Aufgaben:Exercise 4.1: About the Gram-Schmidt Process - Revision history

Guenter: Guenter moved page Aufgaben:Eercise 4.1: About the Gram-Schmidt Process to Aufgaben:Exercise 4.1: About the Gram-Schmidt Process

2022-07-13T12:56:39Z

Guenter moved page Eercise 4.1: About the Gram-Schmidt Process to Exercise 4.1: About the Gram-Schmidt Process

Guenter at 15:01, 7 July 2022

2022-07-07T15:01:47Z

Guenter at 14:41, 7 July 2022

2022-07-07T14:41:08Z

Guenter: Guenter moved page Aufgaben:Eercise 4.1: About the Gram-Schmidt Method to Aufgaben:Eercise 4.1: About the Gram-Schmidt Process

2022-07-07T14:36:54Z

Guenter moved page Eercise 4.1: About the Gram-Schmidt Method to Eercise 4.1: About the Gram-Schmidt Process

Hwang at 13:50, 13 June 2022

2022-06-13T13:50:12Z

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Guenter at 11:52, 8 June 2022

2022-06-08T11:52:23Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.1: Zum Gram-Schmidt-Verfahren to Aufgaben:Eercise 4.1: About the Gram-Schmidt Method

2022-06-08T11:42:05Z

Guenter moved page Aufgabe 4.1: Zum Gram-Schmidt-Verfahren to Eercise 4.1: About the Gram-Schmidt Method

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

2021-03-23T12:52:18Z

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

Javier: Text replacement - "[[Digitalsignalübertragung" to "[[Digital_Signal_Transmission"

2020-07-09T12:50:30Z

Text replacement - "[[Digitalsignalübertragung" to "[[Digital_Signal_Transmission"

Guenter at 17:08, 10 March 2019

2019-03-10T17:08:41Z

@@ Line 59: / Line 59: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; <u>Solutions 2 and 4</u> are correct:
+'''(1)'''&nbsp; <u>Solutions 2 and 4</u>&nbsp; are correct:
-*Every orthonormal basis function should have energy 1, that is, it must hold:
+*Every orthonormal basis function should have energy&nbsp; $1$,&nbsp; that is,&nbsp; it must hold:
 :$$||\varphi_j(t)||^2 =  \int_{-\infty}^{+\infty}\varphi_j(t)^2\,{\rm d}  t = 1
 \hspace{0.05cm}.$$
-*For this condition to be satisfied, the basis function must have unit $\rm \sqrt{\rm s}$. Another equation to be considered is
+*For this condition to be satisfied,&nbsp; the basis function must have unit&nbsp; $\rm \sqrt{\rm s}$.&nbsp;
 :$$s_i(t) = \sum\limits_{j = 1}^{N}s_{ij} \cdot \varphi_j(t).$$
-*Like $A$, the signals themselves have the unit $\rm \sqrt{\rm W}$. Because of the unit $\rm \sqrt{\rm 1/s}$ of $\varphi_{ j}(t)$, this equation can be satisfied with the correct dimension only if the coefficients $s_{\it ij}$ are given with the unit $\rm \sqrt{\rm Ws}$.
+*Like the parameter&nbsp; $A$,&nbsp; the signals themselves have the unit&nbsp; $\rm \sqrt{\rm W}$.&nbsp;
-'''(2)'''&nbsp; The energy of the signal $s_1(t)$ is equal to $E_1 = 2$. It follows for the norm, the basis function $\varphi_1(t)$ and the coefficient $s_{\rm 11}$:
+'''(2)'''&nbsp; The energy of the signal&nbsp; $s_1(t)$ &nbsp;is equal to&nbsp; $E_1 = 2$.&nbsp;
 :$$||s_1(t)|| =  \sqrt{2},\hspace{0.9cm}\varphi_1(t) = \frac{s_1(t)}{||s_1(t)||},\hspace{0.9cm}
 s_{11} = \sqrt{E_1} = \sqrt{2} \hspace{0.1cm}\hspace{0.15cm}\underline { {\approx 1.414} }
 \hspace{0.05cm}.$$
+'''(3)'''&nbsp; Since at most two basis functions are found after considering $s_2(t)$ &nbsp; &rArr; &nbsp; $s_{\rm 23} \hspace{0.15cm} \underline{= 0}$ holds with certainty.&nbsp; On the other hand one obtains
-'''(3)'''&nbsp; Since at most two basis functions are found after considering $s_2(t)$, $s_{\rm 23} \hspace{0.15cm} \underline{= 0}$ holds with certainty. On the other hand one obtains for the coefficient
+*for the coefficient
 :$$||s_1(t)|| =  \sqrt{2},\hspace{0.9cm}\varphi_1(t) = \frac{s_1(t)}{||s_1(t)||},\hspace{0.9cm}
 s_{11} = \sqrt{E_1} = \sqrt{2} \hspace{0.1cm}\hspace{0.15cm}\underline { {\approx 1.414} }
 \hspace{0.05cm};$$
-for the auxiliary function $\theta_2(t)$:
+*for the auxiliary function $\theta_2(t)$:
 :$$\theta_2(t) = s_2(t) - s_{21} \cdot \varphi_1(t) = \left\{ \begin{array}{c} 1 - 0.707 \cdot 0.707 = 0.5\\
 - 0.707 \cdot (-0.707) = 0.5  \end{array} \right.\quad
@@ Line 90: / Line 93: @@
 \hspace{0.05cm}; $$
-for the second basis function:
+*for the second basis function:
 :$$\varphi_2(t) = \frac{\theta_2(t)}{||\theta_2(t)||},\hspace{0.2cm}
 ||\theta_2(t)|| = \sqrt{0.5^2 + 0.5^2} = \sqrt{0.5} \approx 0.707$$
@@ Line 99: / Line 102: @@
 \hspace{0.05cm}; $$
-and finally for the second coefficient
+*and finally for the second coefficient
 :$$s_{22}  = \hspace{0.1cm} < \hspace{-0.1cm} s_2(t), \hspace{0.1cm}\varphi_2(t) \hspace{-0.1cm} > \hspace{0.1cm} = 1 \cdot 0.707 + 0 \cdot 0.707 \hspace{0.1cm}\hspace{0.15cm}\underline {  = 0.707}
 \hspace{0.05cm}.$$
 The calculations are illustrated in the graph below.
-'''(4)'''&nbsp; It can be seen immediately that $s_3(t)$ can be expressed as a linear combination of $s_1(t)$ and $s_2(t)$.
+'''(4)'''&nbsp; It can be seen immediately that&nbsp; $s_3(t)$&nbsp; can be expressed as a linear combination of&nbsp; $s_1(t)$&nbsp; and&nbsp; $s_2(t)$.
-:$$s_{3}(t)  = -s_{1}(t) + s_{2}(t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}s_{31} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - s_{11} + s_{21} = -1.414 + 0.707 = \hspace{0.1cm}\hspace{0.15cm}\underline {-0.707}\hspace{0.05cm},$$
+:$$s_{3}(t)  = -s_{1}(t) + s_{2}(t),$$
 :$$s_{32} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - s_{12} + s_{22} = 0 + 0.707 \hspace{0.1cm}\underline {= 0.707}\hspace{0.05cm},$$
 :$$s_{33} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - s_{13} + s_{23} = 0 + 0 \hspace{0.1cm}\underline {= 0}\hspace{0.05cm}. $$
-'''(5)'''&nbsp; The range $2 &#8804; t &#8804; 3$ is not covered by either $\varphi_1(t)$ or $\varphi_2(t)$. Therefore, $s_4(t)$ provides the new basis function $\varphi_3(t)$. Furthermore, since $s_4(t)$ has components only in the range $2 &#8804; t &#8804; 3$ and $||s_4(t)|| = 1$, we obtain $\varphi_3(t) = s_4(t)$ as well as
+'''(5)'''&nbsp; The range&nbsp; $2 &#8804; t &#8804; 3$&nbsp; is not covered by&nbsp; $\varphi_1(t)$&nbsp; and&nbsp; $\varphi_2(t)$.
 :$$s_{41} \hspace{0.1cm}\hspace{0.15cm}\underline {= 0},  \hspace{0.2cm}s_{42} \hspace{0.1cm}\hspace{0.15cm}\underline {= 0},  \hspace{0.2cm}s_{43} \hspace{0.1cm}\hspace{0.15cm}\underline { = 1} \hspace{0.05cm}.  $$
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 [[File:P_ID1994__Dig_A_4_1.png|right|frame|Specification for the Gram-Schmidt process]]
-For the four signals&nbsp; $s_1(t), \, \text{...} \, , s_4(t)$&nbsp; defined by the figure, the three resulting basis functions&nbsp; $\varphi_1(t)$,&nbsp; $\varphi_2(t)$&nbsp; and&nbsp; $\varphi_3(t)$&nbsp; are to be determined by applying the Gram-Schmidt process, so that for the signals with &nbsp; $i = 1, \, \text{...} \, , 4$&nbsp; can be written:
+For the four signals&nbsp; $s_1(t), \, \text{...} \, , s_4(t)$&nbsp; defined by the figure,&nbsp; the three resulting basis functions&nbsp; $\varphi_1(t)$,&nbsp; $\varphi_2(t)$&nbsp; and&nbsp; $\varphi_3(t)$&nbsp; are to be determined by applying the Gram-Schmidt process,&nbsp; so that for the signals with &nbsp; $i = 1, \, \text{...} \, , 4$&nbsp; can be written:
 :$$s_i(t) = s_{i1} \cdot \varphi_1(t) + s_{i2} \cdot \varphi_2(t) + s_{i3} \cdot \varphi_3(t)\hspace{0.05cm}.$$
-*In subtask '''(1)''', let&nbsp; $A^2 = 1 \ \rm mW$&nbsp; and&nbsp; $T = 1 \ \rm &micro; s$.
+*In subtask&nbsp; '''(1)''',&nbsp; let&nbsp; $A^2 = 1 \ \rm mW$&nbsp; and&nbsp; $T = 1 \ \rm &micro; s$.
-*In the later subtasks, the amplitude and the time are normalized quantities, respectively: &nbsp; $A = 1$,&nbsp; $T = 1$.
-*Thus, both the coefficients&nbsp; $s_{\it ij}$&nbsp; and the basis functions&nbsp; $\varphi_{\it j}(t)$&nbsp; &ndash; with&nbsp; $j = 1, 2, 3$&nbsp; &ndash; are dimensionless quantities.
+*In the later subtasks,&nbsp; the amplitude and the time are normalized quantities: &nbsp; $A = 1$,&nbsp; $T = 1$.
 *Reference is made in particular to the sections&nbsp; [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces#Orthonormal_basis_functions|"Orthonormal basis functions"]]&nbsp; and&nbsp; [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces#The_Gram-Schmidt_process|"Gram-Schmidt process"]].
@@ Line 31: / Line 32: @@
 + The coefficients &nbsp;$s_{\it ij}$&nbsp; have the unit &nbsp;$\rm \sqrt{\rm Ws}$.
-{Perform the first step of the Gram-Schmidt process. As for the other tasks, let &nbsp;$A = 1$&nbsp; and &nbsp;$T = 1$ hold.
+{Perform the first step of the Gram-Schmidt process.&nbsp; As for the other tasks,&nbsp; let &nbsp;$A = 1$&nbsp; and &nbsp;$T = 1$ hold.
 |type="{}"}
 $s_{\rm 11} \ = \ $ { 1.414 3% }

← Older revision		Revision as of 11:52, 8 June 2022
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−	[[Category:Digital Signal Transmission: Exercises\|^4.1 ~~Basisfunktionen~~ & ~~Vektorräume~~^]]	+	[[Category:Digital Signal Transmission: Exercises\|^4.1 Basis Functions & Vector Spaces^]]

← Older revision		Revision as of 12:52, 23 March 2021
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−	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^4.1 Basisfunktionen & Vektorräume^]]	+	[[Category:Digital Signal Transmission: Exercises\|^4.1 Basisfunktionen & Vektorräume^]]

@@ Line 17: / Line 17: @@
 ''Hinweise:''
 *Die Aufgabe gehört zum  Kapitel&nbsp;  [[Digitalsignal%C3%BCbertragung/Signale,_Basisfunktionen_und_Vektorr%C3%A4ume| Signale, Basisfunktionen und Vektorräume]].
-*Bezug genommen wird insbesondere auf die Seiten&nbsp; [[Digitalsignalübertragung/Signale,_Basisfunktionen_und_Vektorräume#Orthonormale_Basisfunktionen|Orthonormale Basisfunktionen]]&nbsp; und&nbsp; [[Digitalsignal%C3%BCbertragung/Signale,_Basisfunktionen_und_Vektorr%C3%A4ume#Das_Verfahren_nach_Gram-Schmidt|Gram&ndash;Schmidt&ndash;Verfahren]].
+*Bezug genommen wird insbesondere auf die Seiten&nbsp; [[Digital_Signal_Transmission/Signale,_Basisfunktionen_und_Vektorräume#Orthonormale_Basisfunktionen|Orthonormale Basisfunktionen]]&nbsp; und&nbsp; [[Digitalsignal%C3%BCbertragung/Signale,_Basisfunktionen_und_Vektorr%C3%A4ume#Das_Verfahren_nach_Gram-Schmidt|Gram&ndash;Schmidt&ndash;Verfahren]].

← Older revision	Revision as of 12:56, 13 July 2022
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← Older revision	Revision as of 14:36, 7 July 2022
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← Older revision	Revision as of 11:42, 8 June 2022
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@@ Line 65: / Line 65: @@
 :$$s_i(t) = \sum\limits_{j = 1}^{N}s_{ij} \cdot \varphi_j(t).$$
 *Die Signale selbst weisen wie $A$ die Einheit $\rm \sqrt{\rm W}$ auf. Wegen der Einheit $\rm \sqrt{\rm 1/s}$ von $\varphi_{ j}(t)$ ist diese Gleichung nur dann mit der richtigen Dimension zu erfüllen, wenn die Koeffizienten $s_{\it ij}$ mit der Einheit $\rm \sqrt{\rm Ws}$ angegeben werden.
 '''(2)'''&nbsp; Die Energie des Signals $s_1(t)$ ist gleich $E_1 = 2$. Daraus folgt für die Norm, die Basisfunktion $\varphi_1(t)$ und den Koeffizienten $s_{\rm 11}$:
-:$$||s_1(t)|| =  \sqrt{2},\hspace{0.2cm}\varphi_1(t) = \frac{s_1(t)}{||s_1(t)||},\hspace{0.2cm}
+:$$||s_1(t)|| =  \sqrt{2},\hspace{0.9cm}\varphi_1(t) = \frac{s_1(t)}{||s_1(t)||},\hspace{0.9cm}
 s_{11} = \sqrt{E_1} = \sqrt{2} \hspace{0.1cm}\hspace{0.15cm}\underline { {\approx 1.414} }
 \hspace{0.05cm}.$$
 Die weiteren Koeffizienten sind $\underline {s_{\rm 12} = s_{\rm 13} = 0}$, da die zugehörigen Basisfunktionen bisher noch gar nicht gefunden wurden, während $\varphi_1(t)$ formgleich mit $s_1(t)$ ist.
 '''(3)'''&nbsp; Da nach Berücksichtigung von $s_2(t)$ höchstens zwei Basisfunktionen gefunden sind, gilt mit Sicherheit $s_{\rm 23} \hspace{0.15cm} \underline{= 0}$. Dagegen erhält man für den Koeffizienten
-:$$||s_1(t)|| =  \sqrt{2},\hspace{0.2cm}\varphi_1(t) = \frac{s_1(t)}{||s_1(t)||},\hspace{0.2cm}
+:$$||s_1(t)|| =  \sqrt{2},\hspace{0.9cm}\varphi_1(t) = \frac{s_1(t)}{||s_1(t)||},\hspace{0.9cm}
 s_{11} = \sqrt{E_1} = \sqrt{2} \hspace{0.1cm}\hspace{0.15cm}\underline { {\approx 1.414} }
-\hspace{0.05cm}.$$
+\hspace{0.05cm};$$
 für die Hilfsfunktion $\theta_2(t)$:
@@ Line 85: / Line 87: @@
 \begin{array}{*{1}c} 0 \le t < 1
 \\  1 \le t < 2 \\ \end{array}
-\hspace{0.05cm}, $$
+\hspace{0.05cm}; $$
 für die zweite Basisfunktion:
@@ Line 94: / Line 96: @@
 \begin{array}{*{1}c} 0 \le t < 2
 \\  2 \le t < 3 \\ \end{array}
-\hspace{0.05cm}, $$
+\hspace{0.05cm}; $$
 und schließlich für den zweiten Koeffizienten
@@ Line 108: / Line 110: @@
 :$$s_{32} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - s_{12} + s_{22} = 0 + 0.707 \hspace{0.1cm}\underline {= 0.707}\hspace{0.05cm},$$
 :$$s_{33} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - s_{13} + s_{23} = 0 + 0 \hspace{0.1cm}\underline {= 0}\hspace{0.05cm}. $$
 '''(5)'''&nbsp; Der Bereich $2 &#8804; t &#8804; 3$ wird weder von $\varphi_1(t)$ noch von $\varphi_2(t)$ abgedeckt. Deshalb liefert $s_4(t)$ die neue Basisfunktion $\varphi_3(t)$. Da außerdem $s_4(t)$ nur Anteile im Bereich $2 &#8804; t &#8804; 3$ aufweist und $||s_4(t)|| = 1$ ist, ergibt sich $\varphi_3(t) = s_4(t)$ sowie