Aufgaben:Exercise 1.5: Cosine-Square Spectrum - Revision history

Guenter at 12:24, 2 May 2022

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Guenter at 15:30, 2 February 2022

2022-02-02T15:30:31Z

Guenter: Guenter moved page Aufgaben:Aufgabe 1.5: Cosinus-Quadrat-Spektrum to Aufgaben:Exercise 1.5: Cosine-Square Spectrum

2022-02-02T15:28:37Z

Guenter moved page Aufgabe 1.5: Cosinus-Quadrat-Spektrum to Exercise 1.5: Cosine-Square Spectrum

Javier: Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

2021-03-23T12:52:19Z

Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

Javier: Text replacement - "[[Digitalsignalübertragung" to "[[Digital_Signal_Transmission"

2020-07-09T12:44:09Z

Text replacement - "[[Digitalsignalübertragung" to "[[Digital_Signal_Transmission"

Guenter at 17:02, 30 January 2019

2019-01-30T17:02:47Z

@@ Line 9: / Line 9: @@
 Accordingly,&nbsp; the associated pulse &nbsp;$g(t)$&nbsp; has zero crossings at multiples of &nbsp;$T$,&nbsp; where &nbsp;$T$&nbsp; remains to be determined.&nbsp; The inverse Fourier  transform of &nbsp;$G(f)$&nbsp; yields the equation for the time course:
 :$$g( t )= g_0 \cdot   \frac{\cos(\pi \cdot t/T)}{1 - (2 \cdot
-t/T)^2}\cdot {\rm sinc}( {t}/{T})\hspace{0.05cm}.$$
+t/T)^2}\cdot {\rm sinc}( {t}/{T})\hspace{0.5cm} \text{with}\hspace{0.5cm} {\rm sinc}(x)=\sin(\pi x)/(\pi x)\hspace{0.05cm}.$$
 The questions for this exercise refer to the following properties:
 *The spectral function &nbsp;$G(f)$&nbsp; is a special case of the cosine rolloff spectrum,&nbsp; which is point symmetric about the Nyquist frequency &nbsp;$f_{\rm Nyq}$.&nbsp;

@@ Line 116: / Line 116: @@
 *A second solution method leads to the expression:
-:$$\frac{g( t )}{g_0}  = {\rm si}(\pi \cdot \frac{t}{T}) \cdot
+:$$\frac{g( t )}{g_0}  = {\rm sinc}( \frac{t}{T}) \cdot
-\frac {\pi}{4} \cdot \big [ {\rm si}(\pi \cdot (t/T + 1/2)) +
+\frac {\pi}{4} \cdot \big [ {\rm sinc}( t/T + 1/2) +
-{\rm si}(\pi \cdot (t/T - 1/2))\big] \hspace{0.05cm}.$$
+{\rm sinc}(t/T - 1/2)\big] \hspace{0.05cm}.$$
 *The second bracket expression can be transformed as follows:
 :$$\frac {\pi}{4} \cdot \bigg [ \hspace{0.1cm}... \hspace{0.1cm}
@@ Line 132: / Line 132: @@
 \cdot t/T)^2}\hspace{0.05cm}.$$
 *It follows that both expressions are actually equal.&nbsp; Thus,&nbsp; for time&nbsp; $t = T/2$,&nbsp; the following is still true:
-:$$\frac{g( t = T/2)}{g_0}  = {\rm si}(  \frac{\pi}{2}) \cdot \frac
+:$$\frac{g( t = T/2)}{g_0}  = {\rm sinc}(  0.5) \cdot \frac
-{\pi}{4} \cdot \left [ {\rm si}(\pi ) + {\rm si}(0)\right]= \frac
+{\pi}{4} \cdot \big [ {\rm sinc}(1 ) + {\rm sinc}(0)\big]= \frac
 {2}{\pi}\cdot \frac {\pi}{4} = 0.5 \hspace{0.05cm}.$$
 {{ML-Fuß}}

@@ Line 49: / Line 49: @@
 + $g(t)$&nbsp; satisfies the first Nyquist criterion because of the &nbsp;$\rm sinc$&ndash;term.
 - $g(t)$&nbsp; has further zero crossings at &nbsp;$\pm 0.5T, &nbsp;\pm 1.5T, &nbsp;\pm 2.5 T, \text{...}$
-+ The &nbsp;$\cos^{2}$ spectrum also satisfies the second Nyquist criterion.
++ The &nbsp;$\cos^{2}$&ndash;spectrum also satisfies the second Nyquist criterion.
@@ Line 62: / Line 62: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The upper corner frequency can be read from the diagram: &nbsp; $f_{2} \underline{= 2 \ \rm MHz}$. Since the spectrum is not constant in any range, $f_{1} \underline {= 0}$.
+'''(1)'''&nbsp; The upper corner frequency can be read from the diagram: &nbsp; $f_{2} \underline{= 2 \ \rm MHz}$.&nbsp; Since the spectrum is not constant in any range, $f_{1} \underline {= 0}$.
@@ Line 75: / Line 75: @@
-'''(4)'''&nbsp; <u>Statements 1 and 3</u> are correct:
+'''(4)'''&nbsp; <u>Statements 1 and 3</u>&nbsp; are correct:
-*The first statement is correct: &nbsp; The function $si(π · t/T)$ leads to zero crossings at $\nu T (\nu \neq 0)$.
+*The first statement is correct: &nbsp; The function&nbsp; ${\rm sinc}(t/T)$&nbsp; leads to zero crossings at&nbsp; $\nu T (\nu \neq 0)$.
-*The last statement is also true: &nbsp;Because of $g(t) = 0$ for $t =\pm 1.5T, \pm 2.5T, \pm 3.5T, ...$ the second Nyquist criterion is also fulfilled.
+*The last statement is also true: &nbsp;Because of&nbsp; $g(t) = 0$&nbsp; for&nbsp; $t =\pm 1.5T, \pm 2.5T, \pm 3.5T, ...$&nbsp; the second Nyquist criterion is also fulfilled.
-*On the other hand, the middle statement is false, since $g(t = T/2) \neq 0$.
+*On the other hand,&nbsp; the middle statement is false,&nbsp; since $g(t = T/2) \neq 0$.
 *The condition for the second Nyquist criterion is in the frequency domain:
 :$$G_{\rm Per}(f) = \sum_{k = -\infty}^{+\infty} \frac {G \left ( f -
-\frac{k}{T} \right)}{\cos(\pi \cdot f \cdot T - k \cdot \pi)}=
+{k}/{T} \right)}{\cos(\pi \cdot f \cdot T - k \cdot \pi)}=
 {\rm const.}$$
-*The condition is indeed fulfilled for the cos$^{2}$ spectrum, as can be shown after a longer calculation. We restrict ourselves here to the frequency range $| f · T | \leq 1$ and set $g_{0} \cdot  T = 1$ for simplicity:
+*The condition is indeed fulfilled for the&nbsp; cos$^{2}$&ndash;spectrum,&nbsp; as can be shown after a longer calculation.
 :$$G_{\rm Per}(f) =  \frac {\cos^2 \left [\pi/2 \cdot   ( f_{\rm Nyq}
 - f) \cdot T \right ]}{\cos \left [\pi \cdot   ( f_{\rm Nyq} - f)
@@ Line 98: / Line 98: @@
 - f) \cdot T \right ]} +1- \frac {1}{\cos \left [\pi \cdot ( f_{\rm Nyq}
 + f) \cdot T \right ]}\right ]\hspace{0.05cm}.$$
-* Because of $\cos \left [ \pi \cdot ( f_{\rm Nyq} \pm f) \cdot T \right] = \cos
+* Because of&nbsp; $\cos \left [ \pi \cdot ( f_{\rm Nyq} \pm f) \cdot T \right] = \cos
 \left (  {\pi}/{2} \pm \pi  f  T \right) =  \sin \left ( \pm
 \pi  f  T \right)\text{:}$
@@ Line 105: / Line 105: @@
-'''(5)'''&nbsp; For $t = T/2$, the given equation yields an indeterminate value (0 divided by 0), which can be determined using l'Hospital's rule.
+'''(5)'''&nbsp; For&nbsp; $t = T/2$,&nbsp; the given equation yields an indeterminate value&nbsp; ("0 divided by 0"),&nbsp; which can be determined using l'Hospital's rule.
-*To do this, form the derivatives of the numerator and denominator and insert the desired time $t = T/2$ into the result:
+*To do this,&nbsp; form the derivatives of the numerator and denominator and insert the desired time&nbsp; $t = T/2$&nbsp; into the result:
-:$$\frac{g( t = T/2)}{g_0}   = \ {{\rm si}(\pi \cdot \frac{t}{T})
+:$$\frac{g( t = T/2)}{g_0}   = \ {{\rm sinc}( \frac{t}{T})
 \cdot \frac{{\rm d}/{\rm d}t \left [ \cos(\pi \cdot
 t/T)\right]}{{\rm d}/{\rm d}t\left [ 1 - (2 \cdot t/T)^2\right]}}
-\bigg |_{t = T/2}  = \ {{\rm si}(\pi \cdot \frac{t}{T}) \cdot \frac{- \pi/T \cdot
+\bigg |_{t = T/2}  = \ {{\rm sinc}( \frac{t}{T}) \cdot \frac{- \pi/T \cdot
   \sin(\pi \cdot t/T)}{-2 \cdot (2\cdot t/T) \cdot (2/T)}} \bigg |_{t = T/2} = \frac {2}{\pi}\cdot
 \frac {\pi}{4}\hspace{0.1cm}\underline { = 0.5}\hspace{0.05cm}.$$
@@ Line 131: / Line 131: @@
 \cdot t/T)(1- 2 \cdot t/T)}= \frac{\cos(\pi \cdot t/T)}{1 - (2
 \cdot t/T)^2}\hspace{0.05cm}.$$
-*It follows that both expressions are actually equal. Thus, for time $t = T/2$, the following is still true:
+*It follows that both expressions are actually equal.&nbsp; Thus,&nbsp; for time&nbsp; $t = T/2$,&nbsp; the following is still true:
 :$$\frac{g( t = T/2)}{g_0}  = {\rm si}(  \frac{\pi}{2}) \cdot \frac
 {\pi}{4} \cdot \left [ {\rm si}(\pi ) + {\rm si}(0)\right]= \frac

@@ Line 5: / Line 5: @@
 [[File:P_ID1282__Dig_A_1_5.png|right|frame|Cosine-square Nyquist spectrum]]
-The spectrum &nbsp;$G(f)$&nbsp; with &nbsp;$\cos^{2}$–shaped course is considered according to the sketch. This satisfies the first Nyquist criterion:
+The spectrum &nbsp;$G(f)$&nbsp; with &nbsp;$\cos^{2}$–shaped course is considered according to the sketch.&nbsp; This satisfies the first Nyquist criterion:
 :$$\sum_{k = -\infty}^{+\infty} G(f -{k}/{T} ) =  {\rm const.}$$
-Accordingly, the associated pulse &nbsp;$g(t)$&nbsp; has zero crossings at multiples of &nbsp;$T$, where &nbsp;$T$&nbsp; remains to be determined. Fourier inverse transformation of &nbsp;$G(f)$&nbsp; yields the equation for the time course:
+Accordingly,&nbsp; the associated pulse &nbsp;$g(t)$&nbsp; has zero crossings at multiples of &nbsp;$T$,&nbsp; where &nbsp;$T$&nbsp; remains to be determined.&nbsp; The inverse Fourier  transform of &nbsp;$G(f)$&nbsp; yields the equation for the time course:
 :$$g( t )= g_0 \cdot   \frac{\cos(\pi \cdot t/T)}{1 - (2 \cdot
-t/T)^2}\cdot {\rm si}(\pi \cdot {t}/{T})\hspace{0.05cm}.$$
+t/T)^2}\cdot {\rm sinc}( {t}/{T})\hspace{0.05cm}.$$
 The questions for this exercise refer to the following properties:
-*The spectral function &nbsp;$G(f)$&nbsp; is a special case of the cosine rolloff spectrum, which is point symmetric about the Nyquist frequency &nbsp;$f_{\rm Nyq}$.&nbsp;
+*The spectral function &nbsp;$G(f)$&nbsp; is a special case of the cosine rolloff spectrum,&nbsp; which is point symmetric about the Nyquist frequency &nbsp;$f_{\rm Nyq}$.&nbsp;
 *The cosine rolloff spectrum is completely characterized by the corner frequencies &nbsp;$f_{1}$&nbsp; and &nbsp;$f_{2}$.&nbsp;
-* For &nbsp;$| f | < f_{1}$,&nbsp; &nbsp;$G(f) = g_{0} \cdot T = \rm const.$, while the spectrum for &nbsp;$| f | > f_{2}$&nbsp; has no components.
+* For &nbsp;$| f | < f_{1}$,&nbsp; &nbsp;$G(f) = g_{0} \cdot T = \rm const.$,&nbsp; while the spectrum for &nbsp;$| f | > f_{2}$&nbsp; has no components.&nbsp;
 *The relation between the Nyquist frequency and the corner frequencies is:
 :$$f_{\rm Nyq}=  \frac{f_1 +f_2 }
 {2 }\hspace{0.05cm}.$$
-*The transition steepness is characterized by the so-called rolloff factor:
+*The edge steepness is characterized by the so-called rolloff factor:
 :$$r = \frac{f_2 -f_1 }
 {f_2 +f_1 }\hspace{0.2cm}(0 \le r \le 1) \hspace{0.05cm}.$$
@@ Line 23: / Line 23: @@
+Note:&nbsp; The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems|"Properties of Nyquist Systems"]].
@@ Line 50: / Line 47: @@
 {Which of the following statements is true?
 |type="[]"}
-+ $g(t)$&nbsp; satisfies the first Nyquist criterion because of the &nbsp;$\rm si$ term.
++ $g(t)$&nbsp; satisfies the first Nyquist criterion because of the &nbsp;$\rm sinc$&ndash;term.
 - $g(t)$&nbsp; has further zero crossings at &nbsp;$\pm 0.5T, &nbsp;\pm 1.5T, &nbsp;\pm 2.5 T, \text{...}$
 + The &nbsp;$\cos^{2}$ spectrum also satisfies the second Nyquist criterion.
-{What is the (normalized) value of the pulse at time &nbsp;$t = T/2$?
+{What is the&nbsp; (normalized)&nbsp; value of the pulse at time &nbsp;$t = T/2$?
 |type="{}"}
 $g(t = T/2)/g_{0} \ = \ $ { 0.5 3% }

← Older revision		Revision as of 15:30, 2 February 2022
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−	[[Category:Digital Signal Transmission: Exercises\|^1.3 ~~Eigenschaften von Nyquistsystemen~~^]]	+	[[Category:Digital Signal Transmission: Exercises\|^1.3 Nyquist System Properties^]]

← Older revision	Revision as of 15:28, 2 February 2022
(No difference)

← Older revision		Revision as of 12:52, 23 March 2021
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−	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^1.3 Eigenschaften von Nyquistsystemen^]]	+	[[Category:Digital Signal Transmission: Exercises\|^1.3 Eigenschaften von Nyquistsystemen^]]

← Older revision		Revision as of 12:44, 9 July 2020
Line 26:		Line 26:

	''Hinweis:''		''Hinweis:''
−	*Die Aufgabe gehört zum Kapitel  [[~~Digitalsignalübertragung~~/Eigenschaften_von_Nyquistsystemen\|Eigenschaften von Nyquistsystemen]].	+	*Die Aufgabe gehört zum Kapitel  [[Digital_Signal_Transmission/Eigenschaften_von_Nyquistsystemen\|Eigenschaften von Nyquistsystemen]].

@@ Line 65: / Line 65: @@
 ===Musterlösung===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Die obere Eckfrequenz kann aus der Grafik abgelesen werden: $f_{2} \underline{= 2 \ \rm MHz}$. Da das Spektrum in keinem Bereich konstant ist, gilt: $f_{1} \underline {= 0}$.
+'''(1)'''&nbsp; Die obere Eckfrequenz kann aus der Grafik abgelesen werden: &nbsp; $f_{2} \underline{= 2 \ \rm MHz}$. Da das Spektrum in keinem Bereich konstant ist, gilt $f_{1} \underline {= 0}$.
@@ Line 71: / Line 72: @@
 :$$f_{\rm Nyq}  = \  \frac{f_1 +f_2 }
 {2 }\hspace{0.1cm}\underline { = 1\,{\rm MHz}}\hspace{0.05cm},\hspace{0.5cm} r = \ \frac{f_2 -f_1 } {f_2 +f_1 }\hspace{0.1cm}\underline { = 1 }\hspace{0.05cm}.$$
 '''(3)'''&nbsp; Der Abstand äquidistanter Nulldurchgänge hängt direkt mit der Nyquistfrequenz zusammen:
-:$$f_{\rm Nyq}= \frac{1}{2T} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} T= \frac{1}{2f_{\rm Nyq}}\hspace{0.1cm}\underline { = 0.5\,{\rm \mu s}}\hspace{0.05cm}.$$
+:$$f_{\rm Nyq}= \frac{1}{2T} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} T= \frac{1}{2f_{\rm Nyq}}\hspace{0.1cm}\underline { = 0.5\,{\rm &micro; s}}\hspace{0.05cm}.$$
 '''(4)'''&nbsp; Richtig sind die <u>Aussagen 1 und 3</u>:
-*Die erste Aussage ist richtig: Die Funktion $si(π · t/T)$ führt zu Nulldurchgängen bei $\nu T (\nu \neq 0)$.
+*Die erste Aussage ist richtig: &nbsp; Die Funktion $si(π · t/T)$ führt zu Nulldurchgängen bei $\nu T (\nu \neq 0)$.
-*Auch die letzte Aussage trifft zu: Wegen $g(t) = 0$ für $t =\pm 1.5T, \pm 2.5T, \pm 3.5T, ...$ wird auch das zweite Nyquistkriterium erfüllt.
+*Auch die letzte Aussage trifft zu: &nbsp;Wegen $g(t) = 0$ für $t =\pm 1.5T, \pm 2.5T, \pm 3.5T, ...$ wird auch das zweite Nyquistkriterium erfüllt.
 *Falsch ist dagegen die mittlere Aussage, da $g(t = T/2) \neq 0$ ist.
+*Die Bedingung für das zweite Nyquistkriterium lautet im Frequenzbereich:
 :$$G_{\rm Per}(f) = \sum_{k = -\infty}^{+\infty} \frac {G \left ( f -
 \frac{k}{T} \right)}{\cos(\pi \cdot f \cdot T - k \cdot \pi)}=
 {\rm const.}$$
-Die Bedingung ist beim cos$^{2}$–Spektrum tatsächlich erfüllt, wie man nach längerer Rechnung zeigen kann. Wir beschränken uns hier auf den Frequenzbereich $| f · T | \leq 1$ und setzen vereinfachend $g_{0} \cdot  T = 1$:
+*Die Bedingung ist beim cos$^{2}$–Spektrum tatsächlich erfüllt, wie man nach längerer Rechnung zeigen kann. Wir beschränken uns hier auf den Frequenzbereich $| f · T | \leq 1$ und setzen vereinfachend $g_{0} \cdot  T = 1$:
 :$$G_{\rm Per}(f) =  \frac {\cos^2 \left [\pi/2 \cdot   ( f_{\rm Nyq}
 - f) \cdot T \right ]}{\cos \left [\pi \cdot   ( f_{\rm Nyq} - f)
@@ Line 91: / Line 93: @@
 + f) \cdot T \right ]}{\cos \left [\pi \cdot   ( f_{\rm Nyq} + f)
 \cdot T \right ]}\hspace{0.05cm}.$$
-Weiter gilt:
+*Weiter gilt:
 :$$\frac {\cos^2 (x)}{\cos(2x)} = {1}/{2} \cdot \frac
 {1+\cos(2x)}{\cos(2x)}= {1}/{2} \cdot \left [1+ \frac
@@ Line 99: / Line 101: @@
 - f) \cdot T \right ]} +1- \frac {1}{\cos \left [\pi \cdot ( f_{\rm Nyq}
 + f) \cdot T \right ]}\right ]\hspace{0.05cm}.$$
-:$$\text{Wegen }\cos \left [ \pi \cdot ( f_{\rm Nyq} \pm f) \cdot T \right] = \cos
+* Wegen $\cos \left [ \pi \cdot ( f_{\rm Nyq} \pm f) \cdot T \right] = \cos
 \left (  {\pi}/{2} \pm \pi  f  T \right) =  \sin \left ( \pm
-\pi  f  T \right)\text{:}$$
+\pi  f  T \right)\text{:}$
 :$$\Rightarrow \hspace{0.3cm} G_{\rm Per}(f) =   2 - \frac {1}{\sin
 (\pi  f  T)} + \frac {1}{\sin (\pi  f  T)} = 2  = {\rm const}\hspace{0.05cm}.$$
 '''(5)'''&nbsp; Für $t = T/2$ liefert die angegebene Gleichung einen unbestimmten Wert (0 geteilt durch 0), der mit der Regel von l'Hospital ermittelt werden kann. Dazu bildet man die Ableitungen von Zähler und Nenner und setzt in das Ergebnis den gewünschten Zeitpunkt $t = T/2$ ein:
 :$$\frac{g( t = T/2)}{g_0}   = \ {{\rm si}(\pi \cdot \frac{t}{T})
@@ Line 114: / Line 118: @@
 \frac {\pi}{4}\hspace{0.1cm}\underline { = 0.5}\hspace{0.05cm}.$$
-Ein zweiter Lösungsweg führt zu der Darstellung:
+*Ein zweiter Lösungsweg führt zu der Darstellung:
 :$$\frac{g( t )}{g_0}  = {\rm si}(\pi \cdot \frac{t}{T}) \cdot
 \frac {\pi}{4} \cdot \big [ {\rm si}(\pi \cdot (t/T + 1/2)) +
 {\rm si}(\pi \cdot (t/T - 1/2))\big] \hspace{0.05cm}.$$
-Der zweite Klammerausdruck kann wie folgt umgeformt werden:
+*Der zweite Klammerausdruck kann wie folgt umgeformt werden:
 :$$\frac {\pi}{4} \cdot \bigg [ \hspace{0.1cm}... \hspace{0.1cm}
 \bigg ]   = \  \frac {\pi}{4} \cdot \left [ \frac {{\rm sin}(\pi
@@ Line 130: / Line 134: @@
 \cdot t/T)(1- 2 \cdot t/T)}= \frac{\cos(\pi \cdot t/T)}{1 - (2
 \cdot t/T)^2}\hspace{0.05cm}.$$
-Daraus folgt, dass beide Ausdrücke tatsächlich gleich sind. Für den Zeitpunkt $t = T/2$ gilt somit weiterhin:
+*Daraus folgt, dass beide Ausdrücke tatsächlich gleich sind. Für den Zeitpunkt $t = T/2$ gilt somit weiterhin:
 :$$\frac{g( t = T/2)}{g_0}  = {\rm si}(  \frac{\pi}{2}) \cdot \frac
 {\pi}{4} \cdot \left [ {\rm si}(\pi ) + {\rm si}(0)\right]= \frac