Aufgaben:Exercise 4.18: Non-Coherent FSK Demodulation - Revision history

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	[[Category:Digital Signal Transmission: Exercises\|^4.5 Non-Coherent Demodulation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.5 Non-Coherent Demodulation^]]
	[[de:Aufgaben:4.~~18_Nichtkohärente_BFSK–Demodulation~~]]		[[de:Aufgaben:Aufgabe 4.18: Nichtkohärente FSK–Demodulation]]

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	For the probability that the decision erroneously chooses the symbol  $m_1$  and,  moreover,  indicates that this decision is to be considered  "reliable"  (particularly reprehensible),  the following applies		For the probability that the decision erroneously chooses the symbol  $m_1$  and,  moreover,  indicates that this decision is to be considered  "reliable"  (particularly reprehensible),  the following applies
	:$${\rm Pr}\{\hat{m} = m_1,\hspace{0.15cm}{\it \Gamma} = {\rm Z} \hspace{0.05cm}\| \hspace{0.05cm}m_0 \} = \frac{1}{1 + \gamma^2} \cdot		:$${\rm Pr}\{\hat{m} = m_1,\hspace{0.15cm}{\it \Gamma} = {\rm Z} \hspace{0.05cm}\| \hspace{0.05cm}m_0 \} = \frac{1}{1 + \gamma^2} \cdot{\rm exp } \left [ - \frac{\gamma^2 \cdot E_{\rm S}}{(1+\gamma^2) \cdot N_{\rm 0}}\right ] \hspace{0.05cm}.$$
	{\rm exp } \left [ - \frac{\gamma^2 \cdot E_{\rm S}}{(1+\gamma^2) \cdot N_{\rm 0}}\right ] \hspace{0.05cm}.$$


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	*It follows that:		*It follows that:
	:$$p_{\rm S} = \frac{1}{1 + \gamma^2} \cdot		:$$p_{\rm S} = \frac{1}{1 + \gamma^2} \cdot{\rm exp } \left [ - \frac{\gamma^2 \cdot E_{\rm S}}{(1+\gamma^2) \cdot N_{\rm 0}}\right ]_{\gamma = 1} ={1}/{2} \cdot {\rm e }^{-E_{\rm S}/(2N_0)}\hspace{0.05cm}.$$
	{\rm exp } \left [ - \frac{\gamma^2 \cdot E_{\rm S}}{(1+\gamma^2) \cdot N_{\rm 0}}\right ]_{\gamma = 1} =
	{1}/{2} \cdot {\rm e }^{-E_{\rm S}/(2N_0)}\hspace{0.05cm}.$$

	*With  $E_{\rm S}/N_0 = 10$,  we get  $p_{\rm S} = 1/2 \cdot {\rm e}^{\rm –5} \approx \underline {0.337 \cdot \%}$.		*With  $E_{\rm S}/N_0 = 10$,  we get  $p_{\rm S} = 1/2 \cdot {\rm e}^{\rm –5} \approx \underline {0.337 \cdot \%}$.
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	'''(4)'''  For symmetry reasons,  this probability results in		'''(4)'''  For symmetry reasons,  this probability results in
	:$${\rm Pr}({\it \Gamma} = {\rm Z}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) =		:$${\rm Pr}({\it \Gamma} = {\rm Z}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) ={1}/{2} \cdot {\rm Pr}\{(\hat{m} = m_1)\hspace{0.05cm}\cap\hspace{0.05cm}{\it \Gamma} = {\rm Z} \hspace{0.05cm}\| \hspace{0.05cm}m_0 \} + {1}/{2}\cdot {\rm Pr}\{(\hat{m} = m_0)\hspace{0.05cm}\cap\hspace{0.05cm}({\it \Gamma} = {\rm Z}) \hspace{0.05cm}\| \hspace{0.05cm}m_1 \} $$
	{1}/{2} \cdot {\rm Pr}\{(\hat{m} = m_1)\hspace{0.05cm}\cap\hspace{0.05cm}{\it \Gamma} = {\rm Z} \hspace{0.05cm}\| \hspace{0.05cm}m_0 \} + {1}/{2}\cdot {\rm Pr}\{(\hat{m} = m_0)\hspace{0.05cm}\cap\hspace{0.05cm}({\it \Gamma} = {\rm Z}) \hspace{0.05cm}\| \hspace{0.05cm}m_1 \} $$		:$$\Rightarrow \hspace{0.3cm} {\rm Pr}({\it \Gamma} = {\rm Z}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) = {\rm Pr}\{(\hat{m} = m_1)\hspace{0.05cm}\cap\hspace{0.05cm}({\it \Gamma} = {\rm Z}) \hspace{0.05cm}\| \hspace{0.05cm}m_0 \} =\frac{1}{1 + 2^2} \cdot{\rm exp } \left [ - \frac{2^2 \cdot E_{\rm S}}{(1+2^2) \cdot N_{\rm 0}}\right ] = {1}/{5} \cdot {\rm e }^{-8} = \underline{6.7 \cdot 10^{-5}}\hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm} {\rm Pr}({\it \Gamma} = {\rm Z}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) = {\rm Pr}\{(\hat{m} = m_1)\hspace{0.05cm}\cap\hspace{0.05cm}({\it \Gamma} = {\rm Z}) \hspace{0.05cm}\| \hspace{0.05cm}m_0 \} =\frac{1}{1 + 2^2} \cdot
	{\rm exp } \left [ - \frac{2^2 \cdot E_{\rm S}}{(1+2^2) \cdot N_{\rm 0}}\right ] = {1}/{5} \cdot {\rm e }^{-8} = \underline{6.7 \cdot 10^{-5}}\hspace{0.05cm}.$$



	'''(5)'''  Since  ${\rm U}$  and  ${\rm Z}$  give a  "complete system"  according to statistics,  with the results of subtasks  '''(3)'''  and  '''(4)''':		'''(5)'''  Since  ${\rm U}$  and  ${\rm Z}$  give a  "complete system"  according to statistics,  with the results of subtasks  '''(3)'''  and  '''(4)''':
	:$${\rm Pr}({\it \Gamma} = {\rm U}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) ={\rm Pr}( {\rm error} ) -		:$${\rm Pr}({\it \Gamma} = {\rm U}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) ={\rm Pr}( {\rm error} ) -{\rm Pr}({\it \Gamma} = {\rm Z}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) = 3.37 \cdot 10^{-3} - 6.7 \cdot 10^{-5} = \underline{3.3 \cdot 10^{-3}}\hspace{0.05cm}.$$
	{\rm Pr}({\it \Gamma} = {\rm Z}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) = 3.37 \cdot 10^{-3} - 6.7 \cdot 10^{-5} = \underline{3.3 \cdot 10^{-3}}
	\hspace{0.05cm}.$$

	*Thus,  the conditional probability,  we are looking for,  is  "${\rm Pr}({\it \Gamma}= {\rm U}\hspace{0.05cm}\|\hspace{0.05cm} {\rm error})$":		*Thus,  the conditional probability,  we are looking for,  is  "${\rm Pr}({\it \Gamma}= {\rm U}\hspace{0.05cm}\|\hspace{0.05cm} {\rm error})$":
	:$${\rm Pr}({\it \Gamma}= {\rm U}\hspace{0.05cm}\|\hspace{0.05cm} {\rm error} ) =		:$${\rm Pr}({\it \Gamma}= {\rm U}\hspace{0.05cm}\|\hspace{0.05cm} {\rm error} ) =\frac{{\rm Pr}({\it \Gamma} = {\rm U}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) }{{\rm Pr}( {\rm error} )} = \frac{3.3 \cdot 10^{-3}}{3.37 \cdot 10^{-3}} \hspace{0.15cm} \underline{\approx 98 \%}\hspace{0.05cm}.$$
	\frac{{\rm Pr}({\it \Gamma} = {\rm U}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) }{{\rm Pr}( {\rm error} )} = \frac{3.3 \cdot 10^{-3}}{3.37 \cdot 10^{-3}} \hspace{0.15cm} \underline{\approx 98 \%}
	\hspace{0.05cm}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Digital Signal Transmission: Exercises\|^4.5 Non-Coherent Demodulation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.5 Non-Coherent Demodulation^]]
			[[de:Aufgaben:4.18_Nichtkohärente_BFSK–Demodulation]]

Guenter at 14:25, 8 October 2022

2022-10-08T14:25:29Z

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	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation}}

	[[File:~~EN_Dig_A_4_18~~.png\|right\|frame\|Block diagram for non-coherent demodulation ~~'''KORREKTUR: amount -> magnitude'''~~]]		[[File:EN_Dig_Z_4_18_neu.png\|right\|frame\|Block diagram for non-coherent demodulation]]
	We consider  "Frequency Shift Keying"  $\rm (FSK)$  with  $M = 2$   ⇒   binary signaling.  The two low-pass basis functions in this case are complex:		We consider  "Frequency Shift Keying"  $\rm (FSK)$  with  $M = 2$   ⇒   binary signaling.  The two low-pass basis functions in this case are complex:
	:$$\xi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},$$		:$$\xi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},$$

Hwang at 03:34, 11 September 2022

2022-09-11T03:34:44Z

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	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation}}

	[[File:EN_Dig_A_4_18.png\|right\|frame\|Block diagram for non-coherent demodulation]]		[[File:EN_Dig_A_4_18.png\|right\|frame\|Block diagram for non-coherent demodulation '''KORREKTUR: amount -> magnitude''']]
	We consider  "Frequency Shift Keying"  $\rm (FSK)$  with  $M = 2$   ⇒   binary signaling.  The two low-pass basis functions in this case are complex:		We consider  "Frequency Shift Keying"  $\rm (FSK)$  with  $M = 2$   ⇒   binary signaling.  The two low-pass basis functions in this case are complex:
	:$$\xi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},$$		:$$\xi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},$$

Guenter at 12:53, 30 August 2022

2022-08-30T12:53:21Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Solutions 1, 2 and 4</u> are correct:		'''(1)'''  <u>Solutions 1, 2 and 4</u>  are correct:
	*In coherent demodulation, orthogonality is given when the modulation index $h$ is a multiple of $0.5$. Binary FSK with $h = 0.5$ is also called ~~<i>~~Minimum Shift Keying~~</i>~~ (MSK).		*In coherent demodulation,  orthogonality is given when the modulation index  $h$  is a multiple of  $0.5$.  Binary FSK with  $h = 0.5$  is also called  "Minimum Shift Keying"  $\rm (MSK)$.
	*Since phase control compensates for the phase shift (delay) on the AWGN transmission channel   ⇒   ${\rm e}^{{\rm j}\phi} \cdot {\rm e}^{-{\rm j}\phi} = 1$, the signals in the equivalent low-pass region are actually:   $r(t) = s(t) + n(t) \hspace{0.05cm}.$
			*Since phase control compensates for the phase shift  ("delay")  on the AWGN transmission channel   ⇒   ${\rm e}^{{\rm j}\phi} \cdot {\rm e}^{-{\rm j}\phi} = 1$,  the signals in the equivalent low-pass region are actually:   $r(t) = s(t) + n(t) \hspace{0.05cm}.$



	'''(2)'''  Here, only the <u>first solution</u> is correct, i.e., now $h = 1, \ 2, \ 3, \text{ ...}$ must be an integer.		'''(2)'''  Here,  only the  <u>first solution</u>  is correct,  i.e., now $h = 1, \ 2, \ 3, \text{ ...}$  must be an integer.
	*So a non-coherent demodulation of FSK is not possible.		*So a non-coherent demodulation of FSK is not possible.
	*Because of the missing phase control, the following is also true:
			*Because of the missing phase control,  the following is also true:
	:$$r(t) = s(t) \cdot {\rm e }^{ - {\rm j }\hspace{0.05cm}\phi} + n(t) \hspace{0.05cm}.$$		:$$r(t) = s(t) \cdot {\rm e }^{ - {\rm j }\hspace{0.05cm}\phi} + n(t) \hspace{0.05cm}.$$

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	:$$p_{\rm S} = {\rm Pr}({\cal{E}}) = {\rm Pr}({\cal{E}}\| \hspace{0.05cm}m = m_0) = {\rm Pr}(\hat{m}= m_1\| \hspace{0.05cm}m = m_0)\hspace{0.05cm}.$$		:$$p_{\rm S} = {\rm Pr}({\cal{E}}) = {\rm Pr}({\cal{E}}\| \hspace{0.05cm}m = m_0) = {\rm Pr}(\hat{m}= m_1\| \hspace{0.05cm}m = m_0)\hspace{0.05cm}.$$

	*This probability is obtained from the given equation with $\gamma = 1$.		*This probability is obtained from the given equation with $\gamma = 1$.
	*In this case, ${\it \Gamma} ={\rm Z}$ is always and the decision rule is:   Decide on the symbol $m_0$ if $y_1 > y_2$.
			*In this case,  ${\it \Gamma} ={\rm Z}$  is always and the decision rule is:   "Decide on the symbol  $m_0$,  if $y_1 > y_2$".

	*It follows that:		*It follows that:
	:$$p_{\rm S} = \frac{1}{1 + \gamma^2} \cdot		:$$p_{\rm S} = \frac{1}{1 + \gamma^2} \cdot
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	{1}/{2} \cdot {\rm e }^{-E_{\rm S}/(2N_0)}\hspace{0.05cm}.$$		{1}/{2} \cdot {\rm e }^{-E_{\rm S}/(2N_0)}\hspace{0.05cm}.$$

	*With $E_{\rm S}/N_0 = 10$, we get $p_{\rm S} = 1/2 \cdot {\rm e}^{\rm –5} \approx \underline {0.337 \cdot \%}$.		*With  $E_{\rm S}/N_0 = 10$,  we get  $p_{\rm S} = 1/2 \cdot {\rm e}^{\rm –5} \approx \underline {0.337 \cdot \%}$.



	'''(4)'''  For symmetry reasons, this probability results in		'''(4)'''  For symmetry reasons,  this probability results in
	:$${\rm Pr}({\it \Gamma} = {\rm Z}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) =		:$${\rm Pr}({\it \Gamma} = {\rm Z}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) =
	{1}/{2} \cdot {\rm Pr}\{(\hat{m} = m_1)\hspace{0.05cm}\cap\hspace{0.05cm}{\it \Gamma} = {\rm Z} \hspace{0.05cm}\| \hspace{0.05cm}m_0 \} + {1}/{2}\cdot {\rm Pr}\{(\hat{m} = m_0)\hspace{0.05cm}\cap\hspace{0.05cm}({\it \Gamma} = {\rm Z}) \hspace{0.05cm}\| \hspace{0.05cm}m_1 \} $$		{1}/{2} \cdot {\rm Pr}\{(\hat{m} = m_1)\hspace{0.05cm}\cap\hspace{0.05cm}{\it \Gamma} = {\rm Z} \hspace{0.05cm}\| \hspace{0.05cm}m_0 \} + {1}/{2}\cdot {\rm Pr}\{(\hat{m} = m_0)\hspace{0.05cm}\cap\hspace{0.05cm}({\it \Gamma} = {\rm Z}) \hspace{0.05cm}\| \hspace{0.05cm}m_1 \} $$
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	'''(5)'''  Since ${\rm U}$ and ${\rm Z}$ give a complete system according to statistics, with the results of subtasks '''(3)''' and '''(4)''':		'''(5)'''  Since  ${\rm U}$  and  ${\rm Z}$  give a  "complete system"  according to statistics,  with the results of subtasks  '''(3)'''  and  '''(4)''':
	:$${\rm Pr}({\it \Gamma} = {\rm U}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) ={\rm Pr}( {\rm error} ) -		:$${\rm Pr}({\it \Gamma} = {\rm U}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) ={\rm Pr}( {\rm error} ) -
	{\rm Pr}({\it \Gamma} = {\rm Z}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) = 3.37 \cdot 10^{-3} - 6.7 \cdot 10^{-5} = \underline{3.3 \cdot 10^{-3}}		{\rm Pr}({\it \Gamma} = {\rm Z}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) = 3.37 \cdot 10^{-3} - 6.7 \cdot 10^{-5} = \underline{3.3 \cdot 10^{-3}}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*Thus, the conditional probability we are looking for is ${\rm Pr}({\it \Gamma}= {\rm U}\hspace{0.05cm}\|\hspace{0.05cm} {\rm error})$:		*Thus,  the conditional probability,  we are looking for,  is  "${\rm Pr}({\it \Gamma}= {\rm U}\hspace{0.05cm}\|\hspace{0.05cm} {\rm error})$":
	:$${\rm Pr}({\it \Gamma}= {\rm U}\hspace{0.05cm}\|\hspace{0.05cm} {\rm error} ) =		:$${\rm Pr}({\it \Gamma}= {\rm U}\hspace{0.05cm}\|\hspace{0.05cm} {\rm error} ) =
	\frac{{\rm Pr}({\it \Gamma} = {\rm U}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) }{{\rm Pr}( {\rm error} )} = \frac{3.3 \cdot 10^{-3}}{3.37 \cdot 10^{-3}} \hspace{0.15cm} \underline{= 98 \%}		\frac{{\rm Pr}({\it \Gamma} = {\rm U}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) }{{\rm Pr}( {\rm error} )} = \frac{3.3 \cdot 10^{-3}}{3.37 \cdot 10^{-3}} \hspace{0.15cm} \underline{\approx 98 \%}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 12:40, 30 August 2022

2022-08-30T12:40:25Z

← Older revision		Revision as of 14:40, 30 August 2022
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	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation}}

	[[File:EN_Dig_A_4_18.png\|right\|frame\|~~Non~~-coherent demodulation]]		[[File:EN_Dig_A_4_18.png\|right\|frame\|Block diagram for non-coherent demodulation]]
	We consider ~~<i>~~Frequency Shift Keying~~</i>~~  (FSK) with  $M = 2$   ⇒   binary signaling. The two low-pass basis functions in this case are complex ~~and are~~:		We consider  "Frequency Shift Keying"  $\rm (FSK)$  with  $M = 2$   ⇒   binary signaling.  The two low-pass basis functions in this case are complex:
	:$$\xi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},$$		:$$\xi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},$$
	:$$ \xi_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}-{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T \hspace{0.05cm}.$$		:$$ \xi_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}-{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T \hspace{0.05cm}.$$

	Then, for the two possible waveforms in the low-pass region, with the mean symbol energy  $E_{\rm S}$:		Then,  for the two possible waveforms in the low-pass region,  with the mean symbol energy  $E_{\rm S}$:
	:$$m_0\text{:}\hspace{0.2cm} s_{\rm TP,\hspace{0.05cm}0} = \sqrt{E_{\rm S}} \cdot \xi_1(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\boldsymbol{ s}_{\rm 0} = (\sqrt{E_{\rm S}}, 0)\hspace{0.05cm},$$		:$$m_0\text{:}\hspace{0.2cm} s_{\rm TP,\hspace{0.05cm}0} = \sqrt{E_{\rm S}} \cdot \xi_1(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\boldsymbol{ s}_{\rm 0} = (\sqrt{E_{\rm S}}, 0)\hspace{0.05cm},$$
	:$$m_1\text{:}\hspace{0.2cm} s_{\rm TP,\hspace{0.05cm}1} = \sqrt{E_{\rm S}} \cdot \xi_2(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\boldsymbol{ s}_{\rm 1} = (0, \sqrt{E_{\rm S}})\hspace{0.05cm}.$$		:$$m_1\text{:}\hspace{0.2cm} s_{\rm TP,\hspace{0.05cm}1} = \sqrt{E_{\rm S}} \cdot \xi_2(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\boldsymbol{ s}_{\rm 1} = (0, \sqrt{E_{\rm S}})\hspace{0.05cm}.$$

	Here,  $h$  indicates the so-called ~~<i>~~modulation index~~</i>~~. This must meet certain criteria to result in orthogonal waveforms. However, these criteria depend on whether a coherent or a non-coherent demodulator is used at the receiver.		Here,  $h$  indicates the so-called  "modulation index".  This must meet certain criteria to result in orthogonal waveforms.  However,  these criteria depend on whether a coherent or a non-coherent demodulator is used at the receiver.

	The diagram shows the non-coherent demodulator for ~~binary <i>~~Frequency Shift Keying~~</i>~~  (~~FSK~~) in the lower section. All complex signals are labeled in blue, complex values in green, and real values in red.		The diagram shows the non-coherent demodulator for  "Binary Frequency Shift Keying"  $\rm (BFSK)$  in the lower section.  All complex signals are labeled in blue,  complex values in green,  and real values in red.

	Compared to the decision process given in the  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation\| "theory part"]],  we now consider a complex decision that outputs a ~~<i>~~safety flag~~</i>~~  ${\it \Gamma} = \{{\rm Z}, \ {\rm U}\}$  in addition to the estimated value. The logical values  ${\rm Z}$  and  ${\rm U}$  stand for a realiable ~~('''Z'''uverlässige) and an '''U'''nreliable~~ decision~~, respectively. Thus, there are four possibilities for the decision, controlled by the parameter~~  ~~$\gamma$~~:		Compared to the decision process given in the  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation\| "theory part"]],  we now consider a complex decision that outputs a  "safety flag"  ${\it \Gamma} = \{{\rm Z}, \ {\rm U}\}$  in addition to the estimated value.  The logical values  ${\rm Z}$  and  ${\rm U}$  stand for
	:~~$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_0,\hspace{0.05cm}{\it \Gamma} = {\rm~~ Z}, ~~\hspace{0.5cm}{\rm falls} \hspace{0.15cm}y_1 > \gamma \cdot y_2\hspace{0.05cm},$$~~		*realiable decision  (German: "zuverlässig"   ⇒   subscript:  "Z"),
	:~~$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_0,\hspace{0.05cm}{\it \Gamma} = {\rm U}, \hspace{0.5cm}{\rm falls} \hspace{0.15cm}y_2 < y_1 < \gamma \cdot y_2\hspace{0.05cm},$$~~		* unreliable decision  (German: "unzuverlässig"   ⇒   subscript:  "U").
	:~~$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_1,\hspace{0.05cm}{\it \Gamma} = {\rm Z}, \hspace{0.5cm}{\rm falls} \hspace{0.15cm}y_2 > \gamma \cdot y_1\hspace{0.05cm},$$~~
	~~:$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_1,\hspace{0.05cm}{\it \Gamma} = {\rm~~ U~~}, \hspace{0.5cm}{\rm falls} \hspace{0~~.~~15cm}y_1 < y_2 < \gamma \cdot y_1\hspace{0.05cm}.$$~~

	~~In the questions for the exercise, the two values  $\gamma = 1$  and  $\gamma = 2$  are considered.~~

	For the probability that the decision erroneously chooses the symbol  $m_1$  and, moreover, indicates that this decision is to be considered reliable (particularly reprehensible), the following applies		Thus,  there are four possibilities for the decision,  controlled by the parameter  $\gamma$:
	:$${\rm Pr}\{\hat{m} = m_1,\hspace{0.~~05cm~~}{\it \Gamma} = {\rm Z} \hspace{0.05cm}\| \hspace{0.05cm}m_0 \} = \frac{1}{1 + \gamma^2} \cdot		:$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_0,\hspace{0.15cm}{\it \Gamma} = {\rm Z}, \hspace{0.5cm}{\rm if} \hspace{0.15cm}y_1 > \gamma \cdot y_2\hspace{0.05cm},$$
			:$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_0,\hspace{0.15cm}{\it \Gamma} = {\rm U}, \hspace{0.5cm}{\rm if} \hspace{0.15cm}y_2 < y_1 < \gamma \cdot y_2\hspace{0.05cm},$$
			:$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_1,\hspace{0.15cm}{\it \Gamma} = {\rm Z}, \hspace{0.5cm}{\rm if} \hspace{0.15cm}y_2 > \gamma \cdot y_1\hspace{0.05cm},$$
			:$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_1,\hspace{0.15cm}{\it \Gamma} = {\rm U}, \hspace{0.5cm}{\rm if} \hspace{0.15cm}y_1 < y_2 < \gamma \cdot y_1\hspace{0.05cm}.$$

			In the questions for the exercise,  the two values  $\gamma = 1$  and  $\gamma = 2$  are considered.

			For the probability that the decision erroneously chooses the symbol  $m_1$  and,  moreover,  indicates that this decision is to be considered  "reliable"  (particularly reprehensible),  the following applies
			:$${\rm Pr}\{\hat{m} = m_1,\hspace{0.15cm}{\it \Gamma} = {\rm Z} \hspace{0.05cm}\| \hspace{0.05cm}m_0 \} = \frac{1}{1 + \gamma^2} \cdot
	{\rm exp } \left [ - \frac{\gamma^2 \cdot E_{\rm S}}{(1+\gamma^2) \cdot N_{\rm 0}}\right ] \hspace{0.05cm}.$$		{\rm exp } \left [ - \frac{\gamma^2 \cdot E_{\rm S}}{(1+\gamma^2) \cdot N_{\rm 0}}\right ] \hspace{0.05cm}.$$

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	''Note:''		Note:  The exercise belongs to the the chapter  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation\|"Carrier Frequency Systems with Non-Coherent Demodulation"]].
	* The exercise belongs to the ~~topic of~~ the chapter  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation\|"Carrier Frequency Systems with Non-Coherent Demodulation"]].


Line 39:		Line 43:
	===Questions===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{Which statements are true in the case of coherent demodulation of the ~~FSK~~?		{Which statements are true in the case of coherent demodulation of the BFSK?
	\|type="[]"}		\|type="[]"}
	+ Orthogonality results if  $h$  is an integer.		+ Orthogonality results if  $h$  is an integer.
	+ Orthogonal waveforms also result for  $h = 0.5$.		+ Orthogonal waveforms also result for  $h = 0.5$.
	- Orthogonality cannot be achieved in principle.		- Orthogonality cannot be achieved in principle.
	+ For the AWGN channel  $r(t) = s(t) + n(t)$ is valid.		+ For the AWGN channel   $r(t) = s(t) + n(t)$   is valid.

	{Which statements are true for the non-coherent demodulation of ~~FSK~~?		{Which statements are true for the non-coherent demodulation of the BFSK?
	\|type="[]"}		\|type="[]"}
	+ Orthogonality results when  $h$  is an integer.		+ Orthogonality results when  $h$  is an integer.
	- Orthogonal waveforms also result for  $h = 0.5$.		- Orthogonal waveforms also result for  $h = 0.5$.
	- Orthogonality cannot be achieved in principle.		- Orthogonality cannot be achieved in principle.
	- For the AWGN channel  $r(t) = s(t) + n(t)$ is valid.		- For the AWGN channel   $r(t) = s(t) + n(t)$   is valid.

	{Let  $E_{\rm S}/N_0 = 10$. What is the error probability, i.e., the probability that the estimated value deviates from the transmitted message?		{Let  $E_{\rm S}/N_0 = 10$.  What is the error probability,  i.e.,  the probability that the estimated value deviates from the transmitted message?
	\|type="{}"}		\|type="{}"}
	$p_{\rm S} \ = \ $ { 0.337 3% } $\ \%$		$p_{\rm S} \ = \ $ { 0.337 3% } $\ \%$

	{Let  $\gamma = 2$  and  $E_{\rm S}/N_0 = 10$. What is the probability that the security feature signals a reliable decision despite the errors?		{Let  $\gamma = 2$  and  $E_{\rm S}/N_0 = 10$.  What is the probability that the security feature signals a reliable decision despite the errors?
	\|type="{}"}		\|type="{}"}
	${\rm Pr}({\it \Gamma} = {\rm Z\hspace{0.05cm} \| \hspace{0.05cm}error}) \ = \ $ { 6.7 3% } $\ \cdot 10^{\rm –5}$		${\rm Pr}({\it \Gamma} = {\rm Z\hspace{0.05cm} \| \hspace{0.05cm}error}) \ = \ $ { 6.7 3% } $\ \cdot 10^{\rm –5}$

	{Furthermore, let  $\gamma = 2$ and  $E_{\rm S}/N_0 = 10$. What is the (conditional) probability that the additional information "unreliable" is displayed in case of an error?		{Furthermore,  let  $\gamma = 2$  and  $E_{\rm S}/N_0 = 10$.  What is the  (conditional)  probability that the additional information  "unreliable"  is displayed in case of an error?
	\|type="{}"}		\|type="{}"}
	${\rm Pr}({\it \Gamma} = {\rm U\hspace{0.05cm} \| \hspace{0.05cm}error}) \ = \ $ { 98 3% } $\ \%$		${\rm Pr}({\it \Gamma} = {\rm U\hspace{0.05cm} \| \hspace{0.05cm}error}) \ = \ $ { 98 3% } $\ \%$

Guenter at 15:04, 24 August 2022

2022-08-24T15:04:21Z

← Older revision		Revision as of 17:04, 24 August 2022
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	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation}}

	[[File:~~P_ID2080__Dig_A_4_18~~.png\|right\|frame\|Non-coherent demodulation]]		[[File:EN_Dig_A_4_18.png\|right\|frame\|Non-coherent demodulation]]
	We consider <i>Frequency Shift Keying</i>  (FSK) with  $M = 2$   ⇒   binary signaling. The two low-pass basis functions in this case are complex and are:		We consider <i>Frequency Shift Keying</i>  (FSK) with  $M = 2$   ⇒   binary signaling. The two low-pass basis functions in this case are complex and are:
	:$$\xi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},$$		:$$\xi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},$$

Hwang at 17:10, 16 August 2022

2022-08-16T17:10:57Z

Show changes

Hwang at 15:01, 16 August 2022

2022-08-16T15:01:10Z

← Older revision		Revision as of 17:01, 16 August 2022
Line 1:		Line 1:

	{{quiz-Header\|Buchseite=~~Digitalsignalübertragung~~/~~Trägerfrequenzsysteme mit nichtkohärenter Demodulation~~}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation}}

	[[File:P_ID2080__Dig_A_4_18.png\|right\|frame\|~~Nichtkohärente Demodulation~~]]		[[File:P_ID2080__Dig_A_4_18.png\|right\|frame\|Non-coherent demodulation]]
	~~Wir betrachten~~ <i>Frequency Shift Keying</i>  (FSK) ~~mit~~  $M = 2$   ⇒   ~~binäre Signalisierung~~. ~~Die beiden Basisfunktionen im Tiefpassbereich sind~~ in ~~diesem Fall komplex und lauten~~:		We consider <i>Frequency Shift Keying</i>  (FSK) with  $M = 2$   ⇒   binary signaling. The two low-pass basis functions in this case are complex and are:
	:$$\xi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},$$		:$$\xi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},$$
	:$$ \xi_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}-{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T \hspace{0.05cm}.$$		:$$ \xi_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}-{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T \hspace{0.05cm}.$$

	~~Für die zwei möglichen Signalformen im Tiefpassbereich gilt dann mit der mittleren Symbolenergie~~  $E_{\rm S}$:		Then, for the two possible waveforms in the low-pass region, with the mean symbol energy  $E_{\rm S}$:
	:$$m_0\text{:}\hspace{0.2cm} s_{\rm TP,\hspace{0.05cm}0} = \sqrt{E_{\rm S}} \cdot \xi_1(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\boldsymbol{ s}_{\rm 0} = (\sqrt{E_{\rm S}}, 0)\hspace{0.05cm},$$		:$$m_0\text{:}\hspace{0.2cm} s_{\rm TP,\hspace{0.05cm}0} = \sqrt{E_{\rm S}} \cdot \xi_1(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\boldsymbol{ s}_{\rm 0} = (\sqrt{E_{\rm S}}, 0)\hspace{0.05cm},$$
	:$$m_1\text{:}\hspace{0.2cm} s_{\rm TP,\hspace{0.05cm}1} = \sqrt{E_{\rm S}} \cdot \xi_2(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\boldsymbol{ s}_{\rm 1} = (0, \sqrt{E_{\rm S}})\hspace{0.05cm}.$$		:$$m_1\text{:}\hspace{0.2cm} s_{\rm TP,\hspace{0.05cm}1} = \sqrt{E_{\rm S}} \cdot \xi_2(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\boldsymbol{ s}_{\rm 1} = (0, \sqrt{E_{\rm S}})\hspace{0.05cm}.$$

	~~Hierbei gibt~~  $h$  ~~den~~ so ~~genannten~~ <i>~~Modulationsindex~~</i> an. ~~Dieser muss gewisse Kriterien erfüllen, damit sich orthogonale Signalformen ergeben~~. ~~Diese Kriterien hängen allerdings davon ab~~, ~~ob beim Empfänger ein kohärenter oder ein nichtkohärenter Demodulator verwendet wird~~.		Here,  $h$  indicates the so-called <i>modulation index</i>. This must meet certain criteria to result in orthogonal waveforms. However, these criteria depend on whether a coherent or a non-coherent demodulator is used at the receiver.

	~~Die Grafik zeigt im unteren Bereich den nichtkohärenten Demodulator für binäres~~ <i>Frequency Shift Keying</i>  (FSK). ~~Alle komplexen Signale sind blau beschriftet~~, ~~komplexe Werte grün und reelle Werte rot~~.		The diagram shows the non-coherent demodulator for binary <i>Frequency Shift Keying</i>  (FSK) in the lower section. All complex signals are labeled in blue, complex values in green, and real values in red.

	~~Gegenüber dem im~~  [[~~Digitalsignal%C3%BCbertragung~~/~~Tr%C3%A4gerfrequenzsysteme_mit_nichtkoh%C3%A4renter_Demodulation~~\| ~~Theorieteil~~]]  ~~angegeben Entscheidungsprozess wird nun ein komplizierter Entscheider betrachtet, der außer dem Schätzwert noch ein~~ <i>~~Sicherheitsflag~~</i>  ${\it \Gamma} = \{{\rm Z}, \ {\rm U}\}$  ~~ausgibt~~. ~~Die logischen Größen~~  ${\rm Z}$  ~~und~~  ${\rm U}$  ~~stehen hierbei für eine~~ '''Z'''uverlässige ~~bzw. eine~~ '''U'''~~nzuverlässige Entscheidung~~. ~~Es gibt also vier Möglichkeiten der Entscheidung~~, ~~gesteuert durch den Parameter~~  $\gamma$:		Compared to the decision process given in the  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation\| "theory part"]],  we now consider a complex decision that outputs a <i>safety flag</i>  ${\it \Gamma} = \{{\rm Z}, \ {\rm U}\}$  in addition to the estimated value. The logical values  ${\rm Z}$  and  ${\rm U}$  stand for a realiable ('''Z'''uverlässige) and an '''U'''nreliable decision, respectively. Thus, there are four possibilities for the decision, controlled by the parameter  $\gamma$:
	:$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_0,\hspace{0.05cm}{\it \Gamma} = {\rm Z}, \hspace{0.5cm}{\rm falls} \hspace{0.15cm}y_1 > \gamma \cdot y_2\hspace{0.05cm},$$		:$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_0,\hspace{0.05cm}{\it \Gamma} = {\rm Z}, \hspace{0.5cm}{\rm falls} \hspace{0.15cm}y_1 > \gamma \cdot y_2\hspace{0.05cm},$$
	:$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_0,\hspace{0.05cm}{\it \Gamma} = {\rm U}, \hspace{0.5cm}{\rm falls} \hspace{0.15cm}y_2 < y_1 < \gamma \cdot y_2\hspace{0.05cm},$$		:$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_0,\hspace{0.05cm}{\it \Gamma} = {\rm U}, \hspace{0.5cm}{\rm falls} \hspace{0.15cm}y_2 < y_1 < \gamma \cdot y_2\hspace{0.05cm},$$
Line 21:		Line 21:
	:$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_1,\hspace{0.05cm}{\it \Gamma} = {\rm U}, \hspace{0.5cm}{\rm falls} \hspace{0.15cm}y_1 < y_2 < \gamma \cdot y_1\hspace{0.05cm}.$$		:$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_1,\hspace{0.05cm}{\it \Gamma} = {\rm U}, \hspace{0.5cm}{\rm falls} \hspace{0.15cm}y_1 < y_2 < \gamma \cdot y_1\hspace{0.05cm}.$$

	In ~~den Fragen zur Aufgabe werden die beiden Werte~~  $\gamma = 1$  ~~und~~  $\gamma = 2$  ~~betrachtet~~.		In the questions for the exercise, the two values  $\gamma = 1$  and  $\gamma = 2$  are considered.

	~~Für die Wahrscheinlichkeit, dass sich der Entscheider fälschlicherweise für das Symbol~~  $m_1$  ~~entscheidet und zudem anzeigt~~, ~~dass diese Entscheidung als zuverlässig zu betrachten ist~~ (~~besonders verwerflich~~), ~~gilt~~		For the probability that the decision erroneously chooses the symbol  $m_1$  and, moreover, indicates that this decision is to be considered reliable (particularly reprehensible), the following applies
	:$${\rm Pr}\{\hat{m} = m_1,\hspace{0.05cm}{\it \Gamma} = {\rm Z} \hspace{0.05cm}\| \hspace{0.05cm}m_0 \} = \frac{1}{1 + \gamma^2} \cdot		:$${\rm Pr}\{\hat{m} = m_1,\hspace{0.05cm}{\it \Gamma} = {\rm Z} \hspace{0.05cm}\| \hspace{0.05cm}m_0 \} = \frac{1}{1 + \gamma^2} \cdot
	{\rm exp } \left [ - \frac{\gamma^2 \cdot E_{\rm S}}{(1+\gamma^2) \cdot N_{\rm 0}}\right ] \hspace{0.05cm}.$$		{\rm exp } \left [ - \frac{\gamma^2 \cdot E_{\rm S}}{(1+\gamma^2) \cdot N_{\rm 0}}\right ] \hspace{0.05cm}.$$
Line 31:		Line 31:


	''~~Hinweis~~:''		''Note:''
	* ~~Die Aufgabe gehört zum Themengebiet des Kapitels~~  [[~~Digitalsignal%C3%BCbertragung~~/~~Tr%C3%A4gerfrequenzsysteme_mit_nichtkoh%C3%A4renter_Demodulation~~\| ~~Trägerfrequenzsysteme mit nichtkohärenter~~ Demodulation]].		* The exercise belongs to the topic of the chapter  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation\|"Carrier Frequency Systems with Non-Coherent Demodulation"]].