Exercise 2.3: Cosine and Sine Components

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Spektrum von Cosinus- und Sinusanteilen

Given is the amplitude spectrum  $X(f)$  of a signal  $x(t)$  according to the graph.

• Let  $f_1 = 4\,\text{kHz}$ be the normalisation frequency.
• Thus the actual frequencies of the signal components are  $0\,\text{kHz}$,  $4\,\text{kHz}$  and  $10\,\text{kHz}$.

This signal  $x(t)$  is at the input of a linear differentiator whose output can be represented with  $\omega_1 = 2\pi f_1$  as follows:

$$y(t)=\frac{1}{\omega_1}\cdot\frac{ {\rm d} x(t)}{{\rm d} t}.$$

Hint:

• This exercise belongs to the chapter  Harmonic Oscillation.

Questions

1

Give  $x(t)$  analytically.  What is the signal value at  $t = 0$?

$x(t=0)\ = \ $   ${\rm V}$

2

What is the period duration of the signal  $x(t)$?

$T_0\ = \ $   ${\rm ms}$

3

Calculate the output signal  $y(t)$  of the differentiator  What is the signal value at time  $t = 0$?

$y(t=0)\ = \ $   ${\rm V}$

4

Which of the following statements are true regarding the signal  $y(t)$  or its spectrum  $Y(f)$ ?

	$y(t)$  has the same period duration as the signal  $x(t)$.
	$Y(f)$  contains a Dirac function at the frequency  $f = 0$.
	$Y(f)$  contains a Dirac function at  $+f_1$  with weight  $\rm{j} · 1\,{\rm V}$.
	$Y(f)$  contains a Dirac function at  $–\hspace{-0.1cm}2.5 \cdot f_1$  with weight  $5\,{\rm V}$.

Solution

Solution

Summensignal aus Cosinus- und Sinusanteilen

(1)  The time signal has the following form:

$$x(t)={\rm 3V}-{\rm 2V}\cdot \cos(\omega_{\rm 1} \cdot t)+{\rm 4V} \cdot \sin(2.5 \cdot \omega_{\rm 1} \cdot t).$$

• Here  $\omega_1 = 2\pi f_1$  denotes the angular frequency of the cosine component.
• At time  $t = 0$  the signal has the value  $x(t=0)\hspace{0.15 cm}\underline{=1\,\rm V}$.

(2)  The base frequency  $f_0$  is the least common divisor

• of $f_1 = 4{\,\rm kHz}$
• and $2.5 · f_1 = 10{\,\rm kHz}$.

From this follows  $f_0 = 2{\,\rm kHz}$   ⇒   period duration $T_0 = 1/f_0 \hspace{0.1cm}\underline{= 0.5 {\,\rm ms}}$.

Spektrum mit diskreten Anteilen

(3)  The following applies to the output signal $y(t)$ of the differentiatior:

$$y(t)=\frac{1}{\omega_1}\cdot\frac{ {\rm d}x(t)}{{\rm d}t}=\frac{ {\rm -2V}}{\omega_1}\cdot\omega_1 \cdot (-\sin(\omega_1 t))+\frac{\rm 4V}{\omega_1}\cdot 2.5\omega_1\cdot {\rm cos}(2.5\omega_1t).$$

• This leads to the solution:

$$y(t)={\rm 2V}\cdot\sin(\omega_1 t)+{\rm 10V}\cdot\cos(2.5\omega_1 t).$$

• For  $t = 0$  the value  $y(t=0)\hspace{0.15cm}\underline{=10\,\rm V}$ follows.
• The spectrum  $Y(f)$  is shown on the right.

(4)  The solutions 1 and 4 are correct:

• The period duration $T_0$ is not changed by the amplitude and phase of the two components.
• This means, that  $T_0 = 0.5 {\,\rm ms}$  still applies.
• The DC component disappears due to the differentiation.
• The component  $f_1$  is sinusoidal. Thus  $X(f)$  has an (imaginary) Dirac at  $f = f_1$, but with a negative sign.
• The cosine component with amplitude  ${10\,\rm V}$  results in the two Dirac functions at  $\pm 2.5 \cdot f_1$ , each with weight  ${5\,\rm V}$ .