Aufgaben:Exercise 4.14: 8-PSK and 16-PSK - Revision history

Hwang at 19:31, 1 September 2022

2022-09-01T19:31:09Z

Hwang at 19:28, 1 September 2022

2022-09-01T19:28:03Z

Hwang at 19:27, 1 September 2022

2022-09-01T19:27:33Z

Hwang at 19:26, 1 September 2022

2022-09-01T19:26:53Z

Guenter at 14:13, 23 August 2022

2022-08-23T14:13:11Z

Guenter at 13:16, 23 August 2022

2022-08-23T13:16:32Z

Guenter at 12:36, 23 August 2022

2022-08-23T12:36:29Z

Guenter at 12:08, 23 August 2022

2022-08-23T12:08:50Z

Hwang at 12:55, 12 August 2022

2022-08-12T12:55:18Z

Show changes

Hwang at 12:05, 12 August 2022

2022-08-12T12:05:55Z

@@ Line 167: / Line 167: @@
 The following statements are valid:
 # The actual error probability&nbsp; $p_{\rm S}$&nbsp; is composed of the components&nbsp; $\rm A$,&nbsp; $\rm B$&nbsp; and&nbsp; $\rm C$.
-# On the other hand,&nbsp; the&nbsp; "Union Bound"&nbsp; results from&nbsp; $\rm A+B$&nbsp; $($falsification into the symbol&nbsp;  $\boldsymbol{s}_1))$&nbsp; plus&nbsp; $\rm C+B$&nbsp; $($falsification into the symbol $\boldsymbol{s}_{M-1})$.
+# On the other hand,&nbsp; the&nbsp; "Union Bound"&nbsp; results from&nbsp; $\rm A+B$&nbsp; $($falsification into the symbol&nbsp;  $\boldsymbol{s}_1)$&nbsp; plus&nbsp; $\rm C+B$&nbsp; $($falsification into the symbol $\boldsymbol{s}_{M-1})$.
 # Thus,&nbsp; $p_{\rm S} = p_{\rm UB}-B$&nbsp; always holds,&nbsp; but the larger&nbsp; $M$&nbsp; is, the smaller the fraction&nbsp; $\rm B$&nbsp; is.

@@ Line 81: / Line 81: @@
 $M = 16 \text{:} \hspace{0.2cm} p_{\rm UB}$ = { 6 3% } $\ \%$
-{Does the statement: &nbsp; "$p_{\rm UB}$&nbsp; approximates &nbsp;$p_{\rm S}$&nbsp; more closely the larger &nbsp;$M$ &nbsp; is"&nbsp; hold?
+{Does the statement &nbsp; "$p_{\rm UB}$&nbsp; approximates &nbsp;$p_{\rm S}$&nbsp; more closely the larger &nbsp;$M$ &nbsp; is"&nbsp; hold?
 |type="()"}
 + YES.

@@ Line 81: / Line 81: @@
 $M = 16 \text{:} \hspace{0.2cm} p_{\rm UB}$ = { 6 3% } $\ \%$
-{Does the statement: &nbsp; "$p_{\rm UB}$&nbsp; approximates &nbsp;$p_{\rm S}$&nbsp; more closely the larger &nbsp;$M$ &nbsp; is"nbsp; hold?
+{Does the statement: &nbsp; "$p_{\rm UB}$&nbsp; approximates &nbsp;$p_{\rm S}$&nbsp; more closely the larger &nbsp;$M$ &nbsp; is"&nbsp; hold?
 |type="()"}
 + YES.

@@ Line 81: / Line 81: @@
 $M = 16 \text{:} \hspace{0.2cm} p_{\rm UB}$ = { 6 3% } $\ \%$
-{Does the statement: nbsp; "$p_{\rm UB}$&nbsp; approximates &nbsp;$p_{\rm S}$&nbsp; more closely the larger &nbsp;$M$ &nbsp; is"nbsp; hold?
+{Does the statement: &nbsp; "$p_{\rm UB}$&nbsp; approximates &nbsp;$p_{\rm S}$&nbsp; more closely the larger &nbsp;$M$ &nbsp; is"nbsp; hold?
 |type="()"}
 + YES.

@@ Line 8: / Line 8: @@
 *This is a&nbsp; "phase modulation"&nbsp; with&nbsp; $M$&nbsp; signal shapes.&nbsp;
-*This modulation process is also called&nbsp; $\rm M&ndash;PSK$.&nbsp; $M$&nbsp; is usually a power of two.
+*This modulation process is also called&nbsp; "$\rm M&ndash;PSK$".&nbsp; $M$&nbsp; is usually a power of two.
@@ Line 97: / Line 97: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; <u>Solutions 2 and 4</u>&nbsp; are correct:
-*The signal set can be represented in the band-pass domain with the given trigonometric transformation as follows:
+*The signal set&nbsp; $\{s_i(t)\}$&nbsp; can be represented in the band-pass domain with the given trigonometric transformation as follows:
 :$$s_i(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} A \cdot \cos \left ( 2\pi f_{\rm T}t + { 2\pi }/{ M} \cdot i \right ) = A \cdot \cos \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \cos \left ( 2\pi f_{\rm T}t  \right )-
   A \cdot \sin \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \sin \left ( 2\pi f_{\rm T}t  \right )\hspace{0.05cm}.$$
@@ Line 158: / Line 158: @@
 '''(5)'''&nbsp; Correct is <u>YES</u>.&nbsp; The graph shows the constellation for&nbsp; "8&ndash;PSK"&nbsp; and&nbsp; "16&ndash;PSK".
-[[File:P_ID2071__Dig_A_4_14e.png|right|frame|For the interpretation of the Union Bound at the&nbsp;  $\rm M–PSK$]]
+[[File:P_ID2071__Dig_A_4_14e.png|right|frame|For the interpretation of the Union Bound at the&nbsp;  "$\rm M–PSK$"]]
 *Each valid under the assumption that&nbsp; $\boldsymbol{s}_0$&nbsp; was transmitted.
@@ Line 168: / Line 168: @@
 # The actual error probability&nbsp; $p_{\rm S}$&nbsp; is composed of the components&nbsp; $\rm A$,&nbsp; $\rm B$&nbsp; and&nbsp; $\rm C$.
 # On the other hand,&nbsp; the&nbsp; "Union Bound"&nbsp; results from&nbsp; $\rm A+B$&nbsp; $($falsification into the symbol&nbsp;  $\boldsymbol{s}_1))$&nbsp; plus&nbsp; $\rm C+B$&nbsp; $($falsification into the symbol $\boldsymbol{s}_{M-1})$.
-# Thus,&nbsp; $p_{\rm S} = p_{\rm UB}-B$ always holds,&nbsp; but the larger&nbsp; $M$&nbsp; is, the smaller the fraction&nbsp; $\rm B$&nbsp; is.
+# Thus,&nbsp; $p_{\rm S} = p_{\rm UB}-B$&nbsp; always holds,&nbsp; but the larger&nbsp; $M$&nbsp; is, the smaller the fraction&nbsp; $\rm B$&nbsp; is.

@@ Line 157: / Line 157: @@
-'''(5)'''&nbsp; Correct is <u>YES</u>.&nbsp; The graph shows the constellation for&nbsp; "8&ndash;PSK"&nbsp; and&nbsp; "16&ndash;PSK",
+'''(5)'''&nbsp; Correct is <u>YES</u>.&nbsp; The graph shows the constellation for&nbsp; "8&ndash;PSK"&nbsp; and&nbsp; "16&ndash;PSK".
 [[File:P_ID2071__Dig_A_4_14e.png|right|frame|For the interpretation of the Union Bound at the&nbsp;  $\rm M–PSK$]]
-*each valid under the assumption that&nbsp; $\boldsymbol{s}_0$&nbsp; was transmitted.
+*Each valid under the assumption that&nbsp; $\boldsymbol{s}_0$&nbsp; was transmitted.
 *The contour lines of the AWGN PDF are then circles around&nbsp; $\boldsymbol{s}_0$.
 The following statements are valid:
-* The actual error probability $p_{\rm S}$ is composed of the components <b>A</b>, <b>B</b> and <b>C</b>.
+# The actual error probability&nbsp; $p_{\rm S}$&nbsp; is composed of the components&nbsp; $\rm A$,&nbsp; $\rm B$&nbsp; and&nbsp; $\rm C$.
-* On the other hand, the "Union Bound" results from <b>A + B</b> (falsification into the symbol $\boldsymbol{s}_1$) plus <b>B + C</b> (falsification into the symbol $\boldsymbol{s}_{M-1}$).
+# On the other hand,&nbsp; the&nbsp; "Union Bound"&nbsp; results from&nbsp; $\rm A+B$&nbsp; $($falsification into the symbol&nbsp;  $\boldsymbol{s}_1))$&nbsp; plus&nbsp; $\rm C+B$&nbsp; $($falsification into the symbol $\boldsymbol{s}_{M-1})$.
-* Thus, $p_{\rm S} = p_{\rm UB}$ &ndash; <b>B</b> always holds, but the larger $M$ is, the smaller the fraction <b>B</b> is.
+# Thus,&nbsp; $p_{\rm S} = p_{\rm UB}-B$ always holds,&nbsp; but the larger&nbsp; $M$&nbsp; is, the smaller the fraction&nbsp; $\rm B$&nbsp; is.
-'''(6)'''&nbsp; For verification we assume, for example, $p_{\rm B} = 10^{\rm &ndash;4}$. From this follows for $M = 2$ and $M = 4$:
+'''(6)'''&nbsp; For verification we assume&nbsp; $p_{\rm B} = 10^{\rm &ndash;4}$.
-:$$p_{\rm B}  ={\rm Q} \left (  \sqrt{ { {2E_{\rm B}}}/{ N_0} }\right ) = 10^{-4}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
   \sqrt{ { {2E_{\rm B}}}/{ N_0} } \approx 3.72
   \hspace{0.05cm}.$$
-On the other hand, with the given equation for $M = 8$:
+*On the other hand,&nbsp; with the given equation for&nbsp; $M = 8$:
 :$$p_{\rm B} \hspace{-0.1cm} \ \le \ \hspace{-0.1cm}  \frac{2}{{\rm log_2} \hspace{0.05cm}(M)} \cdot {\rm Q} \left ( \sqrt{{\rm log_2} \hspace{0.05cm}(M)} \cdot \sin ({ \pi}/{ M}) \cdot \sqrt{ { {2E_{\rm S}}}/{ N_0} }\right )=  {2}/{3} \cdot {\rm Q} \left ( \sqrt{3} \cdot 0.383 \cdot 3.72 \right ) \approx {2}/{3} \cdot {\rm Q} \left ( 2.46 \right ) \approx 0.46 \%
   \hspace{0.05cm}.$$
-Accordingly, for $M = 16$ &#8658; ${\rm log}_2 (M) = 4$ is obtained:
+*Accordingly,&nbsp; for&nbsp; $M = 16$ &#8658; ${\rm log}_2 (M) = 4$&nbsp; is obtained:
 :$$p_{\rm B} =  {2}/{4} \cdot {\rm Q} \left ( \sqrt{4} \cdot 0.195 \cdot 3.72 \right ) \approx {1}/{2} \cdot {\rm Q} \left ( 1.45 \right ) \approx 4.8 \%
   \hspace{0.05cm}.$$
-Thus, <u>statements 1 and 4</u> are correct:
+Thus,&nbsp; The <u>statements 1 and 4</u>&nbsp; are correct:
-*Thus, the main advantage of a higher-level PSK is not the lower bit error rate, but the lower demand on the very expensive resource "bandwidth".
+*Thus,&nbsp; the main advantage of a higher-level PSK is not the lower bit error rate,&nbsp; but the lower demand on the very expensive resource&nbsp; "bandwidth".
-*It should also be noted that the results are completely different when a (highly) distorting channel is present, as is common in <i>wireline transmission technology</i>.
 {{ML-Fuß}}

@@ Line 42: / Line 42: @@
-''Notes:''
+Notes:
 * The exercise belongs to the topic of the chapter&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation|"Carrier Frequency Systems with Coherent Demodulation"]].
@@ Line 96: / Line 96: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; <u>Solutions 2 and 4</u> are correct:
+'''(1)'''&nbsp; <u>Solutions 2 and 4</u>&nbsp; are correct:
 *The signal set can be represented in the band-pass domain with the given trigonometric transformation as follows:
 :$$s_i(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} A \cdot \cos \left ( 2\pi f_{\rm T}t + { 2\pi }/{ M} \cdot i \right ) = A \cdot \cos \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \cos \left ( 2\pi f_{\rm T}t  \right )-
   A \cdot \sin \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \sin \left ( 2\pi f_{\rm T}t  \right )\hspace{0.05cm}.$$
-*The respective first terms in this difference lead to the signal space point $\boldsymbol{s}_i$, the respective second terms to the basis functions $\varphi_1(t)$ and $\varphi_2(t)$. Here it is to be noted that these must be energy normalized in each case:
+*The respective first terms in this difference lead to the signal space point&nbsp; $\boldsymbol{s}_i$,&nbsp; the respective second terms to the basis functions&nbsp; $\varphi_1(t)$&nbsp; and&nbsp; $\varphi_2(t)$.
 :$$\int_{0}^{T} \varphi_1 ( t)^2 \,{\rm d} t = \int_{0}^{T} \left [K \cdot \cos( 2\pi f_{\rm T}t)\right ]^2 \,{\rm d} t = 1\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} \frac{K^2 \cdot T}{2} = 2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} K = \sqrt{2/T } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
   \varphi_1 ( t) = \sqrt{2/T } \cdot \cos( 2\pi f_{\rm T}t)\hspace{0.05cm}.$$
+*Using the same arithmetic,&nbsp; we arrive at the second basis function:
 :$$\varphi_2 ( t) = -\sqrt{2/T } \cdot \sin( 2\pi f_{\rm T}t)\hspace{0.05cm}.$$
-'''(2)'''&nbsp; <u>Solutions 1 and 4</u> are correct:
+'''(2)'''&nbsp; <u>Solutions 1 and 4</u>&nbsp; are correct:
-*Using the basis functions just calculated, the signal set $s_i(t)$ can be represented as follows (again, limited to the range $0 &#8804; t &#8804; T$):
+*Using the basis functions just calculated,&nbsp; the signal set&nbsp; $s_i(t)$&nbsp; can be represented as follows&nbsp; $($again, limited to the range&nbsp; $0 &#8804; t &#8804; T)$:
 :$$s_i(t) = A \cdot \sqrt{T/2 } \cdot \cos ( 2\pi  \cdot i /M ) \cdot \varphi_1 ( t)+
   A \cdot \sqrt{T/2 } \cdot \sin ( 2\pi  \cdot i /M ) \cdot \varphi_2 ( t)
   \hspace{0.05cm}.$$
-*With the energy $E = 0.5 \cdot A^2 \cdot T$, which is the same for all $M$ points and which is also the mean signal energy per symbol ($E_{\rm S}$), the above equation is
+*With the energy&nbsp; $E = 0.5 \cdot A^2 \cdot T$,&nbsp; which is the same for all&nbsp; $M$&nbsp; points and which is also the average signal energy per symbol&nbsp; $(E_{\rm S})$,&nbsp; the above equation is
 :$$s_i(t) = s_{{\rm I}i} \cdot \varphi_1 ( t)+
   s_{{\rm Q}i}  \cdot \varphi_2 ( t)\hspace{0.3cm}
@@ Line 122: / Line 123: @@
   s_{{\rm Q}i} = \sqrt{E_{\rm S} }\cdot \sin ( 2\pi  \cdot i /M )\hspace{0.05cm}. $$
-*The signal space points sketched in the front graph for $M = 8$ and $M = 16$, respectively, can be represented in exactly this way.
+*The signal space points sketched in the front graph for&nbsp; $M = 8$&nbsp; resp.&nbsp; $M = 16$,&nbsp; can be represented in exactly this way.
 :$$d_{\rm norm}^2 \hspace{-0.01cm}=\hspace{-0.01cm} \sin^2(\frac{2\pi}{M}) + [1- \cos(\frac{2\pi}{M})]^2 $$
 :$$\Rightarrow \hspace{0.3cm} d_{\rm norm}^2 \hspace{-0.01cm}=\hspace{-0.01cm} 2 \cdot [1- \cos( \frac{2\pi}{M} )] =  4 \cdot \sin^2(\pi/M )\hspace{0.05cm} $$
@@ Line 133: / Line 136: @@
 \Rightarrow \hspace{0.3cm} d =  2\sqrt {E_{\rm S}} \cdot \sin{\pi}/{M })\hspace{0.05cm}.$$
-This results in the following numerical values:
+*This results in the following numerical values:
 :$$M = 8\hspace{-0.09cm}: \hspace{0.2cm}d  \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \underline{0.765} \cdot \sqrt {E_{\rm S} } \hspace{0.05cm},$$
 :$$M = 16\hspace{-0.09cm}: \hspace{0.2cm} d  \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \underline{0.390} \cdot \sqrt {E_{\rm S} } \hspace{0.05cm}.$$
-'''(4)'''&nbsp; Using the result from '''(3)''', we obtain with the equation given in front:
+'''(4)'''&nbsp; Using the result from subtask&nbsp; '''(3)''',&nbsp; we obtain with the equation given in front:
 :$$p_{\rm S} \le  p_{\rm UB}  =  2 \cdot {\rm Q} \left [ \sin ({ \pi}/{ M}) \cdot \sqrt{ { 2E_{\rm S}}/{ N_0}} \right ]  \hspace{0.05cm}.$$
@@ Line 145: / Line 148: @@
 :$$ M = 16\hspace{-0.09cm}: \hspace{0.2cm} p_{\rm UB}  \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  2 \cdot {\rm Q}  ( 10 \cdot 0.195 )= 2 \cdot {\rm Q} ( 1.95 ) \approx \underline{6 \%}  \hspace{0.05cm}.$$
-*Thus, the error probability becomes larger and larger as $M$ increases, assuming constant $E_{\rm S}/N_0$, as in this case.
+*Thus,&nbsp; the symbol error probability becomes larger and larger as&nbsp; $M$&nbsp; increases,&nbsp; assuming constant&nbsp; $E_{\rm S}/N_0$,&nbsp; as in this case.
 *The most favorable value would result for $M = 2$ (the factor $2$ of the Union Bound is then not necessary) as follows
 :$$M = 2\hspace{-0.09cm}: \hspace{0.2cm}p_{\rm UB} = p_{\rm S}  = {\rm Q}  ( 10  ) \approx { 10^{-23}}.$$
-*Thus, multi-level PSK would not make sense if there were not other reasons for its use, which will be discussed in subtask '''(6)'''.
+*Thus,&nbsp; multi-level PSK would not make sense if there were not other reasons for its use,&nbsp; which will be discussed in subtask&nbsp; '''(6)'''.
-'''(5)'''&nbsp; Correct is <u>YES</u>. The graph shows the constellation for 8&ndash;PSK and 16&ndash;PSK, each valid under the assumption that $\boldsymbol{s}_0$ was transmitted. The contour lines of the AWGN PDF are then circles around $\boldsymbol{s}_0$.
+'''(5)'''&nbsp; Correct is <u>YES</u>.&nbsp; The graph shows the constellation for&nbsp; "8&ndash;PSK"&nbsp; and&nbsp; "16&ndash;PSK",
-[[File:P_ID2071__Dig_A_4_14e.png|center|frame|For the interpretation of the Union Bound at <i>M</i>–PSK]]
+[[File:P_ID2071__Dig_A_4_14e.png|right|frame|For the interpretation of the Union Bound at the&nbsp;  $\rm M–PSK$]]
 The following statements are valid:
 * The actual error probability $p_{\rm S}$ is composed of the components <b>A</b>, <b>B</b> and <b>C</b>.

@@ Line 3: / Line 3: @@
 [[File:P_ID2067__Dig_A_4_14.png|right|frame|Signal space constellations <br>of the 8-PSK and 16-PSK]]
-Now a signal set&nbsp; $\{s_i(t)\}$ is considered, which is limited to the time domain&nbsp; $0 &#8804; t &#8804; T$.&nbsp; The index&nbsp; $i$&nbsp; runs through the values&nbsp; $0, \ \text{...} \ , M-1$:
+Now a signal set&nbsp; $\{s_i(t)\}$&nbsp; is considered,&nbsp; which is limited to the time domain&nbsp; $0 &#8804; t &#8804; T$.&nbsp; The index&nbsp; $i$&nbsp; runs through the values&nbsp; $0, \ \text{...} \ , M-1$:
 :$$s_i(t) = A \cdot \cos \left ( 2\pi f_{\rm T}t + { 2\pi }/{ M} \cdot i \right ) \hspace{0.05cm}.$$
-This is a <i>phase modulation</i>&nbsp; with&nbsp; $M$&nbsp; signal shapes. This modulation process is also called <b><i>M</i>&ndash;PSK</b>. $M$&nbsp; is usually a power of two.
+*This is a&nbsp; "phase modulation"&nbsp; with&nbsp; $M$&nbsp; signal shapes.&nbsp;
-The graphic shows the signal space constellation for $M = 8$ (top) and $M = 16$ (bottom). All signal space points have equal energy $||\boldsymbol{s}_i||^2 = E_{\rm S}$ ("mean symbol energy").
+*This modulation process is also called&nbsp; $\rm M&ndash;PSK$.&nbsp; $M$&nbsp; is usually a power of two.
-The exact calculation of the error probability is difficult for&nbsp; $M &ne; 2$.&nbsp; However, the so-called <i>Union Bound</i>&nbsp; can always be given as an upper bound for the symbol error probability &nbsp;$(p_{\rm UB} &#8805; p_{\rm S})$:
 :$$  p_{\rm UB}  = 2 \cdot {\rm Q} \left ( \frac{ d/2}{ \sigma_n}\right ) = 2 \cdot {\rm Q} \left (\sqrt{ \frac{ d^2}{ 2 N_0}}\right ) \hspace{0.05cm}.$$
 The following quantities are used here:
-* $d$&nbsp; is the distance between two neighboring points, for example between&nbsp; $\boldsymbol{s}_0$&nbsp; and&nbsp; $\boldsymbol{s}_1$. If the decision boundary is exactly centered perpendicular to the line connecting&nbsp; $\boldsymbol{s}_0$&nbsp; and&nbsp; $\boldsymbol{s}_1$,&nbsp; then&nbsp; $d/2$&nbsp; is the distance of&nbsp; $\boldsymbol{s}_0$&nbsp; or&nbsp; $\boldsymbol{s}_1$&nbsp; from this decision boundary.
+* $d$&nbsp; is the distance between two neighboring points,&nbsp; for example between&nbsp; $\boldsymbol{s}_0$&nbsp; and&nbsp; $\boldsymbol{s}_1$.
 * The variance of the AWGN noise is&nbsp; $\sigma_n^2 = N_0/2$.
-* The factor of&nbsp; $2$&nbsp; in the above limit takes into account that for&nbsp; $M > 2$&nbsp; each signal space point can be falsified in two directions, for example, for the 8&ndash;PSK the symbol&nbsp; $\boldsymbol{s}_0$&nbsp; into the symbol&nbsp; $\boldsymbol{s}_1$&nbsp; or into the symbol&nbsp; $\boldsymbol{s}_7$.
 * ${\rm Q}(x)$&nbsp; is the complementary Gaussian error function for which the following approximation holds:
 :$${\rm Q}(x)  \approx   \frac{1}{\sqrt{2\pi} \cdot x} \cdot {\rm e}^{-x^2/2}
 \hspace{0.05cm}.$$
-The last subtask deals with the bit error probability. For this, the following bound was given in the&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation| "theory section"]]&nbsp; under the assumption of a <i>Gray code</i>:&nbsp;
+The last subtask deals with the bit error probability.&nbsp; For this,&nbsp; the following bound was given in the&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation| "theory section"]]&nbsp; under the assumption of a&nbsp; "Gray code":&nbsp;
 :$$p_{\rm B}  \le \frac{2}{{\rm log_2} \hspace{0.05cm}(M)} \cdot {\rm Q} \left ( \sqrt{{\rm log_2} \hspace{0.05cm}(M)} \cdot \sin ({ \pi}/{ M}) \cdot \sqrt{ { {2E_{\rm B}}}/{ N_0} }\right )
   \hspace{0.05cm}.$$
-However, this equation is only applicable for&nbsp; $M > 4$.&nbsp; In contrast, the exact solution
+However,&nbsp; this equation is only applicable for&nbsp; $M > 4$.&nbsp; In contrast,&nbsp; the exact solution
-* for&nbsp; $M = 2$&nbsp; from the identity with the&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Binary_phase_shift_keying_.28BPSK.29| BPSK]], and
+* for&nbsp; $M = 2$ &nbsp; from the identity with the&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Binary_phase_shift_keying_.28BPSK.29| "BPSK"]], and
 * for&nbsp; $M = 4$&nbsp; from the fact that the 4&ndash;PSK is identical with the&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Quadrature_amplitude_modulation_.28M-QAM.29|"4&ndash;QAM"]]:
 :$$p_{\rm B}  ={\rm Q} \left (  \sqrt{ { {2E_{\rm B}}}/{ N_0} }\right )
   \hspace{0.05cm}.$$
@@ Line 35: / Line 44: @@
 ''Notes:''
 * The exercise belongs to the topic of the chapter&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation|"Carrier Frequency Systems with Coherent Demodulation"]].
-*Reference is made in particular to the section&nbsp;   [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Multi-level_phase.E2.80.93shift_keying_.28M.E2.80.93PSK.29| "Multi-level ''Phase Shift Keying''"]].
-* The assignment of the 8 or 16 symbols to binary sequences of length 3 or 4 according to the Gray coding can be taken from the graphic (red labeling).
+*Reference is made in particular to the section&nbsp;   [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Multi-level_phase.E2.80.93shift_keying_.28M.E2.80.93PSK.29| "Multi-level Phase Shift Keying"]].
 * When solving the exercise, you can use the following equations:
 :$$\cos(\alpha + \beta) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \cos(\alpha ) \cdot \cos( \beta) - \sin(\alpha ) \cdot \sin( \beta)\hspace{0.05cm}, \hspace{0.25cm}
@@ Line 45: / Line 57: @@
 ===Questions===
 <quiz display=simple>
-{What are the basis functions in the band-pass representation? Let $\varphi_1(t)$&nbsp; and&nbsp; $\varphi_2(t)$&nbsp; each be limited to the range&nbsp; $0 &#8804; t &#8804; T$.&nbsp;
+{What are the basis functions in the band-pass representation?&nbsp; Let $\varphi_1(t)$&nbsp; and&nbsp; $\varphi_2(t)$&nbsp; each be limited to the range&nbsp; $0 &#8804; t &#8804; T$.&nbsp;
 |type="[]"}
 - $\varphi_1(t) = \cos {(2\pi f_{\rm T}t)}$,
@@ Line 52: / Line 64: @@
 + $\varphi_2(t) = \, &ndash;\sqrt {2/T} \cdot \sin {(2\pi f_{\rm T}t)}$.
-{What are the in-phase and quadrature components of the signal space point&nbsp; $\boldsymbol{s}_i$? Which statements are true?
+{What are the in-phase and quadrature components of the signal space point&nbsp; $\boldsymbol{s}_i$?&nbsp; Which statements are true?
 |type="[]"}
 + $s_{\rm I \it i} = \sqrt {E_{\rm S}} \cdot \cos {(2\pi \cdot i/M)}$,
@@ Line 59: / Line 71: @@
 + $s_{\rm Q \it i} = \sqrt {E_{\rm S}} \cdot \sin {(2\pi \cdot i/M)}$.
-{What is the distance&nbsp; $d$&nbsp; between two adjacent signal space points? What are the values for&nbsp; $M = 8$&nbsp; and &nbsp;$M = 16$, respectively?
+{What is the distance&nbsp; $d$&nbsp; between two adjacent signal space points?&nbsp; What are the values for&nbsp; $M = 8$&nbsp; and &nbsp;$M = 16$, respectively?
 |type="{}"}
 $M = 8 \text{:} \hspace{0.45cm} d \ = \ $ { 0.765 3% } $\ \cdot \sqrt {E_{\rm S}}$
 $M = 16 \text{:} \hspace{0.2cm} d \ = \ $ { 0.39 3% } $\ \cdot \sqrt {E_{\rm S}}$
-{What is the value of Union Bound&nbsp; $(p_{\rm UB})$&nbsp; with &nbsp;$E_{\rm S}/N_0 = 50$.
+{What is the value of the&nbsp; "Union Bound"&nbsp; $(p_{\rm UB})$&nbsp; with &nbsp;$E_{\rm S}/N_0 = 50$.
 |type="{}"}
 $M = 8 \text{:} \hspace{0.45cm} p_{\rm UB}$ = { 0.014 3% } $\ \%$
 $M = 16 \text{:} \hspace{0.2cm} p_{\rm UB}$ = { 6 3% } $\ \%$
-{Does the statement: "$p_{\rm UB}$&nbsp; approximates &nbsp;$p_{\rm S}$&nbsp; more closely the larger &nbsp;$M$ &nbsp; is" hold?
+{Does the statement: nbsp; "$p_{\rm UB}$&nbsp; approximates &nbsp;$p_{\rm S}$&nbsp; more closely the larger &nbsp;$M$ &nbsp; is"nbsp; hold?
 |type="()"}
 + YES.

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Digitalsignalübertragung/Trägerfrequenzsysteme mit kohärenter Demodulation}}
+{{quiz-Header|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation}}
-[[File:P_ID2067__Dig_A_4_14.png|right|frame|Signalraumkonstellationen <br>der 8–PSK und 16–PSK]]
+[[File:P_ID2067__Dig_A_4_14.png|right|frame|Signal space constellations <br>of the 8-PSK and 16-PSK]]
-Betrachtet wird nun eine Signalmenge&nbsp; $\{s_i(t)\}$, die auf den Zeitbereich&nbsp; $0 &#8804; t &#8804; T$&nbsp; begrenzt ist. Der Index&nbsp; $i$&nbsp; durchläuft die Werte&nbsp; $0, \ \text{...} \ , M-1$:
+Now a signal set&nbsp; $\{s_i(t)\}$ is considered, which is limited to the time domain&nbsp; $0 &#8804; t &#8804; T$.&nbsp; The index&nbsp; $i$&nbsp; runs through the values&nbsp; $0, \ \text{...} \ , M-1$:
 :$$s_i(t) = A \cdot \cos \left ( 2\pi f_{\rm T}t + { 2\pi }/{ M} \cdot i \right ) \hspace{0.05cm}.$$
-Es handelt sich hierbei um eine <i>Phasenmodulation</i>&nbsp; mit&nbsp; $M$&nbsp; Signalformen. Man nennt dieses Modulationsverfahren auch <b><i>M</i>&ndash;PSK</b>. $M$&nbsp; ist meist eine Zweierpotenz.
+This is a <i>phase modulation</i>&nbsp; with&nbsp; $M$&nbsp; signal shapes. This modulation process is also called <b><i>M</i>&ndash;PSK</b>. $M$&nbsp; is usually a power of two.
-Die Grafik zeigt die Signalraumkonstellation für $M = 8$ (oben) und $M = 16$ (unten). Alle Signalraumpunkte haben gleiche Energie $||\boldsymbol{s}_i||^2 = E_{\rm S}$ ("mittlere Symbolenergie").
+The graphic shows the signal space constellation for $M = 8$ (top) and $M = 16$ (bottom). All signal space points have equal energy $||\boldsymbol{s}_i||^2 = E_{\rm S}$ ("mean symbol energy").
-Die exakte Berechnung der Fehlerwahrscheinlichkeit ist für&nbsp; $M &ne; 2$&nbsp; schwierig. Angegeben werden kann dagegen stets die sogenannte <i>Union Bound</i>&nbsp; als obere Schranke für die Symbolfehlerwahrscheinlichkeit &nbsp;$(p_{\rm UB} &#8805; p_{\rm S})$:
+The exact calculation of the error probability is difficult for&nbsp; $M &ne; 2$.&nbsp; However, the so-called <i>Union Bound</i>&nbsp; can always be given as an upper bound for the symbol error probability &nbsp;$(p_{\rm UB} &#8805; p_{\rm S})$:
 :$$  p_{\rm UB}  = 2 \cdot {\rm Q} \left ( \frac{ d/2}{ \sigma_n}\right ) = 2 \cdot {\rm Q} \left (\sqrt{ \frac{ d^2}{ 2 N_0}}\right ) \hspace{0.05cm}.$$
-Hierbei sind folgende Größen verwendet:
+The following quantities are used here:
-* $d$&nbsp; ist der Abstand zwischen zwei benachbarten Punkten, zum Beispiel zwischen&nbsp; $\boldsymbol{s}_0$&nbsp; und&nbsp; $\boldsymbol{s}_1$. Verläuft die Entscheidungsgrenze senkrecht zur Verbindungslinie von&nbsp; $\boldsymbol{s}_0$&nbsp; und&nbsp; $\boldsymbol{s}_1$&nbsp; genau mittig, so ist&nbsp; $d/2$&nbsp; der Abstand von&nbsp; $\boldsymbol{s}_0$&nbsp; bzw.&nbsp; $\boldsymbol{s}_1$&nbsp; von dieser Entscheidungsgrenze.
+* $d$&nbsp; is the distance between two neighboring points, for example between&nbsp; $\boldsymbol{s}_0$&nbsp; and&nbsp; $\boldsymbol{s}_1$. If the decision boundary is exactly centered perpendicular to the line connecting&nbsp; $\boldsymbol{s}_0$&nbsp; and&nbsp; $\boldsymbol{s}_1$,&nbsp; then&nbsp; $d/2$&nbsp; is the distance of&nbsp; $\boldsymbol{s}_0$&nbsp; or&nbsp; $\boldsymbol{s}_1$&nbsp; from this decision boundary.
-* Die Varianz des AWGN&ndash;Rauschens ist&nbsp; $\sigma_n^2 = N_0/2$.
+* The variance of the AWGN noise is&nbsp; $\sigma_n^2 = N_0/2$.
-* Der Faktor&nbsp; $2$&nbsp; in obiger Grenze berücksichtigt, dass für&nbsp; $M > 2$&nbsp; jeder Signalraumpunkt in zwei Richtungen verfälscht werden kann, zum Beispiel bei der 8&ndash;PSK das Symbol&nbsp; $\boldsymbol{s}_0$&nbsp; in das Symbol&nbsp; $\boldsymbol{s}_1$&nbsp; oder in das Symbol&nbsp; $\boldsymbol{s}_7$.
+* The factor of&nbsp; $2$&nbsp; in the above limit takes into account that for&nbsp; $M > 2$&nbsp; each signal space point can be falsified in two directions, for example, for the 8&ndash;PSK the symbol&nbsp; $\boldsymbol{s}_0$&nbsp; into the symbol&nbsp; $\boldsymbol{s}_1$&nbsp; or into the symbol&nbsp; $\boldsymbol{s}_7$.
-* ${\rm Q}(x)$&nbsp; ist die komplementäre Gaußsche Fehlerfunktion, für die folgende Näherung gilt:
+* ${\rm Q}(x)$&nbsp; is the complementary Gaussian error function for which the following approximation holds:
 :$${\rm Q}(x)  \approx   \frac{1}{\sqrt{2\pi} \cdot x} \cdot {\rm e}^{-x^2/2}
 \hspace{0.05cm}.$$
-In der letzten Teilaufgabe geht es um die Bitfehlerwahrscheinlichkeit. Für diese wurde im&nbsp; [[Digitalsignal%C3%BCbertragung/Tr%C3%A4gerfrequenzsysteme_mit_koh%C3%A4renter_Demodulation| Theorieteil]]&nbsp; unter der Voraussetzung eines <i>Graycodes</i>&nbsp; folgende Schranke angegeben:
+The last subtask deals with the bit error probability. For this, the following bound was given in the&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation| "theory section"]]&nbsp; under the assumption of a <i>Gray code</i>:&nbsp;
 :$$p_{\rm B}  \le \frac{2}{{\rm log_2} \hspace{0.05cm}(M)} \cdot {\rm Q} \left ( \sqrt{{\rm log_2} \hspace{0.05cm}(M)} \cdot \sin ({ \pi}/{ M}) \cdot \sqrt{ { {2E_{\rm B}}}/{ N_0} }\right )
   \hspace{0.05cm}.$$
@@ Line 34: / Line 34: @@
 ''Hinweise:''
-* Die Aufgabe gehört zum Themengebiet des Kapitels&nbsp; [[Digitalsignal%C3%BCbertragung/Tr%C3%A4gerfrequenzsysteme_mit_koh%C3%A4renter_Demodulation| Trägerfrequenzen mit kohärenter Demodulation]].
+* Die Aufgabe gehört zum Themengebiet des Kapitels&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation|"Carrier Frequency Systems with Coherent Demodulation"]].
 *Bezug genommen eird insbesondere auf die Seite&nbsp;   [[Digitalsignal%C3%BCbertragung/Tr%C3%A4gerfrequenzsysteme_mit_koh%C3%A4renter_Demodulation#Mehrstufiges_Phase.E2.80.93Shift_Keying_.28M.E2.80.93PSK.29| Mehrstufiges ''Phase Shift Keying'']].
 * Die Zuordnung der 8 bzw. 16 Symbole zu Binärfolgen der Länge 3 bzw. 4 nach der Graycodierung kann der Grafik entnommen werden (rote Beschriftung).