[[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:
*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 $j = 1, 2, 3$ – 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% }
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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 $\rm \sqrt{\rm s}$.
*For this condition to be satisfied, the basis function must have unit $\rm \sqrt{\rm s}$. Another equation to be considered is
*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 $s_{\rm 11}$:
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 $s_1(t)$.
'''(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 $s_{\rm 11}$:
*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
'''(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
'''(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 $s_2(t)$.
'''(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
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:
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 $s_{\rm 11}$:
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
