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$\pi f$)