Recently I came across an exercise which I can’t confirm myself If I’m doing it right. The question itself asked to Design a Band Pass Filter


  • Lower -3db
  • cutoff-frequency 500Hz
  • Higher -3db
  • cutoff-freuqency 1000Hz
  • Gain in Pass-Band 20db
  • List item
  • Attenuation: 20bd/Dekade

Below is a picture of what I came up with, this no homework. It just interested me and tried to solve it considering that I don’t know much about band pass filters I came here .

enter image description here

  • \$\begingroup\$ The easy way to check your work is to enter the circuit into your favorite SPICE program and directly check whether it has the properties you designed for. \$\endgroup\$
    – The Photon
    Commented Dec 30, 2019 at 17:15
  • \$\begingroup\$ @The Photon, I'm sorry but I dont have access. I even felt bad for just putting a picture of my attempt solving this task. I tried to design the circuit on my phone but really tricky doing something like this using only but the phone \$\endgroup\$
    – E199504
    Commented Dec 30, 2019 at 17:18
  • \$\begingroup\$ You will need to get access to a computer to do engineering work. I encourage you to keep at it, but Stackexchange isn't here to do your compute work for you. \$\endgroup\$
    – The Photon
    Commented Dec 30, 2019 at 17:19
  • \$\begingroup\$ @The Photon, sorry it came across like that I just thought maybe someone can confirm If I have done it right or not \$\endgroup\$
    – E199504
    Commented Dec 30, 2019 at 17:20
  • \$\begingroup\$ Right? Is it optimal: probably not. Does it meet the requirement: prove this to yourself by using a spice simulator. \$\endgroup\$
    – Andy aka
    Commented Dec 30, 2019 at 17:23

2 Answers 2


1st order filters are pretty easy as in this case such as yours. Good job.

But if you have a dual Op Amp, why stop at 1st order when you can make a second order High and Low Pass filter by combining two Butterworth Bandpass Filters.

Solution in < 5 minutes using Falstad Browser Simulator

  • Compute geometric centre of each -3dB breakpoints
    • \$\sqrt(500^2*1000^2)=707 Hz~~~~~\$ then \$BW = 500 Hz\$ = 1000-500 for -3dB points with -12dB/octave on either side.
  • Let's choose a Butterworth response

enter image description here

Notice (inserted time response in green) how each edge of the 100Hz square wave produces a sinus wave resembling the heart QRST wave but in fact is faster than the rep rate of a muscle AXON wave. This is due to the center/BW ratio . You can change the shape of each hump but this is forward ,backward with +/- edges.


I'm making some assumptions about the filter specification:

  • far below the -3dB lower frequency of 500 Hz, attenuation slope is 20dB/decade

  • far above the -3dB upper frequency of 1000 Hz, attenuation slope is -20dB/decade

Single-pole RC stages, cascaded and buffered can almost meet this spec. Perhaps good enough.

A crude attempt at such a filter might deal with upper and lower -3dB points separately. Very easy to calculate:

  • High pass \$ RC={{1}\over{2 \pi 500}}\$

  • Low pass \$ RC={1\over{2 \pi 1000}}\$

But its wrong...midband gain when cascaded is less than desired. Of course, the opamp can make up this lack of gain. But the two frequencies where the output falls 3dB below midband gain are not 500 Hz, and 1000 Hz. Those two frequencies are just too close to one another. Were they much farther separated, this simple solution could be good enough.

As is, their separate responses interact, making the proper solution somewhat more complex (pun intended). One might question if the OP's circuit topology could achieve the spec...
Suppose we make both corner frequencies the same (at the geometric mean, at 707.107 Hz.). Mid-band gain not including opamp gain is -6dB: -3dB from low-pass and -3dB from high-pass.

Where do the two cascaded -3dB frequencies fall, relative to mid-band gain? Alas, below 500 Hz, and above 1000 Hz. Conclusion - the OP's circuit topology cannot meet the spec. A higher circuit Q is required.


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