Aufgaben:Exercise 1.4: Nyquist Criteria - Revision history

Guenter at 07:16, 20 May 2022

2022-05-20T07:16:42Z

Hwang at 15:52, 9 May 2022

2022-05-09T15:52:08Z

Guenter at 15:01, 1 May 2022

2022-05-01T15:01:37Z

Guenter at 14:43, 1 May 2022

2022-05-01T14:43:40Z

Guenter at 13:28, 1 May 2022

2022-05-01T13:28:59Z

Hwang at 14:53, 7 March 2022

2022-03-07T14:53:06Z

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Guenter at 16:43, 8 February 2022

2022-02-08T16:43:59Z

Guenter at 15:29, 2 February 2022

2022-02-02T15:29:40Z

Guenter: Guenter moved page Aufgaben:Aufgabe 1.4: Nyquistkriterien to Aufgaben:Exercise 1.4: Nyquist Criteria

2022-02-02T15:26:32Z

Guenter moved page Aufgabe 1.4: Nyquistkriterien to Exercise 1.4: Nyquist Criteria

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

2021-03-23T12:47:06Z

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

@@ Line 5: / Line 5: @@
 [[File:P_ID1278__Dig_A_1_4.png|right|frame|Rectangular Nyquist spectrum]]
-Given by the sketch is the spectrum &nbsp;$G(f)$&nbsp; of the basic detection pulse, '''KORREKTUR: basic transmitter pulse''' &nbsp; where the parameter &nbsp;$A$&nbsp; is still to be determined.&nbsp; Among other things it is to be checked whether this basic detection pulse '''KORREKTUR: basic transmitter pulse''' fulfills one of the two Nyquist criteria.&nbsp; These are:
+Given by the sketch is the spectrum &nbsp;$G(f)$&nbsp; of the basic detection pulse,&nbsp; where the parameter &nbsp;$A$&nbsp; is still to be determined.&nbsp; Among other things it is to be checked whether this basic detection pulse fulfills one of the two Nyquist criteria.&nbsp; These are:
 *The&nbsp; '''first Nyquist criterion'''&nbsp; is fulfilled if for the spectral function holds:
 :$$\sum_{k = -\infty}^{+\infty} G(f -

@@ Line 5: / Line 5: @@
 [[File:P_ID1278__Dig_A_1_4.png|right|frame|Rectangular Nyquist spectrum]]
-Given by the sketch is the spectrum &nbsp;$G(f)$&nbsp; of the basic detection pulse,&nbsp; where the parameter &nbsp;$A$&nbsp; is still to be determined.&nbsp; Among other things it is to be checked whether this basic detection pulse fulfills one of the two Nyquist criteria.&nbsp; These are:
+Given by the sketch is the spectrum &nbsp;$G(f)$&nbsp; of the basic detection pulse, '''KORREKTUR: basic transmitter pulse''' &nbsp; where the parameter &nbsp;$A$&nbsp; is still to be determined.&nbsp; Among other things it is to be checked whether this basic detection pulse '''KORREKTUR: basic transmitter pulse''' fulfills one of the two Nyquist criteria.&nbsp; These are:
 *The&nbsp; '''first Nyquist criterion'''&nbsp; is fulfilled if for the spectral function holds:
 :$$\sum_{k = -\infty}^{+\infty} G(f -

@@ Line 113: / Line 113: @@
     )$$
 :$$\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.
 <br clear=all>
@@ Line 128: / Line 128: @@
 '''(5)'''&nbsp; The second Nyquist criterion states that the Nyquist pulse&nbsp; $g(t)$&nbsp; has zero crossings at $\pm 1.5T, \pm 2.5T, \pm 3.5T, ...$
-*According to the result from&nbsp; '''(4)'''&nbsp; this condition is not fulfilled here.
+*According to the result from&nbsp; '''(4)'''&nbsp; this condition is not fulfilled here.&nbsp; Therefore <u>solution 2</u> is correct.

@@ Line 24: / Line 24: @@
 \\   {\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 sinc}(2 f_0 T) \hspace{0.05cm},$$
+=2 \cdot A \cdot f_0 \cdot {\rm si}(2 \pi f_0 T) \hspace{0.05cm},\hspace{0.4cm} {\rm si} (x) = \sin(x)/x\hspace{0.05cm},$$
 :$$\sin(\alpha) \cdot \cos (\beta)  =  \frac{1}{2} \cdot \big[ \sin(\alpha - \beta) + \sin(\alpha + \beta)\big]
   \hspace{0.05cm}.$$
@@ Line 62: / Line 62: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The following diagram shows the spectrum (the index "Per" here stands for "Periodic Continuation"):
+'''(1)'''&nbsp; The following diagram shows the spectrum&nbsp; (the index&nbsp; "Per"&nbsp; here stands for&nbsp; "Periodic Continuation"):
 :$$G_{\rm Per}(f) = \sum_{k = -\infty}^{+\infty} G(f -
 \frac{k}{T} ) \hspace{0.05cm}.$$
-*The indexing variable $k = 0$ indicates the original spectral function $G(f)$. This is filled in gray.
+*The indexing variable&nbsp; $k = 0$&nbsp; indicates the original spectral function&nbsp; $G(f)$.&nbsp; 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&nbsp; $1/T = 10\, \rm kHz$&nbsp; belongs to&nbsp; $k = 1$&nbsp; and is marked in green,&nbsp; while&nbsp; $k =  -1$&nbsp; 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&nbsp; $k = 2$&nbsp; and&nbsp; $k = - 2$.
-[[File:P_ID1280__Dig_A_1_4a.png|center|frame|Illustration of the first Nyquist criterion]]
+It can be seen that&nbsp; $G_{\rm Per}(f)$&nbsp; is constant.&nbsp; From this it follows that the first Nyquist criterion is fulfilled &nbsp; &rArr; &nbsp; <u>solution 1</u>&nbsp; is correct.
 '''(2)'''&nbsp; Due to the Fourier integrals the following relation holds:
 :$$g(t=0) =  \int_{-\infty}^{\infty}G(f) \,{\rm d} f
-  = A \cdot ( 2\,{\rm kHz}+6\,{\rm kHz}+2\,{\rm kHz})= A \cdot 10\,{\rm kHz}\hspace{0.3cm}
+  = A \cdot ( 2\,{\rm kHz}+6\,{\rm kHz}+2\,{\rm kHz})= A \cdot 10\,{\rm kHz}$$
-\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}.$$
-'''(3)'''&nbsp; Let $g(t) = g_{1}(t) +g_{2}(t)$, where
+'''(3)'''&nbsp; Let&nbsp; $g(t) = g_{1}(t) +g_{2}(t)$,&nbsp; where
-*$g_{1}(t)$ contains the spectral components in the interval $\pm 3 \, \rm kHz$ and
+*$g_{1}(t)$&nbsp; contains the spectral components in the interval $\pm 3 \, \rm kHz$&nbsp; and
-*$g_{2}(t)$ those between $13 \, \rm kHz$ and $15 \, \rm kHz$ (and between  $-13 \, \rm kHz$ and $-15 \, \rm kHz$).
+*$g_{2}(t)$&nbsp; those between $13 \, \rm kHz$ and $15 \, \rm kHz$&nbsp; (and between&nbsp;  $-13 \, \rm kHz$&nbsp; and&nbsp; $-15 \, \rm kHz$).
-With the Fourier correspondence given, the two components are:
+With the Fourier correspondence given,&nbsp; the two components are:
 :$$g_1(t)  \ = \ A \cdot 6\,{\rm kHz}  \cdot {\rm si}(\pi \cdot 6\,{\rm kHz} \cdot t)
     \hspace{0.05cm},$$
@@ Line 97: / Line 98: @@
 \\   {\rm{for}}\hspace{0.15cm}|f| > 1\,{\rm kHz}  \hspace{0.05cm}. \\
 \end{array}$$
-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&nbsp; $g(t)$.&nbsp; For the time&nbsp; $t = T = 0.1\, \rm ms$&nbsp; (yellow square)&nbsp; 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)
        =  \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)] $$
 :$$\Rightarrow \hspace{0.3cm} g_2(t = T )  = -\frac{ A \cdot 10\,{\rm kHz}}{ \pi}\cdot {\rm sin}(0.6 \cdot
     \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&nbsp; $g_1(t)$,&nbsp; at time&nbsp; $t = T$:
-   )= \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
 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)'''&nbsp; For $t = 2.5 T$ (green square), the following partial results are obtained:
 :$$g_1(t = 2.5 T ) =  A \cdot 6\,{\rm kHz}  \cdot {\rm si}(1.5 \cdot \pi
@@ Line 121: / Line 123: @@
     )\cdot \cos (7 \cdot \pi)=- \frac{ A \cdot 8\,{\rm kHz}}{ \pi} $$
 :$$\Rightarrow \hspace{0.3cm} g(t = 2.5  T )  = g_1(t = 2.5  T )  +g_2(t = 2.5  T ) =  - \frac{ A \cdot 12\,{\rm kHz}}{ \pi} \hspace{0.05cm}.$$
-Considering $g(t = 0) = A \cdot 10 \ \rm kHz$, we get:
+Considering&nbsp; $g(t = 0) = A \cdot 10 \ \rm kHz$,&nbsp; we get:
 :$$\frac{g(t = 2.5  T )}{g(t = 0)} =  -\frac{ 1.2}{ \pi} \hspace{0.1cm}\underline {= -0.382 } \hspace{0.05cm}.$$
-'''(5)'''&nbsp; 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)'''&nbsp; The second Nyquist criterion states that the Nyquist pulse&nbsp; $g(t)$&nbsp; 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&nbsp; '''(4)'''&nbsp; this condition is not fulfilled here.
 *Therefore <u>solution 2</u> is correct.

@@ Line 5: / Line 5: @@
 [[File:P_ID1278__Dig_A_1_4.png|right|frame|Rectangular Nyquist spectrum]]
-Given by the sketch is the spectrum &nbsp;$G(f)$&nbsp; of the basic detection pulse, where the parameter &nbsp;$A$&nbsp; 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 &nbsp;$G(f)$&nbsp; of the basic detection pulse,&nbsp; where the parameter &nbsp;$A$&nbsp; is still to be determined.&nbsp; Among other things it is to be checked whether this basic detection pulse fulfills one of the two Nyquist criteria.&nbsp; These are:
-*The '''first Nyquist criterion''' is fulfilled if for the spectral function holds:
+*The&nbsp; '''first Nyquist criterion'''&nbsp; is fulfilled if for the spectral function holds:
 :$$\sum_{k = -\infty}^{+\infty} G(f -
 {k}/{T} ) =  {\rm const.}$$
-:In this case, the pulse &nbsp;$g(t)$&nbsp; has zero crossings at &nbsp;$t = &nu; \cdot T$ for all integer values of &nbsp;$&nu;$&nbsp; except &nbsp;$&nu; = 0$&nbsp;. For the entire exercise, &nbsp;$T = 0.1 \, \rm  ms$&nbsp; is assumed.
+:In this case,&nbsp; the pulse &nbsp;$g(t)$&nbsp; has zero crossings at &nbsp;$t = &nu; \cdot T$&nbsp; for all integer values of &nbsp;$&nu;$&nbsp; except &nbsp;$&nu; = 0$.&nbsp; For the entire exercise, &nbsp;$T = 0.1 \, \rm  ms$&nbsp; is assumed.
-*If the '''second Nyquist criterion''' is satisfied, &nbsp;$g(t)$&nbsp; has zero crossings at &nbsp;$\pm 1.5 T$, &nbsp;$\pm 2.5 T$, etc.
+*If the&nbsp; '''second Nyquist criterion'''&nbsp; is satisfied, &nbsp;$g(t)$&nbsp; has zero crossings at &nbsp;$\pm 1.5 T$, &nbsp;$\pm 2.5 T$, etc.
@@ Line 16: / Line 16: @@
-''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 two equations are assumed to be known:
@@ Line 24: / Line 24: @@
 \\   {\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},$$
+=2 \cdot A \cdot f_0 \cdot {\rm sinc}(2 f_0 T) \hspace{0.05cm},$$
 :$$\sin(\alpha) \cdot \cos (\beta)  =  \frac{1}{2} \cdot \big[ \sin(\alpha - \beta) + \sin(\alpha + \beta)\big]
   \hspace{0.05cm}.$$
@@ Line 43: / Line 43: @@
 $A \ = \ $ { 0.2 3% } $ \ \rm mV/Hz$
-{Calculate &nbsp;$g(t)$&nbsp; from &nbsp;$G(f)$&nbsp; by applying the Fourier inverse transformation. <br>What (normalized) function value is obtained at &nbsp;$t = T$?
+{Calculate &nbsp;$g(t)$&nbsp; from &nbsp;$G(f)$&nbsp; by applying the inverse  Fourier transform.&nbsp; What&nbsp; (normalized)&nbsp; function value is obtained at &nbsp;$t = T$?
 |type="{}"}
 $ g(t = T)/g(t = 0) \ = \ $ { 0. }
-{Which (normalized) value results for &nbsp;$t = 2.5T$?
+{Which&nbsp; (normalized)&nbsp; value results for &nbsp;$t = 2.5T$?
 |type="{}"}
 $g(t = 2.5 T)/g(t = 0)\ = \ $ { -0.39346--0.37054 }

@@ Line 111: / Line 111: @@
 :$$\Rightarrow \hspace{0.3cm} g(t = T ) = g_1(t = T ) + g_2(t = T )\hspace{0.1cm}\underline {= 0 } \hspace{0.05cm}.$$
 Dieses Ergebnis ist aufgrund der Nyquisteigenschaft nicht überraschend.
-[[File:P_ID1281__Dig_A_1_4c.png|center|frame|Höherfrequenter Nyquistimpuls]]
+[[File:EN_Dig_A_1_4.png|center|frame|Höherfrequenter Nyquistimpuls]]

← Older revision		Revision as of 15:29, 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 12:47, 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^]]