Inverting band-pass 1st order

Transfer function: $H(j\omega )=A.\frac{j{\displaystyle \frac{\omega }{{\omega }_1}}}{\left(1+j{\displaystyle \frac{\omega }{{\omega }_1}}\right).\left(1+j{\displaystyle \frac{\omega }{{\omega }_2}}\right)}$

Gain in the bandwidth: $A=-\frac{R2}{R1}$

Lower cut-off frequency: $f_1=\frac{1}{2\pi .R_1{C}_1}$

Higher cut-off frequency: $f_2=\frac{1}{2\pi .R_2{C}_2}$