I've thought about designing an active first-order bandpass filter with variable center frequency and fixed bandwidth, using OpAmps, as a project for an Applied Electronics class. My first try was to use the following topology, fixing the capacitor values and changing the frequencies by varying the resistances. But to do this without changes in the bandwidth, the bandwidth's derivative in respect to R should be 0. It happens that the only solution to this equation is to have equal capacitances, what would lead to equal frequencies to both low and high-pass filters. enter image description here Given this mathematical no-go, please give me some advice on how to proceed about this project idea.

Thank you.

EDIT: Thanks to the commenters, I've realised a fixed bandwidth wasn't really necessary for me, and using the State Variable Filter topology solved my problem.

  • \$\begingroup\$ Have you already looked up the approach used in setting up the center frequency of a parametric filter? Terms such as "image pole location" and "all pass filter transformation" or "low pass to band pass transformation?" \$\endgroup\$
    – jonk
    Jun 22, 2018 at 0:50
  • 1
    \$\begingroup\$ Have you looked into a state variable filter? For example: State Variable Filters - "For the values shown, centre frequency is 159Hz. R6 and R7 can be replaced with a dual gang pot, making the filter frequency variable over a wide range." \$\endgroup\$ Jun 22, 2018 at 2:54
  • \$\begingroup\$ I am afraid, you will have problems to find a bandpass topology with a fixed bandwidth for a variable center frequency. That means: The quality factor Q=fo/BW must vary synchronously with the center frequency fo (for BW=const.). More than that, it is even not easy to find a bandpass structure which can be single-element tuned. In most cases, two parts are to be varied at the same time (two resistors or two capacitors). \$\endgroup\$
    – LvW
    Jun 22, 2018 at 8:02
  • \$\begingroup\$ Do you have a good reason for wanting to keep bandwidth constant whilst centre frequency changes. It may be that there is another solution. \$\endgroup\$
    – Andy aka
    Jun 22, 2018 at 9:04
  • \$\begingroup\$ I hadn't, @jonkm but I'll search for the meaning of these terms. \$\endgroup\$ Jun 23, 2018 at 2:55

1 Answer 1



simulate this circuit – Schematic created using CircuitLab

The shown bandpass circuit (GIC bandpass) uses an active simulated inductor. The corresponding equations are:

R1=R, R5=R6, k2=R2/R1, k4=R4/R1, Co=C3=C

Midband gain: Ao=2

Midband frequency: wo=1/[RC*SQRT(k2k4)]

Quality factor: Q=1/[SQRT(k2k4)]=woRC

Because bandwidt BW=wo/Q=1/RC we can tune the frequency with k2 or k4 (single-element tuning) without changing the bandwidth BW.

As an alternative, we can tune the midfrequency wo by varying R1=R without touching the quality factor Q.

  • \$\begingroup\$ is o a variable, or a subscript of w, like this: \$w_o\$? Is w the same as W, or is it \$\omega\$? is k2 the same as 2k (2×k), or is it \$k_2\$? \$\endgroup\$ Jun 23, 2018 at 9:26
  • \$\begingroup\$ Harry - of course, o is a subscript; and k2 is a factor ("2" is a subscript) because it is - as defined - the ratio R2/R1. \$\endgroup\$
    – LvW
    Jun 23, 2018 at 12:14

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