# Exercise 4.3: Different Frequencies

In the diagram  $M = 5$  different signals  $s_i(t)$  are shown.  Contrary to the nomenclature in the theory section,  the indexing variable  $i$  can have the values  $0, \ \text{...} \ , M-1$.

To be noted:

• All signals are time-limited to  $0$  ...   $T$;  thus the energy of all signals is finite.
• The signal  $s_1(t)$  has the period  $T_0 = T$.  The frequency is therefore  $f_0 = 1/T$.
• The signals  $s_i(t)$  with  $i ≠ 0$  are cosine oscillations with frequency  $i \cdot f_0$.
• In contrast,  $s_0(t)$  is constant between  $0$  and  $T$.
• The maximum value of all signals is  $A$  and  $|s_i(t)| ≤ A$ holds.

In this exercise we are looking for the  $N$  basis functions,  which are numbered here with  $j = 0, \ \text{...} \ , N-1$.

Note:  The exercise belongs to the chapter  "Signals, Basis Functions and Vector Spaces".

### Question

1

Describe the signal set  $\{s_i(t)\}$  with  $0 ≤ i ≤ 4$  as compactly as possible.
Which description form is correct?

 $s_i(t) = A \cdot \cos {(2\pi \cdot i \cdot t/T)}$. $s_i(t) = A \cdot \cos {(2\pi \cdot i \cdot t/T)}$  for  $0 ≤ t < T$,  otherwise $0$. $s_i(t) = A \cdot \cos {(2\pi t/T \, – \, i \cdot \pi/2)}$  for  $0 ≤ t < T$,  otherwise $0$.

2

Specify the number  $N$  of basis functions required.

 $N \ = \$

3

What is the basis function  $\varphi_0(t)$ that is equal in form to  $s_0(t)$?

 $\varphi_0(t) = s_0(t)$, $\varphi_0(t) = \sqrt{1/T}$  for  $0 ≤ t < T$,  outside  $0$. $\varphi_0(t) = \sqrt{2/T}$  for  $0 ≤ t < T$,  outside  $0$.

4

What is the basis function  $\varphi_1(t)$  that is equal in form to  $s_1(t)$?

 $\varphi_1(t) = s_1(t)$, $\varphi_1(t) = \sqrt{1/T} \cdot \cos {(2\pi t/T)}$ for $0 ≤ t < T$,  outside $0$. $\varphi_1(t) =\sqrt{2/T} \cdot \cos {(2\pi t/T)}$ for $0 ≤ t < T$,  outside $0$.

### Solution

#### Solution

(1)  Correct is the  solution 2:

• This takes into account the different frequencies and the limitation to the range  $0 ≤ t < T$.
• The signals  $s_i(t)$  according to suggestion 3,  on the other hand,  do not differ with respect to frequency,  but have different phase positions.

(2)  The energy-limited signals   $s_i(t) = A \cdot \cos {(2\pi \cdot i \cdot t/T)}$   are orthogonal to each other   ⇒   the inner product of two signals  $s_i(t)$,  $s_k(t)$  with  $i ≠ k$  is always  $0$:

$$< \hspace{-0.1cm}s_i(t), \hspace{0.1cm} s_k(t)\hspace{-0.1cm} > \hspace{0.1cm} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} A^2 \cdot \int_{0}^{T}\cos(2\pi \cdot i \cdot t/T) \cdot \cos(2\pi \cdot k \cdot t/T)\,{\rm d} t$$
$$\Rightarrow \hspace{0.3cm} < \hspace{-0.1cm}s_i(t), \hspace{0.1cm} s_k(t)\hspace{-0.1cm} > \hspace{0.1cm} \hspace{-0.1cm} {A^2}/{2} \cdot \int_{0}^{T}\cos(2\pi (i-k) t/T) \,{\rm d} t + \frac{A^2}{2} \cdot \int_{0}^{T}\cos(2\pi (i+k) t/T) \,{\rm d} t \hspace{0.05cm}.$$
• With  $i ∈ \{0, \ \text{...} \ , 4\}$  and  $k ∈ \{0, \ \text{...}\ , 4\}$  as well as  $i ≠ j$,  both  $i \, - k$  is integer  $\ne0$,  as is the  sum $i + k$.
• Thus,  both integrals yield the result zero:
$$< \hspace{-0.1cm}s_i(t), \hspace{0.1cm} s_k(t)\hspace{-0.1cm} > \hspace{0.1cm} \hspace{-0.1cm}= 0 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \hspace{0.05cm}\hspace{0.15cm}\underline {N = M = 5} \hspace{0.05cm}.$$

(3)  The energy of the signal  $s_0(t)$,  which is constant within  $T$,  is equal to

$$E_0 = ||s_0(t)||^2 = A^2 \cdot T \hspace{0.3cm}\Rightarrow \hspace{0.3cm} ||s_0(t)|| = A \cdot \sqrt{T} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \varphi_0 (t) = \frac{s_0(t)}{||s_0(t)||} = \left\{ \begin{array}{c} 1/\sqrt{T} \\ 0 \end{array} \right.\quad \begin{array}{*{1}c} 0 \le t < T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$

Therefore,  solution 2  is correct.

(4)  The  last solution  is correct because of

$$E_1 = ||s_1(t)||^2 = \frac{A^2 \cdot T}{2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} ||s_1(t)|| = A \cdot \sqrt{{T}/{2}} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \varphi_1 (t) = \frac{s_1(t)}{||s_1(t)||} = \left\{ \begin{array}{c} \sqrt{2/T} \cdot \cos(2\pi t/T) \\ 0 \end{array} \right.\quad \begin{array}{*{1}c} 0 \le t < T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$