Exercise 2.5: Half-Wave Rectification

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Rectified cosine functions

We are looking for the Fourier coefficients of the signal  $x(t)$, sketched below, which results from the one-way rectification of the sinusoidal signal  $w(t)$  with amplitude  $\pi /2$.

The Fourier series representation of the signal  $u(t)$ sketched above is assumed to be known.  This was already determined in  Exercise 2.4.  Taking into account the amplitude  $\pi /2$  the following applies:

$$u(t)=1+\frac{2}{3} \cdot \cos(\omega_1t)-\frac{2}{15}\cdot \cos(2\omega_1t)+\frac{2}{35}\cdot \cos(3\omega_1t)-\dots$$

It should be noted:

  • The basic circular frequency is denoted by  $\omega_1$ .  But since the period of the signals  $u(t)$  and  $v(t)$  is  $T/2$,   $\omega_1 = 2\pi /(T/2) = 4 \pi /T$.
  • Because in this task the signals  $u(t)$,  $w(t)$  and  $x(t)$  are to be related to each other, the signal  $u(t)$  must also be represented with the period duration  $T$  of the signal  $x(t)$. 
  • With  $\omega_0 = 2\pi /T = \omega_1/2$  the same applies:
$$u(t)=1+\frac{2}{3} \cdot \cos(2\omega_0t)-\frac{2}{15} \cdot \cos(4\omega_0t)+\frac{2}{35} \cdot \cos(6\omega_0t)-\dots$$

For the Fourier coefficients this means:

  • The DC coefficient results in  $A_0 = 1$,
  • All sine coefficients are  $B_n = 0$,
  • The cosine coefficients with odd  $n = 1, \ 3, \ 5, \dots$ are all  $0$,
  • The cosine coefficients with even  $n = 2, \ 4, \ 6, \dots$ are not equal to  $0$ :

This results in the following numerical values:

$$A_2=2/3; \;A_4=-2/15;\;A_6=2/35;\;A_8=-2/63.$$


  • This exercise belongs to the chapter  Fourier Series.
  • You can find a compact summary of the topic in the two learning videos
Zur Berechnung der Fourierkoeffizienten  ⇒   "To calculate the Fourier coefficients",
Eigenschaften der Fourierreihendarstellung   ⇒   "Properties of the Fourier series representation".



Calculate the Fourier coefficients of the signal  $v(t)$.  What is the coefficient  $A_2$?

$v(t)$:   $A_2\ = \ $


Calculate the Fourier coefficients of the signal  $w(t)$.  What is the coefficient  $B_1$?

$w(t)$:  $B_1\ = \ $


How can  $x(t)$  be composed of  $v(t)$  and  $w(t)$?  Give the corresponding Fourier coefficients of the signal  $x(t)$.  In particular:

$x(t)$:  $A_0\ = \ $

$\hspace{1cm}B_1\ = \ $

$\hspace{1cm}A_2\ = \ $


(1)  The shifted signal  $v(t)$  is also even and all sine coefficients are accordingly zero.

  • Nothing changes in the DC signal coefficient:   $A_0 = 1$.
  • From the signal curves it can be seen that  $v(t) = u(t - T/4)$  applies: 
$$v(t)=1+\frac{2}{3}\cdot \cos(2\omega_0(t-\frac{T}{4}))-\frac{2}{15}\cdot \cos(4\omega_0(t-\frac{T}{4}))+\frac{2}{35}\cdot \cos(6\omega_0(t-\frac{T}{4}))-\dots$$
  • The cosine terms can now be transformed with  $\omega_0 \cdot T = 2 \pi$  : 
  • This gives us for the Fourier series:
$$v(t)=1-{2}/{3}\cdot \cos(2\omega_0t)-{2}/{15}\cdot \cos(4\omega_0t)-{2}/{35}\cdot \cos(6\omega_0t)-\dots$$
or for the cosine coefficients with even-numbered  $n$:
$$A_n=\frac{-2}{n^2-1}\hspace{0.5cm}\Rightarrow\hspace{0.5cm}A_2=-\hspace{-0.05cm}2/3 \hspace{0.1cm}\underline{= -\hspace{-0.05cm}0.667}.$$

(2)  Because of  $w(t) = \pi /2 \cdot \sin(\omega_0 t)$  all Fourier coefficients except  $B_1 = \pi /2 \hspace{0.1cm}\underline{=1.571}$  are zero.

(3)  From the graphical representation one can see the relationship  $x(t)={1}/{2} \cdot \big [v(t)+w(t) \big].$ This means:

$$x(t)=\frac{1}{2}+\frac{\pi}{4}\cdot \sin(\omega_0 t)-\frac{1}{3}\cdot \cos(2\omega_0 t)-\frac{1}{15}\cdot \cos(4\omega_0 t)-\frac{1}{35}\cdot \cos(6\omega_0 t)-\ldots$$
  • The Fourier coefficients sought are thus:
$$A_0 \hspace{0.1cm}\underline{=0.5},\hspace{1cm} B_1 = \pi /4 \hspace{0.1cm}\underline{= 0.785},\hspace{1cm} A_2\hspace{0.1cm}\underline{ = -0.333}.$$