Difference between revisions of "Aufgaben:Exercise 4.8Z: BPSK Error Probability"

From LNTwww
 
(19 intermediate revisions by 4 users not shown)
Line 1: Line 1:
  
{{quiz-Header|Buchseite=Modulationsverfahren/Lineare digitale Modulationsverfahren
+
{{quiz-Header|Buchseite=Modulation_Methods/Linear_Digital_Modulation
 
}}
 
}}
  
[[File:P_ID1681__Dig_Z_4_1.png|right|]]
+
[[File:P_ID1681__Dig_Z_4_1.png|right|frame|Table of the Complementary Gaussian Error Function  ${\rm Q}(x)$]]
Wir gehen von dem optimalen Basisbandübertragungssystem für Binärsignale aus mit
+
We assume the optimal baseband transmission system for binary signals with
:* bipolaren Amplitudenkoeffizienten $a_ν$ ∈ {–1, +1},
+
* bipolar amplitude coefficients  $a_ν ∈ \{-1, +1\}$,
:* rechteckförmigem Sendesignal mit den Signalwerten $±s_0$ und der Bitdauer $T_B$,
+
* rectangular transmitted signal  $s(t)$  with signal values  $±s_0$  and bit duration  $T_{\rm B}$,
:* AWGN–Rauschen mit der Rauschleistungsdichte $N_0$,
+
* AWGN noise with the (one-sided) noise power density  $N_0$,
:* Empfangsfilter gemäß dem Matched–Filter–Prinzip,
+
* receiver filter according to the matched filter principle,
:* Entscheider mit optimalem Schwellenwert E = 0.
+
* decision with optimal threshold value  $E = 0$.
  
Wenn nichts anderes angegeben ist, so sollten Sie von den folgenden Zahlenwerten ausgehen:
 
$$s_0 = 4\,{\rm V},\hspace{0.2cm} T_{\rm B} = 1\,{\rm ns},\hspace{0.2cm}N_0 = 2 \cdot 10^{-9}\, {\rm V^2/Hz} \hspace{0.05cm}.$$
 
Die Fehlerwahrscheinlichkeit dieses Basisbandsystems ist
 
$$ p_{\rm BB} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )\hspace{0.2cm}{\rm mit}\hspace{0.2cm}\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm B}}}.$$
 
Hierbei bezeichnet $σ_d$ den Rauscheffektivwert am
 
  
Entscheider und $Q(x)$ die komplementäre Gaußsche Fehlerfunktion, die hier tabellarisch gegeben ist. Diese Fehlerwahrscheinlichkeit kann auch in der Form
+
If nothing else is specified,  you should assume the following numerical values:
$$p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right )$$
+
:$$s_0 = 4\,{\rm V},\hspace{0.2cm} T_{\rm B} = 1\,{\rm ns},\hspace{0.2cm}N_0 = 2 \cdot 10^{-9}\, {\rm V^2/Hz} \hspace{0.05cm}.$$
geschrieben werden, wobei $E_B$ die „Signalenergie pro Bit” angibt. Die Fehlerwahrscheinlichkeit eines vergleichbaren Übertragungssystems mit ''Binary Phase Shift Keying (BPSK)'' lautet:
+
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:
$$ p_{\rm BPSK} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )\hspace{0.2cm}{\rm mit}\hspace{0.2cm}\sigma_d = \sqrt{\frac{N_0}{T_{\rm B}}}.$$
+
:$$ p_{\rm BB} = {\rm Q}\left ( {s_0}/{\sigma_d } \right )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{{N_0}/(2 \cdot T_{\rm B}}).$$
'''Hinweis:''' Die Aufgabe gehört zum Themengebiet von [http://en.lntwww.de/Modulationsverfahren/Lineare_digitale_Modulationsverfahren Kapitel 4.2].
+
This bit error probability can also be expressed in the form
 +
:$$p_{\rm BB} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ),$$
 +
where  $E_{\rm B}$  indicates the  "signal energy per bit".
  
 +
The bit error probability of a comparable transmission system with  "Binary Phase Shift Keying"  $\rm (BPSK)$  is:
 +
:$$ p_{\rm BPSK} = {\rm Q}\left ( {s_0}/{\sigma_d } \right )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{{N_0}/{T_{\rm B}}}.$$
  
===Fragebogen===
+
 
 +
 
 +
 
 +
 
 +
 
 +
Notes:
 +
*The exercise belongs to the chapter  [[Modulation_Methods/Linear_Digital_Modulation|"Linear Digital Modulation"]].
 +
*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  $1 \ \rm \Omega$.
 +
 +
 
 +
 
 +
 
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Wie groß ist die Fehlerwahrscheinlichkeit des Basisbandsystems?
+
{Let &nbsp;$s_0 = 4\,{\rm V},\hspace{0.2cm} T_{\rm B} = 1\,{\rm ns},\hspace{0.2cm}N_0 = 2 \cdot 10^{-9}\, {\rm V^2/Hz} \hspace{0.05cm}.$&nbsp; What is the error probability &nbsp;$p_{\rm BB}$&nbsp; of the baseband system?
 
|type="{}"}
 
|type="{}"}
$s_0 = 4V:  p_{BB}$ = { 0.317 3% } $10^{-4}$
+
$p_{\rm BB} \ = \ $ { 0.317 3% } $\ \cdot 10^{-4}$
  
  
{Wie groß ist die Energie pro Bit beim Basisbandsystem?
+
{For this parameter set,&nbsp; what is the energy per bit &nbsp; &rArr; &nbsp; &nbsp;$E_{\rm B}$&nbsp; for the baseband system?
 
|type="{}"}
 
|type="{}"}
$s_0 = 4V:  E_B$ = { 1.6 3% } $10^{-8}$ $V^2 s$
+
$E_{\rm B}  \ = \ $ { 1.6 3% } $\ \cdot 10^{-8} \ \rm V^2 s$
  
{Welche Fehlerwahrscheinlichkeit ergibt sich bei halber Sendeamplitude?
+
{What is the error probability at half the transmitted amplitude &nbsp; &rArr; &nbsp; &nbsp;$s_0 = 2\,{\rm V}$?
 
|type="{}"}
 
|type="{}"}
$s_0 = 2V:  p_{BB}$ = { 0.227 3% } $10^{-1}$
+
$p_{\rm BB} \ = \ $ { 227 3% } $\ \cdot 10^{-4}$
 
   
 
   
{Geben Sie die Fehlerwahrscheinlichkeit der BPSK abhängig vom Quotienten $E-B/N_0$ an. Welches Ergebnis stimmt?
+
{Give the error probability of the BPSK depending on the quotient &nbsp;$E_{\rm B}/N_0$.&nbsp; Which result is correct?
|type="[]"}  
+
|type="()"}  
- $p_{BPSK} = Q[(E_B/N_0)^{1/2}],$
+
- $p_{\rm BPSK} = {\rm Q}\big[(E_{\rm B}/N_0)^{1/2}\big],$
+ $p_{BPSK} = Q[(2E_B/N_0)^{1/2}],$
+
+ $p_{\rm BPSK} = {\rm Q}\big[(2E_{\rm B}/N_0)^{1/2}\big],$
-  $p_{BPSK} = Q[(4E_B/N_0)^{1/2}].$
+
-  $p_{\rm BPSK} = {\rm Q}\big[(4E_{\rm B}/N_0)^{1/2}\big].$
  
{Welche Fehlerwahrscheinlichkeiten ergeben sich bei der BPSK für $E_B/N_0 = 8$ und $E_B/N_0 = 2$?
+
{What are the error probabilities for BPSK with &nbsp;$E_{\rm B}/N_0 = 8$&nbsp; and &nbsp;$E_{\rm B}/N_0 = 2$?
 
|type="{}"}
 
|type="{}"}
$E_B/N_0 = 8:  p_{BPSK}$ = { 0.317 3% } $10^{-4}$
+
$E_{\rm B}/N_0 = 8\text{:} \ \ \ \   p_{\rm BPSK} \ = \ $ { 0.317 3% } $\ \cdot 10^{-4}$
$E_B/N_0 = 2:  p_{BPSK}$ = { 0.227 3% } $10^{-1}$
+
$E_{\rm B}/N_0 = 2\text{:} \ \ \ \   p_{\rm BPSK} \ = \ $ { 227 3% } $\ \cdot 10^{-4}$
  
  
Line 55: Line 69:
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''1.''' Der Rauscheffektivwert ergibt sich hier zu
+
'''(1)'''&nbsp;  The noise rms value here is given by
$$\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm B}}}= \sqrt{\frac{2 \cdot 10^{-9}\,{\rm V^2/Hz}}{2 \cdot 1\,{\rm ns}}}= 1\,{\rm V}$$
+
:$$\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm B}}}= \sqrt{\frac{2 \cdot 10^{-9}\,{\rm V^2/Hz}}{2 \cdot 1\,{\rm ns}}}= 1\,{\rm V}
$$: \Rightarrow \hspace{0.3cm}p_{\rm BB} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )= {\rm Q}(4)\hspace{0.15cm}\underline {= 0.317 \cdot 10^{-4}}.$$
+
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm BB} = {\rm Q}\left ( {s_0}/{\sigma_d } \right )= {\rm Q}(4)\hspace{0.15cm}\underline {= 0.317 \cdot 10^{-4}}.$$
 +
 
 +
 
 +
'''(2)'''&nbsp;  For the baseband system:
 +
:$$E_{\rm B} = s_0^2 \cdot T_{\rm B}= (4\,{\rm V})^2 \cdot 10^{-9}\,{\rm s}\hspace{0.15cm}\underline {= 1.6 \cdot 10^{-8}\,{\rm V^2s}}.$$
 +
*Of course,&nbsp; the second equation gives the exact same error probability
 +
:$$ p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 16 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\rm V^2/Hz} }} \hspace{0.1cm}\right ) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$
 +
 
 +
 
 +
'''(3)'''&nbsp;  When the transmitted amplitude is half &nbsp; &rArr; &nbsp; $s_0 = 2\,{\rm V}$,&nbsp; the energy per bit decreases to one-fourth and the following equations apply:
 +
:$$ p_{\rm BB} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )= {\rm Q}\left ( \frac{2\,{\rm V}}{1\,{\rm V}} \right )= {\rm Q}(2)= 227 \cdot 10^{-4},$$
 +
:$$ p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 4 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\rm V^2/Hz} }} \hspace{0.1cm}\right ) = {\rm Q}(2)\hspace{0.15cm}\underline {= 227 \cdot 10^{-4}}.$$
  
'''2.''' Beim Basisbandsystem gilt:
 
$$E_{\rm B} = s_0^2 \cdot T_{\rm B}= (4\,{\rm V})^2 \cdot 10^{-9}\,{\rm s}\hspace{0.15cm}\underline {= 1.6 \cdot 10^{-8}\,{\rm V^2s}}.$$
 
Natürlich ergibt sich mit der zweiten angegebenen Gleichung die genau gleiche Fehlerwahrscheinlichkeit
 
$$ p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 16 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\rm V^2/Hz} }} \hspace{0.1cm}\right ) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$
 
  
'''3.''' Bei halber Sendeamplitude $s_0 = 2 V$ sinkt die Energie pro Bit auf ein Viertel und es gelten folgende Gleichungen:
+
'''(4)'''&nbsp;  <u>Answer 2</u>&nbsp; is correct:
$$ p_{\rm BB} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )= {\rm Q}\left ( \frac{2\,{\rm V}}{1\,{\rm V}} \right )= {\rm Q}(2)= 0.227 \cdot 10^{-1},$$
+
*Considering the energy&nbsp; $E_{\rm B} = s_0^2 · T_{\rm B}/2$&nbsp; we obtain
$$ p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 4 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\rm V^2/Hz} }} \hspace{0.1cm}\right ) = {\rm Q}(2)\hspace{0.15cm}\underline {= 0.227 \cdot 10^{-1}}.$$
+
:$$ p_{\rm BPSK} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )= {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }}\hspace{0.1cm}\right ).$$
 +
*Thus,&nbsp; the same result is obtained as for the optimal baseband transmission system.
  
'''4.''' Unter Berücksichtigung der Energie $E_B = s_0^2 · T_B/2$ erhält man mit
 
$$ p_{\rm BPSK} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )= {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }}\hspace{0.1cm}\right )$$
 
das gleiche Ergebnis wie beim optimalen Basisbandübertragungssystem. Richtig ist somit Antwort 2.
 
  
  
'''5.''' Es ergeben sich die genau gleichen Ergebnisse wie bei der Basisbandübertragung:
+
'''(5)'''&nbsp;  Exactly the same results are obtained as for the baseband transmission in questions&nbsp; '''(1)'''&nbsp; and&nbsp; '''(3)''':
$$\frac{ E_{\rm B}}{N_0 }= 8: \hspace{0.2cm}p_{\rm BPSK} = {\rm Q}(\sqrt{16}) = {\rm Q}(4)\hspace{0.15cm}\underline {= 0.317 \cdot 10^{-4}},$$
+
:$${ E_{\rm B}}/{N_0 }= 8: \hspace{0.2cm}p_{\rm BPSK} = {\rm Q}(\sqrt{16}) = {\rm Q}(4)\hspace{0.15cm}\underline {= 0.317 \cdot 10^{-4}},$$
$$ \frac{ E_{\rm B}}{N_0 }= 2: \hspace{0.2cm}p_{\rm BPSK} = {\rm Q}(\sqrt{4}) = {\rm Q}(2)\hspace{0.15cm}\underline {= 0.227 \cdot 10^{-1}}.$$
+
:$$ { E_{\rm B}}/{N_0 }= 2: \hspace{0.2cm}p_{\rm BPSK} = {\rm Q}(\sqrt{4}) = {\rm Q}(2)\hspace{0.15cm}\underline {= 227 \cdot 10^{-4}}.$$
  
 
{{ML-Fuß}}
 
{{ML-Fuß}}
Line 83: Line 102:
  
  
[[Category:Aufgaben zu  Modulationsverfahren|^4.2 Lineare digitale Modulationsverfahren^]]
+
[[Category:Modulation Methods: Exercises|^4.2 Linear Digital Modulation^]]

Latest revision as of 17:26, 15 April 2022

Table of the Complementary Gaussian Error Function  ${\rm Q}(x)$

We assume the optimal baseband transmission system for binary signals with

  • bipolar amplitude coefficients  $a_ν ∈ \{-1, +1\}$,
  • 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  $N_0$,
  • receiver filter according to the matched filter principle,
  • decision with optimal threshold value  $E = 0$.


If nothing else is specified,  you should assume the following numerical values:

$$s_0 = 4\,{\rm V},\hspace{0.2cm} T_{\rm B} = 1\,{\rm ns},\hspace{0.2cm}N_0 = 2 \cdot 10^{-9}\, {\rm V^2/Hz} \hspace{0.05cm}.$$

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:

$$ p_{\rm BB} = {\rm Q}\left ( {s_0}/{\sigma_d } \right )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{{N_0}/(2 \cdot T_{\rm B}}).$$

This bit error probability can also be expressed in the form

$$p_{\rm BB} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ),$$

where  $E_{\rm B}$  indicates the  "signal energy per bit".

The bit error probability of a comparable transmission system with  "Binary Phase Shift Keying"  $\rm (BPSK)$  is:

$$ p_{\rm BPSK} = {\rm Q}\left ( {s_0}/{\sigma_d } \right )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{{N_0}/{T_{\rm B}}}.$$




Notes:



Questions

1

Let  $s_0 = 4\,{\rm V},\hspace{0.2cm} T_{\rm B} = 1\,{\rm ns},\hspace{0.2cm}N_0 = 2 \cdot 10^{-9}\, {\rm V^2/Hz} \hspace{0.05cm}.$  What is the error probability  $p_{\rm BB}$  of the baseband system?

$p_{\rm BB} \ = \ $

$\ \cdot 10^{-4}$

2

For this parameter set,  what is the energy per bit   ⇒    $E_{\rm B}$  for the baseband system?

$E_{\rm B} \ = \ $

$\ \cdot 10^{-8} \ \rm V^2 s$

3

What is the error probability at half the transmitted amplitude   ⇒    $s_0 = 2\,{\rm V}$?

$p_{\rm BB} \ = \ $

$\ \cdot 10^{-4}$

4

Give the error probability of the BPSK depending on the quotient  $E_{\rm B}/N_0$.  Which result is correct?

$p_{\rm BPSK} = {\rm Q}\big[(E_{\rm B}/N_0)^{1/2}\big],$
$p_{\rm BPSK} = {\rm Q}\big[(2E_{\rm B}/N_0)^{1/2}\big],$
$p_{\rm BPSK} = {\rm Q}\big[(4E_{\rm B}/N_0)^{1/2}\big].$

5

What are the error probabilities for BPSK with  $E_{\rm B}/N_0 = 8$  and  $E_{\rm B}/N_0 = 2$?

$E_{\rm B}/N_0 = 8\text{:} \ \ \ \ p_{\rm BPSK} \ = \ $

$\ \cdot 10^{-4}$
$E_{\rm B}/N_0 = 2\text{:} \ \ \ \ p_{\rm BPSK} \ = \ $

$\ \cdot 10^{-4}$


Solution

(1)  The noise rms value here is given by

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


(2)  For the baseband system:

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


(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:

$$ p_{\rm BB} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )= {\rm Q}\left ( \frac{2\,{\rm V}}{1\,{\rm V}} \right )= {\rm Q}(2)= 227 \cdot 10^{-4},$$
$$ p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 4 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\rm V^2/Hz} }} \hspace{0.1cm}\right ) = {\rm Q}(2)\hspace{0.15cm}\underline {= 227 \cdot 10^{-4}}.$$


(4)  Answer 2  is correct:

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


(5)  Exactly the same results are obtained as for the baseband transmission in questions  (1)  and  (3):

$${ E_{\rm B}}/{N_0 }= 8: \hspace{0.2cm}p_{\rm BPSK} = {\rm Q}(\sqrt{16}) = {\rm Q}(4)\hspace{0.15cm}\underline {= 0.317 \cdot 10^{-4}},$$
$$ { E_{\rm B}}/{N_0 }= 2: \hspace{0.2cm}p_{\rm BPSK} = {\rm Q}(\sqrt{4}) = {\rm Q}(2)\hspace{0.15cm}\underline {= 227 \cdot 10^{-4}}.$$