Aufgaben:Exercise 4.15Z: MSK Basic Pulse and MSK Spectrum - Revision history

Guenter at 09:50, 12 April 2022

2022-04-12T09:50:45Z

Reed at 16:54, 21 March 2022

2022-03-21T16:54:59Z

Reed at 16:54, 21 March 2022

2022-03-21T16:54:39Z

Reed: /* Solution */

2022-03-21T16:53:41Z

Solution

Reed at 16:47, 21 March 2022

2022-03-21T16:47:47Z

Reed at 16:47, 21 March 2022

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Reed at 16:46, 21 March 2022

2022-03-21T16:46:53Z

Reed at 16:46, 21 March 2022

2022-03-21T16:46:37Z

Reed at 16:38, 21 March 2022

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Reed at 16:34, 21 March 2022

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 }}
-[[File:P_ID1744__Mod_Z_4_14.png|right|frame|MSK fundamental pulse and its spectrum]]
+[[File:P_ID1744__Mod_Z_4_14.png|right|frame|MSK basic pulse and its spectrum]]
 The fundamental pulse that is always required to &nbsp;[[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK|realize MSK as Offset–QPSK]] &nbsp; has the form shown in the graph above:
 :$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} | t | \le T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$

@@ Line 23: / Line 23: @@
 ''Hints:''
 *This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Nonlinear_Digital_Modulation|Nonlinear Digital Modulation]].
-*Particular reference is made tot he page&nbsp; [[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK|Realizing MSK as Offset–QPSK]].
+*Particular reference is made to the page&nbsp; [[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK|Realizing MSK as Offset–QPSK]].
 *The result obtained here is also used in &nbsp;[[Aufgaben:Exercise_4.15:_MSK_Compared_with_BPSK_and_QPSK|Exercise 4.15]]&nbsp;.

@@ Line 54: / Line 54: @@
-'''(2)'''&nbsp; The spectral function of a rectangular pulse of height&nbsp; $g_0$&nbsp; und duration&nbsp;$ 2T$&nbsp; is:
+'''(2)'''&nbsp; The spectral function of a rectangular pulse of height&nbsp; $g_0$&nbsp; and duration&nbsp;$ 2T$&nbsp; is:
 :$$R(f) = g_0 \cdot 2 T \cdot {\rm si} ( \pi f \cdot 2T )\hspace{0.2cm}{\rm mit}\hspace{0.2cm}{\rm si} (x) = \sin(x)/x \hspace{0.3cm}
   \Rightarrow \hspace{0.3cm}R(f = 0) \hspace{0.15cm}\underline {= 2} \cdot g_0 \cdot T\hspace{0.05cm}.$$

@@ Line 51: / Line 51: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Die Periodendauer des Cosinussignals muss&nbsp; $T_0 = 4T$&nbsp; sein.&nbsp; Damit ist die Frequenz $f_0 = 1/T_0\hspace{0.15cm}\underline {= 0.25} · 1/T$.
+'''(1)'''&nbsp; The period of the cosine signal must be&nbsp; $T_0 = 4T$&nbsp;.&nbsp; Thus, the frequency is $f_0 = 1/T_0\hspace{0.15cm}\underline {= 0.25} · 1/T$.
-'''(2)'''&nbsp; Die Spektralfunktion eines Rechteckimpulses der Höhe&nbsp; $g_0$&nbsp; und der Dauer&nbsp;$ 2T$&nbsp; lautet:
+'''(2)'''&nbsp; The spectral function of a rectangular pulse of height&nbsp; $g_0$&nbsp; und duration&nbsp;$ 2T$&nbsp; is:
 :$$R(f) = g_0 \cdot 2 T \cdot {\rm si} ( \pi f \cdot 2T )\hspace{0.2cm}{\rm mit}\hspace{0.2cm}{\rm si} (x) = \sin(x)/x \hspace{0.3cm}
   \Rightarrow \hspace{0.3cm}R(f = 0) \hspace{0.15cm}\underline {= 2} \cdot g_0 \cdot T\hspace{0.05cm}.$$
-'''(3)'''&nbsp; Aus&nbsp; $g(t) = c(t) · r(t)$&nbsp; folgt nach dem Faltungssatz: &nbsp; $ G(f) = C(f) \star R(f)\hspace{0.05cm}.$
+'''(3)'''&nbsp; When&nbsp; $g(t) = c(t) · r(t)$&nbsp;, it follows from the convolution theorem that: &nbsp; $ G(f) = C(f) \star R(f)\hspace{0.05cm}.$
-*Die Spektralfunktion&nbsp; $C(f)$&nbsp; besteht aus zwei Diracfunktionen bei&nbsp; $± f_0$, jeweils mit dem Gewicht&nbsp; $1/2$.&nbsp; Daraus folgt:
+*The spectral function&nbsp; $C(f)$&nbsp; consists of two  Dirac functions at&nbsp; $± f_0$, each with weight&nbsp; $1/2$.&nbsp; From this follows:
 :$$ G(f)  =  2 \cdot g_0 \cdot T \cdot \big [ 1/2 \cdot \delta (f - f_0 ) + 1/2 \cdot \delta (f + f_0 )\big ] \star {\rm si} ( 2 \pi f T )=  g_0 \cdot T \cdot \big [ {\rm si} ( 2 \pi T \cdot (f - f_0 ) ) + {\rm si} ( 2 \pi T \cdot (f + f_0 ) ) \big ] \hspace{0.05cm}.$$
-*Mit dem Ergebnis&nbsp; $f_0 = 1/(4T)$&nbsp; der Teilaufgabe&nbsp; '''(1)'''&nbsp; gilt weiter:
+*Using the result&nbsp; $f_0 = 1/(4T)$&nbsp; from question&nbsp; '''(1)'''&nbsp;, it further holds that:
 :$$G(f) = g_0 \cdot T \cdot \big [ {\rm si} ( 2 \pi f T - \pi / 2 ) + {\rm si} ( 2 \pi f T + \pi / 2) \big ]$$
 :$$\Rightarrow \hspace{0.3cm} G(f = 0)  =  g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T \hspace{-0.02cm}\cdot\hspace{-0.02cm} \big [ {\rm si} ( - \pi/2 ) + {\rm si} ( +\pi/2 ) \big ] = 2 \cdot g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T \hspace{-0.02cm}\cdot\hspace{-0.02cm} {\rm si} ( \pi/2 ) = 2 \hspace{-0.02cm}\cdot\hspace{-0.02cm} g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T \hspace{-0.02cm}\cdot\hspace{-0.02cm} \frac {{\rm sin}({\pi}/{2}) } { {\pi}/{2} } ={4}/{\pi} \hspace{-0.02cm}\cdot\hspace{-0.02cm} g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T \hspace{0.15cm}\underline {\approx 1.273} \hspace{-0.02cm}\cdot\hspace{-0.02cm} g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T .$$
-'''(4)'''&nbsp; Schreibt man die&nbsp; $\rm si$–Funktion aus, so erhält man mit&nbsp; $\sin (α ± π/2) = ± \cos(α)$:
+'''(4)'''&nbsp; By writing out the&nbsp; $\rm si$–function, with &nbsp; $\sin (α ± π/2) = ± \cos(α)$, one gets:
 :$$G(f)  =  g_0 \cdot T \cdot \left [ \frac{{\rm sin} ( 2 \pi f T - \pi / 2 )}{2 \pi f T - \pi / 2 } + \frac{{\rm sin} ( 2 \pi f T + \pi / 2 )}{2 \pi f T + \pi / 2 } \right ]=  g_0 \cdot T \cdot \frac {2}{\pi}\cdot\left [ \frac{-{\rm cos} ( 2 \pi f T )}{4 f T - 1 } + \frac{{\rm cos} ( 2 \pi f T )}{4 f T + 1 } \right ]$$
 :$$\Rightarrow \hspace{0.3cm} G(f)  =  g_0 \cdot T \cdot \frac {2}{\pi}\cdot \frac{(1+4 f T ) \cdot {\rm cos} ( 2 \pi f T )+ (1-4 f T ) \cdot {\rm cos} ( 2 \pi f T )}{1 - (4 f T)^2 } =   \frac {4}{\pi}\cdot g_0 \cdot T \cdot \frac{ {\rm cos} ( 2 \pi f T )}{1 - (4 f T)^2 }\hspace{0.05cm}.$$
-*Die Nullstellen von&nbsp; $G(f)$&nbsp; werden allein durch die Cosinusfunktion im Zähler bestimmt und würden bei den Frequenzen&nbsp; $f · T = 0.25,\ 0.75,\ 1.25,$&nbsp; ... liegen.
+*The zeroes of&nbsp; $G(f)$&nbsp; are exclusively determined by the cosine function in the numerator, and are found at the frequencies&nbsp; $f · T = 0.25,\ 0.75,\ 1.25,$&nbsp; ...
-*Allerdings wird die erste Nullstelle bei&nbsp; $f · T = 0.25$&nbsp; durch die gleichzeitige Nullstelle des Nenners aufgehoben.&nbsp; Deshalb gilt:
+*However, the first zero at&nbsp; $f · T = 0.25$&nbsp; is cancelled out by the simultaneously occuring zero in the denominator.&nbsp; Therefore:
 :$$f_1 \hspace{0.15cm}\underline {= 0.75} \cdot 1/T \hspace{0.05cm}.$$

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1744__Mod_Z_4_14.png|right|frame|MSK fundamental pulse&nbsp; and&nbsp; its spectrum]]
+[[File:P_ID1744__Mod_Z_4_14.png|right|frame|MSK fundamental pulse and its spectrum]]
 The fundamental pulse that is always required to &nbsp;[[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK|realize MSK as Offset–QPSK]] &nbsp; has the form shown in the graph above:
 :$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} | t | \le T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$

@@ Line 5: / Line 5: @@
 [[File:P_ID1744__Mod_Z_4_14.png|right|frame|MSK&ndash;Grundimpuls&nbsp; und&nbsp; &ndash;Spektrum]]
 The fundamental pulse that is always required to &nbsp;[[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK|realize MSK as Offset–QPSK]] &nbsp; has the form shown in the graph above:
-:$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} | t | \le T \hspace{0.05cm}, \\ {\rm sonst}\hspace{0.05cm}. \\ \end{array}$$
+:$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} | t | \le T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$
 The spectral function &nbsp;$G(f)$ is drawn below, that is, the &nbsp;[[Signal_Representation/Fourier_Transform_and_its_Inverse#The_first_Fourier_integral|Fourier transform]]&nbsp; of &nbsp;$g(t)$.

@@ Line 4: / Line 4: @@
 [[File:P_ID1744__Mod_Z_4_14.png|right|frame|MSK&ndash;Grundimpuls&nbsp; und&nbsp; &ndash;Spektrum]]
-Der zur Realisierung der &nbsp;[[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK|Realizing MSK as Offset–QPSK]] &nbsp; stets erforderliche Grundimpuls hat die in der Grafik oben dargestellte Form:
+The fundamental pulse that is always required to &nbsp;[[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK|realize MSK as Offset–QPSK]] &nbsp; has the form shown in the graph above:
 :$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} | t | \le T \hspace{0.05cm}, \\ {\rm sonst}\hspace{0.05cm}. \\ \end{array}$$
-Darunter gezeichnet ist die Spektralfunktion &nbsp;$G(f)$, also die &nbsp;[[Signal_Representation/Fourier_Transform_and_its_Inverse#The_first_Fourier_integral|Fourier transform]]&nbsp; of &nbsp;$g(t)$.
+The spectral function &nbsp;$G(f)$ is drawn below, that is, the &nbsp;[[Signal_Representation/Fourier_Transform_and_its_Inverse#The_first_Fourier_integral|Fourier transform]]&nbsp; of &nbsp;$g(t)$.
-Die dazugehörige Gleichung soll in dieser Aufgabe ermittelt werden, wobei zu berücksichtigen ist:
+The corresponding equation is to be determined in this task, by considering:
 :$$g(t) = c(t) \cdot r(t)\hspace{0.05cm}.$$
-Hierbei sind folgende  Abkürzungen verwendet:
+The following abbreviations are used here:
-* $c(t)$&nbsp; ist eine Cosinusschwingung mit Amplitude &nbsp;$1$&nbsp; und&nbsp; (noch zu bestimmender)&nbsp; Frequenz &nbsp;$f_0$.
+* $c(t)$&nbsp; is a cosine oscillation with amplitude &nbsp;$1$&nbsp; and&nbsp; frequency &nbsp;$f_0$ (yet to be determined).
-* $r(t)$&nbsp; ist eine Rechteckfunktion mit der Amplitude &nbsp;$g_0$&nbsp; und der Dauer &nbsp;$2T$.
+* $r(t)$&nbsp; is a square wave function with amplitude&nbsp;$g_0$&nbsp; and duration&nbsp;$2T$.
@@ Line 23: / Line 23: @@
 ''Hints:''
 *This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Nonlinear_Digital_Modulation|Nonlinear Digital Modulation]].
-*Bezug genommen wird insbesondere auf den Abschnitt&nbsp; [[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK|Realizing MSK as Offset–QPSK]].
+*Particular reference is made tot he page&nbsp; [[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK|Realizing MSK as Offset–QPSK]].
-*Das hier gewonnene Ergebnis wird auch in der &nbsp;[[Aufgaben:Exercise_4.15:_MSK_Compared_with_BPSK_and_QPSK|Exercise 4.15]]&nbsp;.
+*The result obtained here is also used in &nbsp;[[Aufgaben:Exercise_4.15:_MSK_Compared_with_BPSK_and_QPSK|Exercise 4.15]]&nbsp;.
@@ Line 31: / Line 31: @@
 <quiz display=simple>
-{Wie ist die Frequenz &nbsp;$f_0$&nbsp; der Cosinusschwingung &nbsp;$c(t)$&nbsp; zu wählen, damit &nbsp;$g(t) = c(t) · r(t)$&nbsp; gilt?
+{How should one choose the frequency&nbsp;$f_0$&nbsp; of the cosine oscillation &nbsp;$c(t)$&nbsp; so that  &nbsp;$g(t) = c(t) · r(t)$&nbsp;?
 |type="{}"}
 $f_0 \ = \ $  { 0.25 3% } $\ \cdot 1/T$
-{Wie lautet das Spektrum &nbsp;$R(f)$&nbsp; der Rechteckfunktion &nbsp;$r(t)$?&nbsp; Welcher Spektralwert tritt bei &nbsp;$f = 0$&nbsp; auf?
+{What is the spectrum &nbsp;$R(f)$&nbsp; of the rectangular function &nbsp;$r(t)$?&nbsp; What spectral value occurs when&nbsp;$f = 0$&nbsp;?
 |type="{}"}
 $R(f=0) \ = \ $  { 2 3%  } $\ \cdot g_0 \cdot T$
-{Berechnen Sie das Spektrum &nbsp;$G(f)$&nbsp; des MSK&ndash;Impuses &nbsp;$g(t)$, insbesondere den Spektralwert bei &nbsp;$f = 0$.
+{Calculate the spectrun &nbsp;$G(f)$&nbsp; of the MSK pulse &nbsp;$g(t)$, particularly the spectral value at &nbsp;$f = 0$.
 |type="{}"}
 $G(f=0) \ = \ $ { 1.273 3% } $\ \cdot g_0 \cdot T$
-{Fassen Sie das Ergebnis der Teilaufgabe&nbsp; '''(3)'''&nbsp; in einem Term zusammen.&nbsp; Bei welcher Frequenz &nbsp;$f_1$&nbsp; besitzt &nbsp;$G(f)$&nbsp; seine erste Nullstelle?
+{Summarize the result of question &nbsp; '''(3)'''&nbsp; in one term. At what frequency &nbsp;$f_1$&nbsp; does &nbsp;$G(f)$&nbsp; have its first zero?
 |type="{}"}
 $f_1 \ = \ $ { 0.75 3% } $\ \cdot 1/T$