Aufgaben:Exercise 5.4: Sine Wave Generator - Revision history

Guenter at 17:53, 10 February 2022

2022-02-10T17:53:28Z

Guenter at 17:44, 10 February 2022

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Guenter at 17:38, 10 February 2022

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Guenter at 17:34, 10 February 2022

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Guenter at 17:18, 10 February 2022

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Hwang at 10:52, 17 January 2022

2022-01-17T10:52:01Z

Guenter at 10:36, 12 January 2022

2022-01-12T10:36:03Z

Guenter: Guenter moved page Aufgaben:Aufgabe 5.4: Sinusgenerator to Aufgaben:Exercise 5.4: Sine Wave Generator

2022-01-12T08:59:32Z

Guenter moved page Aufgabe 5.4: Sinusgenerator to Exercise 5.4: Sine Wave Generator

Javier: Text replacement - "”" to """

2021-05-28T14:34:18Z

Text replacement - "”" to """

Javier: Text replacement - "„" to """

2021-05-28T14:32:53Z

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← Older revision		Revision as of 17:53, 10 February 2022
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	*Checking the coefficient  $b_1$  confirms the calculation:		*Checking the coefficient  $b_1$  confirms the calculation:
−	:$$b_1 = 2 \cdot \cos \left( {{{\rm{\pi }}}/{6}} \right) = 2 \cdot c{\sqrt 3 }/{2} = \sqrt 3 .$$	+	:$$b_1 = 2 \cdot \cos \left( {{{\rm{\pi }}}/{6}} \right) = 2 \cdot {\sqrt 3 }/{2} = \sqrt 3 .$$

@@ Line 56: / Line 56: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The "$1$" at the input affects&nbsp;  $($because of&nbsp; $a_0= 0)$&nbsp; at the output only at the time&nbsp; $\nu = 1$:&nbsp;
+'''(1)'''&nbsp; The&nbsp; "$1$"&nbsp; at the input affects&nbsp;  $($because of&nbsp; $a_0= 0)$&nbsp; at the output only at the time&nbsp; $\nu = 1$:&nbsp;
 :$$y_0   \hspace{0.15cm} \underline{= 0},\quad y_1  \hspace{0.15cm} \underline{ = 0.5}.$$
@@ Line 64: / Line 64: @@
-'''(2)'''&nbsp; For&nbsp; $\nu \ge 2$,&nbsp; the filter is purely recursive:
+'''(2)'''&nbsp; For&nbsp; $\nu \ge 2$, &nbsp; the filter is purely recursive:
 :$$y_\nu   = b_1  \cdot y_{\nu  - 1}  - y_{\nu  - 2} .$$
-*In particular, one obtains
+*In particular,&nbsp; one obtains
 :$$y_3  = \sqrt 3  \cdot y_2  - y_1  = \sqrt 3  \cdot {\sqrt 3 }/{2} - {1}/{2} = 1;$$
 :$$y_4  = \sqrt 3  \cdot y_3  - y_2  = \sqrt 3  \cdot 1 - {\sqrt 3 }/{2} = {\sqrt 3 }/{2};$$
@@ Line 86: / Line 86: @@
-'''(4)'''&nbsp; From &nbsp;$f_0 = 10 \hspace{0.05cm} \rm kHz$&nbsp; follows &nbsp;$T_0 = 100 \hspace{0.05cm} \rm &micro; s$&nbsp; or &nbsp;$T_0/T = 100$&nbsp;. This gives:
+'''(4)'''&nbsp; From &nbsp;$f_0 = 10 \hspace{0.15cm} \rm kHz$&nbsp; follows &nbsp;$T_0 = 100 \hspace{0.05cm} \rm &micro; s$&nbsp; or &nbsp;$T_0/T = 100$&nbsp;. This gives:
 :$$a_1  = \sin \left( {2{\rm{\pi }}\cdot{T}/{T_0 }} \right) = \sin \left( {3.6^ \circ  } \right) \hspace{0.15cm} \underline{\approx 0.062},$$
 :$$b_1  = 2 \cdot \cos \left( {2{\rm{\pi }}\cdot{T}/{T_0 }} \right) = 2 \cdot \cos \left( {3.6^ \circ  } \right) \hspace{0.15cm} \underline{\approx 1.996}.$$

@@ Line 18: / Line 18: @@
 Notes:
 *The exercise belongs to the chapter &nbsp; [[Theory_of_Stochastic_Signals/Digital_Filters|Digital Filters]]&nbsp; in this book.
 *The HTML5/JavaScript applet&nbsp; [[Applets:Digital_Filters|"Digital Filters"]]&nbsp; illustrates the subject matter of this chapter.
@@ Line 48: / Line 45: @@
-{Now&nbsp; $T = 1 \hspace{0.05cm} \rm &micro; s$&nbsp; is valid. How must the coefficients&nbsp; $a_1$&nbsp; and&nbsp; $b_1$&nbsp; be selected so that a&nbsp; $\text{10 kHz}$ sine wave is generated?
+{Now&nbsp; $T = 1 \hspace{0.05cm} \rm &micro; s$&nbsp; is valid.&nbsp; How must the coefficients&nbsp; $a_1$&nbsp; and&nbsp; $b_1$&nbsp; be selected so that a&nbsp; $\text{10 kHz}$ sine wave is generated?
 |type="{}"}
 $a_1 \ =  \ $ { 0.062 3% }

@@ Line 4: / Line 4: @@
 [[File:P_ID622__Sto_A_5_4.png|right|frame|Proposed filter structure]]
-The diagram shows a second-order digital filter suitable for generating a discrete-time sinusoidal function on a digital signal processor (DSP), for example:
+The diagram shows a second-order digital filter suitable for generating a discrete-time sinusoidal function on a digital signal processor&nbsp; $\rm (DSP)$,&nbsp; for example:
 :$$\left\langle \hspace{0.05cm}{y_\nu  }\hspace{0.05cm} \right\rangle  = \left\langle {\, \sin ( {\nu T \cdot \omega _0  } )\, }\right\rangle .$$
-*It is assumed that the input sequence&nbsp; $\left\langle \hspace{0.05cm} {x_\nu  } \hspace{0.05cm}\right\rangle$&nbsp; describes a (discrete-time) Dirac function.&nbsp; Thus, all output values&nbsp; $y_\nu$&nbsp; are simultaneously identical to zero for times&nbsp; $\nu\lt 0$.&nbsp;
+*It is assumed that the input sequence&nbsp; $\left\langle \hspace{0.05cm} {x_\nu  } \hspace{0.05cm}\right\rangle$&nbsp; describes a&nbsp; (discrete-time)&nbsp; Dirac delta function.&nbsp; Thus,&nbsp; all output values&nbsp; $y_\nu$&nbsp; are simultaneously identical to zero for times&nbsp; $\nu\lt 0$.&nbsp;
 *The total of five filter coefficients result from the&nbsp; [https://en.wikipedia.org/wiki/Z-transform $Z$-transform]:
 :$$Z \{ {\sin ( {\nu T \omega _0 } )} \} = \frac{{z \cdot \sin \left( {\omega _0  T} \right)}}{{z^2  - 2 \cdot z \cdot \cos \left( {\omega _0   T} \right) + 1}}.$$

@@ Line 23: / Line 23: @@
 ''Notes:''
 *The exercise belongs to the chapter &nbsp; [[Theory_of_Stochastic_Signals/Digital_Filters|Digital Filters]]&nbsp; in this book.
+*The HTML5/JavaScript applet&nbsp; [[Applets:Digital_Filters|"Digital Filters"]]&nbsp; illustrates the subject matter of this chapter.
 *For the subtasks&nbsp; '''(1)'''&nbsp; to&nbsp; '''(3)'''&nbsp; the following apply:
 :$$a_1  = 0.5,\quad b_1  = \sqrt 3 .$$

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Stochastische Signaltheorie/Digitale Filter
+{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Digital_Filters
 }}
-[[File:P_ID622__Sto_A_5_4.png|right|frame|Vorgeschlagene Filterstruktur]]
+[[File:P_ID622__Sto_A_5_4.png|right|frame|Proposed filter structure]]
-Die Grafik zeigt ein digitales Filter zweiter Ordnung, das zum Beispiel zur Erzeugung einer zeitdiskreten Sinusfunktion auf einem digitalen Signalprozessor (DSP) geeignet ist:
+The diagram shows a second-order digital filter suitable for generating a discrete-time sinusoidal function on a digital signal processor (DSP), for example:
 :$$\left\langle \hspace{0.05cm}{y_\nu  }\hspace{0.05cm} \right\rangle  = \left\langle {\, \sin ( {\nu T \cdot \omega _0  } )\, }\right\rangle .$$
-*Vorausgesetzt wird, dass die Eingangsfolge&nbsp; $\left\langle \hspace{0.05cm} {x_\nu  } \hspace{0.05cm}\right\rangle$&nbsp; eine (zeitdiskrete) Diracfunktion beschreibt.&nbsp; Damit sind gleichzeitig alle Ausgangswerte&nbsp; $y_\nu$&nbsp; für Zeiten&nbsp; $\nu\lt 0$&nbsp; identisch Null.
+*It is assumed that the input sequence&nbsp; $\left\langle \hspace{0.05cm} {x_\nu  } \hspace{0.05cm}\right\rangle$&nbsp; describes a (discrete-time) Dirac function.&nbsp; Thus, all output values&nbsp; $y_\nu$&nbsp; are simultaneously identical to zero for times&nbsp; $\nu\lt 0$.&nbsp;
-*Die insgesamt fünf Filterkoeffizienten ergeben sich aus der&nbsp; [https://de.wikipedia.org/wiki/Z-Transformation $Z$-Transformation]:
+*The total of five filter coefficients result from the&nbsp; [https://en.wikipedia.org/wiki/Z-transform $Z$-transform]:
 :$$Z \{ {\sin ( {\nu T \omega _0 } )} \} = \frac{{z \cdot \sin \left( {\omega _0  T} \right)}}{{z^2  - 2 \cdot z \cdot \cos \left( {\omega _0   T} \right) + 1}}.$$
-*Setzt man diese Gleichung durch ein rekursives Filter zweiter Ordnung&nbsp; $(M = 2)$&nbsp; um, so erhält man die folgenden Filterkoeffizienten:
+*Substituting this equation by a recursive second order filter&nbsp; $(M = 2)$,&nbsp; we obtain the following filter coefficients:
 :$$a_0 = 0,\quad a_1  = \sin \left( {\omega _0   T} \right),\quad a_2  = 0, \quad b_1  = 2 \cdot \cos \left( {\omega _0   T} \right),\quad b_2  =  - 1.$$
-In der Grafik ist bereits durch die hellere Umrandung markiert, dass auf die Filterkoeffizienten&nbsp; $a_0$&nbsp; und&nbsp; $a_2$&nbsp; verzichtet werden kann.
+In the diagram it is already marked by the lighter border that the filter coefficients&nbsp; $a_0$&nbsp; and&nbsp; $a_2$&nbsp; can be omitted.
@@ Line 21: / Line 21: @@
-''Hinweise:''
+''Notes:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Digitale_Filter|Digitale Filter]]&nbsp; im vorliegenden Buch.
+*The exercise belongs to the chapter &nbsp; [[Theory_of_Stochastic_Signals/Digital_Filters|Digital Filters]]&nbsp; in this book.
-*Für die Teilaufgaben&nbsp; '''(1)'''&nbsp; bis&nbsp; '''(3)'''&nbsp; gelte:
+*For the subtasks&nbsp; '''(1)'''&nbsp; to&nbsp; '''(3)'''&nbsp; the following apply:
 :$$a_1  = 0.5,\quad b_1  = \sqrt 3 .$$
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Es gelte&nbsp; $a_1  = 0.5$&nbsp; und&nbsp; $b_1  = \sqrt 3 $.&nbsp; Berechnen Sie die Ausgangswerte&nbsp; $y_\nu$&nbsp; zu den Zeitpunkten&nbsp; $\nu = 0$,&nbsp; $\nu = 1$&nbsp; und&nbsp; $\nu = 2$.
+{Let&nbsp; $a_1  = 0.5$&nbsp; and&nbsp; $b_1  = \sqrt 3 $.&nbsp; Calculate the initial values&nbsp; $y_\nu$&nbsp; at time points&nbsp; $\nu = 0$,&nbsp; $\nu = 1$&nbsp; and&nbsp; $\nu = 2$.
 |type="{}"}
 $y_0 \ = \ $ { 0. }
@@ Line 38: / Line 38: @@
-{Wie lautet der Ausgangswert &nbsp;$y_\nu$&nbsp; für &nbsp;$\nu \ge 2$&nbsp; allgemein?&nbsp; Berechnen Sie die Werte &nbsp;$y_3$, ... , $y_7$&nbsp; und geben Sie zur Kontrolle &nbsp;$y_7$&nbsp; ein.
+{What is the initial value &nbsp;$y_\nu$&nbsp; for &nbsp;$\nu \ge 2$&nbsp; in general?&nbsp; Calculate the values &nbsp;$y_3$, ... , $y_7$&nbsp; and enter &nbsp;$y_7$&nbsp; as a check.
 |type="{}"}
 $y_7 \ =  \ $ { -0.515--0.485 }
-{Wie viele Stützstellen&nbsp; $(T_0/T)$&nbsp; stellen eine Periodendauer&nbsp; $(T_0)$&nbsp; dar?
+{How many grid points&nbsp; $(T_0/T)$&nbsp; represent a period&nbsp; $(T_0)$?&nbsp;
 |type="{}"}
 $T_0/T\ =  \ $ { 12 3% }
-{Es gelte nun&nbsp; $T = 1 \hspace{0.05cm} \rm &micro; s$.&nbsp; Wie müssen die Koeffizienten&nbsp; $a_1$&nbsp; und&nbsp; $b_1$&nbsp; gewählt werden, damit eine&nbsp; $\text{10 kHz}$&ndash;Sinusschwingung erzeugt wird?
+{Now&nbsp; $T = 1 \hspace{0.05cm} \rm &micro; s$&nbsp; is valid. How must the coefficients&nbsp; $a_1$&nbsp; and&nbsp; $b_1$&nbsp; be selected so that a&nbsp; $\text{10 kHz}$ sine wave is generated?
 |type="{}"}
 $a_1 \ =  \ $ { 0.062 3% }
@@ Line 57: / Line 57: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Die "$1$" am Eingang wirkt sich&nbsp;  $($wegen&nbsp; $a_0= 0)$&nbsp; am Ausgang erst zum Zeitpunkt&nbsp; $\nu = 1$&nbsp; aus:
+'''(1)'''&nbsp; The "$1$" at the input affects&nbsp;  $($because of&nbsp; $a_0= 0)$&nbsp; at the output only at the time&nbsp; $\nu = 1$:&nbsp;
 :$$y_0   \hspace{0.15cm} \underline{= 0},\quad y_1  \hspace{0.15cm} \underline{ = 0.5}.$$
-*Bei&nbsp; $\nu = 2$&nbsp; wird auch der rekursive Teil des Filters wirksam:
+*At&nbsp; $\nu = 2$,&nbsp; the recursive part of the filter also takes effect:
 :$$y_2  = b_1  \cdot y_1  - y_0  = {\sqrt 3 }/{2} \hspace{0.15cm} \underline{ \approx 0.866}.$$
-'''(2)'''&nbsp; Für&nbsp; $\nu \ge 2$&nbsp; ist das Filter rein rekursiv:
+'''(2)'''&nbsp; For&nbsp; $\nu \ge 2$,&nbsp; the filter is purely recursive:
 :$$y_\nu   = b_1  \cdot y_{\nu  - 1}  - y_{\nu  - 2} .$$
-*Insbesondere erhält man
+*In particular, one obtains
 :$$y_3  = \sqrt 3  \cdot y_2  - y_1  = \sqrt 3  \cdot {\sqrt 3 }/{2} - {1}/{2} = 1;$$
 :$$y_4  = \sqrt 3  \cdot y_3  - y_2  = \sqrt 3  \cdot 1 - {\sqrt 3 }/{2} = {\sqrt 3 }/{2};$$
@@ Line 79: / Line 79: @@
-'''(3)'''&nbsp; Durch Fortsetzung des rekursiven Algorithmuses der Teilaufgabe&nbsp; '''(2)'''&nbsp; erhält man für große&nbsp; $\nu$&ndash;Werte: &nbsp;
+'''(3)'''&nbsp; By continuing the recursive algorithm of subtask&nbsp; '''(2)''',&nbsp; we obtain for large&nbsp; $\nu$&ndash;values: &nbsp;
 :$$y_\nu   = y_{\nu  - 12} .$$
-*Daraus folgt&nbsp; $T_0/T\hspace{0.15cm} \underline{= 12}$.&nbsp; Zum gleichen Ergebnis kommt man durch folgende Überlegungen:
+*From this follows&nbsp; $T_0/T\hspace{0.15cm} \underline{= 12}$.&nbsp; The same result is obtained by the following considerations:
 :$$a_1  = \sin \left( {\omega _0  \cdot T} \right) = \sin \left( {2{\rm{\pi }}\cdot{T}/{T_0 }} \right)\mathop  = \limits^! {1}/{2} = \sin \left( {{{\rm{\pi }}}/{6}} \right) \;\;{\rm \Rightarrow} \;\;{2T}/{T_0 } = {1}/{6}\quad  \Rightarrow \;\;{T_0 }/{T} = 12.$$
-*Die Überprüfung des Koeffizienten&nbsp; $b_1$&nbsp; bestätigt die Rechnung:
+*Checking the coefficient&nbsp; $b_1$&nbsp; confirms the calculation:
 :$$b_1  = 2 \cdot \cos \left( {{{\rm{\pi }}}/{6}} \right) = 2 \cdot c{\sqrt 3 }/{2} = \sqrt 3 .$$
-'''(4)'''&nbsp; Aus &nbsp;$f_0 = 10 \hspace{0.05cm} \rm kHz$&nbsp; folgt &nbsp;$T_0 = 100 \hspace{0.05cm} \rm &micro; s$&nbsp; bzw. &nbsp;$T_0/T = 100$&nbsp;. Damit erhält man:
+'''(4)'''&nbsp; From &nbsp;$f_0 = 10 \hspace{0.05cm} \rm kHz$&nbsp; follows &nbsp;$T_0 = 100 \hspace{0.05cm} \rm &micro; s$&nbsp; or &nbsp;$T_0/T = 100$&nbsp;. This gives:
 :$$a_1  = \sin \left( {2{\rm{\pi }}\cdot{T}/{T_0 }} \right) = \sin \left( {3.6^ \circ  } \right) \hspace{0.15cm} \underline{\approx 0.062},$$
 :$$b_1  = 2 \cdot \cos \left( {2{\rm{\pi }}\cdot{T}/{T_0 }} \right) = 2 \cdot \cos \left( {3.6^ \circ  } \right) \hspace{0.15cm} \underline{\approx 1.996}.$$

← Older revision		Revision as of 10:36, 12 January 2022
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−	[[Category:Theory of Stochastic Signals: Exercises\|^5.2 ~~Digitale Filter~~^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^5.2 Digital Filters^]]

← Older revision	Revision as of 08:59, 12 January 2022
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