I am doing a lot of audio synthesis DSP where I need to use low pass filters to shape the decay of an impulse. I understand that a one pole filter fed a discontinuity, say from a steady state of 1's to a steady state of 0's, will decay with a perfectly exponential curve. I believe the decay will be such that the signal will reach an amplitude of $1/e$ at $time = 1/(2*pi*f)$.

This makes conceptualizing one pole low pass impulse modeling very easy. But I am trying to understand how other filter orders will do the same.

I understand that a second order filter can be approximated running two one pole filters in sequence (ie. each will attenuate by 6 dB/oct, leading to a 12 dB/oct curve). If that is the case, then I would expect the step response of a second order filter without resonance to behave the same as a one pole filter and remain exponential, only that much faster.

So perhaps the time to 1/e would be:

$time = 1/(2*pi*f)^2$

Would that be correct?

I asked here about a lower slope filter and no one has replied so I am guessing it is not common knowledge: A one-pole LPF (6 dB/oct) has a step response to $1/e$ amplitude of $time = 1/(2∗pi∗f)$. What would the response time be of a 3 dB/oct filter?

But I presume the same principle would then apply. A 3 dB/oct low pass filter would have a time constant of:

$time = 1/(2*pi*f)^{0.5}$

And if this is all correct, then we could say the time constants of any non-resonant low pass filter would be roughly:

$time = 1/(2*pi*f)^{filter-order}$

What do you think? Does all this sound correct? If so, there is no difference between low pass filtering with any order of non-resonant filter, as they can all be set up to create the same outcome with different parameters.

Lastly, then the remaining question would be: What does the step response of a resonant second order low pass filter look like, again going from steady 1's to steady 0's? I presume the resonance ruins the precisely exponential decay, but in what manner? Does it create a more compressed curve? Will it resonate creating a wobbling of the output? Will it dip below zero as it tries to settle out?

I tried testing it with a resonant second order LPF but I was just getting almost random maximum amplitudes coming out. Very unpredictable. I'm not sure if I did something wrong or that's to be expected from the resonance.

Thanks for any help understanding all this.

  • $\begingroup$ Your formulation is very uncommon. First off, step or Heaviside function is all zeros followed by all ones and not the other way around. The step response of any system holds the connection $h(t)=s'(t)$ where $h(t)$ is the impulse response and $s(t)$ is the step response. When you write 'first-order system' do you refer the step or the impulse response? Usually, people refer to the impulse response? In any case, try and better formulate your problem, $\endgroup$
    – havakok
    Apr 26, 2020 at 13:30
  • $\begingroup$ The answers to this question should be helpful. $\endgroup$
    – Matt L.
    Apr 26, 2020 at 13:39

1 Answer 1


I can help with some of your multiple questions. First, a cascade of n buffered RC low pass filters (LPFs), a so-called n-th order synchronous LPF, has the impulse response and step response shown in the screenshot below:

Synchronous RC LPF

This is a screenshot from my paper 1 referenced at the bottom. All R values are the same and all C values are the same. Buffering simply means there is a unity gain buffer between each filter and on the first input and final output as well. So no filter is loaded or gets loaded. Notice that the conventional unit step function goes from 0 to 1, so if you want the response to be 1 to 0, just subtract y(t) from 1. Then you can find 1/e times as a function of n, etc.

With regard to the four "time = ..." equations you give, only the first one, for n = 1, is correct. In general, second (and higher) order filters do not all behave the same. The simple RC LPF is the first (primitive) member of several filter families. But things get much more complicated (and useful) as the various properties of the filter families are taken into account and utilized.

Note that a cascade of RC LPFs cannot have overshoot or undershoot. On the other hand, Butterworth LPFs can have "ring" in response to a unit step input. This is nicely illustrated by the following figure from Blinchikoff and Zverev 2:

Butterworth step response


1 E. Voigtman, J.D. Winefordner, “Low-pass filters for signal averaging”, Rev. Sci. Instrum. 57 (1986) 957-966.

2 H.J. Blinchikoff, A.I. Zverev, "Filtering in the Time and Frequency Domains", Wiley-Interscience, John Wiley & Sons, NY, ©1976, p. 114.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.