Aufgaben:Exercise 4.2Z: Eight-level Phase Shift Keying - Revision history

Hwang at 08:41, 12 August 2022

2022-08-12T08:41:28Z

Guenter at 16:26, 13 July 2022

2022-07-13T16:26:31Z

Guenter at 14:32, 13 July 2022

2022-07-13T14:32:09Z

Guenter at 14:18, 13 July 2022

2022-07-13T14:18:59Z

Guenter: Guenter moved page Aufgaben:Exercise 4.2Z: Eight-step Phase Shift Keying to Aufgaben:Exercise 4.2Z: Eight-level Phase Shift Keying

2022-07-07T14:25:50Z

Guenter moved page Exercise 4.2Z: Eight-step Phase Shift Keying to Exercise 4.2Z: Eight-level Phase Shift Keying

Hwang at 19:51, 14 June 2022

2022-06-14T19:51:19Z

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Hwang at 13:45, 14 June 2022

2022-06-14T13:45:04Z

Hwang at 13:11, 14 June 2022

2022-06-14T13:11:22Z

Guenter at 11:55, 8 June 2022

2022-06-08T11:55:31Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.2Z: Achtstufiges Phase Shift Keying to Aufgaben:Exercise 4.2Z: Eight-step Phase Shift Keying

2022-06-08T11:50:06Z

Guenter moved page Aufgabe 4.2Z: Achtstufiges Phase Shift Keying to Exercise 4.2Z: Eight-step Phase Shift Keying

@@ Line 19: / Line 19: @@
   \hspace{0.05cm}, \hspace{0.2cm}i = 0,\text{...} \hspace{0.1cm} , 7 \hspace{0.05cm},$$
-where&nbsp; $a_i$&nbsp; are dimensionless complex  coefficients and the energy of the basic transmission pulse&nbsp; $g_s(t)$&nbsp; is&nbsp; $E_{\it g_s}$&nbsp; in the low-pass region.&nbsp; In the case shown here, &nbsp; $g_s(t)$&nbsp; describes a rectangular pulse,&nbsp;s but any other energy-limited pulse can be used for&nbsp; $g_s(t)$.&nbsp;
+where&nbsp; $a_i$&nbsp; are dimensionless complex  coefficients and the energy of the basic transmission pulse&nbsp; $g_s(t)$&nbsp; is&nbsp; $E_{\it g_s}$&nbsp; in the low-pass region.&nbsp; In the case shown here, &nbsp; $g_s(t)$&nbsp; describes a rectangular pulse,&nbsp; but any other energy-limited pulse can be used for&nbsp; $g_s(t)$.&nbsp;
 The graph shows the&nbsp; 8&ndash;PSK&nbsp; signal space representation

@@ Line 36: / Line 36: @@
 * For abbreviation,&nbsp;  use the energy&nbsp; $E = 1/2 \cdot A^2 \cdot T$.
 * In contrast to the theory section and&nbsp; [[Aufgaben:Exercise_4.2:_AM/PM_Oscillations|"Exercise 4.2"]],&nbsp; here the indexing variable&nbsp; $i$&nbsp; can take the values &nbsp;$0, \ \text{...} \, ,M-1$.&nbsp;
@@ Line 70: / Line 72: @@
 - $\psi_1(t) = {\rm j} \cdot \varphi_1(t)$.
-{Let&nbsp; $s_{\rm TP0}(t) = \sqrt{E}$. Which is true:
+{Let&nbsp; $s_{\rm TP0}(t) = \sqrt{E}$. Which is true?
 |type="()"}
 - The energy&nbsp; $E$&nbsp; refers to the low&ndash;pass signal.

@@ Line 78: / Line 78: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The signal $s_0(t)$ is:
+'''(1)'''&nbsp; The signal&nbsp;  $s_0(t)$&nbsp; is:
 :$$s_0(t)=  A \cdot \cos(2\pi f_{\rm T}t ) = s_{01} \cdot \varphi_1(t) + s_{02} \cdot \varphi_2(t) \hspace{0.05cm}.$$
-*Since this signal has no sinusoidal part, $s_{\rm 02} \hspace{0.15cm}\underline {= 0}$.
+*Since this signal has no sinusoidal part,&nbsp; $s_{\rm 02} \hspace{0.15cm}\underline {= 0}$.
-*Further, with the given abbreviation:
+*Further,&nbsp; with the given abbreviation:
 :$$A = s_{01} \cdot \sqrt{{2}/{T}}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
 s_{01}=\sqrt{1/2 \cdot A^2 \cdot T} =  \sqrt{E}\hspace{0.05cm} \hspace{0.15cm}\underline { = 1 \cdot E^{\hspace{0.05cm}0.5}}\hspace{0.05cm}.$$
-'''(2)'''&nbsp; The signal $s_2(t)$ is with $i = 2$ (note that the second basis function is minus&ndash;sine):
+'''(2)'''&nbsp; The signal&nbsp; $s_2(t)$&nbsp; is with&nbsp; $i = 2$&nbsp; (note that the second basis function is minus&ndash;sine):
 :$$s_2(t)=  A \cdot \cos(2\pi f_{\rm T}t + {\pi}/{2})= -  A \cdot \sin(2\pi f_{\rm T}t )\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} s_{21}\hspace{0.05cm} \underline{= 0}\hspace{0.05cm}, \hspace{0.2cm} s_{22}=  \sqrt{E}  \hspace{0.05cm} \hspace{0.15cm}\underline {=1 \cdot E^{\hspace{0.05cm}0.5}}\hspace{0.05cm}.$$
-'''(3)'''&nbsp; According to the solutions to subtasks '''(1)''' and '''(2)''', the following is now true:
+'''(3)'''&nbsp; According to the solutions to subtasks&nbsp; '''(1)'''&nbsp; and&nbsp; '''(2)''',&nbsp; the following is true:
 :$$s_{51}= s_{52}= - \sqrt{E/2} \hspace{0.05cm} \hspace{0.15cm}\underline { = -0.707 \cdot E^{\hspace{0.05cm}0.5}}$$
 :$$\Rightarrow \hspace{0.3cm} s_{5}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  - {A}/{ \sqrt{2}} \cdot \cos(2\pi f_{\rm T}t ) - {A}/{ \sqrt{2}} \cdot \sin(2\pi f_{\rm T}t )=A \cdot \cos(2\pi f_{\rm T}t + \phi_5)\hspace{0.2cm}{\rm with}\hspace{0.2cm}\phi_5 = -0.75 \cdot \pi
@@ Line 99: / Line 99: @@
-'''(4)'''&nbsp; <u>Solutions 1 and 3</u> are correct.
+'''(4)'''&nbsp; <u>Solutions 1 and 3</u>&nbsp; are correct.&nbsp; The following relation holds: &nbsp;
 :$$\xi_1 (t) = \varphi_1 (t) + {\rm j} \cdot \psi_1 (t)\hspace{0.05cm}.$$
-'''(5)'''&nbsp; Correct are the <u>alternatives 2 and 3</u>:
+'''(5)'''&nbsp; Correct are the&nbsp; <u>alternatives 2 and 3</u>:
-*The basis function must be energy normalized.
+*The basis function must be energy-normalized.
 * Like $\psi_1(t)$, $\varphi_1(t)$ is a real function, not an imaginary one:
 :$$\varphi_1 (t) = \psi_1 (t) =
 \left\{ \begin{array}{c} 1/\sqrt{T} \\
@@ Line 114: / Line 114: @@
-'''(6)'''&nbsp; From the low&ndash;pass signal $s_{\rm TP0}(t)$, one can also calculate the band-pass signal $s_0(t)$.
+'''(6)'''&nbsp; From the low&ndash;pass signal&nbsp; $s_{\rm TP0}(t)$,&nbsp; one can also calculate the band-pass signal&nbsp; $s_0(t)$.
-*In the range $0 &#8804; t &#8804; T$, the result from '''(5)''' gives the same result as in subtask '''(1)''' :
+*In the range&nbsp; $0 &#8804; t &#8804; T$,&nbsp; the result from&nbsp; '''(5)'''&nbsp; gives the same result as in subtask&nbsp; '''(1)''':
 :$$s_0(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Re}\left[s_{{\rm TP}0}(t) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi f_{\rm T}t} \right]
   = {\rm Re}\left[\sqrt{E} /{\sqrt{T}} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi f_{\rm T}t} \right]=  \sqrt{E/T} \cdot \cos(2\pi f_{\rm T}t )
 \hspace{0.05cm},$$
-*It follows: The energy $E$ refers to the band&ndash;pass signal even when considered in the equivalent low&ndash;pass region.
+*It follows:&nbsp; The energy&nbsp; $E$&nbsp; refers to the band&ndash;pass signal even when considered in the equivalent low&ndash;pass region.
-*Accordingly, for the signal $s_2(t)$ marked with blue dot in the region of interest:
+*Accordingly,&nbsp; for the signal&nbsp; $s_2(t)$&nbsp; marked with blue dot in the region of interest:
 :$$s_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Re}\big[\hspace{0.05cm}{\rm j} \cdot \sqrt{E/T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi f_{\rm T}t} \big]
   = {\rm Re}\big[\hspace{0.05cm}{\rm j} \cdot \sqrt{E/T} \cdot \cos(2\pi f_{\rm T}t)- \sqrt{E/T} \cdot \sin(2\pi f_{\rm T}t) \big]
@@ Line 128: / Line 128: @@
 \hspace{0.05cm}.$$
-*Finally, for the (green) signal $s_5(t)$ in the range $0 &#8804; t < T$ can be written:
+*Finally,&nbsp; for the&nbsp; (green)&nbsp; signal&nbsp; $s_5(t)$&nbsp; in the range $0 &#8804; t < T$&nbsp; can be written:
 :$$s_5(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Re}\big[\frac{-1 - {\rm j}}{\sqrt{2}} \cdot \sqrt{{E}/{T}} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi f_{\rm T}t} \big] = \text{...}
   =  - \sqrt{\frac{E}{2T}} \cdot \cos(2\pi f_{\rm T}t)+ \sqrt{\frac{E}{2T}} \cdot \sin(2\pi f_{\rm T}t)=\sqrt{E/T} \cdot \cos(2\pi f_{\rm T}t + 1.25 \cdot \pi)
 \hspace{0.05cm}.$$
-*These results also agree with those of subtasks '''(2)''' and '''(3)''', respectively. Therefore,  <u>solution 2</u> is correct.
+*These results also agree with those of subtasks&nbsp; '''(2)'''&nbsp; resp.&nbsp; '''(3)'''.&nbsp; Therefore,&nbsp;  <u>solution 2</u>&nbsp; is correct.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 [[File:P_ID1998__Dig_Z_4_2.png|right|frame|Signal space points at 8-PSK]]
-The&nbsp; $M = 8$&nbsp; possible transmitted signals at 8&ndash;PSK are with&nbsp; $i = 0, \ \text{...} \ , 7$&nbsp; in the range&nbsp; $0 &#8804; t < T$:
+The&nbsp; $M = 8$&nbsp; possible transmitted signals at&nbsp; "8&ndash;PSK"&nbsp; are with &nbsp; $i = 0, \ \text{...} \ , 7$ &nbsp; in the range&nbsp; $0 &#8804; t < T$:
 :$$s_i(t)=  A \cdot \cos(2\pi f_{\rm T}t + i \cdot {\pi}/{4}) \hspace{0.05cm}.$$
@@ Line 15: / Line 15: @@
 :$$s_i(t)= s_{i1} \cdot \varphi_1(t) + s_{i2} \cdot \varphi_2(t) \hspace{0.05cm}.$$
-The equivalent low-pass representation of the signals&nbsp; $s_i(t)$&nbsp; is according to the section&nbsp; [[Modulation_Methods/Quadrature_Amplitude_Modulation#System_description_using_the_equivalent_low-pass_signal|"System description using the equivalent low-pass signal"]]&nbsp; of the book "Modulation methods":
+The equivalent low-pass representation of the signals&nbsp; $s_i(t)$&nbsp; is according to the section&nbsp; [[Modulation_Methods/Quadrature_Amplitude_Modulation#System_description_using_the_equivalent_low-pass_signal|"System description using the equivalent low-pass signal"]]&nbsp; of the book&nbsp; "Modulation methods":
 :$$s_{{\rm TP}i}(t)= a_{i} \cdot g_s(t) \hspace{0.05cm}, \hspace{0.2cm}a_{i} = a_{{\rm I}i} + {\rm j} \cdot a_{{\rm Q}i}
   \hspace{0.05cm}, \hspace{0.2cm}i = 0,\text{...} \hspace{0.1cm} , 7 \hspace{0.05cm},$$
-where&nbsp; $a_i$&nbsp; are complex dimensionless coefficients and the energy of the basic transmission pulse&nbsp; $g_s(t)$&nbsp; is&nbsp; $E_{\it g_s}$&nbsp; in the low-pass region. In the case shown here, &nbsp; $g_s(t)$&nbsp; describes a rectangular pulse, but any other energy-limited pulse can be used for&nbsp; $g_s(t)$.&nbsp;
+where&nbsp; $a_i$&nbsp; are dimensionless complex  coefficients and the energy of the basic transmission pulse&nbsp; $g_s(t)$&nbsp; is&nbsp; $E_{\it g_s}$&nbsp; in the low-pass region.&nbsp; In the case shown here, &nbsp; $g_s(t)$&nbsp; describes a rectangular pulse,&nbsp;s but any other energy-limited pulse can be used for&nbsp; $g_s(t)$.&nbsp;
-The graph shows the signal space representation of the 8&ndash;PSK for the band-pass signal (top) and for the equivalent low-pass signal (bottom):
+The graph shows the&nbsp; 8&ndash;PSK&nbsp; signal space representation
-*It can be seen from this that the two representations differ only in the basis functions used, with&nbsp; $\varphi_1(t)$&nbsp; representing different functions in the upper and lower graphs.
+#for the band-pass signal&nbsp; $s_{ \it i}(t)$&nbsp; (top),&nbsp; and
-*In the low-pass representation,&nbsp; $\varphi_2(t) = {\rm j} \cdot \varphi_1(t)$ holds.
+#for the equivalent low-pass signal&nbsp; $s_{\rm TP \it i}(t)$&nbsp; (bottom):
-''Notes:''
+Notes:
 *The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces|"Signals, Basis Functions and Vector Spaces"]].
-* For abbreviation, use the energy&nbsp; $E = 1/2 \cdot A^2 \cdot T$.
+* For abbreviation,&nbsp;  use the energy&nbsp; $E = 1/2 \cdot A^2 \cdot T$.
 * In contrast to the theory section and&nbsp; [[Aufgaben:Exercise_4.2:_AM/PM_Oscillations|"Exercise 4.2"]],&nbsp; here the indexing variable&nbsp; $i$&nbsp; can take the values &nbsp;$0, \ \text{...} \, ,M-1$.&nbsp;
@@ Line 55: / Line 57: @@
 $s_{\rm 52} \ = \ $ { -0.72821--0.68579 } $\ \cdot \sqrt{E}$
-{By which basis functions can the low&ndash;pass signals&nbsp; $s_{\rm TP \it i}(t)$&nbsp; be represented? By
+{By which basis functions can the low&ndash;pass signals&nbsp; $s_{\rm TP \it i}(t)$&nbsp; be represented?&nbsp; By
 |type="[]"}
 + one complex basis function&nbsp; $\xi_1(t)$,

@@ Line 12: / Line 12: @@
 :$$\varphi_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - \sqrt{{2}/{T}} \cdot \sin(2\pi f_{\rm T}t )\hspace{0.05cm}$$
-wie folgt dargestellt werden kann&nbsp; $(i = 0, \ \text{...} \ , 7)$:
+which can be represented as follows&nbsp; $(i = 0, \ \text{...} \ , 7)$:
 :$$s_i(t)= s_{i1} \cdot \varphi_1(t) + s_{i2} \cdot \varphi_2(t) \hspace{0.05cm}.$$
-Die äquivalente Tiefpassdarstellung der Signale&nbsp; $s_i(t)$&nbsp; lautet entsprechend dem Abschnitt&nbsp; [[Modulation_Methods/Quadratur%E2%80%93Amplitudenmodulation#Systembeschreibung_durch_das_.C3.A4quivalente_Tiefpass.E2.80.93Signal| Systembeschreibung durch das äquivalente Tiefpass–Signal]]&nbsp; des Buches "Modulationsverfahren":
+The equivalent low-pass representation of the signals&nbsp; $s_i(t)$&nbsp; is according to the section&nbsp; [[Modulation_Methods/Quadrature_Amplitude_Modulation#System_description_using_the_equivalent_low-pass_signal|"System description using the equivalent low-pass signal"]]&nbsp; of the book "Modulation methods":
 :$$s_{{\rm TP}i}(t)= a_{i} \cdot g_s(t) \hspace{0.05cm}, \hspace{0.2cm}a_{i} = a_{{\rm I}i} + {\rm j} \cdot a_{{\rm Q}i}
   \hspace{0.05cm}, \hspace{0.2cm}i = 0,\text{...} \hspace{0.1cm} , 7 \hspace{0.05cm},$$
-wobei&nbsp; $a_i$&nbsp; komplexe dimensionslose Koeffizienten sind und die Energie des Sendegrundimpulses&nbsp; $g_s(t)$&nbsp; im Tiefpassbereich&nbsp; $E_{\it g_s}$&nbsp; beträgt. Im hier dargestellten Fall beschreibt&nbsp; $g_s(t)$&nbsp; einen Rechteckimpuls, doch kann für&nbsp; $g_s(t)$&nbsp; auch ein jeder andere energiebegrenzte Impuls verwendet werden.
+where&nbsp; $a_i$&nbsp; are complex dimensionless coefficients and the energy of the basic transmission pulse&nbsp; $g_s(t)$&nbsp; is&nbsp; $E_{\it g_s}$&nbsp; in the low-pass region. In the case shown here, &nbsp; $g_s(t)$&nbsp; describes a rectangular pulse, but any other energy-limited pulse can be used for&nbsp; $g_s(t)$.&nbsp;
-Die Grafik zeigt die Signalraumdarstellung der 8&ndash;PSK für das Bandpass&ndash;Signal (oben) sowie für das äquivalente Tiefpass&ndash;Signal (unten):
+The graph shows the signal space representation of the 8&ndash;PSK für das Bandpass&ndash;Signal (oben) sowie für das äquivalente Tiefpass&ndash;Signal (unten):
 *Man erkennt daraus, dass sich die beiden Darstellungen nur duch die verwendeten Basisfunktionen unterscheiden, wobei&nbsp; $\varphi_1(t)$&nbsp; in der oberen und der unteren Grafik für unterschiedliche Funktionen steht.
 *In der Tiefpassdarstellung gilt&nbsp; $\varphi_2(t) = {\rm j} \cdot \varphi_1(t)$.

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@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Digitalsignalübertragung/Signale, Basisfunktionen und Vektorräume}}
+{{quiz-Header|Buchseite=Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces}}
-[[File:P_ID1998__Dig_Z_4_2.png|right|frame|Signalraumpunkte bei 8-PSK]]
+[[File:P_ID1998__Dig_Z_4_2.png|right|frame|Signal space points at 8-PSK]]
-Die&nbsp; $M = 8$&nbsp; möglichen Sendesignale bei 8&ndash;PSK lauten mit&nbsp; $i = 0, \ \text{...} \ , 7$&nbsp; im Bereich&nbsp; $0 &#8804; t < T$:
+The&nbsp; $M = 8$&nbsp; possible transmitted signals at 8&ndash;PSK are with&nbsp; $i = 0, \ \text{...} \ , 7$&nbsp; in the range&nbsp; $0 &#8804; t < T$:
 :$$s_i(t)=  A \cdot \cos(2\pi f_{\rm T}t + i \cdot {\pi}/{4}) \hspace{0.05cm}.$$
-Außerhalb der Symboldauer&nbsp; $T$&nbsp; sind die Signale&nbsp; $s_i(t)$&nbsp; alle Null.
+Outside the symbol duration&nbsp; $T$,&nbsp; the signals&nbsp; $s_i(t)$&nbsp; are all zero.
-In der&nbsp; [[Aufgaben:4.2_AM/PM-Schwingungen| Aufgabe 4.2]]&nbsp; wurde gezeigt, dass diese Signalmenge durch die Basisfunktionen
+In &nbsp; [[Aufgaben:Exercise_4.2:_AM/PM_Oscillations|"Exercise 4.2"]]&nbsp; it was shown that this signal set is given by the basis functions
 :$$\varphi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{{2}/{T}} \cdot \cos(2\pi f_{\rm T}t )\hspace{0.05cm},$$
 :$$\varphi_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - \sqrt{{2}/{T}} \cdot \sin(2\pi f_{\rm T}t )\hspace{0.05cm}$$
@@ Line 31: / Line 31: @@
 ''Hinweise:''
-*Die Aufgabe gehört zum  Kapitel&nbsp;  [[Digitalsignal%C3%BCbertragung/Signale,_Basisfunktionen_und_Vektorr%C3%A4ume| Signale, Basisfunktionen und Vektorräume]].
+*Die Aufgabe gehört zum  Kapitel&nbsp;  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces|"Signals, Basis Functions and Vector Spaces"]].
 * Verwenden Sie zur Abkürzung die Energie&nbsp; $E = 1/2 \cdot A^2 \cdot T$.

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−	[[Category:Digital Signal Transmission: Exercises\|^4.1 ~~Basisfunktionen~~ & ~~Vektorräume~~^]]	+	[[Category:Digital Signal Transmission: Exercises\|^4.1 Basis Functions & Vector Spaces^]]

← Older revision	Revision as of 11:50, 8 June 2022
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