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

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:58:05Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:58, 16 March 2026
Line 82:		Line 82:

	'''(3)'''  For a pulse with real low–pass spectrum,  the maximum is always at  $t = 0$  and it holds:		'''(3)'''  For a pulse with real low–pass spectrum,  the maximum is always at  $t = 0$  and it holds:
	:$$g_0 = g(t=0) = \int_{-\infty}^{+\infty}G(f) \,{\rm d} f= A \cdot 2 f_{\rm Nyq} = 10^{-4 }\,\frac{\rm V}{\rm Hz}\cdot 2 \cdot 5 \cdot10^{3} \,{\~~rmHz~~}\hspace{0.1cm}\underline { = 1\,{\rm V}}\hspace{0.05cm}.$$		:$$g_0 = g(t=0) = \int_{-\infty}^{+\infty}G(f) \,{\rm d} f= A \cdot 2 f_{\rm Nyq} = 10^{-4 }\,\frac{\rm V}{\rm Hz}\cdot 2 \cdot 5 \cdot10^{3} \,{\rm Hz}\hspace{0.1cm}\underline { = 1\,{\rm V}}\hspace{0.05cm}.$$

	'''(4)'''  For the Nyquist pulse,  the equidistant zero crossings occur at the interval  $T = 1/(2f_{\rm Nyq}) = 100 \, \rm µ s$.  From this one obtains directly:		'''(4)'''  For the Nyquist pulse,  the equidistant zero crossings occur at the interval  $T = 1/(2f_{\rm Nyq}) = 100 \, \rm µ s$.  From this one obtains directly:
Line 124:		Line 124:

	*In particular,  for time $t = 250 \, \rm µ s$  (green square):		*In particular,  for time $t = 250 \, \rm µ s$  (green square):
	:$$g(t = 2.5 T) = g_{\rm I}(t = 2.5 T) = \ 2 \cdot B \cdot f_0 \cdot{\rm sin} ( 2.5\pi )\cdot {\rm sinc}^2(0.5)= \ \frac{8}{\pi^2} \cdot B \cdot f_0 = \ \frac{8}{\pi^2} \cdot 10^{-4 }\,\frac{\rm V}{\rm Hz}\cdot 2 \cdot 10^{3} \,{\~~rmHz~~}\hspace{0.1cm}\underline {= 0.162\,{\~~rmV~~}} \hspace{0.05cm}.$$		:$$g(t = 2.5 T) = g_{\rm I}(t = 2.5 T) = \ 2 \cdot B \cdot f_0 \cdot{\rm sin} ( 2.5\pi )\cdot {\rm sinc}^2(0.5)= \ \frac{8}{\pi^2} \cdot B \cdot f_0 = \ \frac{8}{\pi^2} \cdot 10^{-4 }\,\frac{\rm V}{\rm Hz}\cdot 2 \cdot 10^{3} \,{\rm Hz}\hspace{0.1cm}\underline {= 0.162\,{\rm V}} \hspace{0.05cm}.$$

	[[File:P_ID1283__Dig_Z_1_4g.png\|right\|frame\|Asymmetric Nyquist pulse   $g(t)= g_{\rm R}(t) + g_{\rm I}(t) $]]		[[File:P_ID1283__Dig_Z_1_4g.png\|right\|frame\|Asymmetric Nyquist pulse   $g(t)= g_{\rm R}(t) + g_{\rm I}(t) $]]
Line 137:		Line 137:

	[[Category:Digital Signal Transmission: Exercises\|^1.3 Nyquist System Properties^]]		[[Category:Digital Signal Transmission: Exercises\|^1.3 Nyquist System Properties^]]
	[[de:Aufgaben:1.~~4Z_Komplexes_Nyquistspektrum~~]]		[[de:Aufgaben:Aufgabe 1.4Z: Komplexes Nyquistspektrum]]

Maintenance script: Join 1 split formula blocks

2026-03-16T14:16:01Z

Join 1 split formula blocks

← Older revision		Revision as of 16:16, 16 March 2026
Line 13:		Line 13:

	It is necessary to check whether the pulse  $g(t)$  satisfies the first Nyquist condition even with complex spectrum:		It is necessary to check whether the pulse  $g(t)$  satisfies the first Nyquist condition even with complex spectrum:
	:$$g(\nu		:$$g(\nuT) = \left\{ \begin{array}{c} g_0 \\0 \\ \end{array} \right.\quad\begin{array}{{1}c} {\rm{for}}\\ {\rm{for}} \\ \end{array}\begin{array}{{20}c}\nu = 0 \hspace{0.05cm}, \\\nu \ne 0 \hspace{0.1cm}. \\\end{array}$$
	T) = \left\{ \begin{array}{c} g_0 \\
	0 \\ \end{array} \right.\quad
	\begin{array}{*{1}c} {\rm{for}}
	\\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c}
	\nu = 0 \hspace{0.05cm}, \\
	\nu \ne 0 \hspace{0.1cm}. \\
	\end{array}$$

	In the course of this exercise,  reference is made to the following descriptive quantities:		In the course of this exercise,  reference is made to the following descriptive quantities:

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:30:25Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:30, 16 March 2026
Line 24:		Line 24:
	In the course of this exercise,  reference is made to the following descriptive quantities:		In the course of this exercise,  reference is made to the following descriptive quantities:
	*The  '''Nyquist frequency'''  gives the symmetry point of the rolloff:		*The  '''Nyquist frequency'''  gives the symmetry point of the rolloff:
	:$$f_{\rm Nyq}= \frac{1}{2T}= \frac{f_1 +f_2 }		:$$f_{\rm Nyq}= \frac{1}{2T}= \frac{f_1 +f_2 }{2 }\hspace{0.05cm}.$$
	{2 }\hspace{0.05cm}.$$
	*The  '''rolloff factor'''  is a measure of the transition steepness:		*The  '''rolloff factor'''  is a measure of the transition steepness:
	:$$r = \frac{f_2 -f_1 }		:$$r = \frac{f_2 -f_1 }{f_2 +f_1 } \hspace{0.05cm}.$$
	{f_2 +f_1 } \hspace{0.05cm}.$$


Line 36:		Line 34:

	*The inverse Fourier transform  $g(t)$  of a trapezoidal Nyquist spectrum with rolloff factor  $r$  can be assumed to be known:		*The inverse Fourier transform  $g(t)$  of a trapezoidal Nyquist spectrum with rolloff factor  $r$  can be assumed to be known:
	:$$g ( t )= g_0 \cdot {\rm sinc} \left ( { t}/{T}\right)\cdot {\rm sinc} \left ( { r \~~cdot~~		:$$g ( t )= g_0 \cdot {\rm sinc} \left ( { t}/{T}\right)\cdot {\rm sinc} \left ( { r \cdott}/{T}\right),\hspace{0.3cm} {\rm sinc}(x)= \sin(\pi \cdot x)/(\pi \cdot x).$$
	t}/{T}\right),\hspace{0.3cm} {\rm sinc}(x)= \sin(\pi \cdot x)/(\pi \cdot x).$$

	*A triangular low–pass spectrum  $G(f)$ limited to  $\| f \| < f_{0}$  and where  $G(f = 0) = B$,  after inverse Fourier transform,  leads to the time function		*A triangular low–pass spectrum  $G(f)$ limited to  $\| f \| < f_{0}$  and where  $G(f = 0) = B$,  after inverse Fourier transform,  leads to the time function
Line 86:		Line 83:
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The Nyquist frequency specifies the symmetry point of the rolloff.  The following holds true:		'''(1)'''  The Nyquist frequency specifies the symmetry point of the rolloff.  The following holds true:
	:$$f_{\rm Nyq}= \frac{f_1 +f_2 }		:$$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}.$$
	{2 }= \frac{3\, {\rm kHz} + 7\, {\rm kHz}} {2 } \hspace{0.1cm}\underline { = 5\, {\rm kHz}}
	\hspace{0.05cm}.$$

	'''(2)'''  The rolloff factor is also determined by the two corner frequencies  $f_{1}$  and  $f_{2}$:		'''(2)'''  The rolloff factor is also determined by the two corner frequencies  $f_{1}$  and  $f_{2}$:
	:$$r = \frac{f_2 -f_1 }		:$$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}.$$
	{f_2 +f_1 } = \frac{7\, {\rm kHz} - 3\, {\rm kHz}} {7\, {\rm kHz}
	+ 3\, {\rm kHz} }\hspace{0.1cm}\underline { = 0.4 }\hspace{0.05cm}.$$

	'''(3)'''  For a pulse with real low–pass spectrum,  the maximum is always at  $t = 0$  and it holds:		'''(3)'''  For a pulse with real low–pass spectrum,  the maximum is always at  $t = 0$  and it holds:
	:$$g_0 = g(t=0) = \int_{-\infty}^{+\infty}G(f) \,{\rm d} f		:$$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} \,{\rmHz}\hspace{0.1cm}\underline { = 1\,{\rm V}}\hspace{0.05cm}.$$
	= A \cdot 2 f_{\rm Nyq} = 10^{-4 }\,\frac{\rm V}{\rm Hz}\cdot 2 \cdot 5 \cdot10^{3} \,{\rm
	Hz}\hspace{0.1cm}\underline { = 1\,{\rm V}}\hspace{0.05cm}.$$

	'''(4)'''  For the Nyquist pulse,  the equidistant zero crossings occur at the interval  $T = 1/(2f_{\rm Nyq}) = 100 \, \rm µ s$.  From this one obtains directly:		'''(4)'''  For the Nyquist pulse,  the equidistant zero crossings occur at the interval  $T = 1/(2f_{\rm Nyq}) = 100 \, \rm µ s$.  From this one obtains directly:
	:$$g(t= 100\,{\rm µ s}) = \ \hspace{0.1cm}\underline { g(T) = 0,}$$		:$$g(t= 100\,{\rm µ s}) = \ \hspace{0.1cm}\underline { g(T) = 0,}$$
	:$$g(t= 200\,{\rm µ s}) = \ \hspace{0.1cm}\underline {g(2T) = 0}		:$$g(t= 200\,{\rm µ s}) = \ \hspace{0.1cm}\underline {g(2T) = 0}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$
	This result also follows from the given equation with  $r = 0.4$:		This result also follows from the given equation with  $r = 0.4$:
	:$$g ( t )= g_0 \cdot {\rm sinc} \left ( {		:$$g ( t )= g_0 \cdot {\rm sinc} \left ( {t}/{T}\right)\cdot {\rm sinc} \left ( {0.4 \cdott}/{T}\right) \hspace{0.05cm}.$$
	t}/{T}\right)\cdot {\rm sinc} \left ( {0.4 \~~cdot~~
	t}/{T}\right) \hspace{0.05cm}.$$
	Responsible for satisfying the first Nyquist condition is the first term.		Responsible for satisfying the first Nyquist condition is the first term.

Line 118:		Line 106:
	'''(6)'''  For the following derivation,  let  $g(t)= g_{\rm R}(t) + g_{\rm I}(t) \hspace{0.05cm},$  where  $g_{\rm R}(t)$  is due to the real part and  $g_{\rm I}(t)$  is due to the imaginary part of  $G(f)$.		'''(6)'''  For the following derivation,  let  $g(t)= g_{\rm R}(t) + g_{\rm I}(t) \hspace{0.05cm},$  where  $g_{\rm R}(t)$  is due to the real part and  $g_{\rm I}(t)$  is due to the imaginary part of  $G(f)$.
	*The first part is calculated exactly as in point  '''(4)''':		*The first part is calculated exactly as in point  '''(4)''':
	:$$g_{\rm R} ( t )= g_0 \cdot {\rm sinc} \left ( {		:$$g_{\rm R} ( t )= g_0 \cdot {\rm sinc} \left ( {t}/{T}\right)\cdot {\rm sinc} \left ( { 0.4 \cdott}/{T}\right) \hspace{0.05cm}.$$
	t}/{T}\right)\cdot {\rm sinc} \left ( { 0.4 \~~cdot~~
	t}/{T}\right) \hspace{0.05cm}.$$
	*To satisfy the first Nyquist criterion, the following must hold for the imaginary part with $1/T = 10 \, \rm kHz$:		*To satisfy the first Nyquist criterion, the following must hold for the imaginary part with $1/T = 10 \, \rm kHz$:
	:$$\sum_{k = -\infty}^{+\infty} {\rm Im}\left[G \left ( f -		:$$\sum_{k = -\infty}^{+\infty} {\rm Im}\left[G \left ( f -{k}/{T} \right)\right]= 0 \hspace{0.05cm}.$$
	{k}/{T} \right)\right]= 0 \hspace{0.05cm}.$$
	*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  $f_{1} = 3 \, \rm kHz$  and  $f_{2} = 7 \ \rm kHz$,  the two triangles are around  $\pm 5\, \rm kHz$,  so the above equation is satisfied.

Line 137:		Line 122:

	'''(7)'''  With the result  $g_{\rm R}(2.5T) = 0$  from  '''(3)''',  it follows  $g(2.5T) = g_{\rm I}(2.5T)$,  where  $g_{\rm I}(t)$  is the inverse Fourier transform of  ${\rm j}\cdot \ G_{\rm I}(f)$.  It holds:		'''(7)'''  With the result  $g_{\rm R}(2.5T) = 0$  from  '''(3)''',  it follows  $g(2.5T) = g_{\rm I}(2.5T)$,  where  $g_{\rm I}(t)$  is the inverse Fourier transform of  ${\rm j}\cdot \ G_{\rm I}(f)$.  It holds:
	:$${\rm j} \cdot G_{\rm I}(f) = {\rm j} \cdot\left[ \delta(f + f_{\rm Nyq}) - \delta(f - f_{\rm Nyq})\right] \star D(f) \hspace{0.3cm}		:$${\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}.$$
	\Rightarrow \hspace{0.3cm} g_{\rm I}(t) = 2 \cdot {\rm sin} ( 2
	\pi\cdot f_{\rm Nyq} \cdot t )\cdot d(t)\hspace{0.05cm}.$$
	*The sine function enforces the required zero crossings at multiples of  $T = 100 \, \rm µ s$.		*The sine function enforces the required zero crossings at multiples of  $T = 100 \, \rm µ s$.

Line 145:		Line 128:

	*For the associated time function we can thus write according to the specification:		*For the associated time function we can thus write according to the specification:
	:$$g_{\rm I}(t ) = 2 \cdot B \cdot f_0 \cdot{\rm sin} ( 2		:$$g_{\rm I}(t ) = 2 \cdot B \cdot f_0 \cdot{\rm sin} ( 2\pi\cdot f_{\rm Nyq} \cdot t)\cdot {\rm sinc}^2( 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 $t = 250 \, \rm µ s$  (green square):		*In particular,  for time $t = 250 \, \rm µ s$  (green square):
	:$$g(t = 2.5 T) = g_{\rm I}(t = 2.5 T) = \ 2 \cdot B \cdot f_0 \cdot{\rm sin} ( 2.5		:$$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 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} \,{\rmHz}\hspace{0.1cm}\underline {= 0.162\,{\rmV}} \hspace{0.05cm}.$$
	\pi )\cdot {\rm sinc}^2(0.5)= \ \frac{8}{\pi^2} \cdot B \cdot f_0 = \ \frac{8}{\pi^2} \cdot 10^{-4 }\,\frac{\rm V}{\rm Hz}\cdot 2 \cdot 10^{3} \,{\rm
	Hz}\hspace{0.1cm}\underline {= 0.162\,{\rm
	V}} \hspace{0.05cm}.$$

	[[File:P_ID1283__Dig_Z_1_4g.png\|right\|frame\|Asymmetric Nyquist pulse   $g(t)= g_{\rm R}(t) + g_{\rm I}(t) $]]		[[File:P_ID1283__Dig_Z_1_4g.png\|right\|frame\|Asymmetric Nyquist pulse   $g(t)= g_{\rm R}(t) + g_{\rm I}(t) $]]

Maintenance script: Add German interlanguage link

2026-03-16T12:30:27Z

Add German interlanguage link

← Older revision		Revision as of 14:30, 16 March 2026
Line 165:		Line 165:

	[[Category:Digital Signal Transmission: Exercises\|^1.3 Nyquist System Properties^]]		[[Category:Digital Signal Transmission: Exercises\|^1.3 Nyquist System Properties^]]
			[[de:Aufgaben:1.4Z_Komplexes_Nyquistspektrum]]

Guenter at 16:31, 1 May 2022

2022-05-01T16:31:57Z

← Older revision		Revision as of 18:31, 1 May 2022
Line 154:		Line 154:
	V}} \hspace{0.05cm}.$$		V}} \hspace{0.05cm}.$$

	[[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   $g(t)= g_{\rm R}(t) + g_{\rm I}(t) $]]
	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  (green time course):
	*The result is now an asymmetric function curve  $g(t)$,  shown in blue.		*The result is now an asymmetric function curve  $g(t)$,  shown in blue.

Guenter at 16:08, 1 May 2022

2022-05-01T16:08:32Z

Guenter at 15:19, 1 May 2022

2022-05-01T15:19:20Z

← Older revision		Revision as of 17:19, 1 May 2022
Line 5:		Line 5:

	[[File:P_ID1279__Dig_Z_1_4.png\|right\|frame\|Complex Nyquist spectrum]]		[[File:P_ID1279__Dig_Z_1_4.png\|right\|frame\|Complex Nyquist spectrum]]
	Consider a pulse  $g(t)$  with spectrum  $G(f)$  according to the sketch. One recognizes from this representation:		Consider a pulse  $g(t)$  with spectrum  $G(f)$  according to the sketch.  One recognizes from this representation:
	*The real part of  $G(f)$  is trapezoidal with the corner frequencies  $f_{1} = 3 \, \rm kHz$  and  $f_{2} = 7 \, \rm kHz$. In the range  $\|f\| < f_{1}$~~  holds~~:		*The real part of  $G(f)$  is trapezoidal with the corner frequencies  $f_{1} = 3 \, \rm kHz$  and  $f_{2} = 7 \, \rm kHz$.  In the range  $\|f\| < f_{1}$:
	:$${\rm Re}\big[G(f)\big] = A = 10^{-4} \, \rm V/Hz.$$		:$${\rm Re}\big[G(f)\big] = A = 10^{-4} \, \rm V/Hz.$$
	*The imaginary part of  $G(f)$  is always assumed to be  ${\rm Im}\big[G(f)\big] =0$  for subtasks '''(1)''' to '''(5)'''. ~~<br>~~In this case  $g(t)$  is certainly a Nyquist pulse.		*The imaginary part of  $G(f)$  is always assumed to be  ${\rm Im}\big[G(f)\big] =0$   for subtasks  '''(1)'''  to  '''(5)'''.  In this case  $g(t)$  is certainly a Nyquist pulse.
	*From subtask '''(6)''', the imaginary part  ${\rm Im}[G(f)]$  in the range  $f_{1} \leq \| f \| \leq f_{2}$  has a triangular shape with the values  $\pm B$  at the triangle peaks.		*From subtask  '''(6)''',  the imaginary part  ${\rm Im}[G(f)]$  in the range  $f_{1} \leq \| f \| \leq f_{2}$  has a triangular shape with the values  $\pm B$  at the triangle peaks.


Line 22:		Line 22:
	\end{array}$$		\end{array}$$

	In the course of this exercise, reference is made to the following descriptive quantities:		In the course of this exercise,  reference is made to the following descriptive quantities:
	*The Nyquist frequency gives the symmetry point of the rolloff:		*The  '''Nyquist frequency'''  gives the symmetry point of the rolloff:
	:$$f_{\rm Nyq}= \frac{1}{2T}= \frac{f_1 +f_2 }		:$$f_{\rm Nyq}= \frac{1}{2T}= \frac{f_1 +f_2 }
	{2 }\hspace{0.05cm}.$$		{2 }\hspace{0.05cm}.$$
	*The rolloff factor is a measure of the transition steepness:		*The  '''rolloff factor'''  is a measure of the transition steepness:
	:$$r = \frac{f_2 -f_1 }		:$$r = \frac{f_2 -f_1 }
	{f_2 +f_1 } \hspace{0.05cm}.$$		{f_2 +f_1 } \hspace{0.05cm}.$$
Line 32:		Line 32:


	''Notes:''		Notes:
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems\|Properties of Nyquist Systems]].		*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems\|"Properties of Nyquist Systems"]].

	*The Fourier ~~inverse transformation~~  $g(t)$  of a trapezoidal Nyquist spectrum with rolloff factor  $r$ can be assumed to be known:		*The inverse Fourier transform  $g(t)$  of a trapezoidal Nyquist spectrum with rolloff factor  $r$  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–pass spectrum  $G(f)$ limited to  $\| f \| < f_{0}$  and where  $G(f = 0) = B$,  after Fourier ~~inverse transformation~~, leads to the time function		*A triangular low–pass spectrum  $G(f)$ limited to  $\| f \| < f_{0}$  and where  $G(f = 0) = B$,  after inverse Fourier transform,  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>		<quiz display=simple>

	{For the first questions, let  $B = 0$. What is the Nyquist frequency?		{For the first questions,  let  $B = 0$.  What is the Nyquist frequency?
	\|type="{}"}		\|type="{}"}
	$f_{\rm Nyq} \ = \ $ { 5 3% } $\ \rm kHz$		$f_{\rm Nyq} \ = \ $ { 5 3% } $\ \rm kHz$
Line 59:		Line 59:
	$g_{0} \ = \ ${ 1 3% } $\ \rm V$		$g_{0} \ = \ ${ 1 3% } $\ \rm V$

	{Further let  $B=0$. Calculate  $g(t)$  for the time points  $t = 100\, µ \rm s$  and  $t = 200\, µ \rm s$.		{Further let  $B=0$.  Calculate  $g(t)$  for the time points  $t = 100\, µ \rm s$  and  $t = 200\, µ \rm s$.
	\|type="{}"}		\|type="{}"}
	$g(t = 100\, µ \rm s) \ = \ $ { 0. } $\ \rm V$		$g(t = 100\, µ \rm s) \ = \ $ { 0. } $\ \rm V$

Hwang at 13:11, 8 March 2022

2022-03-08T13:11:12Z

← Older revision		Revision as of 15:11, 8 March 2022
Line 35:		Line 35:
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems\|Properties of Nyquist Systems]].		*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems\|Properties of Nyquist Systems]].

	*The Fourier ~~retransform~~  $g(t)$  of a trapezoidal Nyquist spectrum with rolloff factor  $r$ can be assumed to be known:		*The Fourier inverse transformation  $g(t)$  of a trapezoidal Nyquist spectrum with rolloff factor  $r$ can be assumed to be known:
	:$$g ( t )= g_0 \cdot {\rm si} \left ( {\pi \cdot		:$$g ( t )= g_0 \cdot {\rm si} \left ( {\pi \cdot
	t}/{T}\right)\cdot {\rm si} \left ( {\pi \cdot r \cdot		t}/{T}\right)\cdot {\rm si} \left ( {\pi \cdot r \cdot
Line 134:		Line 134:


	'''(7)'''  With the result  $g_{\rm R}(2.5T) = 0$  from '''(3)''', it follows  $g(2.5T) = g_{\rm I}(2.5T)$, where  $g_{\rm I}(t)$  is the Fourier ~~retransform~~ of ${\rm j}\cdot \ G_{\rm I}(f)$. It holds:		'''(7)'''  With the result  $g_{\rm R}(2.5T) = 0$  from '''(3)''', it follows  $g(2.5T) = g_{\rm I}(2.5T)$, where  $g_{\rm I}(t)$  is the Fourier inverse transformation of ${\rm j}\cdot \ G_{\rm I}(f)$. It holds:
	:$${\rm j} \cdot G_{\rm I}(f) = {\rm j} \cdot\left[ \delta(f + f_{\rm Nyq}) - \delta(f - f_{\rm Nyq})\right] \star D(f) \hspace{0.3cm}		:$${\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		\Rightarrow \hspace{0.3cm} g_{\rm I}(t) = 2 \cdot {\rm sin} ( 2

Hwang at 13:00, 8 March 2022

2022-03-08T13:00:24Z

Show changes

Guenter at 15:30, 2 February 2022

2022-02-02T15:30:03Z

← Older revision		Revision as of 17:30, 2 February 2022
Line 159:		Line 159:


	[[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 18:08, 1 May 2022
Line 85:		Line 85:
	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The Nyquist frequency specifies the symmetry point of the rolloff. The following holds true:		'''(1)'''  The Nyquist frequency specifies the symmetry point of the rolloff.  The following holds true:
	:$$f_{\rm Nyq}= \frac{f_1 +f_2 }		:$$f_{\rm Nyq}= \frac{f_1 +f_2 }
	{2 }= \frac{3\, {\rm kHz} + 7\, {\rm kHz}} {2 } \hspace{0.1cm}\underline { = 5\, {\rm kHz}}		{2 }= \frac{3\, {\rm kHz} + 7\, {\rm kHz}} {2 } \hspace{0.1cm}\underline { = 5\, {\rm kHz}}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	'''(2)'''  The rolloff factor is also determined by the two corner frequencies $f_{1}$ and $f_{2}$:		'''(2)'''  The rolloff factor is also determined by the two corner frequencies  $f_{1}$  and  $f_{2}$:
	:$$r = \frac{f_2 -f_1 }		:$$r = \frac{f_2 -f_1 }
	{f_2 +f_1 } = \frac{7\, {\rm kHz} - 3\, {\rm kHz}} {7\, {\rm kHz}		{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}.$$		+ 3\, {\rm kHz} }\hspace{0.1cm}\underline { = 0.4 }\hspace{0.05cm}.$$

	'''(3)'''  For a pulse with real low–pass spectrum, the maximum is always at $t = 0$ and it holds:		'''(3)'''  For a pulse with real low–pass spectrum,  the maximum is always at  $t = 0$  and it holds:
	:$$g_0 = g(t=0) = \int_{-\infty}^{+\infty}G(f) \,{\rm d} f		:$$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		= 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}.$$		Hz}\hspace{0.1cm}\underline { = 1\,{\rm V}}\hspace{0.05cm}.$$

	'''(4)'''  For the Nyquist pulse, the equidistant zero crossings occur at the interval $T = 1/(2f_{\rm Nyq}) = 100 \, \rm µ s$. From this one obtains directly:		'''(4)'''  For the Nyquist pulse,  the equidistant zero crossings occur at the interval  $T = 1/(2f_{\rm Nyq}) = 100 \, \rm µ s$.  From this one obtains directly:
	:$$g(t= 100\,{\rm µ s}) = \ \hspace{0.1cm}\underline { g(T) = 0,}$$		:$$g(t= 100\,{\rm µ s}) = \ \hspace{0.1cm}\underline { g(T) = 0,}$$
	:$$g(t= 200\,{\rm µ s}) = \ \hspace{0.1cm}\underline {g(2T) = 0}		:$$g(t= 200\,{\rm µ s}) = \ \hspace{0.1cm}\underline {g(2T) = 0}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	This result also follows from the given equation with $r = 0.4$:		This result also follows from the given equation with  $r = 0.4$:
	:$$g ( t )= g_0 \cdot {\rm si} \left ( {~~\pi \cdot~~		:$$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}.$$		t}/{T}\right) \hspace{0.05cm}.$$
	Responsible for satisfying the first Nyquist condition is the first term.		Responsible for satisfying the first Nyquist condition is the first term.


	'''(5)'''  According to the equation given in '''(4)''':		'''(5)'''  According to the equation given in  '''(4)''':
	:$$g(t= 250\,{\rm µ s})= g_0 \cdot {\rm si} ( {~~2.5 \cdot \pi~~		:$$g(t= 250\,{\rm µ 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  $\nu T$
	This zero is due to the second term and does not lie in the Nyquist time grid $\nu T$


	'''(6)'''  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)'''  For the following derivation,  let  $g(t)= g_{\rm R}(t) + g_{\rm I}(t) \hspace{0.05cm},$  where  $g_{\rm R}(t)$  is due to the real part and  $g_{\rm I}(t)$  is due to the imaginary part of  $G(f)$.
	*The first part is calculated exactly as in point '''(4)''':		*The first part is calculated exactly as in point  '''(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}.$$		t}/{T}\right) \hspace{0.05cm}.$$
	*To satisfy the first Nyquist criterion, the following must hold for the imaginary part with $1/T = 10 \, \rm kHz$:		*To satisfy the first Nyquist criterion, the following must hold for the imaginary part with $1/T = 10 \, \rm kHz$:
	:$$\sum_{k = -\infty}^{+\infty} {\rm Im}\left[G \left ( f -		:$$\sum_{k = -\infty}^{+\infty} {\rm Im}\left[G \left ( f -
	{k}/{T} \right)\right]= 0 \hspace{0.05cm}.$$		{k}/{T} \right)\right]= 0 \hspace{0.05cm}.$$
	*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  $f_{1} = 3 \, \rm kHz$  and  $f_{2} = 7 \ \rm kHz$,  the two triangles are around  $\pm 5\, \rm kHz$,  so the above equation is satisfied.
	*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  $f_{1} = 4.5\, \rm kHz$  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  $\pm 4 \ \rm kHz$.

	*In this case the triangle functions do not cancel due to the periodic continuation and the Nyquist condition is not fulfilled.		*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  <u>1 and 3</u>.


	'''(7)'''  With the result  $g_{\rm R}(2.5T) = 0$  from '''(3)''', it follows  $g(2.5T) = g_{\rm I}(2.5T)$, where  $g_{\rm I}(t)$  is the Fourier ~~inverse transformation~~ of ${\rm j}\cdot \ G_{\rm I}(f)$. It holds:		'''(7)'''  With the result  $g_{\rm R}(2.5T) = 0$  from  '''(3)''',  it follows  $g(2.5T) = g_{\rm I}(2.5T)$,  where  $g_{\rm I}(t)$  is the inverse Fourier transform of  ${\rm j}\cdot \ G_{\rm I}(f)$.  It holds:
	:$${\rm j} \cdot G_{\rm I}(f) = {\rm j} \cdot\left[ \delta(f + f_{\rm Nyq}) - \delta(f - f_{\rm Nyq})\right] \star D(f) \hspace{0.3cm}		:$${\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		\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}.$$		\pi\cdot f_{\rm Nyq} \cdot t )\cdot d(t)\hspace{0.05cm}.$$
	*The sine function enforces the required zero crossings at multiples of $T = 100 \, \rm µ s$.		*The sine function enforces the required zero crossings at multiples of  $T = 100 \, \rm µ s$.
	*$D(f)$ is a triangular function around $f = 0$ with $D(f = 0) = B$ and one-sided width $f_{0}= f_{2} – f_{\rm Nyq} = f_{\rm Nyq} – f_{1} = 2 \, \rm kHz$.
			*$D(f)$  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:		*For the associated time function we can thus write according to the specification:
	:$$g_{\rm I}(t ) = 2 \cdot B \cdot f_0 \cdot{\rm sin} ( 2		:$$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 $t = 250 \, \rm µ s$ (green square):
			*In particular,  for time $t = 250 \, \rm µ s$  (green square):
	:$$g(t = 2.5 T) = g_{\rm I}(t = 2.5 T) = \ 2 \cdot B \cdot f_0 \cdot{\rm sin} ( 2.5		:$$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		Hz}\hspace{0.1cm}\underline {= 0.162\,{\rm
	V}} \hspace{0.05cm}.$$		V}} \hspace{0.05cm}.$$

	[[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]]
	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  (green time course):
	*The result is now an asymmetric function curve $g(t)$, shown in blue.		*The result is now an asymmetric function curve  $g(t)$,  shown in blue.
	*However, the zero crossings of $g_{\rm R}(t)$ at the distance $T$ remain.		*However, the zero crossings of  $g_{\rm R}(t)$  at the distance  $T$  remain.