Exercise 2.4: Rectified Cosine

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Gleichgerichteter Cosinus

A cosine signal  $x(t)$  with the amplitude  $1\,\rm{V}$  and the frequency  $f_0= 10\,\rm{kHz}$  is applied to the input of a half-wave rectifier. At its output, the signal  $y(t)$ results, which is shown in the graph below.

In subtasks  (6)  und  (7)  the error signal  $\varepsilon_3(t) = y_3(t) - y(t)$  is also used. This describes the difference between the Fourier series  ⇒   $y_3(t)$   limited to only  $N = 3$  coefficients and the actual output signal  $y(t)$.

Hints:

• This exercise belongs to the chapter  Fourier Series.

• To solve the problem, you can use the following definite integral   (let $n$ be an integer$)$:

$$\int ^{\pi /2}_{-\pi /2}\cos(u)\cdot\cos(2nu)\,{\rm d}u = (-1)^{n+1}\cdot\frac{2}{4n^2-1}.$$

• You can find a compact summary of the topic in the learning video  Zur Berechnung der Fourierkoeffizienten.

Questions

1

Which of the following statements are true for the signal  $x(t)$  zutreffend?

	The period duration is  $T_0 = 100 \,µ{\rm s}$.
	The DC signal coefficient is  $A_0 = 0$.
	Of all cosine coefficients  $A_n$  exactly one is not equal to zero.
	Of all the sine coefficients  $B_n$  exactly one is not equal to zero.
	The Fourier series  $x_3(t)$  does not deviate from the actual signal  $x(t)$ .

2

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

$T_0\ = \ $   ${\rm µs}$

3

Calculate the DC component of the signal  $y(t)$.

$A_0\ = \ $   ${\rm V}$

4

What are the sine coefficients  $B_n$? Justify your result. Enter the coefficient  $B_2$  as a check.

$B_2\ = \ $   ${\rm V}$

5

Now calculate the cosine coefficient  $A_n$. Enter the coefficient  $A_2$  as a check.

$A_2\ = \ $   ${\rm V}$

6

Specify the Fourier series  $y_3(t)$  analytically (limit to  $N = 3$  sine and cosine coefficients each).
How large is the error between this finite Fourier series and the actual signal value at  $t = 0$?

$\varepsilon_3(t= 0)\ = \ $ ${\rm V}$

7

Now calculate the error  $\varepsilon_3(t= 25 \,µ{\rm s})$. Interpret this value in comparison to the result from  (6).

$\varepsilon_3(t= 25 \,µ{\rm s})\ = \ $ ${\rm V}$

Solution

Solution

(1)  All solutions are correct except the fourth:

• From the signal frequency  $f_0= 10\,\rm{kHz}$  follows  $T_0 = 1/f_0 = 100\,µ\text{s}$.
• The cosine signal is free of equal signals  $(A_0 = 0)$  and is completely described by a single cosine coefficient – namely  $A_1$ .
• All sine coefficients are  $B_n \equiv 0$, since  $x(t)$  is an even function.
• The Fourier series representation  $x_3(t)$  reproduces  $x(t)$  without error.

(2)  Due to the double path rectification, the period duration is now half the value:  $T_0 \hspace{0.1cm}\underline{= 50\,µ\text{s}}$.

• For all subsequent points, the specification  $T_0$  refers to this value, i.e. to the period of the signal  $y(t)$.

(3)  In the range from  $–T_0/2$  to  $+T_0/2 \ (–25\,µ\text{s} \ \text{...} +25\,µ\text{s})$  ist  $y(t) = x(t)$. Mit  $f_x= 10\,\rm{kHz} = 1/(2T_0)$  therefore applies to this section:

$$y(t)={\rm 1V}\cdot\cos(2{\pi} f_0\hspace{0.05cm}t)={\rm 1V}\cdot\cos(\pi \cdot {t}/{T_0}).$$

• This results in the following for the DC signal component:

$$A_0=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2}y(t)\,{\rm d} t=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2}{\rm 1V}\cdot\cos(\pi\cdot {t}/{T_0})\,{\rm d}t.$$

• With the substitution  $u = \pi \cdot t/T_0$  one obtains:

$$A_0=\left. \frac{ {\rm 1V}}{\pi}\int_{-\pi /2}^{\pi/2}\cos(u)\,{\rm d}u=\frac{ {\rm 1V}}{\pi}\sin(u)\; \right| _{-\pi/2}^{\pi/2}=\frac{ {\rm 1V}\cdot 2}{\pi} \hspace{0.15cm}\underline{\approx 0.637\;{\rm V}}.$$

(4)  Since  $y(–t) = y(t)$  holds, all sine coefficients  $B_n = 0$. Thus  $B_2 \hspace{0.1cm}\underline{= 0}$ also holds.

(5)  For the coefficients  $A_n$  applies with the substitution  $u = \pi \cdot t/T_0$  according to the given integral:

$$A_n = \frac{2{\rm V}}{T_0}\int_{-T_0/2}^{T_0/2}\cos(\pi\frac{t}{T_0})\cdot \cos(n\cdot 2\pi\frac{t}{T_0})\,{\rm d}t = \frac{2{\rm V}}{\pi}\int_{-\pi/2}^{\pi/2}\cos(u)\cdot \cos(2n u)\,{\rm d}u \quad \Rightarrow \quad A_n = \left( { - 1} \right)^{n + 1} \frac{{4\;{\rm{V}}}}{{{\rm{\pi }}\left( {4n^2 - 1} \right)}}.$$

The coefficient  $A_2$  is thus equal to  $-4 \,\text{V}/(15\pi) \hspace{0.1cm}\underline{\approx -\hspace{0.05cm}0.085 \, \text{V}}$.

(6)  For the finite Fourier series with  $N = 3$  the following applies in general:

$$y_3(t)=\frac{2{\rm V}}{\pi} \cdot \left [ 1+{2}/{3} \cdot \cos(\omega_0t)-{2}/{15}\cdot \cos(2\omega_0t)+{2}/{35}\cdot \cos(3\omega_0t) \right ].$$

At time  $t = 0$    $y_3(0) \approx 1.0125 \ \rm V$; thus the error is  $\varepsilon_3(t = 0) \hspace{0.15cm}\underline{= 0.0125 \,\text{V}}$ .

(7)  At time  $t = 25\,µ\text{s}$  entspricht der halben Periodendauer des Signals  $y(t)$. Hierfür gilt wegen  $\omega_0 \cdot T_0 = 2\pi$:

$$y_3(T_0/2) = \frac{2{\rm V}}{\pi} \left [1+\frac{2}{3} \cdot \cos({\pi}) -\frac{2}{15}\cdot \cos(2\pi)+\frac{2}{35}\cdot \cos(3\pi)\right ]= \frac{2{\rm V}}{\pi}\left [1-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}\right ] = \frac{2{\rm V}}{7\pi}\approx 0.091{\rm V}.$$

• Da  $y(T_0/2) = 0$  ist, ergibt sich auch  $\varepsilon_3(T_0/2) \hspace{0.15cm}\underline{\approx 0.091\,{\rm V}}$.
• Dieser Fehler ist um mehr als den Faktor $7$ größer als der Fehler bei  $t = 0$, da  $y(t)$  bei  $t = T_0/2$  mehr hochfrequente Anteile besitzt (spitzförmiger Verlauf).
• Wird gefordert, dass der Fehler  $\varepsilon_3(T_0/2)$  kleiner als  $0.01$  sein soll, dann müssten mindestens  $32$  Fourierkoeffizienten berücksichtigt werden.