Passe-bande du 1er ordre non-inverseur

Fonction de transfert: $H(j\omega )=1+\frac{j\omega .R_2{C}_1}{(1+j\frac{\omega }{{\omega }_1})\left(1+j{\displaystyle \frac{\omega }{{\omega }_2}}\right)}$

Si C₁ > 100*C₂: $H(j\omega )\simeq \frac{(1+j\frac{\omega }{{\omega }_0})\left(1+j\frac{\omega }{{\omega }_3}\right)}{(1+j\frac{\omega }{{\omega }_1})\left(1+j{\displaystyle \frac{\omega }{{\omega }_2}}\right)}$

Avec:

$f_0=\frac{1}{2\pi .(R_1+R_2)C_1}$

$f_1=\frac{1}{2\pi .R_1{C}_1}$

$f_2=\frac{1}{2\pi .R_2{C}_2}$

$f_3=\frac{1}{2\pi .(R_1\parallel R_2)C_2}$

Gain quand ω < ω₀: $|H(j\omega )|\stackrel{\omega <{\omega }_0}{\to }1$

Gain dans la bande passante (ω₁ < ω < ω₂): $|H(j\omega )|\simeq \frac{{\omega }_1}{{\omega }_0}=1+\frac{R_2}{R_1}$

Gain quand ω > ω₃: $|H(j\omega )|\stackrel{\omega >{\omega }_3}{\to }\frac{{\omega }_1.{\omega }_2}{{\omega }_0.{\omega }_3}=1$