Difference between revisions of "Aufgaben:Exercise 4.1: About the Gram-Schmidt Process"

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{{quiz-Header|Buchseite=Digitalsignalübertragung/Signale, Basisfunktionen und Vektorräume}}
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{{quiz-Header|Buchseite=Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces}}
  
[[File:P_ID1994__Dig_A_4_1.png|right|frame|Zum Gram-Schmidt-Verfahren]]
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[[File:P_ID1994__Dig_A_4_1.png|right|frame|Specification for the Gram-Schmidt process]]
Für die vier durch die Abbildung definierten Signale $s_1(t), \, ... \, , s_4(t)$ sind durch Anwendung des sog. Gram–Schmidt–Verfahrens die drei sich ergebenden Basisfunktionen $\varphi_1(t)$, $\varphi_2(t)$ und $\varphi_3(t)$ zu ermitteln, so dass für die Signale mit $i = 1, \, ... \, , \ 4$ geschrieben werden kann:
+
For the four signals  $s_1(t), \, \text{...} \, , s_4(t)$  defined by the figure,  the three resulting basis functions  $\varphi_1(t)$,  $\varphi_2(t)$  and  $\varphi_3(t)$  are to be determined by applying the Gram-Schmidt process,  so that for the signals with   $i = 1, \, \text{...} \, , 4$  can be written:
 
:$$s_i(t) = s_{i1} \cdot \varphi_1(t) + s_{i2} \cdot \varphi_2(t) + s_{i3} \cdot \varphi_3(t)\hspace{0.05cm}.$$
 
:$$s_i(t) = s_{i1} \cdot \varphi_1(t) + s_{i2} \cdot \varphi_2(t) + s_{i3} \cdot \varphi_3(t)\hspace{0.05cm}.$$
  
In der Teilaufgabe (1) gelte $A^2 = 1 \ \rm mW$ und $T = 1 \ \rm \mu s$. In den späteren Teilaufgaben sind die Amplitude und die Zeit jeweils normierte Größen:$A = 1$, $T = 1$. Damit sind sowohl die Koeffizienten $s_{\it ij}$ als auch die Basisfunktionen $\varphi_{\it}(t)$ – jeweils mit $j = 1, 2, 3$ – dimensionslose Größen.
+
*In subtask  '''(1)''',  let  $A^2 = 1 \ \rm mW$  and  $T = 1 \ \rm µ s$.
 +
 +
*In the later subtasks,  the amplitude and the time are normalized quantities:   $A = 1$,  $T = 1$.
 +
 +
*Thus,  both the coefficients  $s_{\it ij}$  and the basis functions  $\varphi_{\it j}(t)$  $($with  $j = 1,\ 2,\ 3)$  are dimensionless quantities.
  
''Hinweise:''
 
* Die Aufgabe bezieht sich inhaltlich auf das Kapitel [[Digitalsignal%C3%BCbertragung/Signale,_Basisfunktionen_und_Vektorr%C3%A4ume| Signale, Basisfunktionen und Vektorräume]].
 
* Auf der Seite 3a des Kapitels ist das [[Digitalsignal%C3%BCbertragung/Signale,_Basisfunktionen_und_Vektorr%C3%A4ume#Das_Verfahren_nach_Gram-Schmidt|Gram–Schmidt–Verfahren]] angegeben, auf der Seite 3b finden Sie ein [https://intern.lntwww.de/cgi-bin/extern/uni.pl?uno=hyperlink&due=block&b_id=2735&hyperlink_typ=block_verweis&hyperlink_fenstergroesse=blockverweis_gross| Berechnungsbeispiel] ähnlich zu dieser Aufgabe.
 
  
  
===Fragebogen===
+
 
 +
Notes:
 +
*The exercise belongs to the chapter   [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces|"Signals, Basis Functions and Vector Spaces"]].
 +
 
 +
*Reference is made in particular to the sections  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces#Orthonormal_basis_functions|"Orthonormal basis functions"]]  and  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces#The_Gram-Schmidt_process|"Gram-Schmidt process"]].
 +
 +
 
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Multiple-Choice Frage
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{What are the units of the following quantities with&nbsp; $A^2 = 1 \, \rm mW$&nbsp; and&nbsp; $T = 1 \, {\rm &micro; s}$?
 
|type="[]"}
 
|type="[]"}
- Falsch
+
- The basis functions &nbsp;$\varphi_j(t)$&nbsp; are dimensionless.
+ Richtig
+
+ The basis functions &nbsp;$\varphi_j(t)$&nbsp; have the unit &nbsp;$\rm \sqrt{\rm s}$.
 +
- The coefficients &nbsp;$s_{\it ij}$&nbsp; are dimensionless.
 +
+ The coefficients &nbsp;$s_{\it ij}$&nbsp; have the unit &nbsp;$\rm \sqrt{\rm Ws}$.
 +
 
 +
{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% }
 +
$s_{\rm 12} \ = \ $ { 0. }
 +
$s_{\rm 13} \ = \ $ { 0. }
 +
 
 +
{What are the coefficients of the signal&nbsp; $s_2(t)$&nbsp; with &nbsp;$A = 1$ &nbsp;and&nbsp; $T = 1$?
 +
|type="{}"}
 +
$s_{\rm 21} \ = \ $ { 0.707 3% }
 +
$s_{\rm 22} \ = \ $ { 0.707 3% }
 +
$s_{\rm 23} \ = \ $ { 0. }
 +
 
 +
{What are the coefficients of the signal&nbsp; $s_3(t)$&nbsp; with &nbsp;$A = 1$ &nbsp;and&nbsp; $T = 1$?
 +
|type="{}"}
 +
$s_{\rm 31} \ = \ $ { -0.72821--0.68579 }
 +
$s_{\rm 32} \ = \ $ { 0.707 3% }
 +
$s_{\rm 33} \ = \ $ { 0. }
  
{Input-Box Frage
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{What are the coefficients of the signal&nbsp; $s_4(t)$&nbsp; with &nbsp;$A = 1$ &nbsp;and&nbsp; $T = 1$?
 
|type="{}"}
 
|type="{}"}
$\alpha$ = { 0.3 }
+
$s_{\rm 41} \ = \ $ { 0. }
 +
$s_{\rm 42} \ = \ $ { 0. }
 +
$s_{\rm 43} \ = \ $ { 1 3% }
 
</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp;  
+
'''(1)'''&nbsp; <u>Solutions 2 and 4</u>&nbsp; are correct:
'''(2)'''&nbsp;  
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*Every orthonormal basis function should have energy&nbsp; $1$,&nbsp; that is,&nbsp; it must hold:
'''(3)'''&nbsp;  
+
:$$||\varphi_j(t)||^2 =  \int_{-\infty}^{+\infty}\varphi_j(t)^2\,{\rm d}  t = 1
'''(4)'''&nbsp;  
+
\hspace{0.05cm}.$$
'''(5)'''&nbsp;  
+
*For this condition to be satisfied,&nbsp; the basis function must have unit&nbsp; $\rm \sqrt{\rm s}$.&nbsp;
'''(6)'''&nbsp;  
+
*Another equation to be considered is
 +
:$$s_i(t) = \sum\limits_{j = 1}^{N}s_{ij} \cdot \varphi_j(t).$$
 +
*Like the parameter&nbsp; $A$,&nbsp; the signals themselves have the unit&nbsp; $\rm \sqrt{\rm W}$.&nbsp;
 +
*Because of the unit&nbsp; $\rm \sqrt{\rm 1/s}$&nbsp; of&nbsp; $\varphi_{ j}(t)$,&nbsp; this equation can be satisfied with the correct dimension only if the coefficients&nbsp; $s_{\it ij}$&nbsp; are given with the unit&nbsp; $\rm \sqrt{\rm Ws}$.
 +
 
 +
 
 +
 
 +
'''(2)'''&nbsp; The energy of the signal&nbsp; $s_1(t)$ &nbsp;is equal to&nbsp; $E_1 = 2$.&nbsp;
 +
*It follows for the norm,&nbsp; the basis function&nbsp; $\varphi_1(t)$&nbsp; and the coefficient&nbsp; $s_{\rm 11}$:
 +
:$$||s_1(t)|| =  \sqrt{2},\hspace{0.9cm}\varphi_1(t) = \frac{s_1(t)}{||s_1(t)||},\hspace{0.9cm}
 +
s_{11} = \sqrt{E_1} = \sqrt{2} \hspace{0.1cm}\hspace{0.15cm}\underline { {\approx 1.414} }
 +
\hspace{0.05cm}.$$
 +
*The other coefficients are&nbsp; $\underline {s_{\rm 12} = s_{\rm 13} = 0}$,&nbsp; since the associated basis functions have not been found at all yet,&nbsp; while&nbsp; $\varphi_1(t)$&nbsp; is equal in form to&nbsp; $s_1(t)$.
 +
 
 +
 
 +
 
 +
'''(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
 +
*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 $\theta_2(t)$:
 +
:$$\theta_2(t) = s_2(t) - s_{21} \cdot \varphi_1(t) = \left\{ \begin{array}{c} 1 - 0.707 \cdot 0.707 = 0.5\\
 +
0 - 0.707 \cdot (-0.707) = 0.5  \end{array} \right.\quad
 +
\begin{array}{*{1}c} 0 \le t < 1
 +
\\  1 \le t < 2 \\ \end{array}
 +
\hspace{0.05cm}; $$
 +
 
 +
*for the second basis function:
 +
:$$\varphi_2(t) = \frac{\theta_2(t)}{||\theta_2(t)||},\hspace{0.2cm}
 +
||\theta_2(t)|| = \sqrt{0.5^2 + 0.5^2} = \sqrt{0.5} \approx 0.707$$
 +
:$$\Rightarrow \hspace{0.3cm} \varphi_2(t) = \left\{ \begin{array}{c} 0.5/0.707 = 0.707\\
 +
0  \end{array} \right.\quad
 +
\begin{array}{*{1}c} 0 \le t < 2
 +
\\  2 \le t < 3 \\ \end{array}
 +
\hspace{0.05cm}; $$
 +
 
 +
*and finally for the second coefficient
 +
:$$s_{22}  = \hspace{0.1cm} < \hspace{-0.1cm} s_2(t), \hspace{0.1cm}\varphi_2(t) \hspace{-0.1cm} > \hspace{0.1cm} = 1 \cdot 0.707 + 0 \cdot 0.707 \hspace{0.1cm}\hspace{0.15cm}\underline {  = 0.707}
 +
\hspace{0.05cm}.$$
 +
[[File:P_ID1995__Dig_A_4_1c.png|right|frame|Gram-Schmidt calculations]]
 +
 
 +
The calculations are illustrated in the graph below.
 +
 
 +
 
 +
'''(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; $s_2(t)$.
 +
:$$s_{3}(t)  = -s_{1}(t) + s_{2}(t),$$
 +
:$$s_{31} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - s_{11} + s_{21} = -1.414 + 0.707 = \hspace{0.1cm}\hspace{0.15cm}\underline {-0.707}\hspace{0.05cm},$$
 +
:$$s_{32} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - s_{12} + s_{22} = 0 + 0.707 \hspace{0.1cm}\underline {= 0.707}\hspace{0.05cm},$$
 +
:$$s_{33} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - s_{13} + s_{23} = 0 + 0 \hspace{0.1cm}\underline {= 0}\hspace{0.05cm}. $$
 +
 
 +
 
 +
'''(5)'''&nbsp; The range&nbsp; $2 &#8804; t &#8804; 3$&nbsp; is not covered by&nbsp; $\varphi_1(t)$&nbsp; and&nbsp; $\varphi_2(t)$.
 +
*Therefore,&nbsp; $s_4(t)$&nbsp; provides the new basis function&nbsp; $\varphi_3(t)$.&nbsp;  
 +
*Since $s_4(t)$&nbsp; has components only in the range&nbsp; $2 &#8804; t &#8804; 3$ and $||s_4(t)|| = 1$,&nbsp; we obtain&nbsp; $\varphi_3(t) = s_4(t)$&nbsp; as well as
 +
:$$s_{41} \hspace{0.1cm}\hspace{0.15cm}\underline {= 0},  \hspace{0.2cm}s_{42} \hspace{0.1cm}\hspace{0.15cm}\underline {= 0},  \hspace{0.2cm}s_{43} \hspace{0.1cm}\hspace{0.15cm}\underline { = 1} \hspace{0.05cm}.  $$
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Digitalsignalübertragung|^4.1 Signale, Basisfunktionen, Vektorräume^]]
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[[Category:Digital Signal Transmission: Exercises|^4.1 Basis Functions & Vector Spaces^]]

Latest revision as of 13:56, 13 July 2022

Specification for the Gram-Schmidt process

For the four signals  $s_1(t), \, \text{...} \, , s_4(t)$  defined by the figure,  the three resulting basis functions  $\varphi_1(t)$,  $\varphi_2(t)$  and  $\varphi_3(t)$  are to be determined by applying the Gram-Schmidt process,  so that for the signals with   $i = 1, \, \text{...} \, , 4$  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  $A^2 = 1 \ \rm mW$  and  $T = 1 \ \rm µ s$.
  • In the later subtasks,  the amplitude and the time are normalized quantities:   $A = 1$,  $T = 1$.
  • Thus,  both the coefficients  $s_{\it ij}$  and the basis functions  $\varphi_{\it j}(t)$  $($with  $j = 1,\ 2,\ 3)$  are dimensionless quantities.



Notes:


Questions

1

What are the units of the following quantities with  $A^2 = 1 \, \rm mW$  and  $T = 1 \, {\rm µ s}$?

The basis functions  $\varphi_j(t)$  are dimensionless.
The basis functions  $\varphi_j(t)$  have the unit  $\rm \sqrt{\rm s}$.
The coefficients  $s_{\it ij}$  are dimensionless.
The coefficients  $s_{\it ij}$  have the unit  $\rm \sqrt{\rm Ws}$.

2

Perform the first step of the Gram-Schmidt process.  As for the other tasks,  let  $A = 1$  and  $T = 1$ hold.

$s_{\rm 11} \ = \ $

$s_{\rm 12} \ = \ $

$s_{\rm 13} \ = \ $

3

What are the coefficients of the signal  $s_2(t)$  with  $A = 1$  and  $T = 1$?

$s_{\rm 21} \ = \ $

$s_{\rm 22} \ = \ $

$s_{\rm 23} \ = \ $

4

What are the coefficients of the signal  $s_3(t)$  with  $A = 1$  and  $T = 1$?

$s_{\rm 31} \ = \ $

$s_{\rm 32} \ = \ $

$s_{\rm 33} \ = \ $

5

What are the coefficients of the signal  $s_4(t)$  with  $A = 1$  and  $T = 1$?

$s_{\rm 41} \ = \ $

$s_{\rm 42} \ = \ $

$s_{\rm 43} \ = \ $


Solution

(1)  Solutions 2 and 4  are correct:

  • Every orthonormal basis function should have energy  $1$,  that is,  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
$$s_i(t) = \sum\limits_{j = 1}^{N}s_{ij} \cdot \varphi_j(t).$$
  • Like the parameter  $A$,  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  $\rm \sqrt{\rm Ws}$.


(2)  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  $s_{\rm 11}$:
$$||s_1(t)|| = \sqrt{2},\hspace{0.9cm}\varphi_1(t) = \frac{s_1(t)}{||s_1(t)||},\hspace{0.9cm} s_{11} = \sqrt{E_1} = \sqrt{2} \hspace{0.1cm}\hspace{0.15cm}\underline { {\approx 1.414} } \hspace{0.05cm}.$$
  • The other coefficients are  $\underline {s_{\rm 12} = s_{\rm 13} = 0}$,  since the associated basis functions have not been found at all yet,  while  $\varphi_1(t)$  is equal in form to  $s_1(t)$.


(3)  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
$$||s_1(t)|| = \sqrt{2},\hspace{0.9cm}\varphi_1(t) = \frac{s_1(t)}{||s_1(t)||},\hspace{0.9cm} s_{11} = \sqrt{E_1} = \sqrt{2} \hspace{0.1cm}\hspace{0.15cm}\underline { {\approx 1.414} } \hspace{0.05cm};$$
  • for the auxiliary function $\theta_2(t)$:
$$\theta_2(t) = s_2(t) - s_{21} \cdot \varphi_1(t) = \left\{ \begin{array}{c} 1 - 0.707 \cdot 0.707 = 0.5\\ 0 - 0.707 \cdot (-0.707) = 0.5 \end{array} \right.\quad \begin{array}{*{1}c} 0 \le t < 1 \\ 1 \le t < 2 \\ \end{array} \hspace{0.05cm}; $$
  • for the second basis function:
$$\varphi_2(t) = \frac{\theta_2(t)}{||\theta_2(t)||},\hspace{0.2cm} ||\theta_2(t)|| = \sqrt{0.5^2 + 0.5^2} = \sqrt{0.5} \approx 0.707$$
$$\Rightarrow \hspace{0.3cm} \varphi_2(t) = \left\{ \begin{array}{c} 0.5/0.707 = 0.707\\ 0 \end{array} \right.\quad \begin{array}{*{1}c} 0 \le t < 2 \\ 2 \le t < 3 \\ \end{array} \hspace{0.05cm}; $$
  • and finally for the second coefficient
$$s_{22} = \hspace{0.1cm} < \hspace{-0.1cm} s_2(t), \hspace{0.1cm}\varphi_2(t) \hspace{-0.1cm} > \hspace{0.1cm} = 1 \cdot 0.707 + 0 \cdot 0.707 \hspace{0.1cm}\hspace{0.15cm}\underline { = 0.707} \hspace{0.05cm}.$$
Gram-Schmidt calculations

The calculations are illustrated in the graph below.


(4)  It can be seen immediately that  $s_3(t)$  can be expressed as a linear combination of  $s_1(t)$  and  $s_2(t)$.

$$s_{3}(t) = -s_{1}(t) + s_{2}(t),$$
$$s_{31} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - s_{11} + s_{21} = -1.414 + 0.707 = \hspace{0.1cm}\hspace{0.15cm}\underline {-0.707}\hspace{0.05cm},$$
$$s_{32} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - s_{12} + s_{22} = 0 + 0.707 \hspace{0.1cm}\underline {= 0.707}\hspace{0.05cm},$$
$$s_{33} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - s_{13} + s_{23} = 0 + 0 \hspace{0.1cm}\underline {= 0}\hspace{0.05cm}. $$


(5)  The range  $2 ≤ t ≤ 3$  is not covered by  $\varphi_1(t)$  and  $\varphi_2(t)$.

  • Therefore,  $s_4(t)$  provides the new basis function  $\varphi_3(t)$. 
  • Since $s_4(t)$  has components only in the range  $2 ≤ t ≤ 3$ and $||s_4(t)|| = 1$,  we obtain  $\varphi_3(t) = s_4(t)$  as well as
$$s_{41} \hspace{0.1cm}\hspace{0.15cm}\underline {= 0}, \hspace{0.2cm}s_{42} \hspace{0.1cm}\hspace{0.15cm}\underline {= 0}, \hspace{0.2cm}s_{43} \hspace{0.1cm}\hspace{0.15cm}\underline { = 1} \hspace{0.05cm}. $$