# 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?

• Where did you find that formula (did it come with the typo?)? Most likely: Exactly where you found that formula, you'll find the parametrization methods used there. Point is that $f_1$ and $f_2$ are functions of these components. – Marcus Müller Jan 24 '18 at 11:35
• Please check the derivation I did in a previous post: electronics.stackexchange.com/questions/179356/… – Verbal Kint Jan 24 '18 at 11:44
• That's me copying it out. A booked called RF circuit design. It just states the following formulas are derived from this transfer function. Yeah I gathered, but I cant find a derivation of this anywhere, and would assume all filters where cutoffs are derived the components have to be pulled from somewhere. – user160063 Jan 24 '18 at 11:44

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$\pi f$)