Difference between revisions of "Aufgaben:Exercise 4.2Z: Eight-level Phase Shift Keying"

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{{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 $M = 8$ möglichen Sendesignale bei 8&ndash;PSK lauten mit $i = 0, \ \text{...} \ , 7$ im Bereich $0 &#8804; t < T$:
+
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; $0 &#8804; t < T$:
 
:$$s_i(t)=  A \cdot \cos(2\pi f_{\rm T}t + i \cdot {\pi}/{4}) \hspace{0.05cm}.$$
 
:$$s_i(t)=  A \cdot \cos(2\pi f_{\rm T}t + i \cdot {\pi}/{4}) \hspace{0.05cm}.$$
  
Außerhalb der Symboldauer $T$ sind die Signale $s_i(t)$ alle gleich $0$.
+
Outside the symbol duration&nbsp; $T$,&nbsp; the signals&nbsp; $s_i(t)$&nbsp; are all zero.
  
In der [[Aufgaben:4.2_AM/PM-Schwingungen| Aufgabe 4.2]] 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
 
:$$\varphi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{{2}/{T}} \cdot \cos(2\pi f_{\rm T}t )\hspace{0.05cm},$$
 
:$$\varphi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{{2}/{T}} \cdot \cos(2\pi f_{\rm T}t )\hspace{0.05cm},$$
 
:$$\varphi_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - \sqrt{{2}/{T}} \cdot \sin(2\pi f_{\rm T}t )\hspace{0.05cm}$$
 
:$$\varphi_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - \sqrt{{2}/{T}} \cdot \sin(2\pi f_{\rm T}t )\hspace{0.05cm}$$
  
wie folgt dargestellt werden kann ($i = 0, \ \text{...} \ , 7$):
+
which can be represented as follows&nbsp; $(i = 0, \ \text{...} \ , 7)$:
 
:$$s_i(t)= s_{i1} \cdot \varphi_1(t) + s_{i2} \cdot \varphi_2(t) \hspace{0.05cm}.$$
 
:$$s_i(t)= s_{i1} \cdot \varphi_1(t) + s_{i2} \cdot \varphi_2(t) \hspace{0.05cm}.$$
  
Die äquivalente Tiefpassdarstellung der Signale $s_i(t)$ lautet entsprechend dem Abschnitt [[Modulationsverfahren/Quadratur%E2%80%93Amplitudenmodulation#Systembeschreibung_durch_das_.C3.A4quivalente_Tiefpass.E2.80.93Signal| Systembeschreibung durch das äquivalente Tiefpass–Signal]] des Buches &bdquo;Modulationsverfahren&rdquo;:
+
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}
 
:$$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},$$
 
  \hspace{0.05cm}, \hspace{0.2cm}i = 0,\text{...} \hspace{0.1cm} , 7 \hspace{0.05cm},$$
  
wobei $a_i$ komplexe dimensionslose Koeffizienten sind und die Energie des Sendegrundimpulses $g_s(t)$ im Tiefpassbereich $E_{\it gs}$ beträgt. Im hier dargestellten Fall beschreibt $g_s(t)$ einen Rechteckimpuls, doch kann für $g_s(t)$ auch ein jeder andere energiebegrenzte Impuls verwendet werden.
+
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;
  
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&nbsp; 8&ndash;PSK&nbsp; signal space representation
*Man erkennt daraus, dass sich die beiden Darstellungen nur duch die verwendeten Basisfunktionen unterscheiden, wobei $\varphi_1(t)$ in der oberen und der unteren Grafik für unterschiedliche Funktionen steht.
+
#for the band-pass signal&nbsp; $s_{ \it i}(t)$&nbsp; (top),&nbsp; and
*In der Tiefpassdarstellung gilt $\varphi_2(t) = {\rm j} \cdot \varphi_1(t)$.
+
#for the equivalent low-pass signal&nbsp; $s_{\rm TP \it i}(t)$&nbsp; (bottom):
  
  
''Hinweise:''
+
It can be seen from this that the two representations differ only in the basis functions used,&nbsp; with&nbsp; $\varphi_1(t)$&nbsp; representing different functions in the upper and lower graphs. In the low-pass representation,&nbsp; $\varphi_2(t) = {\rm j} \cdot \varphi_1(t)$&nbsp; holds.
*Die Aufgabe gehört zum  Kapitel  [[Digitalsignal%C3%BCbertragung/Signale,_Basisfunktionen_und_Vektorr%C3%A4ume| Signale, Basisfunktionen und Vektorräume]].
 
*Sollte die Eingabe des Zahlenwertes &bdquo;0&rdquo; erforderlich sein, so geben Sie bitte &bdquo;0.&rdquo; ein.
 
* Verwenden Sie zur Abkürzung die Energie $E = 1/2 \cdot A^2 \cdot T$.
 
* Im Gegensatz zum Theorieteil und zur [[Aufgaben:4.2_AM/PM-Schwingungen| Aufgabe 4.2]] kann hier die Laufvariable $i$ die Werte $0, \ \text{...} \, ,M&ndash;1$ annehmen.  
 
  
  
  
===Fragebogen===
+
 
 +
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,&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;
 +
 
 +
*The color-coded signal space points in the graph&nbsp; (blue,&nbsp; red,&nbsp; green)&nbsp; are referred to in the questionnaire.&nbsp; These look for the signals&nbsp; $s_0(t)$,&nbsp; $s_2(t)$&nbsp; and&nbsp; $s_5(t)$. 
 +
 
 +
 
 +
 
 +
===Questions===
 
<quiz display=simple>
 
<quiz display=simple>
{Wie lauten die Koeffizienten des Signals $s_0(t)$?
+
{What are the coefficients of the signal&nbsp; $s_0(t)$?
 
|type="{}"}
 
|type="{}"}
 
$s_{\rm 01} \ = \ $ { 1 3% } $\ \cdot \sqrt{E}$
 
$s_{\rm 01} \ = \ $ { 1 3% } $\ \cdot \sqrt{E}$
 
$s_{\rm 02} \ = \ $ { 0. } $\ \cdot \sqrt{E}$
 
$s_{\rm 02} \ = \ $ { 0. } $\ \cdot \sqrt{E}$
  
{Wie lauten die Koeffizienten des Signals $s_2(t)$?
+
{What are the coefficients of the signal&nbsp; $s_2(t)$?
 
|type="{}"}
 
|type="{}"}
 
$s_{\rm 21} \ = \ $ { 0. } $\ \cdot \sqrt{E}$
 
$s_{\rm 21} \ = \ $ { 0. } $\ \cdot \sqrt{E}$
 
$s_{\rm 22} \ = \ $ { 1 3% } $\ \cdot \sqrt{E}$
 
$s_{\rm 22} \ = \ $ { 1 3% } $\ \cdot \sqrt{E}$
  
{Wie lauten die Koeffizienten des Signals $s_5(t)$?
+
{What are the coefficients of the signal&nbsp; $s_5(t)$?
 
|type="{}"}
 
|type="{}"}
 
$s_{\rm 51} \ = \ $ { -0.72821--0.68579 } $\ \cdot \sqrt{E}$
 
$s_{\rm 51} \ = \ $ { -0.72821--0.68579 } $\ \cdot \sqrt{E}$
 
$s_{\rm 52} \ = \ $ { -0.72821--0.68579 } $\ \cdot \sqrt{E}$
 
$s_{\rm 52} \ = \ $ { -0.72821--0.68579 } $\ \cdot \sqrt{E}$
  
{Durch welche Basisfunktionen sind die Tiefpass&ndash;Signale $s_{\rm TP \it i}(t)$ darstellbar? Durch
+
{By which basis functions can the low&ndash;pass signals&nbsp; $s_{\rm TP \it i}(t)$&nbsp; be represented?&nbsp; By
 
|type="[]"}
 
|type="[]"}
+ eine komplexe Basisfunktion $\xi_1(t)$,
+
+ one complex basis function&nbsp; $\xi_1(t)$,
- zwei komplexe Basisfunktionen $\xi_1(t)$ und $\xi_2(t)$,
+
- two complex basis functions&nbsp; $\xi_1(t)$&nbsp; and &nbsp;$\xi_2(t)$,
+ zwei reelle Funktionen $\varphi_1(t)$ und $\psi_1(t)$.
+
+ two real functions&nbsp; $\varphi_1(t)$&nbsp; and &nbsp;$\psi_1(t)$.
  
{Wie lauten im vorliegenden Fall die reellen Basisfunktionen?
+
{What are the real basis functions in the present case?
 
|type="[]"}
 
|type="[]"}
 
- $\varphi_1(t) = g_s(t)$,
 
- $\varphi_1(t) = g_s(t)$,
+ $\varphi_1(t) = g_s(t)/\sqrt{E_{\rm gs}}$,
+
+ $\varphi_1(t) = g_s(t)/\sqrt{E_{\rm g_s}}$,
 
+ $\psi_1(t) = \varphi_1(t)$,
 
+ $\psi_1(t) = \varphi_1(t)$,
 
- $\psi_1(t) = {\rm j} \cdot \varphi_1(t)$.
 
- $\psi_1(t) = {\rm j} \cdot \varphi_1(t)$.
  
{Es gelte $s_{\rm TP0}(t) = \sqrt{E}$. Was trifft zu:
+
{Let&nbsp; $s_{\rm TP0}(t) = \sqrt{E}$. Which is true?
 
|type="()"}
 
|type="()"}
- Die Energie $E$ bezieht sich auf das Tiefpass&ndash;Signal.
+
- The energy&nbsp; $E$&nbsp; refers to the low&ndash;pass signal.
+ Die Energie $E$ bezieht sich auf das Bandpass&ndash;Signal.
+
+ The energy&nbsp; $E$&nbsp; refers to the band&ndash;pass signal.
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Das Signal $s_0(t)$ lautet:
+
'''(1)'''&nbsp; The signal&nbsp;  $s_0(t)$&nbsp; is:
 
:$$s_0(t)=  A \cdot \cos(2\pi f_{\rm T}t ) = s_{01} \cdot \varphi_1(t) + s_{02} \cdot \varphi_2(t) \hspace{0.05cm}.$$
 
:$$s_0(t)=  A \cdot \cos(2\pi f_{\rm T}t ) = s_{01} \cdot \varphi_1(t) + s_{02} \cdot \varphi_2(t) \hspace{0.05cm}.$$
  
Da dieses Signal keinen Sinusteil aufweist, ist $s_{\rm 02} \hspace{0.15cm}\underline {= 0}$. Weiter gilt mit der angegebenen Abkürzung:
+
*Since this signal has no sinusoidal part,&nbsp; $s_{\rm 02} \hspace{0.15cm}\underline {= 0}$.  
 +
*Further,&nbsp; with the given abbreviation:
 
:$$A = s_{01} \cdot \sqrt{{2}/{T}}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
 
:$$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}.$$
 
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; Das Signal $s_2(t)$ lautet mit $i = 2$ (beachten Sie, dass die zweite Basisfunktion minus&ndash;sinusförmig ist):
+
 
 +
'''(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}
 
:$$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}.$$
 
\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; Entsprechend den Musterlösungen zu den Teilaufgaben (1) und (2) gilt nun:
+
 
 +
'''(3)'''&nbsp; According to the solutions to subtasks&nbsp; '''(1)'''&nbsp; and&nbsp; '''(2)''',&nbsp; the following is true:
 
:$$s_{51}= s_{52}= - \sqrt{E/2} \hspace{0.05cm} \hspace{0.15cm}\underline { = -0.707 \cdot E^{\hspace{0.05cm}0.5}}$$
 
:$$s_{51}= s_{52}= - \sqrt{E/2} \hspace{0.05cm} \hspace{0.15cm}\underline { = -0.707 \cdot E^{\hspace{0.05cm}0.5}}$$
:$$\Rightarrow \hspace{0.3cm} s_{5}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  - {A}/{ \sqrt{2}} \cdot \cos(2\pi f_{\rm T}t ) - {A}/{ \sqrt{2}} \cdot \sin(2\pi f_{\rm T}t )=A \cdot \cos(2\pi f_{\rm T}t + \phi_5)\hspace{0.2cm}{\rm mit}\hspace{0.2cm}\phi_5 = -0.75 \cdot \pi
+
:$$\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
\hspace{0.2cm}{\rm bzw.}\hspace{0.2cm}\phi_5 = 1.25 \cdot \pi
+
\hspace{0.2cm}{\rm and}\hspace{0.2cm}\phi_5 = 1.25 \cdot \pi
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
'''(4)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 1 und 3</u>. Dabei gilt folgender Zusammenhang: &nbsp; $\xi_1 (t) = \varphi_1 (t) + {\rm j} \cdot \psi_1 (t)\hspace{0.05cm}.$
 
  
'''(5)'''&nbsp; Richtig sind die <u>Alternativen 2 und 3</u>:  
+
'''(4)'''&nbsp; <u>Solutions 1 and 3</u>&nbsp; are correct.&nbsp; The following relation holds: &nbsp;
*Die Basisfunktion muss energienormiert sein.
+
:$$\xi_1 (t) = \varphi_1 (t) + {\rm j} \cdot \psi_1 (t)\hspace{0.05cm}.$$
* $\psi_1(t)$ ist wie $\varphi_1(t)$ eine reelle, nicht etwa eine imaginäre Funktion:
+
 
 +
 
 +
'''(5)'''&nbsp; Correct are the&nbsp; <u>alternatives 2 and 3</u>:  
 +
*The basis function must be energy-normalized.
 +
 
 +
* Like&nbsp; $\psi_1(t)$,&nbsp; the basis&nbsp; $\varphi_1(t)$&nbsp; is a real function,&nbsp; not an imaginary one:
 
:$$\varphi_1 (t) = \psi_1 (t) =
 
:$$\varphi_1 (t) = \psi_1 (t) =
 
\left\{ \begin{array}{c} 1/\sqrt{T} \\
 
\left\{ \begin{array}{c} 1/\sqrt{T} \\
 
  0  \end{array} \right.\quad
 
  0  \end{array} \right.\quad
 
\begin{array}{*{1}c} 0 \le t < T \hspace{0.05cm},
 
\begin{array}{*{1}c} 0 \le t < T \hspace{0.05cm},
\\  {\rm sonst}\hspace{0.05cm}. \\ \end{array}$$
+
\\  {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$
 +
 
  
'''(6)'''&nbsp; Aus dem Tiefpass&ndash;Signal $s_{\rm TP0}(t)$ kann man auch das Bandpass&ndash;Signal $s_0(t)$ berechnen. Im Bereich $0 &#8804; t &#8804; T$ gilt mit dem Ergebnis aus (5):
+
'''(6)'''&nbsp; From the low&ndash;pass signal&nbsp; $s_{\rm TP0}(t)$,&nbsp; one can also calculate the band-pass signal&nbsp; $s_0(t)$.  
:$$s_0(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Re}[s_{{\rm TP}0}(t) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi f_{\rm T}t} ]
+
*In the range&nbsp; $0 &#8804; t &#8804; T$,&nbsp; the result from&nbsp; '''(5)'''&nbsp; gives the same result as in subtask&nbsp; '''(1)''':
  = {\rm Re}[\sqrt{E} \cdot \frac{1}{\sqrt{T}} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi f_{\rm T}t} ]=  \sqrt{E/T} \cdot \cos(2\pi f_{\rm T}t )
+
:$$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},$$
 
\hspace{0.05cm},$$
  
also das gleiche Ergebnis wie in der Teilaufgabe (1). Daraus folgt: Die Energie $E$ bezieht sich auch bei Betrachtung im äquivalenten Tiefpass&ndash;Bereich auf das Bandpass&ndash;Signal.
+
*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.
  
Entsprechend gilt für das mit blauem Punkt markierte Signal $s_2(t)$ im interessierenden Bereich:
+
*Accordingly,&nbsp; for the signal&nbsp; $s_2(t)$&nbsp; marked with blue dot in the region of interest:
:$$s_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Re}[\hspace{0.05cm}{\rm j} \cdot \sqrt{E/T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi f_{\rm T}t} ]
+
:$$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}[\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) ]
+
  = {\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]
 
  = - \sqrt{E/T} \cdot \sin(2\pi f_{\rm T}t)
 
  = - \sqrt{E/T} \cdot \sin(2\pi f_{\rm T}t)
 
\hspace{0.05cm}.$$
 
\hspace{0.05cm}.$$
  
Schließlich kann für das (grüne) Signal $s_5(t)$ im Bereich $0 &#8804; t < T$ geschrieben werden:
+
*Finally,&nbsp; for the&nbsp; (green)&nbsp; signal&nbsp; $s_5(t)$&nbsp; in the range $0 &#8804; t < T$&nbsp; can be written:
:$$s_5(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Re}[\frac{-1 - {\rm j}}{\sqrt{2}} \cdot \sqrt{{E}/{T}} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi f_{\rm T}t} ] = \text{...}  
+
:$$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)
 
  =  - \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}.$$
 
\hspace{0.05cm}.$$
  
Auch diese Ergebnisse stimmen mit denen der Teilaufgaben (2) bzw. (3) überein. Zutreffend ist also der <u>Lösungsvorschlag 2</u>.
+
*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}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Digitalsignalübertragung|^4.1 Basisfunktionen & Vektorräume^]]
+
[[Category:Digital Signal Transmission: Exercises|^4.1 Basis Functions & Vector Spaces^]]

Latest revision as of 09:41, 12 August 2022

Signal space points at 8-PSK

The  $M = 8$  possible transmitted signals at  "8–PSK"  are with   $i = 0, \ \text{...} \ , 7$   in the range  $0 ≤ t < T$:

$$s_i(t)= A \cdot \cos(2\pi f_{\rm T}t + i \cdot {\pi}/{4}) \hspace{0.05cm}.$$

Outside the symbol duration  $T$,  the signals  $s_i(t)$  are all zero.

In   "Exercise 4.2"  it was shown that this signal set is given by the basis functions

$$\varphi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{{2}/{T}} \cdot \cos(2\pi f_{\rm T}t )\hspace{0.05cm},$$
$$\varphi_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - \sqrt{{2}/{T}} \cdot \sin(2\pi f_{\rm T}t )\hspace{0.05cm}$$

which can be represented as follows  $(i = 0, \ \text{...} \ , 7)$:

$$s_i(t)= s_{i1} \cdot \varphi_1(t) + s_{i2} \cdot \varphi_2(t) \hspace{0.05cm}.$$

The equivalent low-pass representation of the signals  $s_i(t)$  is according to the section  "System description using the equivalent low-pass signal"  of the book  "Modulation methods":

$$s_{{\rm TP}i}(t)= a_{i} \cdot g_s(t) \hspace{0.05cm}, \hspace{0.2cm}a_{i} = a_{{\rm I}i} + {\rm j} \cdot a_{{\rm Q}i} \hspace{0.05cm}, \hspace{0.2cm}i = 0,\text{...} \hspace{0.1cm} , 7 \hspace{0.05cm},$$

where  $a_i$  are dimensionless complex coefficients and the energy of the basic transmission pulse  $g_s(t)$  is  $E_{\it g_s}$  in the low-pass region.  In the case shown here,   $g_s(t)$  describes a rectangular pulse,  but any other energy-limited pulse can be used for  $g_s(t)$. 

The graph shows the  8–PSK  signal space representation

  1. for the band-pass signal  $s_{ \it i}(t)$  (top),  and
  2. for the equivalent low-pass signal  $s_{\rm TP \it i}(t)$  (bottom):


It can be seen from this that the two representations differ only in the basis functions used,  with  $\varphi_1(t)$  representing different functions in the upper and lower graphs. In the low-pass representation,  $\varphi_2(t) = {\rm j} \cdot \varphi_1(t)$  holds.



Notes:

  • For abbreviation,  use the energy  $E = 1/2 \cdot A^2 \cdot T$.
  • In contrast to the theory section and  "Exercise 4.2",  here the indexing variable  $i$  can take the values  $0, \ \text{...} \, ,M-1$. 
  • The color-coded signal space points in the graph  (blue,  red,  green)  are referred to in the questionnaire.  These look for the signals  $s_0(t)$,  $s_2(t)$  and  $s_5(t)$.


Questions

1

What are the coefficients of the signal  $s_0(t)$?

$s_{\rm 01} \ = \ $

$\ \cdot \sqrt{E}$
$s_{\rm 02} \ = \ $

$\ \cdot \sqrt{E}$

2

What are the coefficients of the signal  $s_2(t)$?

$s_{\rm 21} \ = \ $

$\ \cdot \sqrt{E}$
$s_{\rm 22} \ = \ $

$\ \cdot \sqrt{E}$

3

What are the coefficients of the signal  $s_5(t)$?

$s_{\rm 51} \ = \ $

$\ \cdot \sqrt{E}$
$s_{\rm 52} \ = \ $

$\ \cdot \sqrt{E}$

4

By which basis functions can the low–pass signals  $s_{\rm TP \it i}(t)$  be represented?  By

one complex basis function  $\xi_1(t)$,
two complex basis functions  $\xi_1(t)$  and  $\xi_2(t)$,
two real functions  $\varphi_1(t)$  and  $\psi_1(t)$.

5

What are the real basis functions in the present case?

$\varphi_1(t) = g_s(t)$,
$\varphi_1(t) = g_s(t)/\sqrt{E_{\rm g_s}}$,
$\psi_1(t) = \varphi_1(t)$,
$\psi_1(t) = {\rm j} \cdot \varphi_1(t)$.

6

Let  $s_{\rm TP0}(t) = \sqrt{E}$. Which is true?

The energy  $E$  refers to the low–pass signal.
The energy  $E$  refers to the band–pass signal.


Solution

(1)  The signal  $s_0(t)$  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,  $s_{\rm 02} \hspace{0.15cm}\underline {= 0}$.
  • Further,  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)  The signal  $s_2(t)$  is with  $i = 2$  (note that the second basis function is minus–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)  According to the solutions to subtasks  (1)  and  (2),  the following is true:

$$s_{51}= s_{52}= - \sqrt{E/2} \hspace{0.05cm} \hspace{0.15cm}\underline { = -0.707 \cdot E^{\hspace{0.05cm}0.5}}$$
$$\Rightarrow \hspace{0.3cm} s_{5}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - {A}/{ \sqrt{2}} \cdot \cos(2\pi f_{\rm T}t ) - {A}/{ \sqrt{2}} \cdot \sin(2\pi f_{\rm T}t )=A \cdot \cos(2\pi f_{\rm T}t + \phi_5)\hspace{0.2cm}{\rm with}\hspace{0.2cm}\phi_5 = -0.75 \cdot \pi \hspace{0.2cm}{\rm and}\hspace{0.2cm}\phi_5 = 1.25 \cdot \pi \hspace{0.05cm}.$$


(4)  Solutions 1 and 3  are correct.  The following relation holds:  

$$\xi_1 (t) = \varphi_1 (t) + {\rm j} \cdot \psi_1 (t)\hspace{0.05cm}.$$


(5)  Correct are the  alternatives 2 and 3:

  • The basis function must be energy-normalized.
  • Like  $\psi_1(t)$,  the basis  $\varphi_1(t)$  is a real function,  not an imaginary one:
$$\varphi_1 (t) = \psi_1 (t) = \left\{ \begin{array}{c} 1/\sqrt{T} \\ 0 \end{array} \right.\quad \begin{array}{*{1}c} 0 \le t < T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$


(6)  From the low–pass signal  $s_{\rm TP0}(t)$,  one can also calculate the band-pass signal  $s_0(t)$.

  • In the range  $0 ≤ t ≤ T$,  the result from  (5)  gives the same result as in subtask  (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–pass signal even when considered in the equivalent low–pass region.
  • Accordingly,  for the signal  $s_2(t)$  marked with blue dot in the region of interest:
$$s_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Re}\big[\hspace{0.05cm}{\rm j} \cdot \sqrt{E/T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}2\pi f_{\rm T}t} \big] = {\rm Re}\big[\hspace{0.05cm}{\rm j} \cdot \sqrt{E/T} \cdot \cos(2\pi f_{\rm T}t)- \sqrt{E/T} \cdot \sin(2\pi f_{\rm T}t) \big] = - \sqrt{E/T} \cdot \sin(2\pi f_{\rm T}t) \hspace{0.05cm}.$$
  • Finally,  for the  (green)  signal  $s_5(t)$  in the range $0 ≤ t < T$  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)  resp.  (3).  Therefore,  solution 2  is correct.