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?

  • \$\begingroup\$ 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. \$\endgroup\$ – Marcus Müller Jan 24 '18 at 11:35
  • 2
    \$\begingroup\$ Please check the derivation I did in a previous post: electronics.stackexchange.com/questions/179356/… \$\endgroup\$ – Verbal Kint Jan 24 '18 at 11:44
  • \$\begingroup\$ 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. \$\endgroup\$ – 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\$)


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