Digital Signal Transmission/Properties of Nyquist Systems - Revision history

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Fix 8 split formula blocks

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	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	'''(1)'''   We start from the first Nyquist condition in the time domain:		'''(1)'''   We start from the first Nyquist condition in the time domain:
	:$$		:$$g_{\hspace{0.05cm}\rm Nyq}(\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}$$
	g_{\hspace{0.05cm}\rm Nyq}(\nuT) = \left\{ \begin{array}{c} g_0 \\0 \\ \end{array} \right.\quad\begin{array}{{1}c} {\rm{for} }\\ {\rm{for} } \\ \end{array}\begin{array}{{20}c}\nu = 0 \hspace{0.05cm}, \\\nu \ne 0 \hspace{0.1cm}. \\\end{array}$$		'''(2)'''   Thus,  from the second Fourier integral,  we obtain for  $\nu \ne 0$::$$g_{\hspace{0.05cm}\rm Nyq}(\nuT) = \int_{-\infty}^{+\infty}G_{\rm Nyq}(f) \cdot {\rm e}^{ \hspace{0.05cm}{\rm j} \hspace{0.05cm}2 \pi f \hspace{0.05cm}\nu\hspace{0.05cm}T}\,{\rm d} f = 0 \hspace{0.05cm}.$$
	'''(2)'''   Thus,  from the second Fourier integral,  we obtain for  $\nu \ne 0$::$$		'''(3)'''   Decomposing the Fourier integral into partial integrals of width  $1/T$,  the conditional equations are::$$\sum_{k = -\infty}^{+\infty} \hspace{0.2cm} \int_{(k-1/2)/T}^{(k+1/2)/T}G_{\rm Nyq}(f) \cdot {\rm e}^{ \hspace{0.05cm}{\rm j} \hspace{0.05cm}2 \pi f \hspace{0.05cm}\nu\hspace{0.05cm}T}\,{\rm d} f = 0 \hspace{0.05cm}.$$
	g_{\hspace{0.05cm}\rm Nyq}(\nuT) = \int_{-\infty}^{+\infty}G_{\rm Nyq}(f) \cdot {\~~rme~~}^{ \hspace{0.05cm}{\rm j} \hspace{0.05cm}2 \pi f \hspace{0.05cm}\nu\hspace{0.05cm}T}\,{\rm d} f = 0 \hspace{0.05cm}.$$		'''(4)'''   With the substitution  $f\hspace{0.08cm}' = f + k/T$  it follows::$$\sum_{k = -\infty}^{+\infty} \hspace{0.2cm} \int_{-1/(2T)}^{1/(2T)}G_{\rm Nyq}(f\hspace{0.08cm}' -\frac{k}{T} ) \cdot {\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}2 \pi \hspace{0.05cm} \cdot \hspace{0.05cm} (f\hspace{0.08cm}'-k/T) \hspace{0.05cm} \cdot \hspace{0.05cm}\nu\hspace{0.05cm}T}\,{\rm d} f \hspace{0.08cm}' = 0 \hspace{0.05cm}.$$
	'''(3)'''   Decomposing the Fourier integral into partial integrals of width  $1/T$,  the conditional equations are::$$		'''(5)'''   For all integer values of  $k$  and  $\nu$  holds::$${\rm e}^{ - {\rm j} \hspace{0.05cm}2 \pi\hspace{0.05cm} k \hspace{0.05cm} \nu } = 1\hspace{0.4cm} \Rightarrow \hspace{0.4cm}\sum_{k = - \infty}^{+\infty} \hspace{0.2cm} \int_{- 1/(2T)}^{1/(2T)}G_{\rm Nyq}(f\hspace{0.08cm}' -\frac{k}{T} ) \cdot {\rm e}^{ {\rm j} \hspace{0.05cm}2 \pi\hspace{0.02cm}f\hspace{0.08cm}' \hspace{0.02cm} \nu \hspace{0.05cm}T}\,{\rm d} f\hspace{0.08cm}' = 0 \hspace{0.05cm}.$$
	\sum_{k = -\infty}^{+\infty} \hspace{0.2cm} \int_{(k-1/2)/T}^{(k+1/2)/T}G_{\rm Nyq}(f) \cdot {\~~rme~~}^{ \hspace{0.05cm}{\rm j} \hspace{0.05cm}2 \pi f \hspace{0.05cm}\nu\hspace{0.05cm}T}\,{\rm d} f = 0 \hspace{0.05cm}.$$		'''(6)'''   Swapping summation and integration and renaming  $f\hspace{0.08cm}'$  to  $f$,  it further follows::$$\int_{-1/(2T)}^{1/(2T)}\hspace{0.2cm} \sum_{k = - \infty}^{+\infty} G_{\rm Nyq}(f -\frac{k}{T} ) \cdot {\rm e}^{ {\rm j} \hspace{0.05cm}2 \pi\hspace{0.02cm}f \hspace{0.02cm} \nu \hspace{0.05cm}T}\,{\rm d} f= 0 \hspace{0.05cm}.$$
	'''(4)'''   With the substitution  $f\hspace{0.08cm}' = f + k/T$  it follows::$$		'''(7)'''   This requirement can be satisfied for all  $\nu \ne 0$  only if the infinite sum is independent of  $f$,  i.e.,  has a constant value::$$\sum_{k = -\infty}^{+\infty} G_{\rm Nyq}(f -\frac{k}{T} ) = K_{\rm Nyq} \hspace{0.05cm}.$$
	\sum_{k = -\infty}^{+\infty} \hspace{0.2cm} \int_{-1/(2T)}^{1/(2T)}G_{\rm Nyq}(f\hspace{0.08cm}' -\frac{k}{T} ) \cdot {\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}2 \pi \hspace{0.05cm} \cdot \hspace{0.05cm} (f\hspace{0.08cm}'-k/T) \hspace{0.05cm} \cdot \hspace{0.05cm}\nu\hspace{0.05cm}T}\,{\rm d} f \hspace{0.08cm}' = 0 \hspace{0.05cm}.$$		'''(8)'''   From the penultimate equation,  we obtain simultaneously for  $\nu = 0$::$$\int_{-1/(2T)}^{1/(2T)}\hspace{0.2cm} \sum_{k = -\infty}^{+\infty} G_{\rm Nyq}(f -\frac{k}{T} ) \,{\rm d} f= K_{\rm Nyq} \cdot \frac{1}{T} = g_0 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}K_{\rm Nyq} = g_0 \cdot T \hspace{0.05cm}.$$
	'''(5)'''   For all integer values of  $k$  and  $\nu$  holds::$$
	{\rm e}^{ - {\rm j} \hspace{0.05cm}2 \pi\hspace{0.05cm} k \hspace{0.05cm} \nu } = 1\hspace{0.4cm} \Rightarrow \hspace{0.4cm}\sum_{k = - \infty}^{+\infty} \hspace{0.2cm} \int_{- 1/(2T)}^{1/(2T)}G_{\rm Nyq}(f\hspace{0.08cm}' -\frac{k}{T} ) \cdot {\rm e}^{ {\rm j} \hspace{0.05cm}2 \pi\hspace{0.02cm}f\hspace{0.08cm}' \hspace{0.02cm} \nu \hspace{0.05cm}T}\,{\rm d} f\hspace{0.08cm}' = 0 \hspace{0.05cm}.$$
	'''(6)'''   Swapping summation and integration and renaming  $f\hspace{0.08cm}'$  to  $f$,  it further follows::$$
	\int_{-1/(2T)}^{1/(2T)}\hspace{0.2cm} \sum_{k = - \infty}^{+\infty} G_{\rm Nyq}(f -\frac{k}{T} ) \cdot {\rm e}^{ {\rm j} \hspace{0.05cm}2 \pi\hspace{0.02cm}f \hspace{0.02cm} \nu \hspace{0.05cm}T}\,{\rm d} f= 0 \hspace{0.05cm}.$$
	'''(7)'''   This requirement can be satisfied for all  $\nu \ne 0$  only if the infinite sum is independent of  $f$,  i.e.,  has a constant value::$$
	\sum_{k = -\infty}^{+\infty} G_{\rm Nyq}(f -\frac{k}{T} ) = K_{\rm Nyq} \hspace{0.05cm}.$$
	'''(8)'''   From the penultimate equation,  we obtain simultaneously for  $\nu = 0$::$$
	\int_{-1/(2T)}^{1/(2T)}\hspace{0.2cm} \sum_{k = -\infty}^{+\infty} G_{\rm Nyq}(f -\frac{k}{T} ) \,{\rm d} f= K_{\rm Nyq} \cdot \frac{1}{T} = g_0 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}K_{\rm Nyq} = g_0 \cdot T \hspace{0.05cm}.$$
	<div align="right">'''q.e.d.'''</div>}}		<div align="right">'''q.e.d.'''</div>}}

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	:$$g_{\rm Nyq-2}( t ) = g_0 \cdot \left [ {\rm sinc}( \frac{t + T/2}{T}) + {\rm sinc}( \frac{t- T/2}{T}) \right] =\frac{2 \cdot g_0}{\pi} \cdot \frac{\cos(\pi \cdot t/T)}{1 - (2 \cdot t/T)^2}\hspace{0.05cm}.$$		:$$g_{\rm Nyq-2}( t ) = g_0 \cdot \left [ {\rm sinc}( \frac{t + T/2}{T}) + {\rm sinc}( \frac{t- T/2}{T}) \right] =\frac{2 \cdot g_0}{\pi} \cdot \frac{\cos(\pi \cdot t/T)}{1 - (2 \cdot t/T)^2}\hspace{0.05cm}.$$
	*Due to the limitation of the spectrum  $G_{\rm Nyq-1}( f)$  to the range  $\vert f \vert \le f_{\rm Nyq} = 1/(2T)$,  the sum in equation  '''(4)'''  is limited to the term with  $k = 0$,  and we obtain:		*Due to the limitation of the spectrum  $G_{\rm Nyq-1}( f)$  to the range  $\vert f \vert \le f_{\rm Nyq} = 1/(2T)$,  the sum in equation  '''(4)'''  is limited to the term with  $k = 0$,  and we obtain:
	:$$G_{\rm Nyq-2}(f) = \left\{ \begin{array}{c} g_0 \cdot T \cdot \cos(\pi/2 \cdot f/f_{\rm Nyq}) \\\\ 0 \\\end{array} \right.\quad\begin{array}{*{1}c} {\rm{for} }\hspace{0.15cm} \vert f \vert < f_{\~~rmNyq~~}\hspace{0.05cm},\\ \\ {\rm{otherwise} }\hspace{0.05cm}. \\\end{array}$$		:$$G_{\rm Nyq-2}(f) = \left\{ \begin{array}{c} g_0 \cdot T \cdot \cos(\pi/2 \cdot f/f_{\rm Nyq}) \\\\ 0 \\\end{array} \right.\quad\begin{array}{*{1}c} {\rm{for} }\hspace{0.15cm} \vert f \vert < f_{\rm Nyq}\hspace{0.05cm},\\ \\ {\rm{otherwise} }\hspace{0.05cm}. \\\end{array}$$
	*This frequency response and the corresponding eye diagram is sketched in the middle column of the above diagram.		*This frequency response and the corresponding eye diagram is sketched in the middle column of the above diagram.
	*From the bottom diagram, one can clearly see the fulfillment of the second Nyquist criterion.}}		*From the bottom diagram, one can clearly see the fulfillment of the second Nyquist criterion.}}

Maintenance script: Fix 1 split formula blocks

2026-03-16T14:23:32Z

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← Older revision		Revision as of 16:23, 16 March 2026
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	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	'''(1)'''   We start from the first Nyquist condition in the time domain:		'''(1)'''   We start from the first Nyquist condition in the time domain:
	:$$g_{\hspace{0.05cm}\rm Nyq}(\nu		:$$
	T) = \left\{ \begin{array}{c} g_0 \\		g_{\hspace{0.05cm}\rm Nyq}(\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}$$
	0 \\ \end{array} \right.\quad		'''(2)'''   Thus,  from the second Fourier integral,  we obtain for  $\nu \ne 0$::$$
	\begin{array}{*{1}c} {\rm{for} }		g_{\hspace{0.05cm}\rm Nyq}(\nuT) = \int_{-\infty}^{+\infty}G_{\rm Nyq}(f) \cdot {\rme}^{ \hspace{0.05cm}{\rm j} \hspace{0.05cm}2 \pi f \hspace{0.05cm}\nu\hspace{0.05cm}T}\,{\rm d} f = 0 \hspace{0.05cm}.$$
	\\ {\rm{for} } \\ \end{array}\begin{array}{*{20}c}		'''(3)'''   Decomposing the Fourier integral into partial integrals of width  $1/T$,  the conditional equations are::$$
	\nu = 0 \hspace{0.05cm}, \\		\sum_{k = -\infty}^{+\infty} \hspace{0.2cm} \int_{(k-1/2)/T}^{(k+1/2)/T}G_{\rm Nyq}(f) \cdot {\rme}^{ \hspace{0.05cm}{\rm j} \hspace{0.05cm}2 \pi f \hspace{0.05cm}\nu\hspace{0.05cm}T}\,{\rm d} f = 0 \hspace{0.05cm}.$$
	\nu \ne 0 \hspace{0.1cm}. \\		'''(4)'''   With the substitution  $f\hspace{0.08cm}' = f + k/T$  it follows::$$
	\end{array}$$'''(2)'''   Thus,  from the second Fourier integral,  we obtain for  $\nu \ne 0$::$$g_{\hspace{0.05cm}\rm Nyq}(\nu		\sum_{k = -\infty}^{+\infty} \hspace{0.2cm} \int_{-1/(2T)}^{1/(2T)}G_{\rm Nyq}(f\hspace{0.08cm}' -\frac{k}{T} ) \cdot {\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}2 \pi \hspace{0.05cm} \cdot \hspace{0.05cm} (f\hspace{0.08cm}'-k/T) \hspace{0.05cm} \cdot \hspace{0.05cm}\nu\hspace{0.05cm}T}\,{\rm d} f \hspace{0.08cm}' = 0 \hspace{0.05cm}.$$
	T) = \int_{-\infty}^{+\infty}G_{\rm Nyq}(f) \cdot {\rm		'''(5)'''   For all integer values of  $k$  and  $\nu$  holds::$$
	e}^{ \hspace{0.05cm}{\rm j} \hspace{0.05cm}2 \pi f \hspace{0.05cm}\nu		{\rm e}^{ - {\rm j} \hspace{0.05cm}2 \pi\hspace{0.05cm} k \hspace{0.05cm} \nu } = 1\hspace{0.4cm} \Rightarrow \hspace{0.4cm}\sum_{k = - \infty}^{+\infty} \hspace{0.2cm} \int_{- 1/(2T)}^{1/(2T)}G_{\rm Nyq}(f\hspace{0.08cm}' -\frac{k}{T} ) \cdot {\rm e}^{ {\rm j} \hspace{0.05cm}2 \pi\hspace{0.02cm}f\hspace{0.08cm}' \hspace{0.02cm} \nu \hspace{0.05cm}T}\,{\rm d} f\hspace{0.08cm}' = 0 \hspace{0.05cm}.$$
	\hspace{0.05cm}T}\,{\rm d} f = 0 \hspace{0.05cm}.$$'''(3)'''   Decomposing the Fourier integral into partial integrals of width  $1/T$,  the conditional equations are::$$\sum_{k = -\infty}^{+\infty} \hspace{0.2cm} \int_{(k-1/2)/T}^{(k+1/2)/T}G_{\rm Nyq}(f) \cdot {\rm		'''(6)'''   Swapping summation and integration and renaming  $f\hspace{0.08cm}'$  to  $f$,  it further follows::$$
	e}^{ \hspace{0.05cm}{\rm j} \hspace{0.05cm}2 \pi f \hspace{0.05cm}\nu		\int_{-1/(2T)}^{1/(2T)}\hspace{0.2cm} \sum_{k = - \infty}^{+\infty} G_{\rm Nyq}(f -\frac{k}{T} ) \cdot {\rm e}^{ {\rm j} \hspace{0.05cm}2 \pi\hspace{0.02cm}f \hspace{0.02cm} \nu \hspace{0.05cm}T}\,{\rm d} f= 0 \hspace{0.05cm}.$$
	\hspace{0.05cm}T}\,{\rm d} f = 0 \hspace{0.05cm}.$$'''(4)'''   With the substitution  $f\hspace{0.08cm}' = f + k/T$  it follows::$$\sum_{k = -\infty}^{+\infty} \hspace{0.2cm} \int_{-1/(2T)}^{1/(2T)}G_{\rm Nyq}(f\hspace{0.08cm}' -		'''(7)'''   This requirement can be satisfied for all  $\nu \ne 0$  only if the infinite sum is independent of  $f$,  i.e.,  has a constant value::$$
	\frac{k}{T} ) \cdot {\rm e}^{\hspace{0.05cm}{\rm j}		\sum_{k = -\infty}^{+\infty} G_{\rm Nyq}(f -\frac{k}{T} ) = K_{\rm Nyq} \hspace{0.05cm}.$$
	\hspace{0.05cm}2 \pi \hspace{0.05cm} \cdot \hspace{0.05cm} (f\hspace{0.08cm}'-		'''(8)'''   From the penultimate equation,  we obtain simultaneously for  $\nu = 0$::$$
	k/T) \hspace{0.05cm} \cdot \hspace{0.05cm}\nu		\int_{-1/(2T)}^{1/(2T)}\hspace{0.2cm} \sum_{k = -\infty}^{+\infty} G_{\rm Nyq}(f -\frac{k}{T} ) \,{\rm d} f= K_{\rm Nyq} \cdot \frac{1}{T} = g_0 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}K_{\rm Nyq} = g_0 \cdot T \hspace{0.05cm}.$$
	\hspace{0.05cm}T}\,{\rm d} f \hspace{0.08cm}' = 0 \hspace{0.05cm}.$$'''(5)'''   For all integer values of  $k$  and  $\nu$  holds::$${\rm e}^{ - {\rm j} \hspace{0.05cm}2 \pi
	\hspace{0.05cm} k \hspace{0.05cm} \nu } = 1
	\hspace{0.4cm} \Rightarrow \hspace{0.4cm}\sum_{k = - \infty}^{+\infty} \hspace{0.2cm} \int_{- 1/(2T)}^{1/(2T)}G_{\rm Nyq}(f\hspace{0.08cm}' -
	\frac{k}{T} ) \cdot {\rm e}^{ {\rm j} \hspace{0.05cm}2 \pi
	\hspace{0.02cm}f\hspace{0.08cm}' \hspace{0.02cm} \nu \hspace{0.05cm}T}\,{\rm d} f
	\hspace{0.08cm}' = 0 \hspace{0.05cm}.$$'''(6)'''   Swapping summation and integration and renaming  $f\hspace{0.08cm}'$  to  $f$,  it further follows::$$\int_{-1/(2T)}^{1/(2T)}\hspace{0.2cm} \sum_{k = - \infty}^{+\infty} G_{\rm Nyq}(f -
	\frac{k}{T} ) \cdot {\rm e}^{ {\rm j} \hspace{0.05cm}2 \pi
	\hspace{0.02cm}f \hspace{0.02cm} \nu \hspace{0.05cm}T}\,{\rm d} f
	= 0 \hspace{0.05cm}.$$'''(7)'''   This requirement can be satisfied for all  $\nu \ne 0$  only if the infinite sum is independent of  $f$,  i.e.,  has a constant value::$$\sum_{k = -\infty}^{+\infty} G_{\rm Nyq}(f -
	\frac{k}{T} ) = K_{\rm Nyq} \hspace{0.05cm}.$$'''(8)'''   From the penultimate equation,  we obtain simultaneously for  $\nu = 0$::$$\int_{-1/(2T)}^{1/(2T)}\hspace{0.2cm} \sum_{k = -\infty}^{+\infty} G_{\rm Nyq}(f -
	\frac{k}{T} ) \,{\rm d} f
	= K_{\rm Nyq} \cdot \frac{1}{T} = g_0 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}K_{\rm Nyq} = g_0 \cdot T \hspace{0.05cm}.$$
	<div align="right">'''q.e.d.'''</div>}}		<div align="right">'''q.e.d.'''</div>}}

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← Older revision		Revision as of 16:15, 16 March 2026
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	The purely real spectrum sketched  $G_1(f)$  on the left satisfies the condition formulated above with the smallest possible bandwidth:		The purely real spectrum sketched  $G_1(f)$  on the left satisfies the condition formulated above with the smallest possible bandwidth:
	[[File:P_ID1273__Dig_T_1_3_S2_v1.png\|right\|frame\|Illustration of the first Nyquist criterion\|class=fit]]		[[File:P_ID1273__Dig_T_1_3_S2_v1.png\|right\|frame\|Illustration of the first Nyquist criterion\|class=fit]]
	:$$G_1(f) \hspace{-0.05cm}=\hspace{-0.05cm} \left\{ \begin{array}{c} g_0 \cdot T \\		:$$G_1(f) \hspace{-0.05cm}=\hspace{-0.05cm} \left\{ \begin{array}{c} g_0 \cdot T \\0 \\ \end{array} \right.\begin{array}{{1}c} \text{for}\\ \text{for} \\ \end{array}\begin{array}{{20}c}\vert f \vert \hspace{-0.08cm}<\hspace{-0.08cm} {1}/(2T), \\\vert f \vert \hspace{-0.08cm}>\hspace{-0.08cm} {1}/(2T), \\\end{array}$$
	0 \\ \end{array} \right.
	\begin{array}{*{1}c} \text{for}
	\\ \text{for} \\ \end{array}\begin{array}{*{20}c}
	\vert f \vert \hspace{-0.08cm}<\hspace{-0.08cm} {1}/(2T), \\
	\vert f \vert \hspace{-0.08cm}>\hspace{-0.08cm} {1}/(2T), \\
	\end{array}$$
	But the associated Nyquist pulse  $g_1(t) = g_0 \cdot {\rm sinc}(t/T)$  decays very slowly, asymptotically with  $1/t$.		But the associated Nyquist pulse  $g_1(t) = g_0 \cdot {\rm sinc}(t/T)$  decays very slowly, asymptotically with  $1/t$.

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← Older revision		Revision as of 14:26, 16 March 2026
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	{{Display}}		{{Display}}
			[[de:Digitalsignalübertragung/Eigenschaften_von_Nyquistsystemen]]

Hwang at 23:20, 12 November 2022

2022-11-12T23:20:58Z

← Older revision		Revision as of 01:20, 13 November 2022
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	:$$g_{\hspace{0.05cm}\rm Nyq} ( t )= g_0 \cdot {\rm sinc} \left ( \frac{t}{T}\right)\cdot {\rm sinc} \left ( \frac{ t}{2 \cdot		:$$g_{\hspace{0.05cm}\rm Nyq} ( t )= g_0 \cdot {\rm sinc} \left ( \frac{t}{T}\right)\cdot {\rm sinc} \left ( \frac{ t}{2 \cdot
	T}\right),$$		T}\right),$$
	which has already been discussed on the ~~page~~  [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Trapezoidal_low-pass_filter\|"Trapezoidal low-pass filter"]]  of the book  "Linear Time Invariant Systems".}}		which has already been discussed in the section  [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Trapezoidal_low-pass_filter\|"Trapezoidal low-pass filter"]]  of the book  "Linear Time Invariant Systems".}}


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	$\text{Note}$:  The proof follows on the  [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems#Proof_of_the_first_Nyquist_criterion\|"next ~~page~~"]].}}		$\text{Note}$:  The proof follows in the  [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems#Proof_of_the_first_Nyquist_criterion\|"next section"]].}}


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	#In the limiting case  $r \to 0$,  both cases yield the rectangular Nyquist spectrum and the  $\rm sinc$–shaped Nyquist pulse,   which decays asymptotically with  $1/t$  (thin green curves).		#In the limiting case  $r \to 0$,  both cases yield the rectangular Nyquist spectrum and the  $\rm sinc$–shaped Nyquist pulse,   which decays asymptotically with  $1/t$  (thin green curves).
	#For an average rolloff  $(r \approx 0.5)$,  the first overshoots are smaller for the trapezoidal spectrum than for the CRO spectrum,  because here the Nyquist slope is flatter for a given  $r$  (blue curves).		#For an average rolloff  $(r \approx 0.5)$,  the first overshoots are smaller for the trapezoidal spectrum than for the CRO spectrum,  because here the Nyquist slope is flatter for a given  $r$  (blue curves).
	#With the rolloff factor  $r = 1$,  the trapezoid becomes a triangle in the frequency domain and the CRO spectrum becomes the  "cosine–square spectrum".  In the diagrams on the  [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems#1.2FT_Nyquist_spectra\|"last ~~page~~"]]  these spectral functions are drawn in red.		#With the rolloff factor  $r = 1$,  the trapezoid becomes a triangle in the frequency domain and the CRO spectrum becomes the  "cosine–square spectrum".  In the diagrams in the  [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems#1.2FT_Nyquist_spectra\|"last section"]]  these spectral functions are drawn in red.
	#With  $r = 1$  the asymptotically decay of the upper time function  (according to the trapezoidal spectrum)  occurs with  $1/t^2$,  the decay of the lower function  (according to the CRO spectrum)  with  $1/t^3$.		#With  $r = 1$  the asymptotically decay of the upper time function  (according to the trapezoidal spectrum)  occurs with  $1/t^2$,  the decay of the lower function  (according to the CRO spectrum)  with  $1/t^3$.
	#This means:   After a longer time,  the CRO Nyquist pulse has settled better than the trapezoidal Nyquist pulse.		#This means:   After a longer time,  the CRO Nyquist pulse has settled better than the trapezoidal Nyquist pulse.

Guenter at 11:37, 30 May 2022

2022-05-30T11:37:42Z

← Older revision		Revision as of 13:37, 30 May 2022
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	$\text{Example 1:}$  The diagram shows the detection signal  $d(t)$  of such a Nyquist system.  Dotted in red are the  (weighted and shifted)  Nyquist pulses  $a_\nu \cdot g_{\hspace{0.05cm}\rm Nyq}(t -\nu \cdot T)$.		$\text{Example 1:}$  The diagram shows the detection signal  $d(t)$  of such a Nyquist system.  Dotted in red are the  (weighted and shifted)  Nyquist pulses  $a_\nu \cdot g_{\hspace{0.05cm}\rm Nyq}(t -\nu \cdot T)$.

	Please note: ~~'''Bitte checken'''~~		Please note:
	*At the detection times hold   $d(\nu \cdot T) = a_\nu \cdot g_{\hspace{0.05cm}\rm Nyq}(0)$,  as shown by the blue circles and the green grid.		*At the detection times hold   $d(\nu \cdot T) = a_\nu \cdot g_{\hspace{0.05cm}\rm Nyq}(0)$,  as shown by the blue circles and the green grid.
	*The trailing edge   ⇒   "trailers"  of preceding pulses  $(\nu < 0)$  as well as the rising edge   ⇒   "precursors"  of following pulses  $(\nu > 0)$  do not affect the detection of the symbol  $a_0$  in the Nyquist system.		*The trailing edge   ⇒   "trailers"  of preceding pulses  $(\nu < 0)$  as well as the rising edge   ⇒   "precursors"  of following pulses  $(\nu > 0)$  do not affect the detection of the symbol  $a_0$  in the Nyquist system.
	~~'''Ende Check 1'''~~

Guenter at 07:15, 20 May 2022

2022-05-20T07:15:40Z

← Older revision		Revision as of 09:15, 20 May 2022
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	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Definition:}$  One denotes a basic detection pulse ~~'''KORREKTUR: basic transmitter pulse'''~~ with the properties		$\text{Definition:}$  One denotes a basic detection pulse with the properties
	:$$g_d ( t = \nu T)= 0 \hspace{0.3cm}{\rm{for} }\hspace{0.3cm}		:$$g_d ( t = \nu T)= 0 \hspace{0.3cm}{\rm{for} }\hspace{0.3cm}
	\nu = \pm 1, \pm 2,\pm 3,\hspace{0.05cm}\text{...}$$		\nu = \pm 1, \pm 2,\pm 3,\hspace{0.05cm}\text{...}$$

Guenter at 07:13, 20 May 2022

2022-05-20T07:13:41Z

Hwang at 15:51, 10 May 2022

2022-05-10T15:51:04Z

← Older revision		Revision as of 17:51, 10 May 2022
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	Before the exact mathematical definition,  the significance of the  '''second Nyquist criterion'''  for the evaluation of a digital system is illustrated by means of diagrams. The graphic shows three examples of Nyquist systems,  in each case:		Before the exact mathematical definition,  the significance of the  '''second Nyquist criterion'''  for the evaluation of a digital system is illustrated by means of diagrams. The graphic shows three examples of Nyquist systems,  in each case:
	*at the top,  the Nyquist spectrum  $G_d(f)$,		*at the top,  the Nyquist spectrum  $G_d(f)$,
	*below the corresponding  "eye diagram"  referring to the  [[Digital_Signal_Transmission/~~Fehlerwahrscheinlichkeit_unter_Berücksichtigung_von_Impulsinterferenzen~~#~~Definition_und_Aussagen_des_Augendiagramms~~\|"third main chapter"]].		*below the corresponding  "eye diagram"  referring to the  [[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Definition_and_statements_of_the_eye_diagram\|"third main chapter"]].

	[[File:EN_Dig_T_1_3_S4_v3.png\|right\|frame\|Clarification of first and second Nyquist criterion]]		[[File:EN_Dig_T_1_3_S4_v3.png\|right\|frame\|Clarification of first and second Nyquist criterion]]

← Older revision		Revision as of 09:13, 20 May 2022
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	For the entire first main chapter it was assumed that the detection of a symbol should not be affected by neighboring pulses.  This is achieved by the detection of the signal		For the entire first main chapter it was assumed that the detection of a symbol should not be affected by neighboring pulses.  This is achieved by the detection of the signal
	:$$d(t) = d_{\rm S}(t) = \sum \limits_{\it (\nu)} a_\nu \cdot g_d ( t - \nu T)$$		:$$d(t) = d_{\rm S}(t) = \sum \limits_{\it (\nu)} a_\nu \cdot g_d ( t - \nu T)$$
	at the detection times  $(\nu \cdot T)$  whenever the basic detection pulse ~~'''KORREKTUR: basic transmitter pulse'''~~ $g_d(t)$		at the detection times  $(\nu \cdot T)$  whenever the basic detection pulse  $g_d(t)$
	*is restricted to the range  $\|t\| < T$,  which was assumed for the last chapter  [[Digital_Signal_Transmission/Fehlerwahrscheinlichkeit_bei_Basisbandübertragung\|"Error Probability for Baseband Transmission"]],   or		*is restricted to the range  $\|t\| < T$,  which was assumed for the last chapter  [[Digital_Signal_Transmission/Fehlerwahrscheinlichkeit_bei_Basisbandübertragung\|"Error Probability for Baseband Transmission"]],   or
	*has equidistant zero crossings at the detection times  $\nu \cdot T$.		*has equidistant zero crossings at the detection times  $\nu \cdot T$.
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	It should be mentioned that this graph is valid for the basic detection pulse ~~'''KORREKTUR: basic transmitter pulse'''~~ with trapezoidal spectrum and rolloff factor  $r = 0.5$:		It should be mentioned that this graph is valid for the basic detection pulse with trapezoidal spectrum and rolloff factor  $r = 0.5$:
	:$$g_{\hspace{0.05cm}\rm Nyq} ( t )= g_0 \cdot {\rm sinc} \left ( \frac{t}{T}\right)\cdot {\rm sinc} \left ( \frac{ t}{2 \cdot		:$$g_{\hspace{0.05cm}\rm Nyq} ( t )= g_0 \cdot {\rm sinc} \left ( \frac{t}{T}\right)\cdot {\rm sinc} \left ( \frac{ t}{2 \cdot
	T}\right),$$		T}\right),$$
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	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{First Nyquist criterion:}$  If the spectrum  $G_d(f)$  of the basic detection pulse ~~'''KORREKTUR: basic transmitter pulse'''~~  $g_d(t)$  fulfills the condition		$\text{First Nyquist criterion:}$  If the spectrum  $G_d(f)$  of the basic detection pulse  $g_d(t)$  fulfills the condition
	:$$\sum \limits_{\it k = -\infty}^{+\infty} G_d \left ( f - {k}/{T} \right)=		:$$\sum \limits_{\it k = -\infty}^{+\infty} G_d \left ( f - {k}/{T} \right)=
	g_0 \cdot T = {\rm const.} \hspace{0.05cm}, $$		g_0 \cdot T = {\rm const.} \hspace{0.05cm}, $$
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	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Summary of Nyquist criteria:}$<br><br>		$\text{Summary of Nyquist criteria:}$<br><br>
	'''(1)'''   In memory of the physicist  [https://en.wikipedia.org/wiki/Harry_Nyquist "Harry Nyquist"]  we denote a basic detection pulse ~~'''KORREKTUR: basic transmitter pulse'''~~  $g_d( t)$  with the properties		'''(1)'''   In memory of the physicist  [https://en.wikipedia.org/wiki/Harry_Nyquist "Harry Nyquist"]  we denote a basic detection pulse  $g_d( t)$  with the properties
	::$$g_d ( t= 0) \ne 0, \hspace{1cm} g_d ( t)= 0 \hspace{0.3cm}{\rm{for} }\hspace{0.3cm}		::$$g_d ( t= 0) \ne 0, \hspace{1cm} g_d ( t)= 0 \hspace{0.3cm}{\rm{for} }\hspace{0.3cm}
	t = \pm T, \pm 2T,\pm 3T,\hspace{0.05cm}\text{...}$$		t = \pm T, \pm 2T,\pm 3T,\hspace{0.05cm}\text{...}$$
	:as the  '''Nyquist–1 pulse'''  $g_{\hspace{0.05cm}\rm Nyq-1}(t)$.  This satisfies the first Nyquist criterion and leads to the maximum vertical eye opening.		:as the  '''Nyquist–1 pulse'''  $g_{\hspace{0.05cm}\rm Nyq-1}(t)$.  This satisfies the first Nyquist criterion and leads to the maximum vertical eye opening.
	'''(2)'''   A basic detection pulse ~~'''KORREKTUR: basic transmitter pulse'''~~  $g_d( t)$  satisfying the second Nyquist criterion must have zero crossings at  $t = \pm 1.5T$,  $t = \pm 2.5T$, ...  have:		'''(2)'''   A basic detection pulse  $g_d( t)$  satisfying the second Nyquist criterion must have zero crossings at  $t = \pm 1.5T$,  $t = \pm 2.5T$, ...  have:
	::$$g_d ( t= 0.5) \ne 0, \hspace{0.8cm} g_d ( t)= 0 \hspace{0.3cm}{\rm{for} }\hspace{0.3cm}		::$$g_d ( t= 0.5) \ne 0, \hspace{0.8cm} g_d ( t)= 0 \hspace{0.3cm}{\rm{for} }\hspace{0.3cm}
	t = \pm 1.5T, \ \pm 2.5T,\ \pm 3.5T,\hspace{0.05cm}\text{...}$$		t = \pm 1.5T, \ \pm 2.5T,\ \pm 3.5T,\hspace{0.05cm}\text{...}$$