The expression derived by LvW is correct but there are no steps showing how he got there. Of course, if you compute the gain \$H(s)=-\frac{Z_1(s)}{Z_2(s)}\$ in which \$Z_1(s)=R_1||\frac{1}{sC_1}\$ and \$Z_2(s)=R_2+\frac{1}{sC_2}\$ you'll get there but a) you can make mistakes while developing the expression b) you will need to rearrange the formula to unveil the needed gain in the flat portion. I will show how the fast analytical techniques or FACTs (see this [book][1]) will get you there just by inspecting simple schematics, no algebra at all. Look at the below schematic:

[![enter image description here][2]][2]

You start by determining the transfer function in dc, for \$s=0\$ and you do that by opening all caps. You find \$H_0=0\$ because the first cap blocks the dc and creates a zero at the origin. Then, you reduce the excitation to 0 V and "look" at the resistance offered by the capacitor connection terminals when they are temporarily removed from the circuit. Associating the resistance with the involved capacitor gives you the corresponding time constants. Considering the virtual ground, it is easy to find \$\tau_1=R_1C_1\$ and \$\tau_2=R_2C_2\$. You can sum these two guys to form \$b_1\$ in the denominator. \$b_2\$ is is simply obtained by combining \$\tau_1\$ with a time constant obtained when \$C_1\$ is replaced by a short circuit: \$\tau_{12}=C_2R_2\$: \$b_2=\tau_1\tau_{12}=R_1C_1R_2C_2\$. There is redundancy meaning \$b_2=\tau_2\tau_{21}\$ in which you combine \$\tau_2\$ with a time constant obtained when \$C_2\$ is replaced by a short circuit. Results are identical. The denominator equals:

\$D(s)=1+sb_1+s^2b_2=1+s^2(R_1C_1+R_2C_2)+s^2R_1C_1R_2C_2\$.

From this second-order polynomial form, we can define a quality coefficient \$Q=\frac{\sqrt{b_2}}{b_1}\$ and a resonant frequency \$\omega_0=\frac{1}{\sqrt{b_2}}\$.

The numerator can be derived using the generalized transfer function form obtained by calculating three gains when each capacitor is alternatively set in its high-frequency state (a short circuit) and when both caps are replaced by a short circuit. You can see from the sketches that a gain \$H_2\$ exists only when \$C_2\$ is replaced by a short circuit. The gain in this case is \$H^2=-\frac{R_1}{R_2}\$. The numerator is defined as:

\$N(s)=H_0+s(H^1\tau_1+H^2\tau_2)+s^2H^{12}\tau_1\tau_{12}=-sR_1C_2=-\frac{s}{\omega_z}\$ where \$\omega_z=\frac{1}{R_1C_2}\$.

With a proper factorization, you can rework this transfer function in a *low-entropy* form in which you immediately see the gain you wanted:

\$H(s)=-H_{bp}\frac{1}{1+Q(\frac{s}{\omega_0}+\frac{\omega_0}{s})}\$ in which the bandpass gain is simply \$H_{bp}=\frac{R_1C_2}{R_1C_1+R_2C_2}\$. With the given values, that bandpass gain is exactly 1.9988 or 6.015 dB. All calculations appear in the below Mathcad screenshots:

[![enter image description here][3]][3]

[![enter image description here][4]][4]

You can apply the FACTs to passive or active circuits. The nice thing is that you individually determine the coefficients of the numerator and the denominator via small sketches. That way, if a deviation exists between the raw formula and what you have derived, it is easy to solve the guilty intermediate step and correct it. With a brute-force analysis, you would have to restart from scratch. An introduction to the FACTs can be found [here][5]. I encourage students and engineers to acquire that skill given the ease and speed it provides when analysing transfer functions.


  [1]: https://www.amazon.com/Linear-Circuit-Transfer-Functions-Introduction/dp/1119236371/ref=asap_bc?ie=UTF8
  [2]: https://i.sstatic.net/27bvX.png
  [3]: https://i.sstatic.net/bzvnX.png
  [4]: https://i.sstatic.net/SIpnT.png
  [5]: http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf