Aufgaben:Exercise 1.4Z: Complex Nyquist Spectrum - Revision history

Guenter at 16:31, 1 May 2022

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Guenter at 16:08, 1 May 2022

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Guenter at 15:19, 1 May 2022

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Guenter at 15:30, 2 February 2022

2022-02-02T15:30:03Z

Guenter: Guenter moved page Aufgaben:Aufgabe 1.4Z: Komplexes Nyquistspektrum to Aufgaben:Exercise 1.4Z: Complex Nyquist Spectrum

2022-02-02T15:27:32Z

Guenter moved page Aufgabe 1.4Z: Komplexes Nyquistspektrum to Exercise 1.4Z: Complex Nyquist Spectrum

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

2021-03-23T12:49:05Z

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

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

2020-07-09T12:45:07Z

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

Guenter at 16:47, 30 January 2019

2019-01-30T16:47:46Z

@@ Line 154: / Line 154: @@
   V}} \hspace{0.05cm}.$$
-[[File:P_ID1283__Dig_Z_1_4g.png|right|frame|Asymmetric Nyquist pulse]]
+[[File:P_ID1283__Dig_Z_1_4g.png|right|frame|Asymmetric Nyquist pulse &nbsp; $g(t)= g_{\rm R}(t) +  g_{\rm I}(t) $]]
 The diagram shows the change of the time function due to the imaginary part&nbsp; (green time course):
 *The result is now an asymmetric function curve&nbsp; $g(t)$,&nbsp; shown in blue.

@@ Line 85: / Line 85: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The Nyquist frequency specifies the symmetry point of the rolloff. The following holds true:
+'''(1)'''&nbsp; The Nyquist frequency specifies the symmetry point of the rolloff.&nbsp; The following holds true:
 :$$f_{\rm Nyq}=  \frac{f_1 +f_2 }
 {2 }= \frac{3\, {\rm kHz} + 7\, {\rm kHz}} {2 } \hspace{0.1cm}\underline { = 5\, {\rm kHz}}
 \hspace{0.05cm}.$$
-'''(2)'''&nbsp; The rolloff factor is also determined by the two corner frequencies $f_{1}$ and $f_{2}$:
+'''(2)'''&nbsp; The rolloff factor is also determined by the two corner frequencies&nbsp; $f_{1}$&nbsp; and&nbsp; $f_{2}$:
 :$$r = \frac{f_2 -f_1 }
 {f_2 +f_1 } = \frac{7\, {\rm kHz} - 3\, {\rm kHz}} {7\, {\rm kHz}
 + 3\, {\rm kHz} }\hspace{0.1cm}\underline { = 0.4 }\hspace{0.05cm}.$$
-'''(3)'''&nbsp; For a pulse with real low–pass spectrum, the maximum is always at $t = 0$ and it holds:
+'''(3)'''&nbsp; For a pulse with real low–pass spectrum,&nbsp; the maximum is always at&nbsp; $t = 0$&nbsp; and it holds:
 :$$g_0 = g(t=0) =  \int_{-\infty}^{+\infty}G(f) \,{\rm d} f
   = A \cdot 2  f_{\rm Nyq} = 10^{-4 }\,\frac{\rm V}{\rm Hz}\cdot 2 \cdot 5 \cdot10^{3} \,{\rm
   Hz}\hspace{0.1cm}\underline { = 1\,{\rm V}}\hspace{0.05cm}.$$
-'''(4)'''&nbsp; For the Nyquist pulse, the equidistant zero crossings occur at the interval $T = 1/(2f_{\rm Nyq}) = 100 \, \rm &micro; s$. From this one obtains directly:
+'''(4)'''&nbsp; For the Nyquist pulse,&nbsp; the equidistant zero crossings occur at the interval&nbsp; $T = 1/(2f_{\rm Nyq}) = 100 \, \rm &micro; s$.&nbsp; From this one obtains directly:
 :$$g(t= 100\,{\rm &micro; s}) = \ \hspace{0.1cm}\underline { g(T) = 0,}$$
 :$$g(t= 200\,{\rm &micro; s}) = \  \hspace{0.1cm}\underline {g(2T) = 0}
   \hspace{0.05cm}.$$
-This result also follows from the given equation with $r = 0.4$:
+This result also follows from the given equation with&nbsp; $r = 0.4$:
-:$$g ( t )= g_0 \cdot {\rm si} \left ( {\pi \cdot
+:$$g ( t )= g_0 \cdot {\rm sinc} \left ( {
-t}/{T}\right)\cdot {\rm si} \left ( {\pi \cdot 0.4 \cdot
+t}/{T}\right)\cdot {\rm sinc} \left ( {0.4 \cdot
 t}/{T}\right) \hspace{0.05cm}.$$
 Responsible for satisfying the first Nyquist condition is the first term.
-'''(5)'''&nbsp; According to the equation given in '''(4)''':
+'''(5)'''&nbsp; According to the equation given in&nbsp; '''(4)''':
-:$$g(t= 250\,{\rm &micro; s})= g_0 \cdot {\rm si}  ( {2.5 \cdot \pi
+:$$g(t= 250\,{\rm &micro; s})= g_0 \cdot {\rm sinc}  ( {1 })\cdot {\rm sinc}  ( 2.5 )\hspace{0.1cm}\underline { = 0} \hspace{0.05cm}.$$
-})\cdot {\rm si}  ( \pi )\hspace{0.1cm}\underline { = 0} \hspace{0.05cm}.$$
+This zero is due to the second term and does not lie in the Nyquist time grid&nbsp; $\nu T$
-'''(6)'''&nbsp; For the following derivation, let $g(t)= g_{\rm R}(t) +  g_{\rm I}(t) \hspace{0.05cm},$ where $g_{\rm R}(t)$ is the real part and $g_{\rm I}(t)$ is the imaginary part of $G(f)$.
+'''(6)'''&nbsp; For the following derivation,&nbsp; let&nbsp; $g(t)= g_{\rm R}(t) +  g_{\rm I}(t) \hspace{0.05cm},$&nbsp; where&nbsp; $g_{\rm R}(t)$&nbsp; is due to the real part and&nbsp; $g_{\rm I}(t)$&nbsp; is due to the imaginary part of&nbsp; $G(f)$.
-*The first part is calculated exactly as in point '''(4)''':
+*The first part is calculated exactly as in point&nbsp; '''(4)''':
-:$$g_{\rm R} ( t )= g_0 \cdot {\rm si} \left ( {\pi \cdot
+:$$g_{\rm R} ( t )= g_0 \cdot {\rm sinc} \left ( {
-t}/{T}\right)\cdot {\rm si} \left ( {\pi \cdot 0.4 \cdot
+t}/{T}\right)\cdot {\rm sinc} \left ( { 0.4 \cdot
 t}/{T}\right) \hspace{0.05cm}.$$
 *To satisfy the first Nyquist criterion, the following must hold for the imaginary part with $1/T = 10 \, \rm kHz$:
 :$$\sum_{k = -\infty}^{+\infty} {\rm Im}\left[G \left ( f -
 {k}/{T} \right)\right]= 0 \hspace{0.05cm}.$$
-*With the given corner frequencies $f_{1} = 3 \, \rm kHz$ and $f_{2} = 7 \ \rm kHz$, the two triangles are around $\pm 5\, \rm kHz$, so the above equation is satisfied.
+*With the given corner frequencies&nbsp; $f_{1} = 3 \, \rm kHz$&nbsp; and&nbsp; $f_{2} = 7 \ \rm kHz$,&nbsp; the two triangles are around&nbsp; $\pm 5\, \rm kHz$,&nbsp; so the above equation is satisfied.
 *The same is true for $f_{1} = 4.5\, \rm kHz$ and $f_{2} = 5.5 \, \rm kHz$.
-*In contrast, the triangle peaks with $f_{1} = 3\, \rm kHz$ and $f_{2} = 5 \, \rm kHz$ are at $\pm 4 \ \rm kHz$.
+*The same is true for&nbsp; $f_{1} = 4.5\, \rm kHz$&nbsp; and&nbsp; $f_{2} = 5.5 \, \rm kHz$.
 *In this case the triangle functions do not cancel due to the periodic continuation and the Nyquist condition is not fulfilled.
-Therefore the correct solutions are <u>1 and 3</u>.
+Therefore the correct solutions are&nbsp; <u>1 and 3</u>.
-'''(7)'''&nbsp; With the result &nbsp;$g_{\rm R}(2.5T) = 0$&nbsp; from '''(3)''', it follows &nbsp;$g(2.5T) = g_{\rm I}(2.5T)$, where &nbsp;$g_{\rm I}(t)$&nbsp; is the Fourier inverse transformation of  ${\rm j}\cdot \ G_{\rm I}(f)$. It holds:
+'''(7)'''&nbsp; With the result &nbsp;$g_{\rm R}(2.5T) = 0$&nbsp; from&nbsp; '''(3)''',&nbsp; it follows &nbsp;$g(2.5T) = g_{\rm I}(2.5T)$,&nbsp; where &nbsp;$g_{\rm I}(t)$&nbsp; is the inverse Fourier transform of&nbsp;  ${\rm j}\cdot \ G_{\rm I}(f)$.&nbsp; It holds:
 :$${\rm j} \cdot G_{\rm I}(f)  = {\rm j} \cdot\left[ \delta(f + f_{\rm Nyq}) - \delta(f - f_{\rm Nyq})\right] \star D(f) \hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} g_{\rm I}(t)  = 2 \cdot {\rm sin}  ( 2
   \pi\cdot f_{\rm Nyq} \cdot t )\cdot  d(t)\hspace{0.05cm}.$$
-*The sine function enforces the required zero crossings at multiples of $T = 100 \, \rm &micro; s$.
+*The sine function enforces the required zero crossings at multiples of&nbsp; $T = 100 \, \rm &micro; s$.
-*$D(f)$ is a triangular function around $f = 0$ with $D(f = 0) = B$ and one-sided width $f_{0}= f_{2} – f_{\rm Nyq} = f_{\rm Nyq} – f_{1} = 2 \, \rm kHz$.
 *For the associated time function we can thus write according to the specification:
 :$$g_{\rm I}(t )  = 2 \cdot B \cdot f_0 \cdot{\rm sin}  ( 2
-  \pi\cdot f_{\rm Nyq} \cdot t)\cdot {\rm si}^2(\pi\cdot f_{\rm 0} \cdot t) \hspace{0.05cm}.$$
+  \pi\cdot f_{\rm Nyq} \cdot t)\cdot {\rm sinc}^2( f_{\rm 0} \cdot t) \hspace{0.05cm}.$$
 *In particular, for time $t = 250 \, \rm &micro; s$  (green square):
 :$$g(t = 2.5 T)  = g_{\rm I}(t = 2.5 T)  = \ 2 \cdot B \cdot f_0 \cdot{\rm sin}  ( 2.5
-  \pi )\cdot {\rm si}^2(\frac{\pi}{2}) = \  \frac{8}{\pi^2} \cdot 10^{-4 }\,\frac{\rm V}{\rm Hz}\cdot 2 \cdot 10^{3} \,{\rm
+  \pi )\cdot {\rm sinc}^2(0.5)= \  \frac{8}{\pi^2} \cdot B \cdot f_0 = \  \frac{8}{\pi^2} \cdot 10^{-4 }\,\frac{\rm V}{\rm Hz}\cdot 2 \cdot 10^{3} \,{\rm
   Hz}\hspace{0.1cm}\underline {= 0.162\,{\rm
   V}} \hspace{0.05cm}.$$
 [[File:P_ID1283__Dig_Z_1_4g.png|right|frame|Asymmetric Nyquist pulse]]
-The diagram shows the change of the time function due to the imaginary part (green time course):
+The diagram shows the change of the time function due to the imaginary part&nbsp; (green time course):
-*The result is now an asymmetric function curve $g(t)$, shown in blue.
+*The result is now an asymmetric function curve&nbsp; $g(t)$,&nbsp; shown in blue.
-*However, the zero crossings of $g_{\rm R}(t)$ at the distance $T$ remain.
+*However, the zero crossings of&nbsp; $g_{\rm R}(t)$&nbsp; at the distance&nbsp; $T$&nbsp; remain.

@@ Line 5: / Line 5: @@
 [[File:P_ID1279__Dig_Z_1_4.png|right|frame|Complex Nyquist spectrum]]
-Consider a pulse &nbsp;$g(t)$&nbsp; with spectrum &nbsp;$G(f)$&nbsp; according to the sketch. One recognizes from this representation:
+Consider a pulse &nbsp;$g(t)$&nbsp; with spectrum &nbsp;$G(f)$&nbsp; according to the sketch.&nbsp; One recognizes from this representation:
-*The real part of &nbsp;$G(f)$&nbsp; is trapezoidal with the corner frequencies &nbsp;$f_{1} = 3 \, \rm kHz$&nbsp; and &nbsp;$f_{2} = 7 \, \rm kHz$. In the range &nbsp;$|f| < f_{1}$&nbsp; holds:
+*The real part of &nbsp;$G(f)$&nbsp; is trapezoidal with the corner frequencies &nbsp;$f_{1} = 3 \, \rm kHz$&nbsp; and &nbsp;$f_{2} = 7 \, \rm kHz$.&nbsp; In the range &nbsp;$|f| < f_{1}$:
 :$${\rm Re}\big[G(f)\big] = A = 10^{-4} \, \rm V/Hz.$$
-*The imaginary part of &nbsp;$G(f)$&nbsp; is always assumed to be &nbsp;${\rm Im}\big[G(f)\big] =0$&nbsp; for subtasks '''(1)''' to '''(5)'''. <br>In this case &nbsp;$g(t)$&nbsp; is certainly a Nyquist pulse.
+*The imaginary part of &nbsp;$G(f)$&nbsp; is always assumed to be &nbsp;${\rm Im}\big[G(f)\big] =0$ &nbsp; for subtasks&nbsp; '''(1)'''&nbsp; to&nbsp; '''(5)'''.&nbsp;  In this case &nbsp;$g(t)$&nbsp; is certainly a Nyquist pulse.
-*From subtask '''(6)''', the imaginary part &nbsp;${\rm Im}[G(f)]$&nbsp; in the range &nbsp;$f_{1} \leq | f | \leq f_{2}$&nbsp; has a triangular shape with the values &nbsp;$\pm B$&nbsp; at the triangle peaks.
+*From subtask&nbsp; '''(6)''',&nbsp; the imaginary part &nbsp;${\rm Im}[G(f)]$&nbsp; in the range &nbsp;$f_{1} \leq | f | \leq f_{2}$&nbsp; has a triangular shape with the values &nbsp;$\pm B$&nbsp; at the triangle peaks.
@@ Line 22: / Line 22: @@
 \end{array}$$
-In the course of this exercise, reference is made to the following descriptive quantities:
+In the course of this exercise,&nbsp;  reference is made to the following descriptive quantities:
-*The Nyquist frequency gives the symmetry point of the rolloff:
+*The&nbsp;  '''Nyquist frequency'''&nbsp;  gives the symmetry point of the rolloff:
 :$$f_{\rm Nyq}= \frac{1}{2T}= \frac{f_1 +f_2 }
 {2 }\hspace{0.05cm}.$$
-*The rolloff factor is a measure of the transition steepness:
+*The&nbsp;  '''rolloff factor'''&nbsp;  is a measure of the transition steepness:
 :$$r = \frac{f_2 -f_1 }
 {f_2 +f_1 } \hspace{0.05cm}.$$
@@ Line 32: / Line 32: @@
-''Notes:''
+Notes:
-*The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems|Properties of Nyquist Systems]].
+*The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems|"Properties of Nyquist Systems"]].
-*The Fourier inverse transformation &nbsp;$g(t)$&nbsp; of a trapezoidal Nyquist spectrum with rolloff factor &nbsp;$r$ can be assumed to be known:
+*The inverse Fourier transform &nbsp;$g(t)$&nbsp; of a trapezoidal Nyquist spectrum with rolloff factor &nbsp;$r$&nbsp;  can be assumed to be known:
-:$$g ( t )= g_0 \cdot {\rm si} \left ( {\pi \cdot
+:$$g ( t )= g_0 \cdot {\rm sinc} \left ( { t}/{T}\right)\cdot {\rm sinc} \left ( { r \cdot
-t}/{T}\right)\cdot {\rm si} \left ( {\pi \cdot r \cdot
+t}/{T}\right),\hspace{0.3cm} {\rm sinc}(x)= \sin(\pi \cdot x)/(\pi \cdot x).$$
 t}/{T}\right).$$
-*A triangular low&ndash;pass spectrum &nbsp;$G(f)$ limited to &nbsp;$| f | < f_{0}$&nbsp; and where &nbsp;$G(f = 0) = B$,&nbsp; after Fourier inverse transformation, leads to the time function
+*A triangular low&ndash;pass spectrum &nbsp;$G(f)$ limited to &nbsp;$| f | < f_{0}$&nbsp; and where &nbsp;$G(f = 0) = B$,&nbsp; after inverse Fourier transform,&nbsp;  leads to the time function
-:$$g ( t )= B \cdot f_0 \cdot {\rm si}^2 \left ( {\pi f_0 t}\right)\hspace{0.05cm}.$$
+:$$g ( t )= B \cdot f_0 \cdot {\rm sinc}^2 \left ( { f_0 \cdot t}\right)\hspace{0.05cm},\hspace{0.3cm} {\rm sinc}(x)= \sin(\pi \cdot x)/(\pi \cdot x).$$
@@ Line 47: / Line 47: @@
 <quiz display=simple>
-{For the first questions, let &nbsp;$B = 0$. What is the Nyquist frequency?
+{For the first questions,&nbsp; let &nbsp;$B = 0$.&nbsp; What is the Nyquist frequency?
 |type="{}"}
 $f_{\rm Nyq} \ =  \ $ { 5 3% } $\ \rm kHz$
@@ Line 59: / Line 59: @@
 $g_{0} \ =  \ ${ 1 3% } $\ \rm V$
-{Further let &nbsp;$B=0$. Calculate &nbsp;$g(t)$&nbsp; for the time points &nbsp;$t = 100\, &micro; \rm s$&nbsp; and &nbsp;$t = 200\, &micro; \rm s$.
+{Further let &nbsp;$B=0$.&nbsp; Calculate &nbsp;$g(t)$&nbsp; for the time points &nbsp;$t = 100\, &micro; \rm s$&nbsp; and &nbsp;$t = 200\, &micro; \rm s$.
 |type="{}"}
 $g(t = 100\, &micro; \rm s) \ =  \ $ { 0. } $\ \rm V$

@@ Line 35: / Line 35: @@
 *The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems|Properties of Nyquist Systems]].
-*The Fourier retransform &nbsp;$g(t)$&nbsp; of a trapezoidal Nyquist spectrum with rolloff factor &nbsp;$r$ can be assumed to be known:
+*The Fourier inverse transformation &nbsp;$g(t)$&nbsp; of a trapezoidal Nyquist spectrum with rolloff factor &nbsp;$r$ can be assumed to be known:
 :$$g ( t )= g_0 \cdot {\rm si} \left ( {\pi \cdot
 t}/{T}\right)\cdot {\rm si} \left ( {\pi \cdot r \cdot
@@ Line 134: / Line 134: @@
-'''(7)'''&nbsp; With the result &nbsp;$g_{\rm R}(2.5T) = 0$&nbsp; from '''(3)''', it follows &nbsp;$g(2.5T) = g_{\rm I}(2.5T)$, where &nbsp;$g_{\rm I}(t)$&nbsp; is the Fourier retransform of  ${\rm j}\cdot \ G_{\rm I}(f)$. It holds:
+'''(7)'''&nbsp; With the result &nbsp;$g_{\rm R}(2.5T) = 0$&nbsp; from '''(3)''', it follows &nbsp;$g(2.5T) = g_{\rm I}(2.5T)$, where &nbsp;$g_{\rm I}(t)$&nbsp; is the Fourier inverse transformation of  ${\rm j}\cdot \ G_{\rm I}(f)$. It holds:
 :$${\rm j} \cdot G_{\rm I}(f)  = {\rm j} \cdot\left[ \delta(f + f_{\rm Nyq}) - \delta(f - f_{\rm Nyq})\right] \star D(f) \hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} g_{\rm I}(t)  = 2 \cdot {\rm sin}  ( 2

← 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 15:27, 2 February 2022
(No difference)

← Older revision		Revision as of 12:49, 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^]]

@@ Line 33: / Line 33: @@
 ''Hinweise:''
-*Die Aufgabe gehört zum  Kapitel&nbsp;  [[Digitalsignalübertragung/Eigenschaften_von_Nyquistsystemen|Eigenschaften von Nyquistsystemen]].
+*Die Aufgabe gehört zum  Kapitel&nbsp;  [[Digital_Signal_Transmission/Eigenschaften_von_Nyquistsystemen|Eigenschaften von Nyquistsystemen]].
 *Als bekannt vorausgesetzt werden kann die Fourierrücktransformierte &nbsp;$g(t)$&nbsp; eines trapezförmigen Nyquistspektrums mit Rolloff&ndash;Faktor &nbsp;$r$:

@@ Line 100: / Line 100: @@
   Hz}\hspace{0.1cm}\underline { = 1\,{\rm V}}\hspace{0.05cm}.$$
-'''(4)'''&nbsp; Beim Nyquistimpuls treten die äquidistanten Nulldurchgänge im Abstand $T = 1/(2f_{\rm Nyq}) = 100 \, \rm \mu s$ auf. Daraus erhält man direkt:
+'''(4)'''&nbsp; Beim Nyquistimpuls treten die äquidistanten Nulldurchgänge im Abstand $T = 1/(2f_{\rm Nyq}) = 100 \, \rm &micro; s$ auf. Daraus erhält man direkt:
-:$$g(t= 100\,{\rm \mu s}) = \ \hspace{0.1cm}\underline { g(T) = 0,}$$
+:$$g(t= 100\,{\rm &micro; s}) = \ \hspace{0.1cm}\underline { g(T) = 0,}$$
-:$$g(t= 200\,{\rm \mu s}) = \  \hspace{0.1cm}\underline {g(2T) = 0}
+:$$g(t= 200\,{\rm &micro; s}) = \  \hspace{0.1cm}\underline {g(2T) = 0}
   \hspace{0.05cm}.$$
 Dieses Ergebnis folgt auch aus der angegebenen Gleichung mit $r = 0.4$:
@@ Line 111: / Line 111: @@
-'''(5)'''&nbsp; Entsprechend der unter (4) angegebenen Gleichung gilt:
+'''(5)'''&nbsp; Entsprechend der unter '''(4)''' angegebenen Gleichung gilt:
-:$$g(t= 250\,{\rm \mu s})= g_0 \cdot {\rm si}  ( {2.5 \cdot \pi
+:$$g(t= 250\,{\rm &micro; s})= g_0 \cdot {\rm si}  ( {2.5 \cdot \pi
 })\cdot {\rm si}  ( \pi )\hspace{0.1cm}\underline { = 0} \hspace{0.05cm}.$$
 Diese Nullstelle ist auf den zweiten Term zurückzuführen und liegt nicht im Nyquist–Zeitraster $\nu T$
-'''(6)'''&nbsp; Für die folgende Herleitung gelte $g(t)= g_{\rm R}(t) +  g_{\rm I}(t) \hspace{0.05cm},$ wobei $g_{\rm R}(t)$ auf den Realteil und $g_{\rm I}(t)$ auf den Imaginärteil von $G(f)$ zurückgeht. Der erste Anteil ist dabei genau wie unter Punkt (4) berechnet:
+'''(6)'''&nbsp; Für die folgende Herleitung gelte $g(t)= g_{\rm R}(t) +  g_{\rm I}(t) \hspace{0.05cm},$ wobei $g_{\rm R}(t)$ auf den Realteil und $g_{\rm I}(t)$ auf den Imaginärteil von $G(f)$ zurückgeht.
 :$$g_{\rm R} ( t )= g_0 \cdot {\rm si} \left ( {\pi \cdot
 t}/{T}\right)\cdot {\rm si} \left ( {\pi \cdot 0.4 \cdot
 t}/{T}\right) \hspace{0.05cm}.$$
-Zur Erfüllung des ersten Nyquistkriteriums muss für den Imaginärteil mit $1/T = 10 \, \rm kHz$ gelten:
+*Zur Erfüllung des ersten Nyquistkriteriums muss für den Imaginärteil mit $1/T = 10 \, \rm kHz$ gelten:
 :$$\sum_{k = -\infty}^{+\infty} {\rm Im}\left[G \left ( f -
 {k}/{T} \right)\right]= 0 \hspace{0.05cm}.$$
-*Mit den gegebenen Eckfrequenzen $f_{1} = 3 \, \rm kHz$ und $f_{2} = 7 \ \rm kHz$ liegen die beiden Dreiecke um $\pm 5\, \rm kHz$, so dass obige Gleichung erfüllt ist. Gleiches gilt für $f_{1} = 4.5\, \rm kHz$ und $f_{2} = 5.5 \, \rm kHz$.
+*Mit den gegebenen Eckfrequenzen $f_{1} = 3 \, \rm kHz$ und $f_{2} = 7 \ \rm kHz$ liegen die beiden Dreiecke um $\pm 5\, \rm kHz$, so dass obige Gleichung erfüllt ist.
-*Dagegen liegen die Dreieckspitzen mit $f_{1} = 3\, \rm kHz$ und $f_{2} = 5 \, \rm kHz$ bei $\pm 4 \ \rm kHz$. In diesem Fall löschen sich die Dreieckfunktionen durch die periodische Fortsetzung nicht aus und die Nyquistbedingung ist nicht erfüllt.
+*Gleiches gilt für $f_{1} = 4.5\, \rm kHz$ und $f_{2} = 5.5 \, \rm kHz$.
 Richtig sind somit die <u>Lösungsvorschläge 1 und 3</u>.
-'''(7)'''&nbsp; Mit dem Ergebnis $g_{\rm R}(2.5T) = 0$ aus  (3) folgt $g(2.5T) = g_{\rm I}(2.5T)$, wobei $g_{\rm I}(t)$ die Fourierrücktransformierte von  ${\rm j}\cdot \ G_{\rm I}(f)$ ist. Es gilt:
+'''(7)'''&nbsp; Mit dem Ergebnis &nbsp;$g_{\rm R}(2.5T) = 0$&nbsp; aus  '''(3)''' folgt &nbsp;$g(2.5T) = g_{\rm I}(2.5T)$, wobei &nbsp;$g_{\rm I}(t)$&nbsp; die Fourierrücktransformierte von  ${\rm j}\cdot \ G_{\rm I}(f)$ ist. Es gilt:
 :$${\rm j} \cdot G_{\rm I}(f)  = {\rm j} \cdot\left[ \delta(f + f_{\rm Nyq}) - \delta(f - f_{\rm Nyq})\right] \star D(f) \hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} g_{\rm I}(t)  = 2 \cdot {\rm sin}  ( 2
   \pi\cdot f_{\rm Nyq} \cdot t )\cdot  d(t)\hspace{0.05cm}.$$
-Die Sinusfunktion erzwingt die erforderlichen Nulldurchgänge bei Vielfachen von $T = 100 \, \rm \mu s$. $D(f)$ ist eine Dreieckfunktion um $f = 0$ mit $D(f = 0) = B$ und der einseitigen Breite $f_{0}= f_{2} – f_{\rm Nyq} = f_{\rm Nyq} – f_{1} = 2 \, \rm kHz$. Für die dazugehörige Zeitfunktion kann somit entsprechend der Angabe geschrieben werden:
+*Die Sinusfunktion erzwingt die erforderlichen Nulldurchgänge bei Vielfachen von $T = 100 \, \rm &micro; s$.
 :$$g_{\rm I}(t )  = 2 \cdot B \cdot f_0 \cdot{\rm sin}  ( 2
   \pi\cdot f_{\rm Nyq} \cdot t)\cdot {\rm si}^2(\pi\cdot f_{\rm 0} \cdot t) \hspace{0.05cm}.$$
-Insbesondere gilt für den Zeitpunkt $t = 250 \, \rm \mu s$  (grünes Quadrat):
+*Insbesondere gilt für den Zeitpunkt $t = 250 \, \rm &micro; s$  (grünes Quadrat):
 :$$g(t = 2.5 T)  = g_{\rm I}(t = 2.5 T)  = \ 2 \cdot B \cdot f_0 \cdot{\rm sin}  ( 2.5
   \pi )\cdot {\rm si}^2(\frac{\pi}{2}) = \  \frac{8}{\pi^2} \cdot 10^{-4 }\,\frac{\rm V}{\rm Hz}\cdot 2 \cdot 10^{3} \,{\rm
   Hz}\hspace{0.1cm}\underline {= 0.162\,{\rm
   V}} \hspace{0.05cm}.$$
 [[File:P_ID1283__Dig_Z_1_4g.png|right|frame|Unsymmetrischer Nyquistimpuls]]
 Die Grafik zeigt die Veränderung der Zeitfunktion aufgrund des Imaginärteils (grüner Zeitverlauf):