Difference between revisions of "Aufgaben:Exercise 4.18: Non-Coherent FSK Demodulation"

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===Fragebogen===
+
===Questions===
 
<quiz display=simple>
 
<quiz display=simple>
{Welche Aussagen sind bei kohärenter Demodulation der FSK zutreffend?
+
{Which statements are true in the case of coherent demodulation of the FSK?
 
|type="[]"}
 
|type="[]"}
+ Orthogonalität ergibt sich, wenn&nbsp; $h$&nbsp; ganzzahlig ist.
+
+ Orthogonality results if&nbsp; $h$&nbsp; is an integer.
+ Auch für&nbsp; $h = 0.5$&nbsp; ergeben sich orthogonale Signalformen.
+
+ Orthogonal waveforms also result for&nbsp; $h = 0.5$.&nbsp;  
- Orthogonalität ist grundsätzlich nicht zu erreichen.
+
- Orthogonality cannot be achieved in principle.
+ Beim AWGN&ndash;Kanal gilt&nbsp; $r(t) = s(t) + n(t)$.
+
+ For the AWGN channel&nbsp; $r(t) = s(t) + n(t)$ is valid.
  
{Welche Aussagen sind bei nichtkohärenter Demodulation der FSK zutreffend?
+
{Which statements are true for the non-coherent demodulation of FSK?
 
|type="[]"}
 
|type="[]"}
+ Orthogonalität ergibt sich, wenn&nbsp; $h$&nbsp; ganzzahlig ist.
+
+ Orthogonality results when&nbsp; $h$&nbsp; is an integer.
- Auch für&nbsp; $h = 0.5$&nbsp; ergeben sich orthogonale Signalformen.
+
- Orthogonal waveforms also result for&nbsp; $h = 0.5$.&nbsp;  
- Orthogonalität ist grundsätzlich nicht zu erreichen.
+
- Orthogonality cannot be achieved in principle.
- Beim AWGN&ndash;Kanal gilt&nbsp; $r(t) = s(t) + n(t)$.
+
- For the AWGN channel&nbsp; $r(t) = s(t) + n(t)$ is valid.
  
{Es sei&nbsp; $E_{\rm S}/N_0 = 10$. Wie groß ist die Fehlerwahrscheinlichkeit, also die Wahrscheinlichkeit, dass der Schätzwert von der gesendeten Nachricht abweicht?  
+
{Let&nbsp; $E_{\rm S}/N_0 = 10$. 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% } $\ \%$
  
{Es sei&nbsp; $\gamma = 2$&nbsp; und&nbsp; $E_{\rm S}/N_0 = 10$. Wie groß ist die Wahrscheinlichkeit, dass trotz Fehler das Sicherheitsflag eine zuverlässige Entscheidung signalisiert?
+
{Let&nbsp; $\gamma = 2$&nbsp; and&nbsp; $E_{\rm S}/N_0 = 10$. 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}Fehler}) \ = \ $ { 6.7 3% } $\ \cdot 10^{\rm &ndash;5}$
+
${\rm Pr}({\it \Gamma} = {\rm Z\hspace{0.05cm}  | \hspace{0.05cm}error}) \ = \ $ { 6.7 3% } $\ \cdot 10^{\rm &ndash;5}$
  
{Es sei weiterhin&nbsp; $\gamma = 2$ und&nbsp; $E_{\rm S}/N_0 = 10$. Wie groß ist die (bedingte) Wahrscheinlichkeit, dass bei einem Fehler die Zusatzinformation "unzuverlässig" angezeigt wird?  
+
{Furthermore, let&nbsp; $\gamma = 2$ and&nbsp; $E_{\rm S}/N_0 = 10$. 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}Fehler}) \ = \ $ { 98 3% } $\ \%$
+
${\rm Pr}({\it \Gamma} = {\rm U\hspace{0.05cm}  | \hspace{0.05cm}error}) \ = \ $ { 98 3% } $\ \%$
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Richtig sind die <u>Lösungsvorschlag 1, 2 und 4</u>:  
+
'''(1)'''&nbsp; <u>Solutions 1, 2 and 4</u> are correct:  
*Bei kohärenter Demodulation ist Orthogonalität dann gegeben, wenn der Modulationsindex $h$ ein Vielfaches von $0.5$ ist. Binäre FSK mit $h = 0.5$ nennt man auch <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 <i>Minimum Shift Keying</i> (MSK).  
*Da durch die Phasenregelung die Phasenverschiebung (Laufzeit) auf dem AWGN&ndash;Übertragungskanal ausgeglichen wird &nbsp; &#8658; &nbsp; ${\rm e}^{{\rm j}\phi} \cdot {\rm e}^{-{\rm j}\phi} = 1$, gilt tatsächlich für die Signale im äquivalenten Tiefpassbereich: &nbsp; $r(t) = s(t) + n(t)  \hspace{0.05cm}.$
+
*Since phase control compensates for the phase shift (delay) on the AWGN transmission channel &nbsp; &#8658; &nbsp; ${\rm e}^{{\rm j}\phi} \cdot {\rm e}^{-{\rm j}\phi} = 1$, the signals in the equivalent low-pass region are actually: &nbsp; $r(t) = s(t) + n(t)  \hspace{0.05cm}.$
  
  
  
'''(2)'''&nbsp; Hier ist nur der <u>erste Lösungsvorschlag</u> richtig, das heißt, es muss nun $h = 1, \ 2, \ 3, \text{ ...}$ ganzzahlig sein.  
+
'''(2)'''&nbsp; Here, only the <u>first solution</u> is correct, i.e., now $h = 1, \ 2, \ 3, \text{ ...}$ must be an integer.
*Nichtkohärente Demodulation von FSK ist somit nicht möglich.  
+
*So a non-coherent demodulation of FSK is not possible.
*Wegen der fehlenden Phasenregelung gilt außerdem:
+
*Because of the missing phase control, the following is also true:
 
:$$r(t) = s(t) \cdot {\rm e }^{ - {\rm j }\hspace{0.05cm}\phi} + n(t)  \hspace{0.05cm}.$$
 
:$$r(t) = s(t) \cdot {\rm e }^{ - {\rm j }\hspace{0.05cm}\phi} + n(t)  \hspace{0.05cm}.$$
  
  
'''(3)'''&nbsp; Bei gleichwahrscheinlichen Nachrichten gilt:
+
'''(3)'''&nbsp; For equally probable messages applies:
 
:$$p_{\rm S} = {\rm Pr}({\cal{E}}) = {\rm Pr}({\cal{E}}| \hspace{0.05cm}m = m_0) =    {\rm Pr}(\hat{m}= m_1| \hspace{0.05cm}m = m_0)\hspace{0.05cm}.$$
 
:$$p_{\rm S} = {\rm Pr}({\cal{E}}) = {\rm Pr}({\cal{E}}| \hspace{0.05cm}m = m_0) =    {\rm Pr}(\hat{m}= m_1| \hspace{0.05cm}m = m_0)\hspace{0.05cm}.$$
  
*Diese Wahrscheinlichkeit ergibt sich aus der vorgegebenen Gleichung mit $\gamma = 1$.  
+
*This probability is obtained from the given equation with $\gamma = 1$.  
*In diesem Fall ist stets ${\it \Gamma} ={\rm Z}$ und die Entscheidungsregel lautet: &nbsp; Entscheide auf das Symbol $m_0$, falls $y_1 > y_2$.
+
*In this case, ${\it \Gamma} ={\rm Z}$ is always and the decision rule is: &nbsp; Decide on the symbol $m_0$ if $y_1 > y_2$.
*Daraus folgt:
+
*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} =   
 
  {\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}.$$
 
  {1}/{2} \cdot {\rm e }^{-E_{\rm S}/(2N_0)}\hspace{0.05cm}.$$
  
*Mit $E_{\rm S}/N_0 = 10$ erhält man $p_{\rm S} = 1/2 \cdot {\rm e}^{\rm &ndash;5} \approx \underline {0.337 \cdot \%}$.
+
*With $E_{\rm S}/N_0 = 10$, we get $p_{\rm S} = 1/2 \cdot {\rm e}^{\rm &ndash;5} \approx \underline {0.337 \cdot \%}$.
  
  
  
'''(4)'''&nbsp; Diese Wahrscheinlichkeit ergibt sich aus Symmetriegründen zu
+
'''(4)'''&nbsp; For symmetry reasons, this probability results in
:$${\rm Pr}({\it \Gamma} = {\rm Z}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm Fehler}  ) =  
+
:$${\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 Fehler}  ) = {\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
+
:$$\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}.$$
 
  {\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)'''&nbsp; Da ${\rm U}$ und ${\rm Z}$ nach der Statistik ein vollständiges System ergeben, gilt mit den Ergebnissen der Teilaufgaben '''(3)''' und '''(4)''':
+
'''(5)'''&nbsp; Since ${\rm U}$ and ${\rm Z}$ give a complete system according to statistics, with the results of subtasks '''(3)''' and '''(4)''':
:$${\rm Pr}({\it \Gamma} = {\rm U}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm Fehler}  ) ={\rm Pr}( {\rm Fehler}  ) -
+
:$${\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 Fehler}  ) = 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}.$$
  
*Damit ist die gesuchte bedingte Wahrscheinlichkeit ${\rm Pr}({\it \Gamma}= {\rm U}\hspace{0.05cm}|\hspace{0.05cm} {\rm Fehler})$:
+
*Thus, the conditional probability we are looking for is ${\rm Pr}({\it \Gamma}= {\rm U}\hspace{0.05cm}|\hspace{0.05cm} {\rm error})$:
:$${\rm Pr}({\it \Gamma}= {\rm U}\hspace{0.05cm}|\hspace{0.05cm} {\rm Fehler}  )  =
+
:$${\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 Fehler}  ) }{{\rm Pr}( {\rm Fehler}  )} = \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{= 98 \%}
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
 
{{ML-Fuß}}
 
{{ML-Fuß}}

Revision as of 19:10, 16 August 2022

Non-coherent demodulation

We consider Frequency Shift Keying  (FSK) with  $M = 2$   ⇒   binary signaling. The two low-pass basis functions in this case are complex and are:

$$\xi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},$$
$$ \xi_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{1/T} \cdot {\rm e}^{\hspace{0.05cm}-{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T \hspace{0.05cm}.$$

Then, for the two possible waveforms in the low-pass region, with the mean symbol energy  $E_{\rm S}$:

$$m_0\text{:}\hspace{0.2cm} s_{\rm TP,\hspace{0.05cm}0} = \sqrt{E_{\rm S}} \cdot \xi_1(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\boldsymbol{ s}_{\rm 0} = (\sqrt{E_{\rm S}}, 0)\hspace{0.05cm},$$
$$m_1\text{:}\hspace{0.2cm} s_{\rm TP,\hspace{0.05cm}1} = \sqrt{E_{\rm S}} \cdot \xi_2(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\boldsymbol{ s}_{\rm 1} = (0, \sqrt{E_{\rm S}})\hspace{0.05cm}.$$

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 Frequency Shift Keying  (FSK) 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  "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 a realiable (Zuverlässige) and an Unreliable decision, respectively. Thus, there are four possibilities for the decision, controlled by the parameter  $\gamma$:

$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_0,\hspace{0.05cm}{\it \Gamma} = {\rm Z}, \hspace{0.5cm}{\rm falls} \hspace{0.15cm}y_1 > \gamma \cdot y_2\hspace{0.05cm},$$
$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_0,\hspace{0.05cm}{\it \Gamma} = {\rm U}, \hspace{0.5cm}{\rm falls} \hspace{0.15cm}y_2 < y_1 < \gamma \cdot y_2\hspace{0.05cm},$$
$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_1,\hspace{0.05cm}{\it \Gamma} = {\rm Z}, \hspace{0.5cm}{\rm falls} \hspace{0.15cm}y_2 > \gamma \cdot y_1\hspace{0.05cm},$$
$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{0.1cm} m_1,\hspace{0.05cm}{\it \Gamma} = {\rm U}, \hspace{0.5cm}{\rm falls} \hspace{0.15cm}y_1 < y_2 < \gamma \cdot y_1\hspace{0.05cm}.$$

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

$${\rm Pr}\{\hat{m} = m_1,\hspace{0.05cm}{\it \Gamma} = {\rm Z} \hspace{0.05cm}| \hspace{0.05cm}m_0 \} = \frac{1}{1 + \gamma^2} \cdot {\rm exp } \left [ - \frac{\gamma^2 \cdot E_{\rm S}}{(1+\gamma^2) \cdot N_{\rm 0}}\right ] \hspace{0.05cm}.$$



Note:



Questions

1

Which statements are true in the case of coherent demodulation of the FSK?

Orthogonality results if  $h$  is an integer.
Orthogonal waveforms also result for  $h = 0.5$. 
Orthogonality cannot be achieved in principle.
For the AWGN channel  $r(t) = s(t) + n(t)$ is valid.

2

Which statements are true for the non-coherent demodulation of FSK?

Orthogonality results when  $h$  is an integer.
Orthogonal waveforms also result for  $h = 0.5$. 
Orthogonality cannot be achieved in principle.
For the AWGN channel  $r(t) = s(t) + n(t)$ is valid.

3

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?

$p_{\rm S} \ = \ $

$\ \%$

4

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?

${\rm Pr}({\it \Gamma} = {\rm Z\hspace{0.05cm} | \hspace{0.05cm}error}) \ = \ $

$\ \cdot 10^{\rm –5}$

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?

${\rm Pr}({\it \Gamma} = {\rm U\hspace{0.05cm} | \hspace{0.05cm}error}) \ = \ $

$\ \%$


Solution

(1)  Solutions 1, 2 and 4 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 Minimum Shift Keying (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:   $r(t) = s(t) + n(t) \hspace{0.05cm}.$


(2)  Here, only the first solution is correct, i.e., now $h = 1, \ 2, \ 3, \text{ ...}$ must be an integer.

  • So a non-coherent demodulation of FSK is not possible.
  • Because of the missing phase control, the following is also true:
$$r(t) = s(t) \cdot {\rm e }^{ - {\rm j }\hspace{0.05cm}\phi} + n(t) \hspace{0.05cm}.$$


(3)  For equally probable messages applies:

$$p_{\rm S} = {\rm Pr}({\cal{E}}) = {\rm Pr}({\cal{E}}| \hspace{0.05cm}m = m_0) = {\rm Pr}(\hat{m}= m_1| \hspace{0.05cm}m = m_0)\hspace{0.05cm}.$$
  • This probability is obtained from the given equation with $\gamma = 1$.
  • 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:
$$p_{\rm S} = \frac{1}{1 + \gamma^2} \cdot {\rm exp } \left [ - \frac{\gamma^2 \cdot E_{\rm S}}{(1+\gamma^2) \cdot N_{\rm 0}}\right ]_{\gamma = 1} = {1}/{2} \cdot {\rm e }^{-E_{\rm S}/(2N_0)}\hspace{0.05cm}.$$
  • 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

$${\rm Pr}({\it \Gamma} = {\rm Z}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) = {1}/{2} \cdot {\rm Pr}\{(\hat{m} = m_1)\hspace{0.05cm}\cap\hspace{0.05cm}{\it \Gamma} = {\rm Z} \hspace{0.05cm}| \hspace{0.05cm}m_0 \} + {1}/{2}\cdot {\rm Pr}\{(\hat{m} = m_0)\hspace{0.05cm}\cap\hspace{0.05cm}({\it \Gamma} = {\rm Z}) \hspace{0.05cm}| \hspace{0.05cm}m_1 \} $$
$$\Rightarrow \hspace{0.3cm} {\rm Pr}({\it \Gamma} = {\rm Z}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) = {\rm Pr}\{(\hat{m} = m_1)\hspace{0.05cm}\cap\hspace{0.05cm}({\it \Gamma} = {\rm Z}) \hspace{0.05cm}| \hspace{0.05cm}m_0 \} =\frac{1}{1 + 2^2} \cdot {\rm exp } \left [ - \frac{2^2 \cdot E_{\rm S}}{(1+2^2) \cdot N_{\rm 0}}\right ] = {1}/{5} \cdot {\rm e }^{-8} = \underline{6.7 \cdot 10^{-5}}\hspace{0.05cm}.$$


(5)  Since ${\rm U}$ and ${\rm Z}$ give a complete system according to statistics, with the results of subtasks (3) and (4):

$${\rm Pr}({\it \Gamma} = {\rm U}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) ={\rm Pr}( {\rm error} ) - {\rm Pr}({\it \Gamma} = {\rm Z}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) = 3.37 \cdot 10^{-3} - 6.7 \cdot 10^{-5} = \underline{3.3 \cdot 10^{-3}} \hspace{0.05cm}.$$
  • Thus, the conditional probability we are looking for is ${\rm Pr}({\it \Gamma}= {\rm U}\hspace{0.05cm}|\hspace{0.05cm} {\rm error})$:
$${\rm Pr}({\it \Gamma}= {\rm U}\hspace{0.05cm}|\hspace{0.05cm} {\rm error} ) = \frac{{\rm Pr}({\it \Gamma} = {\rm U}\hspace{0.05cm}\cap\hspace{0.05cm} {\rm error} ) }{{\rm Pr}( {\rm error} )} = \frac{3.3 \cdot 10^{-3}}{3.37 \cdot 10^{-3}} \hspace{0.15cm} \underline{= 98 \%} \hspace{0.05cm}.$$