Exercise 2.6: Complex Fourier Series

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Various periodic triangular signals

We consider the signal  $x(t)$,  defined by the two parameters  $T_0$  and  $T_1$  where  $T_1 \leq T_0$  should always apply.  For the complex Fourier coefficients

$$D_n=\frac{1}{T_0} \cdot \int_0^{T_0}x(t)\cdot\rm e^{-\rm j\it n\omega_0t}\,{\rm d} \it t$$

of this signal are obtained after mathematical transformations:

$$D_n=\frac{T_0/T_1} {(2\pi n)^2} \cdot \big(1-{\rm e}^{-{\rm j} 2\pi nT_1/T_0}\big)-\frac{\rm j}{2\pi n}.$$

• The parameter set dealt with in subtasks  (1)  and  (3)    $($with $T_1 = T_0/2)$  is represented as the signal  $x(t)$ .
• For  $T_1 = T_0$  ⇒   subtask  (2)  the function  $y(t)$ results.
• In subtask  (4)  the signal  $z(t)$  is considered.  Its Fourier coefficients are:

$$A_0=1/4,\hspace{1cm} A_n=\left\{ \begin{array}{cl} {\frac{\displaystyle-2}{\displaystyle(\pi n)^2}} & {\rm for\; even\; \it n \rm ,} \\ 0 & {\rm for\; odd\; \it n,} \end{array}\right. $$

$$B_n=0\; \;\; \rm{ f\ddot{u}r\; alle\; \it n.}$$

Hints:

• This exercise refers to the page  Complex Fourier series.

Questions

1

Calculate the coefficient  $D_0$  and show that it is always real.  What value results for  $T_1 = T_0/2$, i.e. for the signal  $x(t)$?

$D_0^{(x)}\ = \ $

2

Calculate the complex Fourier coefficients  $D_n^{(y)}$  for  $n \neq 0$  for the special case   $T_1 = T_0$  corresponding to the signal  $y(t)$ .
What are the coefficients  $A_n^{(y)}$  and  $B_n^{(y)}$,  especially for  $n = 1$?

$A_1^{(y)}\ = \ $

$B_1^{(y)}\ = \ $

3

Now calculate the coefficients  $A_n^{(x)}$  and  $B_n^{(x)}$  for the signal  $x(t)$  with  $T_1 = T_0/2$  for  $n \neq 0$.  What are the values for  $A_1^{(x)}$  and  $B_1^{(x)}$?

$A_1^{(x)}\ = \ $

$B_1^{(x)}\ = \ $

4

Which of the following statements are true regarding  $x(t)$,  $y(t)$  and  $z(t)$ ?

	It is true that  $x(t) = y(t) + z(t)$.
	It is true that  $x(t) = y(t) - z(t)$.
	The cosine coefficients  $A_n$  of  $x(t)$  und  $z(t)$  are identical.
	The cosine coefficients  $A_n$  of  $x(t)$  und  $z(t)$  are equal in magnitude.
	The sine coefficients  $B_n$  of  $y(t)$  und  $z(t)$  are identical.

Solution

Solution

(1)  With Euler's theorem, the complex Fourier coefficient  $D_n$ can be represented as follows:

$${\rm Re} [D_n] =\frac{T_0/T_1}{(2\pi n)^2}\cdot(1-\cos(2\pi nT_1/T_0)),$$

$${\rm Im}[D_n] =\frac{T_0/T_1}{(2\pi n)^2} \cdot \sin(2\pi nT_1/T_0)-\frac{1}{2\pi n}.$$

• With the approximation  $\text{sin}(\alpha ) \approx \alpha$  valid for small  $\alpha$–values one obtains for the imaginary part:

$${\rm Im}[D_n] =\frac{T_0/T_1}{(2\pi n)^2}\cdot(2\pi nT_1/T_0)-\frac{1}{2\pi n}=0.$$

• For the real part one obtains with  $\text{cos}(\alpha) \approx 1 – \alpha^{2}/2$:

$${\rm Re}[D_n] =\frac{T_0/T_1}{(2\pi n)^2}\frac{(2\pi nT_1/T_0)^2}{2}=\frac{T_1/T_0}{2}.$$

• For  $T_1 = T_0/2$  it follows that the DC signal coefficient  $D_0^{(x)} \hspace{0.1cm}\underline{= 0.25}$.
• With  $T_1 = T_0$  it results in  $D_0^{(y)} = 0.5$.
• A comparison with the signals   $x(t)$  and  $y(t)$ on the data page show the correctness of these results.

(2)  It is now assumed  $n \neq 0$ . With  $T_1 = T_0$  one obtains for the real part because of  $\text{cos}(2\pi n) = 1$:

$${\rm Re}[D_n^{(y)}] =\frac{1}{(2\pi n)^2}\cdot(1-\cos(2\pi n))=0.$$

• The imagnary part is:

$${\rm Im}[D_n^{(y)}] =\frac{1}{(2\pi n)^2}\cdot(\sin(2\pi n))-\frac{1}{2\pi n}.$$

• Because  $\text{sin}(2\pi n) = 0$  it follows that   ${\rm Im}[D_n] =-{1}/({2\pi n}).$ Thus

$$D_n^{(y)}=\frac{-\rm j}{2\pi n}={1}/{2} \cdot (A_n- {\rm j} \cdot B_n).$$

• The coefficient comparison yields  $A_n^{(y)} = 0$  and  $B_n^{(y)} = 1/(\pi n)$.  In particular  $A_1^{(y)} \hspace{0.1cm}\underline{= 0}$  und  $B_1^{(y)}\hspace{0.1cm}\underline{ \approx 0.318}$.

• As expected, $B_{-n}^{(y)} = -B_n^{(y)}$ always holds.

(3)  From the general equation calculated in subtask  (1)  it follows with  $T_1/T_0 = 1/2$:

$$D_n^{(x)}=\frac{2}{(2\pi n)^2}(1-\cos(\pi n))+{\rm j}\cdot \left[\frac{2\sin(\pi n)}{(2\pi n)^2}-\frac{1}{(2\pi n)}\right].$$

• From this one obtains the cosine coefficients

$$A_n^{(x)}={2}\cdot{\rm Re}[D_n] =\left\{ \begin{array}{cl} {\frac{\displaystyle 2}{\displaystyle(\pi n)^2}} & {\rm for\; odd\; \it n ,} \\ 0 & {\rm for\; even\;\it n.} \end{array}\right. $$

• The sine coefficients are:

$$B_n^{(x)}=-2\cdot{\rm Im}[D_n] =\frac{1}{\pi n}.$$

• Here it is taken into account that for all integer values of  $n$  the function  $\text{sin}(n\pi ) = 0$ . The first real coefficients are as follows

$$A_1^{(x)} = 2/\pi^{2} \hspace{0.1cm}\underline{\approx 0.203},$$

$$B_1 = 1/\pi \hspace{0.1cm}\underline{\approx 0.318}.$$

(4)  The correct solutions are 2, 4 and 5:

• The signal  $x(t)$  is equal to the difference between  $y(t)$  and  $z(t)$.  Since  $z(t)$  is an even and  $y(t)$  an odd function, the cosine coefficients  $A_n$  are determined by the coefficients of the signal  $z(t)$  alone, but with negative signs.
• The sine coefficients  $B_n$  completely agree with those of  $y(t)$.
• The DC component of  $x(t)$  results from the difference of the two DC components of  $y(t)$  und  $z(t)$:  

$$A_0 = 0.5 - 0.25 = 0.25.$$