Aufgaben:Exercise 1.08Z: BPSK Error Probability - Revision history

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	:$$E_{\rm B} = s_0^2 \cdot T_{\rm B}= (4\,{\rm V})^2 \cdot10^{-9}\,{\rm s}\hspace{0.1cm}\underline {= 1.6 \cdot 10^{-8}\,{\rm V^2s}}.$$		:$$E_{\rm B} = s_0^2 \cdot T_{\rm B}= (4\,{\rm V})^2 \cdot10^{-9}\,{\rm s}\hspace{0.1cm}\underline {= 1.6 \cdot 10^{-8}\,{\rm V^2s}}.$$
	*Of course, the additional given equation gives the exact same error probability:		*Of course, the additional given equation gives the exact same error probability:
	:$$p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 16 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\~~rmV~~^2/Hz} }} \hspace{0.1cm}\right) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$		:$$p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 16 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\rm V^2/Hz} }} \hspace{0.1cm}\right) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$

	*A comparison with question  '''(4)'''  of  [[Aufgaben:Exercise_1.08:_Comparison_of_ASK_and_BPSK\|"Exercise 1.8"]]  shows that  $E_{\rm B}/N_{0} = 8$  is not  (exactly)  equal to  $10 \cdot \lg E_{\rm B}/N_{0} = 9 \ \rm dB$.		*A comparison with question  '''(4)'''  of  [[Aufgaben:Exercise_1.08:_Comparison_of_ASK_and_BPSK\|"Exercise 1.8"]]  shows that  $E_{\rm B}/N_{0} = 8$  is not  (exactly)  equal to  $10 \cdot \lg E_{\rm B}/N_{0} = 9 \ \rm dB$.
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	[[Category:Digital Signal Transmission: Exercises\|^1.5 Linear Digital Modulation^]]		[[Category:Digital Signal Transmission: Exercises\|^1.5 Linear Digital Modulation^]]
	[[de:Aufgaben:1.~~08Z_BPSK~~-Fehlerwahrscheinlichkeit]]		[[de:Aufgaben:Aufgabe 1.08Z: BPSK-Fehlerwahrscheinlichkeit]]

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	'''(1)'''  The noise rms value is given here by		'''(1)'''  The noise rms value is given here by
	:$$\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm B}}}= \sqrt{\frac{2 \cdot 10^{-9}\,{\rm V^2/Hz}}{2 \cdot 1\,{\rm ns}}}= 1\,{\rm V}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}p_{\rm BB} = {\rm Q}\left ({s_0}/{\sigma_d } \right		:$$\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)= 0.317 \cdot 10^{-4}\hspace{0.1cm}\underline {= 0.00317 \%}.$$
	)= {\rm Q}(4)= 0.317 \cdot 10^{-4}\hspace{0.1cm}\underline {= 0.00317 \%}.$$


	'''(2)'''  For the baseband system:		'''(2)'''  For the baseband system:
	:$$E_{\rm B} = s_0^2 \cdot T_{\rm B}= (4\,{\rm V})^2 \~~cdot~~		:$$E_{\rm B} = s_0^2 \cdot T_{\rm B}= (4\,{\rm V})^2 \cdot10^{-9}\,{\rm s}\hspace{0.1cm}\underline {= 1.6 \cdot 10^{-8}\,{\rm V^2s}}.$$
	10^{-9}\,{\rm s}\hspace{0.1cm}\underline {= 1.6 \cdot 10^{-8}\,{\rm V^2s}}.$$
	*Of course, the additional given equation gives the exact same error probability:		*Of course, the additional given equation gives the exact same error probability:
	:$$p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right		:$$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}\, {\rmV^2/Hz} }} \hspace{0.1cm}\right) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$
	) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 16 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\rm
	V^2/Hz} }} \hspace{0.1cm}\right
	) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$

	*A comparison with question  '''(4)'''  of  [[Aufgaben:Exercise_1.08:_Comparison_of_ASK_and_BPSK\|"Exercise 1.8"]]  shows that  $E_{\rm B}/N_{0} = 8$  is not  (exactly)  equal to  $10 \cdot \lg E_{\rm B}/N_{0} = 9 \ \rm dB$.		*A comparison with question  '''(4)'''  of  [[Aufgaben:Exercise_1.08:_Comparison_of_ASK_and_BPSK\|"Exercise 1.8"]]  shows that  $E_{\rm B}/N_{0} = 8$  is not  (exactly)  equal to  $10 \cdot \lg E_{\rm B}/N_{0} = 9 \ \rm dB$.

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	[[Category:Digital Signal Transmission: Exercises\|^1.5 Linear Digital Modulation^]]		[[Category:Digital Signal Transmission: Exercises\|^1.5 Linear Digital Modulation^]]
			[[de:Aufgaben:1.08Z_BPSK-Fehlerwahrscheinlichkeit]]

Hwang at 15:12, 10 May 2022

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	Notes:		Notes:
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation\|Linear Digital Modulation - Coherent Demodulation]].		*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation\|"Linear Digital Modulation - Coherent Demodulation"]].

	*Reference is also made to the chapter  [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission\|"Error Probability for Baseband Transmission"]].		*Reference is also made to the chapter  [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission\|"Error Probability for Baseband Transmission"]].
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	) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$		) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$

	*A comparison with question  '''(4)'''  of  [[Aufgaben:Exercise_1.08:_Comparison_of_ASK_and_BPSK\|Exercise 1.8]]  shows that  $E_{\rm B}/N_{0} = 8$  is not  (exactly)  equal to  $10 \cdot \lg E_{\rm B}/N_{0} = 9 \ \rm dB$.		*A comparison with question  '''(4)'''  of  [[Aufgaben:Exercise_1.08:_Comparison_of_ASK_and_BPSK\|"Exercise 1.8"]]  shows that  $E_{\rm B}/N_{0} = 8$  is not  (exactly)  equal to  $10 \cdot \lg E_{\rm B}/N_{0} = 9 \ \rm dB$.
	*In the first case  $p_{\rm BB} = 0.317 \cdot 10^{–4}$ is obtained,  in the second  $p_{\rm BB} = 0.336 \cdot 10^{-4}$.		*In the first case  $p_{\rm BB} = 0.317 \cdot 10^{–4}$ is obtained,  in the second  $p_{\rm BB} = 0.336 \cdot 10^{-4}$.

Guenter at 15:24, 6 May 2022

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Guenter at 14:25, 6 May 2022

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	}}		}}

	[[File:P_ID1681__Dig_Z_4_1.png\|right\|frame\|Numerical values of ~~the~~ function  ${\rm Q}(x)$]]		[[File:P_ID1681__Dig_Z_4_1.png\|right\|frame\|Numerical values of function  ${\rm Q}(x)$]]
	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  $a_{\nu} \in \{–1, +1\}$,		*bipolar amplitude coefficients  $a_{\nu} \in \{–1, +1\}$,

	*rectangular transmitted signal with signal values  $\pm s_{0}$  and bit duration  $T_{\rm B}$,		*rectangular transmitted signal with signal values  $\pm s_{0}$  and bit duration  $T_{\rm B}$,

	*AWGN noise with noise power density  $N_{0}$,		*AWGN noise with noise power density  $N_{0}$,

	*receiver filter according to the matched filter principle,		*receiver filter according to the matched filter principle,

	*decision with the optimal threshold  $E = 0$.		*decision with the optimal threshold  $E = 0$.


	Unless otherwise specified, you should also assume the following numerical values:		Unless otherwise specified,  you should also assume the following numerical values:
	:$$ s_0 = 4\,{\rm V},\hspace{0.2cm} T_{\rm B} = 1\,{\rm ns},\hspace{0.2cm}N_0 = 2 \cdot 10^{-9}\, {\rm V^2/Hz} \hspace{0.05cm}.$$		:$$ s_0 = 4\,{\rm V},\hspace{0.2cm} T_{\rm B} = 1\,{\rm ns},\hspace{0.2cm}N_0 = 2 \cdot 10^{-9}\, {\rm V^2/Hz} \hspace{0.05cm}.$$

	The bit error probability of this "baseband system" has already been given in the chapter  [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission\|Error Probability for Baseband Transmission]]  (Index:  BB):		The bit error probability of this  "baseband system"  has already been given in the chapter  [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission\|"Error Probability for Baseband Transmission"]]  $($Index:  $\rm BB)$:
	:$$p_{\rm BB} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm B}}}.$$		:$$p_{\rm BB} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm B}}}.$$

	Here,  $\sigma_{d}$  denotes the noise rms value at the decision and  ${\rm Q}(x)$  denotes the complementary Gaussian error function, which is given here in tabular form.		Here,  $\sigma_{d}$  denotes the noise rms value at the decision device and  ${\rm Q}(x)$  denotes the complementary Gaussian error function,  which is given here in tabular form.  This error probability can also be expressed in the form
			:$$p_{\rm BB} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ),$$
			where  $E_{\rm B}$  denotes  "energy per bit."

	~~This error probability can also be expressed in the form~~		The error probability of a comparable transmission system with  "Binary Phase Shift Keying" is  $($Index:  $\rm BPSK)$:
	~~:$$p_{\rm BB} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right )$$~~
	~~where  $E_{\rm B}$  denotes "energy per bit."~~

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

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			Notes:
			*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation\|Linear Digital Modulation - Coherent Demodulation]].

			*Reference is also made to the chapter  [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission\|"Error Probability for Baseband Transmission"]].

			*You can check the results with the HTML5/JavaScript applet  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|"Complementary Gaussian Error Functions"]].

	~~''Notes:''~~		*Since the signal value  $s_{0}$  is specified here in  "volts"  and no specification is made for the reference resistance,  $E_{\rm B}$  has the unit "$\rm V^{2}/Hz$".
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation\|Linear Digital Modulation - Coherent Demodulation]].
	*Reference is also made to the chapter  [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission\|Error Probability for Baseband Transmission]].

	*You can check the results with the applet  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|Complementary Gaussian Error Functions]].
	*Since the signal value  $s_{0}$  is specified here in "volts" and no specification is made for the reference resistance,  $E_{\rm B}$  has the unit "$\rm V^{2}/Hz$".


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	<quiz display=simple>		<quiz display=simple>

	{Let  $s_{0} = 4 \, \rm V$. What is the error probability  $p_{\rm BB}$  of the baseband system?		{Let  $s_{0} = 4 \, \rm V$.  What is the error probability  $p_{\rm BB}$  of the baseband system?
	\|type="{}"}		\|type="{}"}
	$p_{\rm BB} \ = \ $ { 0.00317 3% } $\ \% $		$p_{\rm BB} \ = \ $ { 0.00317 3% } $\ \% $
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	$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 $

	{What is the error probability  $p_{\rm BB}$  at half the transmission amplitude  $(s_{0} = 2 \, \rm V)$?		{What is the error probability  $p_{\rm BB}$  at half the transmission amplitude   $(s_{0} = 2 \, \rm V)$?
	\|type="{}"}		\|type="{}"}
	$p_{\rm BB} \ = \ $ { 2.27 3% } $\ \% $		$p_{\rm BB} \ = \ $ { 2.27 3% } $\ \% $
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	-$p_{\rm BPSK} = {\rm Q}\big [(4\cdot E_{\rm B}/N_{0})^{1/2}\big ]$.		-$p_{\rm BPSK} = {\rm Q}\big [(4\cdot E_{\rm B}/N_{0})^{1/2}\big ]$.

	{What are the error probabilities for the BPSK and  $E_{\rm B}/N_{0} = 8$  ~~and~~  $E_{\rm B}/N_{0} = 2$?		{What are the error probabilities for the BPSK and  $E_{\rm B}/N_{0} = 8$  resp.  $E_{\rm B}/N_{0} = 2$?
	\|type="{}"}		\|type="{}"}
	$E_{\rm B}/N_{0} = 8\text{:}\hspace{0.4cm} p_{\rm BPSK} \ = \ $ { 0.00317 3% } $\ \% $		$E_{\rm B}/N_{0} = 8\text{:}\hspace{0.4cm} p_{\rm BPSK} \ = \ $ { 0.00317 3% } $\ \% $

Hwang at 13:56, 29 March 2022

2022-03-29T13:56:04Z

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Guenter at 14:04, 19 March 2022

2022-03-19T14:04:52Z

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	[[Category:Digital Signal Transmission: Exercises\|^1.5 ~~Lineare digitale~~ Modulation^]]		[[Category:Digital Signal Transmission: Exercises\|^1.5 Linear Digital Modulation^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 1.08Z: BPSK-Fehlerwahrscheinlichkeit to Aufgaben:Exercise 1.08Z: BPSK Error Probability

2022-03-19T14:00:54Z

Guenter moved page Aufgaben:Aufgabe 1.08Z: BPSK-Fehlerwahrscheinlichkeit to Aufgaben:Exercise 1.08Z: BPSK Error Probability

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	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^1.5 Lineare digitale Modulation^]]		[[Category:Digital Signal Transmission: Exercises\|^1.5 Lineare digitale Modulation^]]

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The noise rms value ~~here~~ is given by		'''(1)'''  The noise rms value is given here by
	:$$\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm B}}}= \sqrt{\frac{2 \cdot 10^{-9}\,{\rm V^2/Hz}}{2 \cdot 1\,{\rm ns}}}= 1\,{\rm V}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}p_{\rm BB} = {\rm Q}\left ({s_0}/{\sigma_d } \right		:$$\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)= 0.317 \cdot 10^{-4}\hspace{0.1cm}\underline {= 0.00317 \%}.$$		)= {\rm Q}(4)= 0.317 \cdot 10^{-4}\hspace{0.1cm}\underline {= 0.00317 \%}.$$
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	:$$E_{\rm B} = s_0^2 \cdot T_{\rm B}= (4\,{\rm V})^2 \cdot		:$$E_{\rm B} = s_0^2 \cdot T_{\rm B}= (4\,{\rm V})^2 \cdot
	10^{-9}\,{\rm s}\hspace{0.1cm}\underline {= 1.6 \cdot 10^{-8}\,{\rm V^2s}}.$$		10^{-9}\,{\rm s}\hspace{0.1cm}\underline {= 1.6 \cdot 10^{-8}\,{\rm V^2s}}.$$
	Of course, the additional equation ~~given~~ gives the exact same error probability:		*Of course, the additional given equation gives the exact same error probability:
	:$$p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right		:$$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		) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 16 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\rm
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	) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$		) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$

	A comparison with question '''(4)''' of [[Aufgaben:Exercise_1.08:_Comparison_of_ASK_and_BPSK\|Exercise 1.8]] shows that  $E_{\rm B}/N_{0} = 8$  is not (exactly) equal to  $10 \cdot \lg E_{\rm B}/N_{0} = 9 \ \rm dB$.  In the first case  $p_{\rm BB} = 0.317 \cdot 10^{–4}$ is obtained, in the second  $p_{\rm BB} = 0.336 \cdot 10^{-4}$.		*A comparison with question  '''(4)'''  of  [[Aufgaben:Exercise_1.08:_Comparison_of_ASK_and_BPSK\|Exercise 1.8]]  shows that  $E_{\rm B}/N_{0} = 8$  is not  (exactly)  equal to  $10 \cdot \lg E_{\rm B}/N_{0} = 9 \ \rm dB$.
			*In the first case  $p_{\rm BB} = 0.317 \cdot 10^{–4}$ is obtained,  in the second  $p_{\rm BB} = 0.336 \cdot 10^{-4}$.


	'''(3)'''  At half the transmission amplitude $s_{0} = 2 \ \rm V$, the energy per bit decreases to a quarter and the following equations apply:		'''(3)'''  At half the transmission amplitude  $s_{0} = 2 \ \rm V$,  the energy per bit decreases to a quarter and the following equations apply:
	:$$p_{\rm BB} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )= {\rm Q}\left ( \frac{2\,{\rm V}}{1\,{\rm V}} \right )\hspace{0.1cm}\underline {= {\rm Q}(2)= 2.27 \%},$$		:$$p_{\rm BB} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )= {\rm Q}\left ( \frac{2\,{\rm V}}{1\,{\rm V}} \right )\hspace{0.1cm}\underline {= {\rm Q}(2)= 2.27 \%},$$
	:$$p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 4 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\rm V^2/Hz} }} \hspace{0.1cm}\right ) = {\rm Q}(2)= 2.27 \%.$$		:$$p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 4 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\rm V^2/Hz} }} \hspace{0.1cm}\right ) = {\rm Q}(2)= 2.27 \%.$$


	'''(4)'''  Considering only half the energy $E_{\rm B} = s^{2}_{0} \cdot T_{\rm B}/2$, we obtain with $\sigma^{2}_{d} = N_{0}/T_{\rm B}$ and		'''(4)'''  Considering only half the energy  $E_{\rm B} = s^{2}_{0} \cdot T_{\rm B}/2$,  we obtain with  $\sigma^{2}_{d} = N_{0}/T_{\rm B}$  and
	:$$p_{\rm BPSK} = {\rm Q}\left ( {s_0}/{\sigma_d } \right )= {\rm Q}\left ( \sqrt{{s_0^2 \cdot T_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }}\hspace{0.1cm}\right )$$		:$$p_{\rm BPSK} = {\rm Q}\left ( {s_0}/{\sigma_d } \right )= {\rm Q}\left ( \sqrt{{s_0^2 \cdot T_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }}\hspace{0.1cm}\right )$$
	the exact same result as for the optimal baseband system   ⇒   <u>solution 2</u>.		the exact same result as for the optimal baseband system   ⇒   <u>solution 2</u>.


	'''(5)'''  Of course, this also gives the exact same results as for the baseband transmission:		'''(5)'''  Of course,  this also gives the exact same results as for the baseband transmission:
	:$${ E_{\rm B}}/{N_0 }= 8{\rm :} \hspace{0.2cm}p_{\rm BPSK} = {\rm Q}(\sqrt{16}) = {\rm Q}(4)\hspace{0.1cm}\underline {= 0.00317 \%},$$		:$${ E_{\rm B}}/{N_0 }= 8{\rm :} \hspace{0.2cm}p_{\rm BPSK} = {\rm Q}(\sqrt{16}) = {\rm Q}(4)\hspace{0.1cm}\underline {= 0.00317 \%},$$
	:$${ E_{\rm B}}/{N_0 }= 2{\rm :} \hspace{0.2cm}p_{\rm BPSK} = {\rm Q}(\sqrt{4}) = {\rm Q}(2) \hspace{0.1cm}\underline {= 2.27 \%}.$$		:$${ E_{\rm B}}/{N_0 }= 2{\rm :} \hspace{0.2cm}p_{\rm BPSK} = {\rm Q}(\sqrt{4}) = {\rm Q}(2) \hspace{0.1cm}\underline {= 2.27 \%}.$$