68 lines
2.2 KiB
Markdown
68 lines
2.2 KiB
Markdown
|
![[TD_FGE_2025.pdf#page=20]]
|
||
|
|
|
||
|
|
# Exercise 1. Composed Transfer Function
|
||
|
|
![[Pasted image 20251009091732.png]]![[Pasted image 20251009091748.png]]
|
||
|
|
|
||
|
|
1.
|
||
|
|
Étude asymptotique de $\underline{G}=1+j\tau_i\omega$:
|
||
|
|
$G_{dB}=||20log(\underline{G})||=20log(\sqrt{1^2+\tau^2\omega^2})$
|
||
|
|
$\phi=arg(\underline{G})=\arctan(\frac{\tau\omega}{1})$
|
||
|
|
|
||
|
|
Étude en basses fréquences:
|
||
|
|
Pour $\omega\rightarrow∞$:
|
||
|
|
$G_{dB}\approx20log(1)=0$
|
||
|
|
$\phi\approx\arctan{(0)}=0$
|
||
|
|
|
||
|
|
Étude en hautes fréquences:
|
||
|
|
Pour $\omega>>1$
|
||
|
|
$G_{dB}\approx20log(\tau\omega)=20log(\tau)+20log(\omega)$
|
||
|
|
$\phi\approx\arctan{(∞)}=\frac{\pi}{2}$
|
||
|
|
|
||
|
|
On cherche le croisement des deux asymptotes.
|
||
|
|
On cherche donc $20log(\tau\omega)=0$, ce qui implique que $\tau\omega=1$ et par conséquent que $\omega=\frac{1}{\tau}$.
|
||
|
|
Alors, $G_{dB}(\frac{1}{\tau})20log(\sqrt{2})\approx3$
|
||
|
|
$\phi(\frac{1}{\tau})=\arctan{(1)}=\frac{\pi}{4}$
|
||
|
|
|
||
|
|
![[Drawing 2025-10-09 09.27.54.excalidraw]]
|
||
|
|
|
||
|
|
2.
|
||
|
|
$G_2=\frac{1}{G}$, donc
|
||
|
|
$G_{dB_2}=20log(1)-20log(|\underline{G}|)=-G_{dB}$
|
||
|
|
|
||
|
|
$\phi_2=-\phi$
|
||
|
|
|
||
|
|
![[Drawing 2025-10-09 09.53.13.excalidraw]]
|
||
|
|
|
||
|
|
3.
|
||
|
|
$G_{tot}=\frac{(1+j\tau_1\omega)(1+j\tau_2\omega)}{(1+j\tau_3\omega)(1+j\tau_4\omega)}=\frac{G_1G_2}{G_3G_4}$
|
||
|
|
|
||
|
|
# Exercise 2: .Bandpass filter 2nd order
|
||
|
|
|
||
|
|
![[Pasted image 20251009105118.png]]
|
||
|
|
|
||
|
|
1.
|
||
|
|
![[Pasted image 20251009105236.png]]
|
||
|
|
|
||
|
|
On cherche $\underline{V_S}$.
|
||
|
|
|
||
|
|
$Z_{eq}$ correspond à $L$ et $C$, donc $\frac{1}{Z_{eq}}=\frac{1}{Z_L}+\frac{1}{Z_C}=jC\omega-j\frac{1}{L\omega}$
|
||
|
|
Par le pont diviseur, $\underline{V_S}=\frac{Z_{eq}}{R+Z_{eq}}V_e=\frac{1}{\frac{R}{Z_{eq}+1}}V_e$
|
||
|
|
$\implies H=\frac{V_S}{V_e}=\frac{1}{1+jR(C\omega-\frac{1}{L\omega})}$
|
||
|
|
Je cherche $\alpha$ et $\beta$ tel que $C\omega-\frac{1}{L\omega}=\alpha(\frac{\omega}{\beta}-\frac{\beta}{\omega})$
|
||
|
|
$\frac{\alpha}{\beta}=C$ et $\alpha\beta=\frac{1}{L}\implies\begin{cases} \alpha=\beta C\implies\alpha=C\sqrt{\frac{1}{LC}}=\sqrt{\frac{C}{L}} \\ \beta^2C=\frac{1}{L}\implies\beta=\sqrt{\frac{1}{LC}} \end{cases}$
|
||
|
|
|
||
|
|
|
||
|
|
Donc $H(j\omega)=\frac{1}{1+jR\sqrt{\frac{C}{L}}(\frac{\omega}{\sqrt{\frac{1}{LC}}}-\frac{\sqrt{\frac{1}{LC}}}{\omega})}$
|
||
|
|
|
||
|
|
# Exercise 3:
|
||
|
|
![[Pasted image 20251009115819.png]]
|
||
|
|
## Part I: Impedance analyses
|
||
|
|
1.
|
||
|
|
$Z_{eq}=\frac{RjL\omega}{R+jL\omega}+\frac{\frac{R}{jC\omega}}{R+\frac{1}{jC\omega}}=$
|
||
|
|
![[IMG_7440.jpeg]]
|
||
|
|
|
||
|
|
2.
|
||
|
|
|
||
|
|
|
||
|
|
|