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

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	[[Category:Digital Signal Transmission: Exercises\|^4.1 Basis Functions & Vector Spaces^]]		[[Category:Digital Signal Transmission: Exercises\|^4.1 Basis Functions & Vector Spaces^]]
	[[de:Aufgaben:4.~~1_Gram~~-Schmidt-Verfahren]]		[[de:Aufgaben:Aufgabe 4.1: Zum Gram-Schmidt-Verfahren]]

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	'''(1)'''  <u>Solutions 2 and 4</u>  are correct:		'''(1)'''  <u>Solutions 2 and 4</u>  are correct:
	*Every orthonormal basis function should have energy  $1$,  that is,  it must hold:		*Every orthonormal basis function should have energy  $1$,  that is,  it must hold:
	:$$\|\|\varphi_j(t)\|\|^2 = \int_{-\infty}^{+\infty}\varphi_j(t)^2\,{\rm d} t = 1		:$$\|\|\varphi_j(t)\|\|^2 = \int_{-\infty}^{+\infty}\varphi_j(t)^2\,{\rm d} t = 1\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$
	*For this condition to be satisfied,  the basis function must have unit  $\rm \sqrt{\rm s}$.		*For this condition to be satisfied,  the basis function must have unit  $\rm \sqrt{\rm s}$.
	*Another equation to be considered is		*Another equation to be considered is
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	'''(2)'''  The energy of the signal  $s_1(t)$  is equal to  $E_1 = 2$.		'''(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  $s_{\rm 11}$:		*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_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}.$$
	s_{11} = \sqrt{E_1} = \sqrt{2} \hspace{0.1cm}\hspace{0.15cm}\underline { {\approx 1.414} }
	\hspace{0.05cm}.$$
	*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  $s_1(t)$.		*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  $s_1(t)$.

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	'''(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		'''(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		*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_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};$$
	s_{11} = \sqrt{E_1} = \sqrt{2} \hspace{0.1cm}\hspace{0.15cm}\underline { {\approx 1.414} }
	\hspace{0.05cm};$$

	*for the auxiliary function $\theta_2(t)$:		*for the auxiliary function $\theta_2(t)$:
	:$$\theta_2(t) = s_2(t) - s_{21} \cdot \varphi_1(t) = \left\{ \begin{array}{c} 1 - 0.707 \cdot 0.707 = 0.5\\		:$$\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}; $$
	0 - 0.707 \cdot (-0.707) = 0.5 \end{array} \right.\quad
	\begin{array}{*{1}c} 0 \le t < 1
	\\ 1 \le t < 2 \\ \end{array}
	\hspace{0.05cm}; $$

	*for the second basis function:		*for the second basis function:
	:$$\varphi_2(t) = \frac{\theta_2(t)}{\|\|\theta_2(t)\|\|},\hspace{0.2cm}		:$$\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$$
	\|\|\theta_2(t)\|\| = \sqrt{0.5^2 + 0.5^2} = \sqrt{0.5} \approx 0.707$$		:$$\Rightarrow \hspace{0.3cm} \varphi_2(t) = \left\{ \begin{array}{c} 0.5/0.707 = 0.707\\0 \end{array} \right.\quad\begin{array}{*{1}c} 0 \le t < 2\\ 2 \le t < 3 \\ \end{array}\hspace{0.05cm}; $$
	:$$\Rightarrow \hspace{0.3cm} \varphi_2(t) = \left\{ \begin{array}{c} 0.5/0.707 = 0.707\\
	0 \end{array} \right.\quad
	\begin{array}{*{1}c} 0 \le t < 2
	\\ 2 \le t < 3 \\ \end{array}
	\hspace{0.05cm}; $$

	*and finally for the second coefficient		*and finally for the second coefficient
	:$$s_{22} = \hspace{0.1cm} < \hspace{-0.1cm} s_2(t), \hspace{0.1cm}\varphi_2(t) \hspace{-0.1cm} > \hspace{0.1cm} = 1 \cdot 0.707 + 0 \cdot 0.707 \hspace{0.1cm}\hspace{0.15cm}\underline { = 0.707}		:$$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}.$$
	\hspace{0.05cm}.$$
	[[File:P_ID1995__Dig_A_4_1c.png\|right\|frame\|Gram-Schmidt calculations]]		[[File:P_ID1995__Dig_A_4_1c.png\|right\|frame\|Gram-Schmidt calculations]]

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	[[Category:Digital Signal Transmission: Exercises\|^4.1 Basis Functions & Vector Spaces^]]		[[Category:Digital Signal Transmission: Exercises\|^4.1 Basis Functions & Vector Spaces^]]
			[[de:Aufgaben:4.1_Gram-Schmidt-Verfahren]]

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

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Guenter at 15:01, 7 July 2022

2022-07-07T15:01:47Z

← Older revision		Revision as of 17:01, 7 July 2022
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Solutions 2 and 4</u> are correct:		'''(1)'''  <u>Solutions 2 and 4</u>  are correct:
	*Every orthonormal basis function should have energy 1, that is, it must hold:		*Every orthonormal basis function should have energy  $1$,  that is,  it must hold:
	:$$\|\|\varphi_j(t)\|\|^2 = \int_{-\infty}^{+\infty}\varphi_j(t)^2\,{\rm d} t = 1		:$$\|\|\varphi_j(t)\|\|^2 = \int_{-\infty}^{+\infty}\varphi_j(t)^2\,{\rm d} t = 1
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	*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,  the basis function must have unit  $\rm \sqrt{\rm s}$.
			*Another equation to be considered is
	:$$s_i(t) = \sum\limits_{j = 1}^{N}s_{ij} \cdot \varphi_j(t).$$		:$$s_i(t) = \sum\limits_{j = 1}^{N}s_{ij} \cdot \varphi_j(t).$$
	*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  $A$,  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 $s_{\rm 11}$:		'''(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_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} }		s_{11} = \sqrt{E_1} = \sqrt{2} \hspace{0.1cm}\hspace{0.15cm}\underline { {\approx 1.414} }
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
			*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  $s_1(t)$.

	~~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
	'''(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		*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_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} }		s_{11} = \sqrt{E_1} = \sqrt{2} \hspace{0.1cm}\hspace{0.15cm}\underline { {\approx 1.414} }
	\hspace{0.05cm};$$		\hspace{0.05cm};$$

	for the auxiliary function $\theta_2(t)$:		*for the auxiliary function $\theta_2(t)$:
	:$$\theta_2(t) = s_2(t) - s_{21} \cdot \varphi_1(t) = \left\{ \begin{array}{c} 1 - 0.707 \cdot 0.707 = 0.5\\		:$$\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		0 - 0.707 \cdot (-0.707) = 0.5 \end{array} \right.\quad
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	\hspace{0.05cm}; $$		\hspace{0.05cm}; $$

	for the second basis function:		*for the second basis function:
	:$$\varphi_2(t) = \frac{\theta_2(t)}{\|\|\theta_2(t)\|\|},\hspace{0.2cm}		:$$\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$$		\|\|\theta_2(t)\|\| = \sqrt{0.5^2 + 0.5^2} = \sqrt{0.5} \approx 0.707$$
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	\hspace{0.05cm}; $$		\hspace{0.05cm}; $$

	and finally for the second coefficient		*and finally for the second coefficient
	:$$s_{22} = \hspace{0.1cm} < \hspace{-0.1cm} s_2(t), \hspace{0.1cm}\varphi_2(t) \hspace{-0.1cm} > \hspace{0.1cm} = 1 \cdot 0.707 + 0 \cdot 0.707 \hspace{0.1cm}\hspace{0.15cm}\underline { = 0.707}		:$$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}.$$		\hspace{0.05cm}.$$
			[[File:P_ID1995__Dig_A_4_1c.png\|right\|frame\|Gram-Schmidt calculations]]

	The calculations are illustrated in the graph below.		The calculations are illustrated in the graph below.

	~~[[File:P_ID1995__Dig_A_4_1c.png\|center\|frame\|Gram-Schmidt calculations]]~~

	'''(4)'''  It can be seen immediately that $s_3(t)$ can be expressed as a linear combination of $s_1(t)$ and $s_2(t)$.		'''(4)'''  It can be seen immediately that  $s_3(t)$  can be expressed as a linear combination of  $s_1(t)$  and  $s_2(t)$.
	:$$s_{3}(t) = -s_{1}(t) + s_{2}(t)~~\hspace{0.3cm}\Rightarrow \hspace{0.3cm}~~s_{31} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - s_{11} + s_{21} = -1.414 + 0.707 = \hspace{0.1cm}\hspace{0.15cm}\underline {-0.707}\hspace{0.05cm},$$		:$$s_{3}(t) = -s_{1}(t) + s_{2}(t),$$
			:$$s_{31} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - s_{11} + s_{21} = -1.414 + 0.707 = \hspace{0.1cm}\hspace{0.15cm}\underline {-0.707}\hspace{0.05cm},$$
	:$$s_{32} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - s_{12} + s_{22} = 0 + 0.707 \hspace{0.1cm}\underline {= 0.707}\hspace{0.05cm},$$		:$$s_{32} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - s_{12} + s_{22} = 0 + 0.707 \hspace{0.1cm}\underline {= 0.707}\hspace{0.05cm},$$
	:$$s_{33} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - s_{13} + s_{23} = 0 + 0 \hspace{0.1cm}\underline {= 0}\hspace{0.05cm}. $$		:$$s_{33} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} - s_{13} + s_{23} = 0 + 0 \hspace{0.1cm}\underline {= 0}\hspace{0.05cm}. $$


	'''(5)'''  The range $2 ≤ t ≤ 3$ is not covered by ~~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 $2 ≤ t ≤ 3$ and $\|\|s_4(t)\|\| = 1$, we obtain $\varphi_3(t) = s_4(t)$ as well as		'''(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
	:$$s_{41} \hspace{0.1cm}\hspace{0.15cm}\underline {= 0}, \hspace{0.2cm}s_{42} \hspace{0.1cm}\hspace{0.15cm}\underline {= 0}, \hspace{0.2cm}s_{43} \hspace{0.1cm}\hspace{0.15cm}\underline { = 1} \hspace{0.05cm}. $$		:$$s_{41} \hspace{0.1cm}\hspace{0.15cm}\underline {= 0}, \hspace{0.2cm}s_{42} \hspace{0.1cm}\hspace{0.15cm}\underline {= 0}, \hspace{0.2cm}s_{43} \hspace{0.1cm}\hspace{0.15cm}\underline { = 1} \hspace{0.05cm}. $$
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 14:41, 7 July 2022

2022-07-07T14:41:08Z

← Older revision		Revision as of 16:41, 7 July 2022
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	[[File:P_ID1994__Dig_A_4_1.png\|right\|frame\|Specification for the Gram-Schmidt process]]		[[File:P_ID1994__Dig_A_4_1.png\|right\|frame\|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:		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 subtask '''(1)''', let  $A^2 = 1 \ \rm mW$  and  $T = 1 \ \rm µ s$.		*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~~, respectively~~:   $A = 1$,  $T = 1$.
	*Thus, both the coefficients  $s_{\it ij}$  and the basis functions  $\varphi_{\it j}(t)$  ~~–~~ with  $j = 1, 2, 3$&nbsp~~; &ndash~~; are dimensionless quantities.		*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:
			*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces\|"Signals, Basis Functions and Vector Spaces"]].


	~~''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"]].		*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"]].

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	+ The coefficients  $s_{\it ij}$  have the unit  $\rm \sqrt{\rm Ws}$.		+ The coefficients  $s_{\it ij}$  have the unit  $\rm \sqrt{\rm Ws}$.

	{Perform the first step of the Gram-Schmidt process. As for the other tasks, let  $A = 1$  and  $T = 1$ hold.		{Perform the first step of the Gram-Schmidt process.  As for the other tasks,  let  $A = 1$  and  $T = 1$ hold.
	\|type="{}"}		\|type="{}"}
	$s_{\rm 11} \ = \ $ { 1.414 3% }		$s_{\rm 11} \ = \ $ { 1.414 3% }

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

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Guenter at 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^]]

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

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