Exercise 2.6Z: Magnitude and Phase

From LNTwww

Signal  $x(t)$  to be analysed

The aim is to show the connection between

  • the real Fourier coefficients  $A_n$  und  $B_n$,
  • the complex coefficients  $D_n$, and
  • the magnitude or phase coefficients  $(C_n$,  $\varphi_n)$.


For this we consider the periodic signal

$$x(t)=1{\rm V+2V}\cdot\cos(\omega_0 t) +{\rm 2V}\cdot\cos(2\omega_0 t)- \ {\rm 1V}\cdot\sin(2\omega_0 t)-{\rm 1V}\cdot\sin(3\omega_0 t).$$

This signal is shown in the graph in the range from  $–2T_0$  to  $+2T_0$.



Hints:

  • 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".


Questions

1

What are the coefficients  $A_0$,  $D_0$,  $C_0$ and  $\varphi_0$?

$A_0\ = \ $

 $\text{V}$
$D_0\ = \ $

 $\text{V}$
$C_0\ = \ $

 $\text{V}$
$\varphi_0\ = \ $

 $\text{deg}$

2

Which of the cosine and sine coefficients are not equal to zero?

$\ A_1$,
$\ B_1$,
$\ A_2$,
$\ B_2$,
$\ A_3$,
$\ B_3$.

3

What are the coefficients  $\varphi_1$,  $C_1$  and  $D_1$?

$\varphi_1\ = \ $

 $\text{deg}$
$C_1\ = \ $

 $\text{V}$
$\text{Re}[D_1]\ = \ $

 $\text{V}$
$\text{Im}[D_1] \ = \ $

 $\text{V}$

4

What are the coefficients  $\varphi_2$,  $C_2$  and  $D_2$?

$\varphi_2\ = \ $

 $\text{deg}$
$\text{Re}[D_2]\ = \ $

 $\text{V}$
$\text{Im}[D_2]\ = \ $

 $\text{V}$

5

What are the coefficients  $\varphi_3$  and  $C_3$?

$\varphi_3\ = \ $

 $\text{deg}$
$C_3\ = \ $

 $\text{V}$

6

What is the complex Fourier coefficient  $D_\text{–3}$?

$\text{Re}[D_{-3}]\ = \ $

 $\text{V}$
$\text{Im}[D_{-3}]\ = \ $

 $\text{V}$


Solution

(1)  The DC signal coefficient is  $A_0 = 1\,{\rm V}$.

  • At the same time,  $C_0 = D_0 = A_0 \hspace{0.1cm}\Rightarrow \hspace{0.1cm} C_0 \hspace{0.1cm}\underline{= 1\,{\rm V}}, \varphi_0 \hspace{0.1cm}\underline{= 0}$.


(2)  The correct answers are 1, 3, 4 and 6:

  • There are no components with  $\sin(\omega_0t)$  and  $\cos(3\omega_0t)$.
  • It follows directly that  $B_1 = A_3 = 0$.
  • All other coefficients listed here are non-zero.


(3)  In general:

$$\varphi_n=\arctan\left({B_n}/{A_n}\right),\hspace{0.5cm}C_n=\sqrt{A_n^2+B_n^2},\hspace{0.5cm}D_n={1}/{2} \cdot (A_n-{\rm j}\cdot B_n).$$
  • Because  $B_1 = 0$  we get  $\varphi_1 \hspace{0.1cm}\underline{= 0}, \ C_1 = A_1 \hspace{0.1cm}\underline{= 2 \,{\rm V}}$  and  $D_1 = A_1/2 \hspace{0.1cm}\underline{= 1 \,{\rm V}}$.


(4)  With  $A_2 = 2\,{\rm V}$  and  $B_2 = -1\,{\rm V}$  one obtains:

$$\varphi_2=\arctan(-0.5)\hspace{0.15cm}\underline{=-26.56^{\circ}},\hspace{0.5cm}C_2=\sqrt{A_2^2+B_2^2}\hspace{0.15cm}\underline{=2.236 \; \rm V},$$
$$D_2={1}/{2} \cdot (A_2-{\rm j}\cdot B_2)=1\;\rm V+{\rm j}\cdot 0.5\, {\rm V} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Re}[D_2]\hspace{0.15cm}\underline{ = 1 \,{\rm V}}, \hspace{0.2cm}{\rm Im}[D_2]\hspace{0.15cm}\underline{ = 0.5\, {\rm V}} .$$


(5)  It is  $\varphi_3 \hspace{0.15cm}\underline{=\hspace{0.1cm}-90^{\circ}}$  and  $C_3 = |B_3| \hspace{0.15cm}\underline{ = 1 \,{\rm V}}$.


(6)  It is  $D_3 = -{\rm j} · B_3/2 ={\rm j}· 0.5 \,{\rm V}$  and  $D_\text{–3} = D_3^{\star} ={\rm j}· B_3/2 = {- {\rm j} · 0.5 \,{\rm V}}$

$$\Rightarrow \hspace{0.3cm} \text{Re}[D_{-3}]\hspace{0.15cm}\underline{=0}, \hspace{0.5cm}\text{Im}[D_{-3}]\hspace{0.15cm}\underline{=\hspace{0.1cm}- 0.5 \,{\rm V}}.$$