Deriving The transfer Function of a high Pass active second Order filter
Regarding to derive the transfer function of this second order high pass active filter consider, for example, to apply the KCL to nodes A and B on the figure below (assuming an ideal op. amp.):



$$
\left\{\begin{matrix}
\frac{V_A-V_i}{\frac{1}{sC_1}}+ \frac{V_A-V_O}{R_2}+\frac{V_A-V_B}{\frac{1}{sC_2}}=0 &amp; (1)\\ 
\frac{V_B-V_A}{\frac{1}{sC_2}}+ \frac{V_B}{R_1}+0= 0 &amp; (2)
\end{matrix}\right.
$$

Replacing:

$$ V_B=\frac{V_oR_B}{R_A+R_B} $$

in (1) and (2):

$$\left\{\begin{matrix}
(V_A-V_i)sC_1+ \frac{V_A-V_O}{R_2}+(V_A-\frac{V_oR_B}{R_A+R_B})sC_2=0 &amp; (3)\\ 
(\frac{V_oR_B}{R_A+R_B}-V_A)sC_2+\frac{V_oR_B}{(R_A+R_B)R_1}=0  &amp; (4)
\end{matrix}\right.$$

From (4):
$$
V_A=\frac{V_oR_B}{R_A+R_B}\left ( 1+\frac{1}{sR_1C_2} \right )
$$

Returning with this expression on (3) and, after a long algebraic manipulation, it's possible to obtain \$\frac{V_o}{V_i}\$ as:

$$
\frac{V_o(s)}{V_i(s)}=\frac{\left( 1+\frac{R_A}{R_B}\right)R_1R_2C_1C_2s^2}{R_1R_2C_1C_2s^2+\left[ R_2C_1+R_2C_2 - \frac{R_A}{R_B}R_1C_2 \right]s+1}
$$

The cut-off frequency (in Hz) is given by:

$$ f_c=\frac{1}{2\pi\sqrt{R_1R_2C_1C_2}} $$