# High pass filter: Overshooting before attenuation

Working with a 2nd order Sallen-Key high pass Butterworth filter with a cutoff frequency of $$\f_c=0.5\$$ Hz I am having difficulties making sense of a result that I got:

After applying at its input a square signal with its first harmonic having a frequency of $$\0.1\$$ Hz < $$\f_c\$$. First the output signal surpass the input signal (overshoot) and then gets attenuated:

Blue: Input signal, Yellow: Output signal of the filter

What's confusing me is that this overshooting may be related to the first harmonic (along with the higher frequency ones). But in that case, shouldn't this first harmonic be attenuated in the first place so that this overshooting can't happen?

Edit: The filter schematic is the following:

• This is expected for high Q filters. Butterworth with at least 2 order overshoots. But you can reduce the Q sightly. Use this calc: sim.okawa-denshi.jp/en/OPseikiHikeisan.htm – tobalt Jun 9 at 23:17
• A step response has no harmonic. It has a continuous spectrum. The square wave frequency is not relevant here. You will get the overshoot at each isolated step – tobalt Jun 9 at 23:33
• It would help if you posted a schematic of your filter. – qrk Jun 10 at 0:11
• That doesn't look like an overshoot to me, it looks like three squares amplitude, which is exactly the amplitude of the input step. Your highpass filter has a gain of one at high frequencies, that is the frequencies that define the leading edge of the step. – Neil_UK Jun 10 at 5:47
• If the "overshoot" (undershoot) is a problem, redesign filter reducing Q slightly. – user_1818839 Jun 10 at 12:50

First the output signal surpass the input signal (overshoot)

Initially the output signal follows the input exactly and goes up by the same amount relative to their current baselines. However the output signal quickly 'sags' back to the baseline as each capacitor in the filter charges, with a small undershoot caused by the second capacitor charging less due to the reducing output voltage from the first capacitor.

You could argue that this is 'overshoot' because an AC square wave goes above and below its true baseline and so has a lower amplitude, but in this time frame we (and the filter) cannot know that. It could just be a single step going from low to high voltage, and the filter has neither the 'clairvoyance' to see into the future nor sufficient 'memory' to account for what might have happened in the distant past.

What's confusing me is that this overshooting may be related to the first harmonic (along with the higher frequency ones).

Frequencies below the cutoff frequency have less effect on the output because they are attenuated. What you are seeing is mostly the effect of higher frequencies, with the fundamental frequency determining the 'repetition rate'. Here are the results (simulated with LTspice) for fundamental frequencies of 1 Hz, 10 Hz, and 100 Hz:-

It makes virtually no difference to the impulse waveform whether the lowest harmonic is 0.1Hz, 1Hz, 10Hz or 100Hz. You could remove those lower frequencies entirely and the waveform wouldn't change much. It's not just some effect of being an 'active' op amp based filter either, a similar waveform is produced even with a passive RC filter:-

• Thanks a lot for the help and your time, it has been very clarifying – Javier TG Jun 10 at 8:40

You are looking at the typical step response of a highpass filter. You can start with the prototype Butterworth highpass transfer function and determine its step response:

\begin{align} H(s)&=\dfrac{s^2}{s^2+\sqrt2 s+1} \tag{1} \\ h(t)&=\mathrm{e}^{-\frac{t}{\sqrt2}}\left(\cos\left(\dfrac{t}{\sqrt2}\right)-\sin\left(\dfrac{t}{\sqrt2}\right)\right) \tag{2} \end{align}

And this is how it looks like when plotted (does it look familiar?):

• Crystal clear, thanks a lot. If it was possible I would accept both answers. – Javier TG Jun 10 at 8:39