Aufgaben:Exercise 1.4: Nyquist Criteria - Revision history

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	[[Category:Digital Signal Transmission: Exercises\|^1.3 Nyquist System Properties^]]		[[Category:Digital Signal Transmission: Exercises\|^1.3 Nyquist System Properties^]]
	[[de:Aufgaben:1.~~4_Nyquistkriterien~~]]		[[de:Aufgaben:Aufgabe 1.4: Nyquistkriterien]]

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	Given by the sketch is the spectrum  $G(f)$  of the basic detection pulse,  where the parameter  $A$  is still to be determined.  Among other things it is to be checked whether this basic detection pulse fulfills one of the two Nyquist criteria.  These are:		Given by the sketch is the spectrum  $G(f)$  of the basic detection pulse,  where the parameter  $A$  is still to be determined.  Among other things it is to be checked whether this basic detection pulse fulfills one of the two Nyquist criteria.  These are:
	*The  '''first Nyquist criterion'''  is fulfilled if for the spectral function holds:		*The  '''first Nyquist criterion'''  is fulfilled if for the spectral function holds:
	:$$\sum_{k = -\infty}^{+\infty} G(f -		:$$\sum_{k = -\infty}^{+\infty} G(f -{k}/{T} ) = {\rm const.}$$
	{k}/{T} ) = {\rm const.}$$
	:In this case,  the pulse  $g(t)$  has zero crossings at  $t = ν \cdot T$  for all integer values of  $ν$  except  $ν = 0$.  For the entire exercise,  $T = 0.1 \, \rm ms$  is assumed.		:In this case,  the pulse  $g(t)$  has zero crossings at  $t = ν \cdot T$  for all integer values of  $ν$  except  $ν = 0$.  For the entire exercise,  $T = 0.1 \, \rm ms$  is assumed.
	*If the  '''second Nyquist criterion'''  is satisfied,  $g(t)$  has zero crossings at  $\pm 1.5 T$,  $\pm 2.5 T$, etc.		*If the  '''second Nyquist criterion'''  is satisfied,  $g(t)$  has zero crossings at  $\pm 1.5 T$,  $\pm 2.5 T$, etc.
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	*The two equations are assumed to be known:		*The two equations are assumed to be known:
	:$$X(f) = \left\{ \begin{array}{c} A \\ 0 \\\end{array} \right.\quad		:$$X(f) = \left\{ \begin{array}{c} A \\ 0 \\\end{array} \right.\quad\begin{array}{*{1}c} {\rm{for}}\hspace{0.15cm}\|f\| < f_0 \hspace{0.05cm},\\ {\rm{for}}\hspace{0.15cm}\|f\| > f_0 \hspace{0.08cm} \\\end{array} \hspace{0.4cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ \hspace{0.4cm} x(t)=2 \cdot A \cdot f_0 \cdot {\rm si}(2 \pi f_0 T) \hspace{0.05cm},\hspace{0.4cm} {\rm si} (x) = \sin(x)/x\hspace{0.05cm},$$
	\begin{array}{*{1}c} {\rm{for}}\hspace{0.15cm}\|f\| < f_0 \hspace{0.05cm},		:$$\sin(\alpha) \cdot \cos (\beta) = \frac{1}{2} \cdot \big[ \sin(\alpha - \beta) + \sin(\alpha + \beta)\big]\hspace{0.05cm}.$$
	\\ {\rm{for}}\hspace{0.15cm}\|f\| > f_0 \hspace{0.08cm} \\
	\end{array} \hspace{0.4cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ \hspace{0.4cm} x(t)
	=2 \cdot A \cdot f_0 \cdot {\rm si}(2 \pi f_0 T) \hspace{0.05cm},\hspace{0.4cm} {\rm si} (x) = \sin(x)/x\hspace{0.05cm},$$
	:$$\sin(\alpha) \cdot \cos (\beta) = \frac{1}{2} \cdot \big[ \sin(\alpha - \beta) + \sin(\alpha + \beta)\big]
	\hspace{0.05cm}.$$


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	[[File:P_ID1280__Dig_A_1_4a.png\|right\|frame\|Illustration of the first Nyquist criterion]]		[[File:P_ID1280__Dig_A_1_4a.png\|right\|frame\|Illustration of the first Nyquist criterion]]

	:$$G_{\rm Per}(f) = \sum_{k = -\infty}^{+\infty} G(f -		:$$G_{\rm Per}(f) = \sum_{k = -\infty}^{+\infty} G(f -\frac{k}{T} ) \hspace{0.05cm}.$$
	\frac{k}{T} ) \hspace{0.05cm}.$$
	*The indexing variable  $k = 0$  indicates the original spectral function  $G(f)$.  This is filled in gray.		*The indexing variable  $k = 0$  indicates the original spectral function  $G(f)$.  This is filled in gray.
	*The spectrum shifted to the right by the value  $1/T = 10\, \rm kHz$  belongs to  $k = 1$  and is marked in green,  while  $k = -1$  leads to the function highlighted in yellow.		*The spectrum shifted to the right by the value  $1/T = 10\, \rm kHz$  belongs to  $k = 1$  and is marked in green,  while  $k = -1$  leads to the function highlighted in yellow.
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	'''(2)'''  Due to the Fourier integrals the following relation holds:		'''(2)'''  Due to the Fourier integrals the following relation holds:
	:$$g(t=0) = \int_{-\infty}^{\infty}G(f) \,{\rm d} f		:$$g(t=0) = \int_{-\infty}^{\infty}G(f) \,{\rm d} f= A \cdot ( 2\,{\rm kHz}+6\,{\rm kHz}+2\,{\rm kHz})= A \cdot 10\,{\rm kHz}$$
	= A \cdot ( 2\,{\rm kHz}+6\,{\rm kHz}+2\,{\rm kHz})= A \cdot 10\,{\rm kHz}$$
	:$$\Rightarrow \hspace{0.3cm}A = \frac{g(t=0)}{10\,{\rm kHz}} = \frac{2\,{\rm V}}{10\,{\rm kHz}} \hspace{0.1cm}\underline {= 0.2 \, {\rm mV/Hz}} \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm}A = \frac{g(t=0)}{10\,{\rm kHz}} = \frac{2\,{\rm V}}{10\,{\rm kHz}} \hspace{0.1cm}\underline {= 0.2 \, {\rm mV/Hz}} \hspace{0.05cm}.$$

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	With the Fourier correspondence given,  the two components are:		With the Fourier correspondence given,  the two components are:
	:$$g_1(t) \ = \ A \cdot 6\,{\rm kHz} \cdot {\rm si}(\pi \cdot 6\,{\rm kHz} \cdot t)		:$$g_1(t) \ = \ A \cdot 6\,{\rm kHz} \cdot {\rm si}(\pi \cdot 6\,{\rm kHz} \cdot t)\hspace{0.05cm},$$
	\hspace{0.05cm},$$		:$$g_2(t) \ = \ A \cdot 2\,{\rm kHz}\cdot{\rm si}(\pi \cdot 2\,{\rm kHz} \cdot t)\cdot 2 \cdot {\rm cos}(2 \pi \cdot 14\,{\rm kHz} \cdot t)\hspace{0.05cm}.$$
	:$$g_2(t) \ = \ A \cdot 2\,{\rm kHz}
	\cdot{\rm si}(\pi \cdot 2\,{\rm kHz} \cdot t)
	\cdot 2 \cdot {\rm cos}(2 \pi \cdot 14\,{\rm kHz} \cdot t)
	\hspace{0.05cm}.$$
	The second equation follows from the relation:		The second equation follows from the relation:
	:$$G_2(f) = \left[ \delta(f + 14\,{\rm kHz}) + \delta(f - 14\,{\rm kHz})\right] \star \left\{ \begin{array}{c} A \\ 0 \\\end{array} \right.\quad		:$$G_2(f) = \left[ \delta(f + 14\,{\rm kHz}) + \delta(f - 14\,{\rm kHz})\right] \star \left\{ \begin{array}{c} A \\ 0 \\\end{array} \right.\quad\begin{array}{*{1}c} {\rm{for}}\hspace{0.15cm}\|f\| < 1\,{\rm kHz} \hspace{0.05cm},\\ {\rm{for}}\hspace{0.15cm}\|f\| > 1\,{\rm kHz} \hspace{0.05cm}. \\\end{array}$$
	\begin{array}{*{1}c} {\rm{for}}\hspace{0.15cm}\|f\| < 1\,{\rm kHz} \hspace{0.05cm},
	\\ {\rm{for}}\hspace{0.15cm}\|f\| > 1\,{\rm kHz} \hspace{0.05cm}. \\
	\end{array}$$
	The bottom diagram shows the numerically determined time history  $g(t)$.  For the time  $t = T = 0.1\, \rm ms$  (yellow square)  we get:		The bottom diagram shows the numerically determined time history  $g(t)$.  For the time  $t = T = 0.1\, \rm ms$  (yellow square)  we get:
	:$$g_2(t = T ) = 2A \cdot 2\,{\rm kHz} \cdot {\rm si}(0.2 \cdot \pi)\cdot \cos (2.8 \cdot \pi)		:$$g_2(t = T ) = 2A \cdot 2\,{\rm kHz} \cdot {\rm si}(0.2 \cdot \pi)\cdot \cos (2.8 \cdot \pi)= \frac{ A \cdot 4\,{\rm kHz}}{0.2 \cdot \pi}\cdot {\rm sin}(0.2 \cdot \pi)\cdot\cos (0.8 \cdot \pi) = \frac{ A \cdot 10\,{\rm kHz}}{ \pi}\cdot [{\rm sin}(-0.6 \cdot\pi)+ {\rm sin}(\pi)] $$
	= \frac{ A \cdot 4\,{\rm kHz}}{0.2 \cdot \pi}\cdot {\rm sin}(0.2 \cdot \pi
	)\cdot\cos (0.8 \cdot \pi) = \frac{ A \cdot 10\,{\rm kHz}}{ \pi}\cdot [{\rm sin}(-0.6 \cdot
	\pi)+ {\rm sin}(\pi)] $$
	[[File:EN_Dig_A_1_4.png\|right\|frame\|Higher frequency Nyquist pulse]]		[[File:EN_Dig_A_1_4.png\|right\|frame\|Higher frequency Nyquist pulse]]
	:$$\Rightarrow \hspace{0.3cm} g_2(t = T ) = -\frac{ A \cdot 10\,{\rm kHz}}{ \pi}\cdot {\rm sin}(0.6 \cdot		:$$\Rightarrow \hspace{0.3cm} g_2(t = T ) = -\frac{ A \cdot 10\,{\rm kHz}}{ \pi}\cdot {\rm sin}(0.6 \cdot\pi).$$
	\pi).$$

	For the first component  $g_1(t)$,  at time  $t = T$:		For the first component  $g_1(t)$,  at time  $t = T$:
	:$$g_1(t = T ) = A \cdot 6\,{\rm kHz} \cdot {\rm sinc}(0.6		:$$g_1(t = T ) = A \cdot 6\,{\rm kHz} \cdot {\rm sinc}(0.6)$$
	)$$		:$$\Rightarrow \hspace{0.3cm}g_1(t = T ) = \frac{ A \cdot 6\,{\rm kHz}}{0.6 \cdot \pi}\cdot {\rm sin}(0.6 \cdot \pi)$$
	:$$\Rightarrow \hspace{0.3cm}g_1(t = T ) = \frac{ A \cdot 6\,{\rm kHz}}{0.6 \cdot \pi}\cdot {\rm sin}(0.6 \cdot \pi
	)$$
	:$$\Rightarrow \hspace{0.3cm}g_1(t = T ) = - g_2(t = T )$$		:$$\Rightarrow \hspace{0.3cm}g_1(t = T ) = - g_2(t = T )$$
	:$$\Rightarrow \hspace{0.3cm} g(t = T ) = g_1(t = T ) + g_2(t = T )\hspace{0.1cm}\underline {= 0 } \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} g(t = T ) = g_1(t = T ) + g_2(t = T )\hspace{0.1cm}\underline {= 0 } \hspace{0.05cm}.$$
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	<br clear=all>		<br clear=all>
	'''(4)'''  For $t = 2.5 T$ (green square), the following partial results are obtained:		'''(4)'''  For $t = 2.5 T$ (green square), the following partial results are obtained:
	:$$g_1(t = 2.5 T ) = A \cdot 6\,{\rm kHz} \cdot {\rm si}(1.5 \cdot \pi		:$$g_1(t = 2.5 T ) = A \cdot 6\,{\rm kHz} \cdot {\rm si}(1.5 \cdot \pi)= \frac{ A \cdot 6\,{\rm kHz}}{1.5 \cdot \pi}\cdot {\rm sin}(1.5 \cdot \pi)= - \frac{ A \cdot 4\,{\rm kHz}}{ \pi}\hspace{0.05cm},$$
	)= \frac{ A \cdot 6\,{\rm kHz}}{1.5 \cdot \pi}\cdot {\rm sin}(1.5 \cdot \pi		:$$g_2(t = 2.5 T ) = 2A \cdot 2\,{\rm kHz} \cdot {\rm si}(0.5 \cdot \pi)\cdot \cos (7 \cdot \pi)=- \frac{ A \cdot 8\,{\rm kHz}}{ \pi} $$
	)= - \frac{ A \cdot 4\,{\rm kHz}}{ \pi}\hspace{0.05cm},$$
	:$$g_2(t = 2.5 T ) = 2A \cdot 2\,{\rm kHz} \cdot {\rm si}(0.5 \cdot \pi
	)\cdot \cos (7 \cdot \pi)=- \frac{ A \cdot 8\,{\rm kHz}}{ \pi} $$
	:$$\Rightarrow \hspace{0.3cm} g(t = 2.5 T ) = g_1(t = 2.5 T ) +g_2(t = 2.5 T ) = - \frac{ A \cdot 12\,{\rm kHz}}{ \pi} \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} g(t = 2.5 T ) = g_1(t = 2.5 T ) +g_2(t = 2.5 T ) = - \frac{ A \cdot 12\,{\rm kHz}}{ \pi} \hspace{0.05cm}.$$
	Considering  $g(t = 0) = A \cdot 10 \ \rm kHz$,  we get:		Considering  $g(t = 0) = A \cdot 10 \ \rm kHz$,  we get:

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

Guenter at 07:16, 20 May 2022

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	[[File:P_ID1278__Dig_A_1_4.png\|right\|frame\|Rectangular Nyquist spectrum]]		[[File:P_ID1278__Dig_A_1_4.png\|right\|frame\|Rectangular Nyquist spectrum]]
	Given by the sketch is the spectrum  $G(f)$  of the basic detection pulse, ~~'''KORREKTUR: basic transmitter pulse'''~~   where the parameter  $A$  is still to be determined.  Among other things it is to be checked whether this basic detection pulse ~~'''KORREKTUR: basic transmitter pulse'''~~ fulfills one of the two Nyquist criteria.  These are:		Given by the sketch is the spectrum  $G(f)$  of the basic detection pulse,  where the parameter  $A$  is still to be determined.  Among other things it is to be checked whether this basic detection pulse fulfills one of the two Nyquist criteria.  These are:
	*The  '''first Nyquist criterion'''  is fulfilled if for the spectral function holds:		*The  '''first Nyquist criterion'''  is fulfilled if for the spectral function holds:
	:$$\sum_{k = -\infty}^{+\infty} G(f -		:$$\sum_{k = -\infty}^{+\infty} G(f -

Hwang at 15:52, 9 May 2022

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	[[File:P_ID1278__Dig_A_1_4.png\|right\|frame\|Rectangular Nyquist spectrum]]		[[File:P_ID1278__Dig_A_1_4.png\|right\|frame\|Rectangular Nyquist spectrum]]
	Given by the sketch is the spectrum  $G(f)$  of the basic detection pulse,  where the parameter  $A$  is still to be determined.  Among other things it is to be checked whether this basic detection pulse fulfills one of the two Nyquist criteria.  These are:		Given by the sketch is the spectrum  $G(f)$  of the basic detection pulse, '''KORREKTUR: basic transmitter pulse'''   where the parameter  $A$  is still to be determined.  Among other things it is to be checked whether this basic detection pulse '''KORREKTUR: basic transmitter pulse''' fulfills one of the two Nyquist criteria.  These are:
	*The  '''first Nyquist criterion'''  is fulfilled if for the spectral function holds:		*The  '''first Nyquist criterion'''  is fulfilled if for the spectral function holds:
	:$$\sum_{k = -\infty}^{+\infty} G(f -		:$$\sum_{k = -\infty}^{+\infty} G(f -

Guenter at 15:01, 1 May 2022

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	)$$		)$$
	:$$\Rightarrow \hspace{0.3cm}g_1(t = T ) = - g_2(t = T )$$		:$$\Rightarrow \hspace{0.3cm}g_1(t = T ) = - g_2(t = T )$$
	$$\Rightarrow \hspace{0.3cm} g(t = T ) = g_1(t = T ) + g_2(t = T )\hspace{0.1cm}\underline {= 0 } \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} g(t = T ) = g_1(t = T ) + g_2(t = T )\hspace{0.1cm}\underline {= 0 } \hspace{0.05cm}.$$
	This result is not surprising because of the Nyquist property.		This result is not surprising because of the Nyquist property.
	<br clear=all>		<br clear=all>
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	'''(5)'''  The second Nyquist criterion states that the Nyquist pulse  $g(t)$  has zero crossings at $\pm 1.5T, \pm 2.5T, \pm 3.5T, ...$		'''(5)'''  The second Nyquist criterion states that the Nyquist pulse  $g(t)$  has zero crossings at $\pm 1.5T, \pm 2.5T, \pm 3.5T, ...$
	*According to the result from  '''(4)'''  this condition is not fulfilled here.		*According to the result from  '''(4)'''  this condition is not fulfilled here.  Therefore <u>solution 2</u> is correct.
	*Therefore <u>solution 2</u> is correct.

Guenter at 14:43, 1 May 2022

2022-05-01T14:43:40Z

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	\\ {\rm{for}}\hspace{0.15cm}\|f\| > f_0 \hspace{0.08cm} \\		\\ {\rm{for}}\hspace{0.15cm}\|f\| > f_0 \hspace{0.08cm} \\
	\end{array} \hspace{0.4cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ \hspace{0.4cm} x(t)		\end{array} \hspace{0.4cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ \hspace{0.4cm} x(t)
	=2 \cdot A \cdot f_0 \cdot {\rm ~~sinc~~}(2 f_0 T) \hspace{0.05cm},$$		=2 \cdot A \cdot f_0 \cdot {\rm si}(2 \pi f_0 T) \hspace{0.05cm},\hspace{0.4cm} {\rm si} (x) = \sin(x)/x\hspace{0.05cm},$$
	:$$\sin(\alpha) \cdot \cos (\beta) = \frac{1}{2} \cdot \big[ \sin(\alpha - \beta) + \sin(\alpha + \beta)\big]		:$$\sin(\alpha) \cdot \cos (\beta) = \frac{1}{2} \cdot \big[ \sin(\alpha - \beta) + \sin(\alpha + \beta)\big]
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The following diagram shows the spectrum (the index "Per" here stands for "Periodic Continuation"):		'''(1)'''  The following diagram shows the spectrum  (the index  "Per"  here stands for  "Periodic Continuation"):
			[[File:P_ID1280__Dig_A_1_4a.png\|right\|frame\|Illustration of the first Nyquist criterion]]

	:$$G_{\rm Per}(f) = \sum_{k = -\infty}^{+\infty} G(f -		:$$G_{\rm Per}(f) = \sum_{k = -\infty}^{+\infty} G(f -
	\frac{k}{T} ) \hspace{0.05cm}.$$		\frac{k}{T} ) \hspace{0.05cm}.$$
	*The indexing variable $k = 0$ indicates the original spectral function $G(f)$. This is filled in gray.		*The indexing variable  $k = 0$  indicates the original spectral function  $G(f)$.  This is filled in gray.
	*The spectrum shifted to the right by the value $1/T = 10\, \rm kHz$ belongs to $k = 1$ and is marked in green, while $k = -1$ leads to the function highlighted in yellow.		*The spectrum shifted to the right by the value  $1/T = 10\, \rm kHz$  belongs to  $k = 1$  and is marked in green,  while  $k = -1$  leads to the function highlighted in yellow.
	*The red and blue areas mark the contributions of the indexing variables $k = 2$ and $k = - 2$.		*The red and blue areas mark the contributions of the indexing variables  $k = 2$  and  $k = - 2$.


	~~[[File:P_ID1280__Dig_A_1_4a.png\|center\|frame\|Illustration of the first Nyquist criterion]]~~		It can be seen that  $G_{\rm Per}(f)$  is constant.  From this it follows that the first Nyquist criterion is fulfilled   ⇒   <u>solution 1</u>  is correct.
	It can be seen that $G_{\rm Per}(f)$ is constant. From this it follows that the first Nyquist criterion is fulfilled~~. Consequently,~~ <u>solution 1</u> is correct.


	'''(2)'''  Due to the Fourier integrals the following relation holds:		'''(2)'''  Due to the Fourier integrals the following relation holds:
	:$$g(t=0) = \int_{-\infty}^{\infty}G(f) \,{\rm d} f		:$$g(t=0) = \int_{-\infty}^{\infty}G(f) \,{\rm d} f
	= A \cdot ( 2\,{\rm kHz}+6\,{\rm kHz}+2\,{\rm kHz})= A \cdot 10\,{\rm kHz}~~\hspace{0.3cm}~~		= A \cdot ( 2\,{\rm kHz}+6\,{\rm kHz}+2\,{\rm kHz})= A \cdot 10\,{\rm kHz}$$
	\Rightarrow \hspace{0.3cm}A = \frac{g(t=0)}{10\,{\rm kHz}} = \frac{2\,{\rm V}}{10\,{\rm kHz}} \hspace{0.1cm}\underline {= 0.2 \, {\rm mV/Hz}} \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm}A = \frac{g(t=0)}{10\,{\rm kHz}} = \frac{2\,{\rm V}}{10\,{\rm kHz}} \hspace{0.1cm}\underline {= 0.2 \, {\rm mV/Hz}} \hspace{0.05cm}.$$


	'''(3)'''  Let $g(t) = g_{1}(t) +g_{2}(t)$, where		'''(3)'''  Let  $g(t) = g_{1}(t) +g_{2}(t)$,  where
	*$g_{1}(t)$ contains the spectral components in the interval $\pm 3 \, \rm kHz$ and		*$g_{1}(t)$  contains the spectral components in the interval $\pm 3 \, \rm kHz$  and
	*$g_{2}(t)$ those between $13 \, \rm kHz$ and $15 \, \rm kHz$ (and between $-13 \, \rm kHz$ and $-15 \, \rm kHz$).		*$g_{2}(t)$  those between $13 \, \rm kHz$ and $15 \, \rm kHz$  (and between  $-13 \, \rm kHz$  and  $-15 \, \rm kHz$).


	With the Fourier correspondence given, the two components are:		With the Fourier correspondence given,  the two components are:
	:$$g_1(t) \ = \ A \cdot 6\,{\rm kHz} \cdot {\rm si}(\pi \cdot 6\,{\rm kHz} \cdot t)		:$$g_1(t) \ = \ A \cdot 6\,{\rm kHz} \cdot {\rm si}(\pi \cdot 6\,{\rm kHz} \cdot t)
	\hspace{0.05cm},$$		\hspace{0.05cm},$$
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	\\ {\rm{for}}\hspace{0.15cm}\|f\| > 1\,{\rm kHz} \hspace{0.05cm}. \\		\\ {\rm{for}}\hspace{0.15cm}\|f\| > 1\,{\rm kHz} \hspace{0.05cm}. \\
	\end{array}$$		\end{array}$$
	The bottom diagram shows the numerically determined time history $g(t)$. For the time $t = T = 0.1\, \rm ms$ (yellow square) we get:		The bottom diagram shows the numerically determined time history  $g(t)$.  For the time  $t = T = 0.1\, \rm ms$  (yellow square)  we get:
	:$$g_2(t = T ) = 2A \cdot 2\,{\rm kHz} \cdot {\rm si}(0.2 \cdot \pi		:$$g_2(t = T ) = 2A \cdot 2\,{\rm kHz} \cdot {\rm si}(0.2 \cdot \pi)\cdot \cos (2.8 \cdot \pi)
	)\cdot \cos (2.8 \cdot \pi)
	= \frac{ A \cdot 4\,{\rm kHz}}{0.2 \cdot \pi}\cdot {\rm sin}(0.2 \cdot \pi		= \frac{ A \cdot 4\,{\rm kHz}}{0.2 \cdot \pi}\cdot {\rm sin}(0.2 \cdot \pi
	)\cdot\cos (0.8 \cdot \pi) = \frac{ A \cdot 10\,{\rm kHz}}{ \pi}\cdot [{\rm sin}(-0.6 \cdot		)\cdot\cos (0.8 \cdot \pi) = \frac{ A \cdot 10\,{\rm kHz}}{ \pi}\cdot [{\rm sin}(-0.6 \cdot
	\pi)+ {\rm sin}(\pi)] $$		\pi)+ {\rm sin}(\pi)] $$
			[[File:EN_Dig_A_1_4.png\|right\|frame\|Higher frequency Nyquist pulse]]
	:$$\Rightarrow \hspace{0.3cm} g_2(t = T ) = -\frac{ A \cdot 10\,{\rm kHz}}{ \pi}\cdot {\rm sin}(0.6 \cdot		:$$\Rightarrow \hspace{0.3cm} g_2(t = T ) = -\frac{ A \cdot 10\,{\rm kHz}}{ \pi}\cdot {\rm sin}(0.6 \cdot
	\pi).$$		\pi).$$
	For the first component $g_1(t)$, at time $t = T$:
	:$$g_1(t = T ) = A \cdot 6\,{\rm kHz} \cdot {\rm si}(0.6 \~~cdot~~ \pi		For the first component  $g_1(t)$,  at time  $t = T$:
	)= \frac{ A \cdot 6\,{\rm kHz}}{0.6 \cdot \pi}\cdot {\rm sin}(0.6 \cdot \pi		:$$g_1(t = T ) = A \cdot 6\,{\rm kHz} \cdot {\rm sinc}(0.6
	)= - g_2(t = T )$$		)$$
	:$$\Rightarrow \hspace{0.3cm} g(t = T ) = g_1(t = T ) + g_2(t = T )\hspace{0.1cm}\underline {= 0 } \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm}g_1(t = T ) = \frac{ A \cdot 6\,{\rm kHz}}{0.6 \cdot \pi}\cdot {\rm sin}(0.6 \cdot \pi
			)$$
			:$$\Rightarrow \hspace{0.3cm}g_1(t = T ) = - g_2(t = T )$$
			$$\Rightarrow \hspace{0.3cm} g(t = T ) = g_1(t = T ) + g_2(t = T )\hspace{0.1cm}\underline {= 0 } \hspace{0.05cm}.$$
	This result is not surprising because of the Nyquist property.		This result is not surprising because of the Nyquist property.
	~~[[File:EN_Dig_A_1_4.png\|center\|frame\|Higher frequency Nyquist pulse]]~~		<br clear=all>


	'''(4)'''  For $t = 2.5 T$ (green square), the following partial results are obtained:		'''(4)'''  For $t = 2.5 T$ (green square), the following partial results are obtained:
	:$$g_1(t = 2.5 T ) = A \cdot 6\,{\rm kHz} \cdot {\rm si}(1.5 \cdot \pi		:$$g_1(t = 2.5 T ) = A \cdot 6\,{\rm kHz} \cdot {\rm si}(1.5 \cdot \pi
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	)\cdot \cos (7 \cdot \pi)=- \frac{ A \cdot 8\,{\rm kHz}}{ \pi} $$		)\cdot \cos (7 \cdot \pi)=- \frac{ A \cdot 8\,{\rm kHz}}{ \pi} $$
	:$$\Rightarrow \hspace{0.3cm} g(t = 2.5 T ) = g_1(t = 2.5 T ) +g_2(t = 2.5 T ) = - \frac{ A \cdot 12\,{\rm kHz}}{ \pi} \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} g(t = 2.5 T ) = g_1(t = 2.5 T ) +g_2(t = 2.5 T ) = - \frac{ A \cdot 12\,{\rm kHz}}{ \pi} \hspace{0.05cm}.$$
	Considering $g(t = 0) = A \cdot 10 \ \rm kHz$, we get:		Considering  $g(t = 0) = A \cdot 10 \ \rm kHz$,  we get:
	:$$\frac{g(t = 2.5 T )}{g(t = 0)} = -\frac{ 1.2}{ \pi} \hspace{0.1cm}\underline {= -0.382 } \hspace{0.05cm}.$$		:$$\frac{g(t = 2.5 T )}{g(t = 0)} = -\frac{ 1.2}{ \pi} \hspace{0.1cm}\underline {= -0.382 } \hspace{0.05cm}.$$


	'''(5)'''  The second Nyquist criterion states that the Nyquist pulse $g(t)$ has zero crossings at $\pm 1.5T, \pm 2.5T, \pm 3.5T, ...$		'''(5)'''  The second Nyquist criterion states that the Nyquist pulse  $g(t)$  has zero crossings at $\pm 1.5T, \pm 2.5T, \pm 3.5T, ...$
	*According to the result from '''(4)''' this condition is not fulfilled here.		*According to the result from  '''(4)'''  this condition is not fulfilled here.
	*Therefore <u>solution 2</u> is correct.		*Therefore <u>solution 2</u> is correct.

Guenter at 13:28, 1 May 2022

2022-05-01T13:28:59Z

← Older revision		Revision as of 15:28, 1 May 2022
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	[[File:P_ID1278__Dig_A_1_4.png\|right\|frame\|Rectangular Nyquist spectrum]]		[[File:P_ID1278__Dig_A_1_4.png\|right\|frame\|Rectangular Nyquist spectrum]]
	Given by the sketch is the spectrum  $G(f)$  of the basic detection pulse, where the parameter  $A$  is still to be determined. Among other things it is to be checked whether this basic detection pulse fulfills one of the two Nyquist criteria. These are:		Given by the sketch is the spectrum  $G(f)$  of the basic detection pulse,  where the parameter  $A$  is still to be determined.  Among other things it is to be checked whether this basic detection pulse fulfills one of the two Nyquist criteria.  These are:
	*The '''first Nyquist criterion''' is fulfilled if for the spectral function holds:		*The  '''first Nyquist criterion'''  is fulfilled if for the spectral function holds:
	:$$\sum_{k = -\infty}^{+\infty} G(f -		:$$\sum_{k = -\infty}^{+\infty} G(f -
	{k}/{T} ) = {\rm const.}$$		{k}/{T} ) = {\rm const.}$$
	:In this case, the pulse  $g(t)$  has zero crossings at  $t = ν \cdot T$ for all integer values of  $ν$  except  $ν = 0$ . For the entire exercise,  $T = 0.1 \, \rm ms$  is assumed.		:In this case,  the pulse  $g(t)$  has zero crossings at  $t = ν \cdot T$  for all integer values of  $ν$  except  $ν = 0$.  For the entire exercise,  $T = 0.1 \, \rm ms$  is assumed.
	*If the '''second Nyquist criterion''' is satisfied,  $g(t)$  has zero crossings at  $\pm 1.5 T$,  $\pm 2.5 T$, etc.		*If the  '''second Nyquist criterion'''  is satisfied,  $g(t)$  has zero crossings at  $\pm 1.5 T$,  $\pm 2.5 T$, etc.


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	''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 two equations are assumed to be known:		*The two equations are assumed to be known:
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	\\ {\rm{for}}\hspace{0.15cm}\|f\| > f_0 \hspace{0.08cm} \\		\\ {\rm{for}}\hspace{0.15cm}\|f\| > f_0 \hspace{0.08cm} \\
	\end{array} \hspace{0.4cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ \hspace{0.4cm} x(t)		\end{array} \hspace{0.4cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ \hspace{0.4cm} x(t)
	=2 \cdot A \cdot f_0 \cdot {\rm si}(2 ~~\pi~~ f_0 T) \hspace{0.05cm},$$		=2 \cdot A \cdot f_0 \cdot {\rm sinc}(2 f_0 T) \hspace{0.05cm},$$
	:$$\sin(\alpha) \cdot \cos (\beta) = \frac{1}{2} \cdot \big[ \sin(\alpha - \beta) + \sin(\alpha + \beta)\big]		:$$\sin(\alpha) \cdot \cos (\beta) = \frac{1}{2} \cdot \big[ \sin(\alpha - \beta) + \sin(\alpha + \beta)\big]
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
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	$A \ = \ $ { 0.2 3% } $ \ \rm mV/Hz$		$A \ = \ $ { 0.2 3% } $ \ \rm mV/Hz$

	{Calculate  $g(t)$  from  $G(f)$  by applying the Fourier ~~inverse transformation~~. ~~<br>~~What (normalized) function value is obtained at  $t = T$?		{Calculate  $g(t)$  from  $G(f)$  by applying the inverse Fourier transform.  What  (normalized)  function value is obtained at  $t = T$?
	\|type="{}"}		\|type="{}"}
	$ g(t = T)/g(t = 0) \ = \ $ { 0. }		$ g(t = T)/g(t = 0) \ = \ $ { 0. }

	{Which (normalized) value results for  $t = 2.5T$?		{Which  (normalized)  value results for  $t = 2.5T$?
	\|type="{}"}		\|type="{}"}
	$g(t = 2.5 T)/g(t = 0)\ = \ $ { -0.39346--0.37054 }		$g(t = 2.5 T)/g(t = 0)\ = \ $ { -0.39346--0.37054 }

Hwang at 14:53, 7 March 2022

2022-03-07T14:53:06Z

Show changes

Guenter at 16:43, 8 February 2022

2022-02-08T16:43:59Z

← Older revision		Revision as of 18:43, 8 February 2022
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	:$$\Rightarrow \hspace{0.3cm} g(t = T ) = g_1(t = T ) + g_2(t = T )\hspace{0.1cm}\underline {= 0 } \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} g(t = T ) = g_1(t = T ) + g_2(t = T )\hspace{0.1cm}\underline {= 0 } \hspace{0.05cm}.$$
	Dieses Ergebnis ist aufgrund der Nyquisteigenschaft nicht überraschend.		Dieses Ergebnis ist aufgrund der Nyquisteigenschaft nicht überraschend.
	[[File:~~P_ID1281__Dig_A_1_4c~~.png\|center\|frame\|Höherfrequenter Nyquistimpuls]]		[[File:EN_Dig_A_1_4.png\|center\|frame\|Höherfrequenter Nyquistimpuls]]