )\hspace{0.05cm},\hspace{0.2cm}
	s_{{\rm Q}i} = \sqrt{E_{\rm S} }\cdot \sin ( 2\pi \cdot i /M )\hspace{0.05cm}. $$

	*The signal space points sketched in the front graph for  $M = 8$  resp.  $M = 16$,  can be represented in exactly this way.		*The signal space points sketched in the front graph for  $M = 8$  resp.  $M = 16$,  can be represented in exactly this way.
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	:$$d_{\rm norm}^2 \hspace{-0.01cm}=\hspace{-0.01cm} \sin^2(\frac{2\pi}{M}) + [1- \cos(\frac{2\pi}{M})]^2 $$		:$$d_{\rm norm}^2 \hspace{-0.01cm}=\hspace{-0.01cm} \sin^2(\frac{2\pi}{M}) + [1- \cos(\frac{2\pi}{M})]^2 $$
	:$$\Rightarrow \hspace{0.3cm} d_{\rm norm}^2 \hspace{-0.01cm}=\hspace{-0.01cm} 2 \cdot [1- \cos( \frac{2\pi}{M} )] = 4 \cdot \sin^2(\pi/M )\hspace{0.05cm} $$		:$$\Rightarrow \hspace{0.3cm} d_{\rm norm}^2 \hspace{-0.01cm}=\hspace{-0.01cm} 2 \cdot [1- \cos( \frac{2\pi}{M} )] = 4 \cdot \sin^2(\pi/M )\hspace{0.05cm} $$
	:$$\Rightarrow \hspace{0.3cm} d_{\rm norm}= 2 \cdot \sin({\pi}/{M }) \hspace{0.3cm}		:$$\Rightarrow \hspace{0.3cm} d_{\rm norm}= 2 \cdot \sin({\pi}/{M }) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} d = 2\sqrt {E_{\rm S}} \cdot \sin{\pi}/{M })\hspace{0.05cm}.$$
	\Rightarrow \hspace{0.3cm} d = 2\sqrt {E_{\rm S}} \cdot \sin{\pi}/{M })\hspace{0.05cm}.$$

	*This results in the following numerical values:		*This results in the following numerical values:
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	*From this follows for  $M = 2$  and  $M = 4$:		*From this follows for  $M = 2$  and  $M = 4$:
	:$$p_{\rm B} ={\rm Q} \left ( \sqrt{ { {2E_{\rm B}}}/{ N_0} }\right ) = 10^{-4}$$		:$$p_{\rm B} ={\rm Q} \left ( \sqrt{ { {2E_{\rm B}}}/{ N_0} }\right ) = 10^{-4}$$
	:$$\Rightarrow \hspace{0.3cm}		:$$\Rightarrow \hspace{0.3cm}\sqrt{ { {2E_{\rm B}}}/{ N_0} } \approx 3.72\hspace{0.05cm}.$$
	\sqrt{ { {2E_{\rm B}}}/{ N_0} } \approx 3.72
	\hspace{0.05cm}.$$

	*On the other hand,  with the given equation for  $M = 8$:		*On the other hand,  with the given equation for  $M = 8$:
	:$$p_{\rm B} \hspace{-0.1cm} \ \le \ \hspace{-0.1cm} \frac{2}{{\rm log_2} \hspace{0.05cm}(M)} \cdot {\rm Q} \left ( \sqrt{{\rm log_2} \hspace{0.05cm}(M)} \cdot \sin ({ \pi}/{ M}) \cdot \sqrt{ { {2E_{\rm S}}}/{ N_0} }\right )= {2}/{3} \cdot {\rm Q} \left ( \sqrt{3} \cdot 0.383 \cdot 3.72 \right ) \approx {2}/{3} \cdot {\rm Q} \left ( 2.46 \right ) \approx 0.46 \%		:$$p_{\rm B} \hspace{-0.1cm} \ \le \ \hspace{-0.1cm} \frac{2}{{\rm log_2} \hspace{0.05cm}(M)} \cdot {\rm Q} \left ( \sqrt{{\rm log_2} \hspace{0.05cm}(M)} \cdot \sin ({ \pi}/{ M}) \cdot \sqrt{ { {2E_{\rm S}}}/{ N_0} }\right )= {2}/{3} \cdot {\rm Q} \left ( \sqrt{3} \cdot 0.383 \cdot 3.72 \right ) \approx {2}/{3} \cdot {\rm Q} \left ( 2.46 \right ) \approx 0.46 \%\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*Accordingly,  for  $M = 16$ ⇒ ${\rm log}_2 (M) = 4$  is obtained:		*Accordingly,  for  $M = 16$ ⇒ ${\rm log}_2 (M) = 4$  is obtained:
	:$$p_{\rm B} = {2}/{4} \cdot {\rm Q} \left ( \sqrt{4} \cdot 0.195 \cdot 3.72 \right ) \approx {1}/{2} \cdot {\rm Q} \left ( 1.45 \right ) \approx 4.8 \%		:$$p_{\rm B} = {2}/{4} \cdot {\rm Q} \left ( \sqrt{4} \cdot 0.195 \cdot 3.72 \right ) \approx {1}/{2} \cdot {\rm Q} \left ( 1.45 \right ) \approx 4.8 \%\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	Thus,  The <u>statements 1 and 4</u>  are correct:		Thus,  The <u>statements 1 and 4</u>  are correct:

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	[[Category:Digital Signal Transmission: Exercises\|^4.4 Coherent Demodulation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.4 Coherent Demodulation^]]
			[[de:Aufgaben:4.14_8-PSK_und_16-PSK]]

Hwang at 19:31, 1 September 2022

2022-09-01T19:31:09Z

← Older revision		Revision as of 21:31, 1 September 2022
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	The following statements are valid:		The following statements are valid:
	# The actual error probability  $p_{\rm S}$  is composed of the components  $\rm A$,  $\rm B$  and  $\rm C$.		# The actual error probability  $p_{\rm S}$  is composed of the components  $\rm A$,  $\rm B$  and  $\rm C$.
	# On the other hand,  the  "Union Bound"  results from  $\rm A+B$  $($falsification into the symbol  $\boldsymbol{s}_1))$  plus  $\rm C+B$  $($falsification into the symbol $\boldsymbol{s}_{M-1})$.		# On the other hand,  the  "Union Bound"  results from  $\rm A+B$  $($falsification into the symbol  $\boldsymbol{s}_1)$  plus  $\rm C+B$  $($falsification into the symbol $\boldsymbol{s}_{M-1})$.
	# Thus,  $p_{\rm S} = p_{\rm UB}-B$  always holds,  but the larger  $M$  is, the smaller the fraction  $\rm B$  is.		# Thus,  $p_{\rm S} = p_{\rm UB}-B$  always holds,  but the larger  $M$  is, the smaller the fraction  $\rm B$  is.

Hwang at 19:28, 1 September 2022

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← Older revision		Revision as of 21:28, 1 September 2022
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	$M = 16 \text{:} \hspace{0.2cm} p_{\rm UB}$ = { 6 3% } $\ \%$		$M = 16 \text{:} \hspace{0.2cm} p_{\rm UB}$ = { 6 3% } $\ \%$

	{Does the statement:   "$p_{\rm UB}$  approximates  $p_{\rm S}$  more closely the larger  $M$   is"  hold?		{Does the statement   "$p_{\rm UB}$  approximates  $p_{\rm S}$  more closely the larger  $M$   is"  hold?
	\|type="()"}		\|type="()"}
	+ YES.		+ YES.

Hwang at 19:27, 1 September 2022

2022-09-01T19:27:33Z

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	$M = 16 \text{:} \hspace{0.2cm} p_{\rm UB}$ = { 6 3% } $\ \%$		$M = 16 \text{:} \hspace{0.2cm} p_{\rm UB}$ = { 6 3% } $\ \%$

	{Does the statement:   "$p_{\rm UB}$  approximates  $p_{\rm S}$  more closely the larger  $M$   is"nbsp; hold?		{Does the statement:   "$p_{\rm UB}$  approximates  $p_{\rm S}$  more closely the larger  $M$   is"  hold?
	\|type="()"}		\|type="()"}
	+ YES.		+ YES.

Hwang at 19:26, 1 September 2022

2022-09-01T19:26:53Z

← Older revision		Revision as of 21:26, 1 September 2022
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	$M = 16 \text{:} \hspace{0.2cm} p_{\rm UB}$ = { 6 3% } $\ \%$		$M = 16 \text{:} \hspace{0.2cm} p_{\rm UB}$ = { 6 3% } $\ \%$

	{Does the statement: nbsp; "$p_{\rm UB}$  approximates  $p_{\rm S}$  more closely the larger  $M$   is"nbsp; hold?		{Does the statement:   "$p_{\rm UB}$  approximates  $p_{\rm S}$  more closely the larger  $M$   is"nbsp; hold?
	\|type="()"}		\|type="()"}
	+ YES.		+ YES.

Guenter at 14:13, 23 August 2022

2022-08-23T14:13:11Z

Guenter at 13:16, 23 August 2022

2022-08-23T13:16:32Z

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	'''(5)'''  Correct is <u>YES</u>.  The graph shows the constellation for  "8–PSK"  and  "16–PSK",		'''(5)'''  Correct is <u>YES</u>.  The graph shows the constellation for  "8–PSK"  and  "16–PSK".
	[[File:P_ID2071__Dig_A_4_14e.png\|right\|frame\|For the interpretation of the Union Bound at the  $\rm M–PSK$]]		[[File:P_ID2071__Dig_A_4_14e.png\|right\|frame\|For the interpretation of the Union Bound at the  $\rm M–PSK$]]

	*~~each~~ valid under the assumption that  $\boldsymbol{s}_0$  was transmitted.		*Each valid under the assumption that  $\boldsymbol{s}_0$  was transmitted.

	*The contour lines of the AWGN PDF are then circles around  $\boldsymbol{s}_0$.		*The contour lines of the AWGN PDF are then circles around  $\boldsymbol{s}_0$.


	The following statements are valid:		The following statements are valid:
	* The actual error probability $p_{\rm S}$ is composed of the components ~~<b>~~A~~</b>~~, ~~<b>~~B~~</b>~~ and ~~<b>~~C~~</b>~~.		# The actual error probability  $p_{\rm S}$  is composed of the components  $\rm A$,  $\rm B$  and  $\rm C$.
	* On the other hand, the "Union Bound" results from ~~<b>~~A + B~~</b>~~ (falsification into the symbol $\boldsymbol{s}_1$) plus ~~<b>~~B ~~+ C</b>~~ (falsification into the symbol $\boldsymbol{s}_{M-1}$).		# On the other hand,  the  "Union Bound"  results from  $\rm A+B$  $($falsification into the symbol  $\boldsymbol{s}_1))$  plus  $\rm C+B$  $($falsification into the symbol $\boldsymbol{s}_{M-1})$.
	* Thus, $p_{\rm S} = p_{\rm UB}$ ~~– <b>B</b>~~ always holds, but the larger $M$ is, the smaller the fraction ~~<b>~~B~~</b>~~ is.		# Thus,  $p_{\rm S} = p_{\rm UB}-B$ always holds,  but the larger  $M$  is, the smaller the fraction  $\rm B$  is.



	'''(6)'''  For verification we assume~~, for example,~~ $p_{\rm B} = 10^{\rm –4}$. From this follows for $M = 2$ and $M = 4$:		'''(6)'''  For verification we assume  $p_{\rm B} = 10^{\rm –4}$.
	:$$p_{\rm B} ={\rm Q} \left ( \sqrt{ { {2E_{\rm B}}}/{ N_0} }\right ) = 10^{-4}~~\hspace{0.3cm}~~\Rightarrow \hspace{0.3cm}
			*From this follows for  $M = 2$  and  $M = 4$:
			:$$p_{\rm B} ={\rm Q} \left ( \sqrt{ { {2E_{\rm B}}}/{ N_0} }\right ) = 10^{-4}$$
			:$$\Rightarrow \hspace{0.3cm}
	\sqrt{ { {2E_{\rm B}}}/{ N_0} } \approx 3.72		\sqrt{ { {2E_{\rm B}}}/{ N_0} } \approx 3.72
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	On the other hand, with the given equation for $M = 8$:		*On the other hand,  with the given equation for  $M = 8$:
	:$$p_{\rm B} \hspace{-0.1cm} \ \le \ \hspace{-0.1cm} \frac{2}{{\rm log_2} \hspace{0.05cm}(M)} \cdot {\rm Q} \left ( \sqrt{{\rm log_2} \hspace{0.05cm}(M)} \cdot \sin ({ \pi}/{ M}) \cdot \sqrt{ { {2E_{\rm S}}}/{ N_0} }\right )= {2}/{3} \cdot {\rm Q} \left ( \sqrt{3} \cdot 0.383 \cdot 3.72 \right ) \approx {2}/{3} \cdot {\rm Q} \left ( 2.46 \right ) \approx 0.46 \%		:$$p_{\rm B} \hspace{-0.1cm} \ \le \ \hspace{-0.1cm} \frac{2}{{\rm log_2} \hspace{0.05cm}(M)} \cdot {\rm Q} \left ( \sqrt{{\rm log_2} \hspace{0.05cm}(M)} \cdot \sin ({ \pi}/{ M}) \cdot \sqrt{ { {2E_{\rm S}}}/{ N_0} }\right )= {2}/{3} \cdot {\rm Q} \left ( \sqrt{3} \cdot 0.383 \cdot 3.72 \right ) \approx {2}/{3} \cdot {\rm Q} \left ( 2.46 \right ) \approx 0.46 \%
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	Accordingly, for $M = 16$ ⇒ ${\rm log}_2 (M) = 4$ is obtained:		*Accordingly,  for  $M = 16$ ⇒ ${\rm log}_2 (M) = 4$  is obtained:
	:$$p_{\rm B} = {2}/{4} \cdot {\rm Q} \left ( \sqrt{4} \cdot 0.195 \cdot 3.72 \right ) \approx {1}/{2} \cdot {\rm Q} \left ( 1.45 \right ) \approx 4.8 \%		:$$p_{\rm B} = {2}/{4} \cdot {\rm Q} \left ( \sqrt{4} \cdot 0.195 \cdot 3.72 \right ) \approx {1}/{2} \cdot {\rm Q} \left ( 1.45 \right ) \approx 4.8 \%
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	Thus, <u>statements 1 and 4</u> are correct:		Thus,  The <u>statements 1 and 4</u>  are correct:
	*Thus, the main advantage of a higher-level PSK is not the lower bit error rate, but the lower demand on the very expensive resource "bandwidth".		*Thus,  the main advantage of a higher-level PSK is not the lower bit error rate,  but the lower demand on the very expensive resource  "bandwidth".
	*It should also be noted that the results are completely different when a (highly) distorting channel is present, as is common in ~~<i>~~wireline transmission technology~~</i>~~.
			*It should also be noted that the results are completely different when a  (highly)  distorting channel is present,  as is common in  "wireline transmission technology".
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 12:36, 23 August 2022

2022-08-23T12:36:29Z

← Older revision		Revision as of 14:36, 23 August 2022
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	''Notes:''		Notes:
	* The exercise belongs to the topic of the chapter  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation\|"Carrier Frequency Systems with Coherent Demodulation"]].		* The exercise belongs to the topic of the chapter  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation\|"Carrier Frequency Systems with Coherent Demodulation"]].

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Solutions 2 and 4</u> are correct:		'''(1)'''  <u>Solutions 2 and 4</u>  are correct:
	*The signal set can be represented in the band-pass domain with the given trigonometric transformation as follows:		*The signal set can be represented in the band-pass domain with the given trigonometric transformation as follows:
	:$$s_i(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} A \cdot \cos \left ( 2\pi f_{\rm T}t + { 2\pi }/{ M} \cdot i \right ) = A \cdot \cos \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \cos \left ( 2\pi f_{\rm T}t \right )-		:$$s_i(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} A \cdot \cos \left ( 2\pi f_{\rm T}t + { 2\pi }/{ M} \cdot i \right ) = A \cdot \cos \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \cos \left ( 2\pi f_{\rm T}t \right )-
	A \cdot \sin \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \sin \left ( 2\pi f_{\rm T}t \right )\hspace{0.05cm}.$$		A \cdot \sin \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \sin \left ( 2\pi f_{\rm T}t \right )\hspace{0.05cm}.$$

	*The respective first terms in this difference lead to the signal space point $\boldsymbol{s}_i$, the respective second terms to the basis functions $\varphi_1(t)$ and $\varphi_2(t)$. Here it is to be noted that these must be energy normalized in each case:		*The respective first terms in this difference lead to the signal space point  $\boldsymbol{s}_i$,  the respective second terms to the basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$.

			*Here it is to be noted that these must be energy normalized in each case:
	:$$\int_{0}^{T} \varphi_1 ( t)^2 \,{\rm d} t = \int_{0}^{T} \left [K \cdot \cos( 2\pi f_{\rm T}t)\right ]^2 \,{\rm d} t = 1\hspace{0.3cm}		:$$\int_{0}^{T} \varphi_1 ( t)^2 \,{\rm d} t = \int_{0}^{T} \left [K \cdot \cos( 2\pi f_{\rm T}t)\right ]^2 \,{\rm d} t = 1\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} \frac{K^2 \cdot T}{2} = 2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} K = \sqrt{2/T } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}		\Rightarrow \hspace{0.3cm} \frac{K^2 \cdot T}{2} = 2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} K = \sqrt{2/T } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	\varphi_1 ( t) = \sqrt{2/T } \cdot \cos( 2\pi f_{\rm T}t)\hspace{0.05cm}.$$		\varphi_1 ( t) = \sqrt{2/T } \cdot \cos( 2\pi f_{\rm T}t)\hspace{0.05cm}.$$
			*Using the same arithmetic,  we arrive at the second basis function:
	*Using the same arithmetic, we arrive at the second basis function:
	:$$\varphi_2 ( t) = -\sqrt{2/T } \cdot \sin( 2\pi f_{\rm T}t)\hspace{0.05cm}.$$		:$$\varphi_2 ( t) = -\sqrt{2/T } \cdot \sin( 2\pi f_{\rm T}t)\hspace{0.05cm}.$$


	'''(2)'''  <u>Solutions 1 and 4</u> are correct:		'''(2)'''  <u>Solutions 1 and 4</u>  are correct:
	*Using the basis functions just calculated, the signal set $s_i(t)$ can be represented as follows (again, limited to the range $0 ≤ t ≤ T$):		*Using the basis functions just calculated,  the signal set  $s_i(t)$  can be represented as follows  $($again, limited to the range  $0 ≤ t ≤ T)$:
	:$$s_i(t) = A \cdot \sqrt{T/2 } \cdot \cos ( 2\pi \cdot i /M ) \cdot \varphi_1 ( t)+		:$$s_i(t) = A \cdot \sqrt{T/2 } \cdot \cos ( 2\pi \cdot i /M ) \cdot \varphi_1 ( t)+
	A \cdot \sqrt{T/2 } \cdot \sin ( 2\pi \cdot i /M ) \cdot \varphi_2 ( t)		A \cdot \sqrt{T/2 } \cdot \sin ( 2\pi \cdot i /M ) \cdot \varphi_2 ( t)
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*With the energy $E = 0.5 \cdot A^2 \cdot T$, which is the same for all $M$ points and which is also the ~~mean~~ signal energy per symbol ($E_{\rm S}$), the above equation is		*With the energy  $E = 0.5 \cdot A^2 \cdot T$,  which is the same for all  $M$  points and which is also the average signal energy per symbol  $(E_{\rm S})$,  the above equation is
	:$$s_i(t) = s_{{\rm I}i} \cdot \varphi_1 ( t)+		:$$s_i(t) = s_{{\rm I}i} \cdot \varphi_1 ( t)+
	s_{{\rm Q}i} \cdot \varphi_2 ( t)\hspace{0.3cm}		s_{{\rm Q}i} \cdot \varphi_2 ( t)\hspace{0.3cm}
Line 122:		Line 123:
	s_{{\rm Q}i} = \sqrt{E_{\rm S} }\cdot \sin ( 2\pi \cdot i /M )\hspace{0.05cm}. $$		s_{{\rm Q}i} = \sqrt{E_{\rm S} }\cdot \sin ( 2\pi \cdot i /M )\hspace{0.05cm}. $$

	*The signal space points sketched in the front graph for $M = 8$ ~~and~~ $M = 16$, ~~respectively,~~ can be represented in exactly this way.		*The signal space points sketched in the front graph for  $M = 8$  resp.  $M = 16$,  can be represented in exactly this way.



			'''(3)'''  Since the distance from one to the nearest point is the same for all  $i$,  we can calculate  $d$, for example,  from the signal space points  $\boldsymbol{s}_0$  and  $\boldsymbol{s}_1$.
			*In doing  so,  consider the sketch below.  According to the  "Pythagorean theorem":
			[[File:P_ID2070__Dig_A_4_14c.png\|right\|frame\|Distance calculation at the  "8-PSK"]]

	~~[[File:P_ID2070__Dig_A_4_14c.png\|right\|frame\|For the distance calculation at 8-PSK]]~~
	'''(3)'''  Since the distance from one to the nearest point is the same for all $i$, we can calculate $d$, for example, from the signal space points $\boldsymbol{s}_0$ and $\boldsymbol{s}_1$. In doing so, consider the sketch below. According to the <i>Pythagorean theorem</i>:
	:$$d_{\rm norm}^2 \hspace{-0.01cm}=\hspace{-0.01cm} \sin^2(\frac{2\pi}{M}) + [1- \cos(\frac{2\pi}{M})]^2 $$		:$$d_{\rm norm}^2 \hspace{-0.01cm}=\hspace{-0.01cm} \sin^2(\frac{2\pi}{M}) + [1- \cos(\frac{2\pi}{M})]^2 $$
	:$$\Rightarrow \hspace{0.3cm} d_{\rm norm}^2 \hspace{-0.01cm}=\hspace{-0.01cm} 2 \cdot [1- \cos( \frac{2\pi}{M} )] = 4 \cdot \sin^2(\pi/M )\hspace{0.05cm} $$		:$$\Rightarrow \hspace{0.3cm} d_{\rm norm}^2 \hspace{-0.01cm}=\hspace{-0.01cm} 2 \cdot [1- \cos( \frac{2\pi}{M} )] = 4 \cdot \sin^2(\pi/M )\hspace{0.05cm} $$
Line 133:		Line 136:
	\Rightarrow \hspace{0.3cm} d = 2\sqrt {E_{\rm S}} \cdot \sin{\pi}/{M })\hspace{0.05cm}.$$		\Rightarrow \hspace{0.3cm} d = 2\sqrt {E_{\rm S}} \cdot \sin{\pi}/{M })\hspace{0.05cm}.$$

	This results in the following numerical values:		*This results in the following numerical values:
	:$$M = 8\hspace{-0.09cm}: \hspace{0.2cm}d \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \underline{0.765} \cdot \sqrt {E_{\rm S} } \hspace{0.05cm},$$		:$$M = 8\hspace{-0.09cm}: \hspace{0.2cm}d \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \underline{0.765} \cdot \sqrt {E_{\rm S} } \hspace{0.05cm},$$
	:$$M = 16\hspace{-0.09cm}: \hspace{0.2cm} d \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \underline{0.390} \cdot \sqrt {E_{\rm S} } \hspace{0.05cm}.$$		:$$M = 16\hspace{-0.09cm}: \hspace{0.2cm} d \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \underline{0.390} \cdot \sqrt {E_{\rm S} } \hspace{0.05cm}.$$


	'''(4)'''  Using the result from '''(3)''', we obtain with the equation given in front:		'''(4)'''  Using the result from subtask  '''(3)''',  we obtain with the equation given in front:
	:$$p_{\rm S} \le p_{\rm UB} = 2 \cdot {\rm Q} \left [ \sin ({ \pi}/{ M}) \cdot \sqrt{ { 2E_{\rm S}}/{ N_0}} \right ] \hspace{0.05cm}.$$		:$$p_{\rm S} \le p_{\rm UB} = 2 \cdot {\rm Q} \left [ \sin ({ \pi}/{ M}) \cdot \sqrt{ { 2E_{\rm S}}/{ N_0}} \right ] \hspace{0.05cm}.$$

Line 145:		Line 148:
	:$$ M = 16\hspace{-0.09cm}: \hspace{0.2cm} p_{\rm UB} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 2 \cdot {\rm Q} ( 10 \cdot 0.195 )= 2 \cdot {\rm Q} ( 1.95 ) \approx \underline{6 \%} \hspace{0.05cm}.$$		:$$ M = 16\hspace{-0.09cm}: \hspace{0.2cm} p_{\rm UB} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 2 \cdot {\rm Q} ( 10 \cdot 0.195 )= 2 \cdot {\rm Q} ( 1.95 ) \approx \underline{6 \%} \hspace{0.05cm}.$$

	*Thus, the error probability becomes larger and larger as $M$ increases, assuming constant $E_{\rm S}/N_0$, as in this case.		*Thus,  the symbol error probability becomes larger and larger as  $M$  increases,  assuming constant  $E_{\rm S}/N_0$,  as in this case.
	*The most favorable value would result for $M = 2$ (the factor $2$ of the Union Bound is then not necessary) as follows
			*The most favorable value would result for  $M = 2$  (the factor  $2$  of the Union Bound is then not necessary)  as follows
	:$$M = 2\hspace{-0.09cm}: \hspace{0.2cm}p_{\rm UB} = p_{\rm S} = {\rm Q} ( 10 ) \approx { 10^{-23}}.$$		:$$M = 2\hspace{-0.09cm}: \hspace{0.2cm}p_{\rm UB} = p_{\rm S} = {\rm Q} ( 10 ) \approx { 10^{-23}}.$$

	*Thus, multi-level PSK would not make sense if there were not other reasons for its use, which will be discussed in subtask '''(6)'''.		*Thus,  multi-level PSK would not make sense if there were not other reasons for its use,  which will be discussed in subtask  '''(6)'''.



	'''(5)'''  Correct is <u>YES</u>. The graph shows the constellation for 8–PSK and 16–PSK, each valid under the assumption that $\boldsymbol{s}_0$ was transmitted. The contour lines of the AWGN PDF are then circles around $\boldsymbol{s}_0$.		'''(5)'''  Correct is <u>YES</u>.  The graph shows the constellation for  "8–PSK"  and  "16–PSK",
	~~[[File:P_ID2071__Dig_A_4_14e.png\|center\|frame\|For the interpretation of the Union Bound at <i>M</i>–PSK]]~~		[[File:P_ID2071__Dig_A_4_14e.png\|right\|frame\|For the interpretation of the Union Bound at the  $\rm M–PSK$]]

			*each valid under the assumption that  $\boldsymbol{s}_0$  was transmitted.

			*The contour lines of the AWGN PDF are then circles around  $\boldsymbol{s}_0$.

	The following statements are valid:		The following statements are valid:
	* The actual error probability $p_{\rm S}$ is composed of the components <b>A</b>, <b>B</b> and <b>C</b>.		* The actual error probability $p_{\rm S}$ is composed of the components <b>A</b>, <b>B</b> and <b>C</b>.

← Older revision		Revision as of 16:13, 23 August 2022
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	*This is a  "phase modulation"  with  $M$  signal shapes.		*This is a  "phase modulation"  with  $M$  signal shapes.

	*This modulation process is also called  $\rm M–PSK$.  $M$  is usually a power of two.		*This modulation process is also called  "$\rm M–PSK$".  $M$  is usually a power of two.


Line 97:		Line 97:
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Solutions 2 and 4</u>  are correct:		'''(1)'''  <u>Solutions 2 and 4</u>  are correct:
	*The signal set can be represented in the band-pass domain with the given trigonometric transformation as follows:		*The signal set  $\{s_i(t)\}$  can be represented in the band-pass domain with the given trigonometric transformation as follows:
	:$$s_i(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} A \cdot \cos \left ( 2\pi f_{\rm T}t + { 2\pi }/{ M} \cdot i \right ) = A \cdot \cos \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \cos \left ( 2\pi f_{\rm T}t \right )-		:$$s_i(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} A \cdot \cos \left ( 2\pi f_{\rm T}t + { 2\pi }/{ M} \cdot i \right ) = A \cdot \cos \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \cos \left ( 2\pi f_{\rm T}t \right )-
	A \cdot \sin \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \sin \left ( 2\pi f_{\rm T}t \right )\hspace{0.05cm}.$$		A \cdot \sin \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \sin \left ( 2\pi f_{\rm T}t \right )\hspace{0.05cm}.$$
Line 158:		Line 158:

	'''(5)'''  Correct is <u>YES</u>.  The graph shows the constellation for  "8–PSK"  and  "16–PSK".		'''(5)'''  Correct is <u>YES</u>.  The graph shows the constellation for  "8–PSK"  and  "16–PSK".
	[[File:P_ID2071__Dig_A_4_14e.png\|right\|frame\|For the interpretation of the Union Bound at the  $\rm M–PSK$]]		[[File:P_ID2071__Dig_A_4_14e.png\|right\|frame\|For the interpretation of the Union Bound at the  "$\rm M–PSK$"]]

	*Each valid under the assumption that  $\boldsymbol{s}_0$  was transmitted.		*Each valid under the assumption that  $\boldsymbol{s}_0$  was transmitted.
Line 168:		Line 168:
	# The actual error probability  $p_{\rm S}$  is composed of the components  $\rm A$,  $\rm B$  and  $\rm C$.		# The actual error probability  $p_{\rm S}$  is composed of the components  $\rm A$,  $\rm B$  and  $\rm C$.
	# On the other hand,  the  "Union Bound"  results from  $\rm A+B$  $($falsification into the symbol  $\boldsymbol{s}_1))$  plus  $\rm C+B$  $($falsification into the symbol $\boldsymbol{s}_{M-1})$.		# On the other hand,  the  "Union Bound"  results from  $\rm A+B$  $($falsification into the symbol  $\boldsymbol{s}_1))$  plus  $\rm C+B$  $($falsification into the symbol $\boldsymbol{s}_{M-1})$.
	# Thus,  $p_{\rm S} = p_{\rm UB}-B$ always holds,  but the larger  $M$  is, the smaller the fraction  $\rm B$  is.		# Thus,  $p_{\rm S} = p_{\rm UB}-B$  always holds,  but the larger  $M$  is, the smaller the fraction  $\rm B$  is.