# Exercise 4.1: About the Gram-Schmidt Process

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

#### 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
$$s_{41} \hspace{0.1cm}\hspace{0.15cm}\underline {= 0}, \hspace{0.2cm}s_{42} \hspace{0.1cm}\hspace{0.15cm}\underline {= 0}, \hspace{0.2cm}s_{43} \hspace{0.1cm}\hspace{0.15cm}\underline { = 1} \hspace{0.05cm}.$$