# Exercise 2.4: Rectified Cosine

Cosine and rectified cosine

A cosine signal  $x(t)$  with amplitude  $1\,\rm{V}$  and 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)  and  (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:

• To solve the problem, you can use the following definite integral   $($let $n$ be an integer$)$:
$$\int ^{\pi /2}_{-\pi /2}\cos(u)\cdot\cos(2nu)\,{\rm d}u = (-1)^{n+1}\cdot\frac{2}{4n^2-1}.$$

### Questions

1

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

 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 mean–free  $(A_0 = 0)$  and it 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  $+T_0/2 \ (–25\,µ\text{s} \ \text{...} +25\,µ\text{s})$  is  $y(t) = x(t)$. With  $f_x= 10\,\rm{kHz} = 1/(2T_0)$  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 coefficient:
$$A_0=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2}y(t)\,{\rm d} t=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2}{\rm 1V}\cdot\cos(\pi\cdot {t}/{T_0})\,{\rm d}t.$$
• 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)  The time  $t = 25\,µ\text{s}$  corresponds to half the period of the signal  $y(t)$.  The following applies here because of  $\omega_0 \cdot T_0 = 2\pi$:

$$y_3(T_0/2) = \frac{2{\rm V}}{\pi} \left [1+\frac{2}{3} \cdot \cos({\pi}) -\frac{2}{15}\cdot \cos(2\pi)+\frac{2}{35}\cdot \cos(3\pi)\right ]= \frac{2{\rm V}}{\pi}\left [1-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}\right ] = \frac{2{\rm V}}{7\pi}\approx 0.091{\rm V}.$$
• Since  $y(T_0/2) = 0$  this also results in  $\varepsilon_3(T_0/2) \hspace{0.15cm}\underline{\approx 0.091\,{\rm V}}$.
• This error is larger than the error at  $t = 0$ by more than a factor of  $7$, since  $y(t)$  has more high-frequency components at  $t = T_0/2$  $($peak-shaped course$)$.
• If it is required that the error  $\varepsilon_3(T_0/2)$  be smaller than  $0.01$  then at least  $32$  Fourier coefficients would have to be taken into account.