band pass - Deriving Bandpass Transfer Function

I'm designing a bandpass filter and have been doing some reading on it. I found a transfer function describing the circuit (which apparently all formulas describing this circuit are derived from):

$$\frac{V_{\text{o}}}{V_{\text{i}}} = -\frac{Kj2\pi f}{\left(1+\frac{jf}{f_1}\right)\left(1+\frac{jf}{f_2}\right)}$$

^{simulate this circuit – Schematic created using CircuitLab}

Circuit Diagram of Inverting Bandpass Filter

Where do they get \$C_1\$, \$C_2\$, \$R_1\$, \$R_2\$ from?

Answer

The TF of the circuit is -\$\dfrac{Z_f}{Z_i}\$ where \$Z_f\$ = \$R_2\$||X\$_{C_2}\$ and \$Z_i\$ = \$R_1\$ +X\$_{C_1}\$.

I.e. \$Z_f\$ is the feedback impedance and \$Z_i\$ is the input impedance.

In terms of the s-plane operator: -

\$Z_f\$ = \$\dfrac{R_2\cdot \frac{1}{sC_2}}{R_2 + \frac{1}{sC_2}}\$ and \$Z_i\$ = \$R_1+\frac{1}{sC_1}\$

The TF then becomes \$-\dfrac{\dfrac{R_2\cdot \frac{1}{sC_2}}{R_2 + \frac{1}{sC_2}}}{R_1+\frac{1}{sC_1}}\$

If you work this down you get TF = \$\dfrac{-R_2}{1+sC_2R_2}\cdot\dfrac{sC_1}{1+sC_1R_1}\$

I think you can see that this pretty much aligns with the TF at the top of the question (when s is replaced by jw where w = 2\$\pi f\$)