I have designed a Sallen-Key bandpass filter with a centre frequency of 5000 Hz, a Q factor of 2 and an H0 of 5.

I am trying to figure out why it responds as seen here to square wave with a frequency of 200 Hz response to 200 Hz square wave

My interpretation is something like this enter image description here but I'm really struggling to figure why it ripples instead of just spiking and why it looks more like an impulse response


1 Answer 1


A square wave of 200 Hz contains odd harmonics all the way to infinity. The 25th harmonic of 200 Hz is 5 kHz hence you see the band pass filter extracting this harmonic and greatly amplifying it.

Here's a picture of the spectrum of a band pass filter on the 17th harmonic (nearest I could find): -

enter image description here

Picture taken from Acoustics and Psychoacoustics: Introduction to sound - Part 8

So, if you look at the step response of a band pass filter you will see waveforms like this: -

enter image description here

Your circuit (Q = 2) will have a zeta of 0.25 (pretty close the the green trace) and each time the square wave pattern changes you get the step response repeating again and again.

Your interpretation is flawed - a sallen key filter isn't the same as two cascaded 1st order filters because two cascaded filters cannot, by mathematical or practical definition, produce a Q greater than 0.5 and therefore cannot generate decaying sinewave responses as you see.

  • \$\begingroup\$ How is the unit step response combined with the amplification of frequencies within the bandwidth to give the resulting outcome? Also why do we get the step response at low frequencies like this one but not at high frequencies? What happens with high frequencies above the bandwidth? \$\endgroup\$
    – Alura
    Commented Oct 28, 2016 at 20:29
  • \$\begingroup\$ Are you asking about how a band-pass filter does what it does? The math is quite tricky in the time domain because it involves convolution - are you familiar with convolution? I don't understand your 2nd question. Above the bandwidth, higher frequencies are present on the output but they get progressively smaller due to the shape of the filter - if you did an FFT of the output of the filter you would see them just like you can see remnants of the 33rd harmonic (not to scale of course) in my 1st picture. \$\endgroup\$
    – Andy aka
    Commented Oct 28, 2016 at 20:44
  • \$\begingroup\$ OK I think I follow your 2nd question. As the input square wave gets closer to resonance the overshoots combine to produce a reasonably clean sinewave but the higher frequency stuff is still present as I answered to your 3rd question. Look closely at the waveform and you will see it but it's masked by the resonant peak at 5 kHz. \$\endgroup\$
    – Andy aka
    Commented Oct 28, 2016 at 20:46
  • \$\begingroup\$ I am a second year engineering student and I understand convolution.I also understand how the filter responds to frequencies near the center frequency - the first harmonic is the most pronounced due to it's amplitude and gain. I just do not understand why at 200 Hz, there is a step response but there isn't a step response at, say, 10000 Hz. Or is there one? I was trying to ask how the step response combines with the overshoots to produce the output. At 5000 Hz my maximum gain is 5 and the 25th harmonic has an amplitude of about 0.051, yet the output at 200 Hz shoots up to about 3.5. \$\endgroup\$
    – Alura
    Commented Oct 28, 2016 at 21:04
  • \$\begingroup\$ At 10 kHz there is no lower frequency to excite the filter at 5 kHz. \$\endgroup\$
    – Andy aka
    Commented Oct 28, 2016 at 22:02

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