Aufgaben:Exercise 4.8Z: BPSK Error Probability - Revision history

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	[[Category:Modulation Methods: Exercises\|^4.2 Linear Digital Modulation^]]		[[Category:Modulation Methods: Exercises\|^4.2 Linear Digital Modulation^]]
	[[de:Aufgaben:4.~~7Z_BPSK–Fehlerwahrscheinlichkeit~~]]		[[de:Aufgaben:Aufgabe 4.8Z: BPSK–Fehlerwahrscheinlichkeit]]

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	'''(1)'''  The noise rms value here is given by		'''(1)'''  The noise rms value here is given by
	:$$\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm B}}}= \sqrt{\frac{2 \cdot 10^{-9}\,{\rm V^2/Hz}}{2 \cdot 1\,{\rm ns}}}= 1\,{\rm V}		:$$\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm B}}}= \sqrt{\frac{2 \cdot 10^{-9}\,{\rm V^2/Hz}}{2 \cdot 1\,{\rm ns}}}= 1\,{\rm V}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm BB} = {\rm Q}\left ( {s_0}/{\sigma_d } \right )= {\rm Q}(4)\hspace{0.15cm}\underline {= 0.317 \cdot 10^{-4}}.$$
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm BB} = {\rm Q}\left ( {s_0}/{\sigma_d } \right )= {\rm Q}(4)\hspace{0.15cm}\underline {= 0.317 \cdot 10^{-4}}.$$

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	[[Category:Modulation Methods: Exercises\|^4.2 Linear Digital Modulation^]]		[[Category:Modulation Methods: Exercises\|^4.2 Linear Digital Modulation^]]
			[[de:Aufgaben:4.7Z_BPSK–Fehlerwahrscheinlichkeit]]

Guenter at 16:26, 15 April 2022

2022-04-15T16:26:38Z

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	'''(2)'''  For the baseband system:		'''(2)'''  For the baseband system:
	:$$E_{\rm B} = s_0^2 \cdot T_{\rm B}= (4\,{\rm V})^2 \cdot 10^{-9}\,{\rm s}\hspace{0.15cm}\underline {= 1.6 \cdot 10^{-8}\,{\rm V^2s}}.$$		:$$E_{\rm B} = s_0^2 \cdot T_{\rm B}= (4\,{\rm V})^2 \cdot 10^{-9}\,{\rm s}\hspace{0.15cm}\underline {= 1.6 \cdot 10^{-8}\,{\rm V^2s}}.$$
	*Of course, the second equation ~~given~~ gives the exact same error probability		*Of course,  the second equation gives the exact same error probability
	:$$ p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 16 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\rm V^2/Hz} }} \hspace{0.1cm}\right ) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$		:$$ p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 16 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\rm V^2/Hz} }} \hspace{0.1cm}\right ) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$


	'''(3)'''  When the transmitted amplitude is half  $s_0 = 2\,{\rm V}$  the energy per bit decreases to one-fourth and the following equations apply:		'''(3)'''  When the transmitted amplitude is half   ⇒   $s_0 = 2\,{\rm V}$,  the energy per bit decreases to one-fourth and the following equations apply:
	:$$ p_{\rm BB} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )= {\rm Q}\left ( \frac{2\,{\rm V}}{1\,{\rm V}} \right )= {\rm Q}(2)= 227 \cdot 10^{-4},$$		:$$ p_{\rm BB} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )= {\rm Q}\left ( \frac{2\,{\rm V}}{1\,{\rm V}} \right )= {\rm Q}(2)= 227 \cdot 10^{-4},$$
	:$$ p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 4 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\rm V^2/Hz} }} \hspace{0.1cm}\right ) = {\rm Q}(2)\hspace{0.15cm}\underline {= 227 \cdot 10^{-4}}.$$		:$$ p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 4 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\rm V^2/Hz} }} \hspace{0.1cm}\right ) = {\rm Q}(2)\hspace{0.15cm}\underline {= 227 \cdot 10^{-4}}.$$


	'''(4)'''  <u>Answer 2</u> is correct:		'''(4)'''  <u>Answer 2</u>  is correct:
	*Considering the energy  $E_{\rm B} = s_0^2 · T_{\rm B}/2$  we obtain		*Considering the energy  $E_{\rm B} = s_0^2 · T_{\rm B}/2$  we obtain
	:$$ p_{\rm BPSK} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )= {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }}\hspace{0.1cm}\right ).$$		:$$ p_{\rm BPSK} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )= {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }}\hspace{0.1cm}\right ).$$
	*Thus, the same result is obtained as for the optimal baseband transmission system.		*Thus,  the same result is obtained as for the optimal baseband transmission system.

Guenter at 16:21, 15 April 2022

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	We assume the optimal baseband transmission system for binary signals with		We assume the optimal baseband transmission system for binary signals with
	* bipolar amplitude coefficients  $a_ν ∈ \{-1, +1\}$,		* bipolar amplitude coefficients  $a_ν ∈ \{-1, +1\}$,
	* rectangular transmitted signal with signal values  $±s_0$  and bit duration  $T_{\rm B}$,		* rectangular transmitted signal  $s(t)$  with signal values  $±s_0$  and bit duration  $T_{\rm B}$,
	* AWGN noise with the (one-sided) noise power density  $N_0$,		* AWGN noise with the (one-sided) noise power density  $N_0$,
	* receiver filter according to the matched filter principle,		* receiver filter according to the matched filter principle,
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	If nothing else is specified, you should assume the following numerical values:		If nothing else is specified,  you should assume the following numerical values:
	:$$s_0 = 4\,{\rm V},\hspace{0.2cm} T_{\rm B} = 1\,{\rm ns},\hspace{0.2cm}N_0 = 2 \cdot 10^{-9}\, {\rm V^2/Hz} \hspace{0.05cm}.$$		:$$s_0 = 4\,{\rm V},\hspace{0.2cm} T_{\rm B} = 1\,{\rm ns},\hspace{0.2cm}N_0 = 2 \cdot 10^{-9}\, {\rm V^2/Hz} \hspace{0.05cm}.$$
	Using the noise rms value  $σ_d$  at the decision and the complementary Gaussian error function  ${\rm Q}(x)$ the bit error probability of this baseband system is   ⇒   see table:		Using the noise rms value  $σ_d$  at the decision and the complementary Gaussian error function  ${\rm Q}(x)$,  the bit error probability of this baseband  $\rm (BB)$  system is   ⇒   see table:
	:$$ p_{\rm BB} = {\rm Q}\left ( {s_0}/{\sigma_d } \right )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{{N_0}/(2 \cdot T_{\rm B}}).$$		:$$ p_{\rm BB} = {\rm Q}\left ( {s_0}/{\sigma_d } \right )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{{N_0}/(2 \cdot T_{\rm B}}).$$
	This bit error probability can also be expressed in the form		This bit error probability can also be expressed in the form
	:$$p_{\rm BB} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right )$$		:$$p_{\rm BB} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ),$$
	where  $E_{\rm B}$  indicates the "signal energy per bit".		where  $E_{\rm B}$  indicates the  "signal energy per bit".

	The bit error probability of a comparable transmission system with  ''Binary Phase Shift Keying''  (BPSK) is:		The bit error probability of a comparable transmission system with  "Binary Phase Shift Keying"  $\rm (BPSK)$  is:
	:$$ p_{\rm BPSK} = {\rm Q}\left ( {s_0}/{\sigma_d } \right )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{{N_0}/{T_{\rm B}}}.$$		:$$ p_{\rm BPSK} = {\rm Q}\left ( {s_0}/{\sigma_d } \right )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{{N_0}/{T_{\rm B}}}.$$

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			Notes:
			*The exercise belongs to the chapter  [[Modulation_Methods/Linear_Digital_Modulation\|"Linear Digital Modulation"]].
	''Notes:''		*Reference is made in particular to the section  [[Modulation_Methods/Linear_Digital_Modulation#Error_probabilities_-_a_brief_overview\|"Error probabilities - a brief overview"]].
	*The exercise belongs to the chapter  [[Modulation_Methods/Linear_Digital_Modulation\|Linear Digital Modulation]].		*The derivations can be found in the chapter  [[Digital_Signal_Transmission/Lineare_digitale_Modulation_–_Kohärente_Demodulation\|"Linear Digital Modulation - Coherent Demodulation"]]  of the book "Digital Signal Transmission".
	*Reference is made in particular to the section  [[Modulation_Methods/Linear_Digital_Modulation#Error_probabilities_-_a_brief_overview\|Error probabilities - a brief overview]].
	*The derivations can be found in the chapter  [[Digital_Signal_Transmission/Lineare_digitale_Modulation_–_Kohärente_Demodulation\|Linear Digital Modulation - Coherent Demodulation]]  of the book "Digital Signal Transmission".
	*The specification of a power in  $\rm V^2$  or an energy in  $\rm V^2 s$  means a conversion to the reference resistance  $1 \ \rm \Omega$.		*The specification of a power in  $\rm V^2$  or an energy in  $\rm V^2 s$  means a conversion to the reference resistance  $1 \ \rm \Omega$.

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	{For this set ~~of parameters~~, what is the energy per bit   ⇒    $E_{\rm B}$  for the baseband system?		{For this parameter set,  what is the energy per bit   ⇒    $E_{\rm B}$  for the baseband system?
	\|type="{}"}		\|type="{}"}
	$E_{\rm B} \ = \ $ { 1.6 3% } $\ \cdot 10^{-8} \ \rm V^2 s$		$E_{\rm B} \ = \ $ { 1.6 3% } $\ \cdot 10^{-8} \ \rm V^2 s$
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	- $p_{\rm BPSK} = {\rm Q}\big[(4E_{\rm B}/N_0)^{1/2}\big].$		- $p_{\rm BPSK} = {\rm Q}\big[(4E_{\rm B}/N_0)^{1/2}\big].$

	{What are the error probabilities for BPSK ~~for~~  $E_{\rm B}/N_0 = 8$  and  $E_{\rm B}/N_0 = 2$?		{What are the error probabilities for BPSK with  $E_{\rm B}/N_0 = 8$  and  $E_{\rm B}/N_0 = 2$?
	\|type="{}"}		\|type="{}"}
	$E_{\rm B}/N_0 = 8\text{:} \ \ \ \ p_{\rm BPSK} \ = \ $ { 0.317 3% } $\ \cdot 10^{-4}$		$E_{\rm B}/N_0 = 8\text{:} \ \ \ \ p_{\rm BPSK} \ = \ $ { 0.317 3% } $\ \cdot 10^{-4}$

Hwang at 17:04, 21 March 2022

2022-03-21T17:04:28Z

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Guenter at 14:03, 16 March 2022

2022-03-16T14:03:04Z

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	[[Category:Modulation Methods: Exercises\|^4.2 ~~Lineare digitale~~ Modulation^]]		[[Category:Modulation Methods: Exercises\|^4.2 Linear Digital Modulation^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 4.8Z: BPSK–Fehlerwahrscheinlichkeit to Aufgaben:Exercise 4.8Z: BPSK Error Probability

2022-03-16T13:59:24Z

Guenter moved page Aufgaben:Aufgabe 4.8Z: BPSK–Fehlerwahrscheinlichkeit to Aufgaben:Exercise 4.8Z: BPSK Error Probability

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	[[Category:~~Aufgaben zu Modulationsverfahren~~\|^4.2 Lineare digitale Modulation^]]		[[Category:Modulation Methods: Exercises\|^4.2 Lineare digitale Modulation^]]

Javier: Text replacement - "[[Digitalsignalübertragung" to "[[Digital_Signal_Transmission"

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	*Die Aufgabe gehört zum Kapitel  [[Modulation_Methods/Lineare_digitale_Modulation\|Lineare digitale Modulation]].		*Die Aufgabe gehört zum Kapitel  [[Modulation_Methods/Lineare_digitale_Modulation\|Lineare digitale Modulation]].
	*Bezug genommen wird insbesondere auf die Seite  [[Modulation_Methods/Lineare_digitale_Modulation#Fehlerwahrscheinlichkeiten_-_ein_kurzer_.C3.9Cberblick\|Fehlerwahrscheinlichkeiten - ein kurzer Überblick]].		*Bezug genommen wird insbesondere auf die Seite  [[Modulation_Methods/Lineare_digitale_Modulation#Fehlerwahrscheinlichkeiten_-_ein_kurzer_.C3.9Cberblick\|Fehlerwahrscheinlichkeiten - ein kurzer Überblick]].
	*Die Herleitungen finden Sie im Kapitel  [[~~Digitalsignalübertragung~~/Lineare_digitale_Modulation_–_Kohärente_Demodulation\|Lineare digitale Modulation – Kohärente Demodulation]]  des Buches „Digitalsignalübertragung”.		*Die Herleitungen finden Sie im Kapitel  [[Digital_Signal_Transmission/Lineare_digitale_Modulation_–_Kohärente_Demodulation\|Lineare digitale Modulation – Kohärente Demodulation]]  des Buches „Digitalsignalübertragung”.
	*Die Angabe einer Leistung in  $\rm V^2$  bzw. einer Energie in  $\rm V^2 s$  bedeutet eine Umrechnung auf den Bezugswiderstand  $1 \ \rm \Omega$.		*Die Angabe einer Leistung in  $\rm V^2$  bzw. einer Energie in  $\rm V^2 s$  bedeutet eine Umrechnung auf den Bezugswiderstand  $1 \ \rm \Omega$.