If you have a One-Pole LPF, and set the cutoff frequency to $\frac{1}{2πt}$ where $t$ is the time to $\frac{1}{e}$ amplitude, it will provide an exponential decay in response to a step function (eg. going from 1 to 0) following the curve $y=e^{\frac{-x}{t}}$.

I am wondering how to similarly predict the decay rate of two One-Pole LPF's cascaded in series.

Ie. If LPF1 is filtering the signal at $t_1 = 2.4$, and then you run the output of LPF1 into LPF2 which is filtering the signal at $t_2 = 0.15$, in a steady state decay (eg. with a step from 1 to 0), what would be the effective "$t$" of this cascaded system?

I presume the cascade would still follow a nice predictable exponential decay just at a slightly different rate? If so, is there an equation that can predict the final "$t$" based on $t_1$ and $t_2$?


  • $\begingroup$ pick which LPF has the dominant pole (the slowest decay). asymptotically that is the decay rate of the whole thing. $\endgroup$ Commented Dec 29, 2020 at 19:19
  • $\begingroup$ Thanks rob. Yeah I noticed it seems to be primarily determined by the slowest one and it's not far off at all from it. $\endgroup$
    – mike
    Commented Dec 29, 2020 at 23:06

1 Answer 1


If both filters have the same time constant, i.e., if their individual impulse responses are

$$h(t)=e^{-\alpha t}u(t),\qquad \alpha>0\tag{1}$$

then the impulse response of the cascade is

$$h_{tot}(t)=te^{-\alpha t}u(t)\tag{2}$$

If both filter have different time constants with individual impulse responses given by

$$h_i(t)=e^{-\alpha_it}u(t),\qquad \alpha_i>0\tag{3}$$

the total impulse response is given by


I don't think it makes much sense to ask for an "effective" time constant of the cascaded systems. How should one define such a time constant? Of course, for large $t$ the impulse response $(2)$ decays as $e^{-\alpha t}$, and the impulse response $(4)$ decays as $e^{-\min(\alpha_1,\alpha_2)t}$, but that doesn't really say much about the systems' behavior.

  • $\begingroup$ Thanks Matt. It's actually relevant to me because I'm using the LPF's to process single sample impulses to create synthesizer envelopes. I know there are other ways to make synth envelopes but I enjoy this method for a few reasons and I'm just trying to standardize what is happening in a predictable way. So I guess in short, as rob said in the comment, the slower LPF mostly dictates the impulse response decay. Thanks for those equations. I will need to play with #4 a bit to see how it all works out. $\endgroup$
    – mike
    Commented Dec 29, 2020 at 23:12
  • $\begingroup$ mike, are you trying to make a sorta smooth ADSR outa this? you want a decay rate and to be able to place the envelope peak where you want it? $\endgroup$ Commented Dec 29, 2020 at 23:42
  • $\begingroup$ Yeah rob. That's the idea. I've got it working and it's pretty nice based on a series of a fast One-Pole LPF (to give an "attack" phase) and slow One-Pole LPF (for the decay). I'm iterating through to find the peak value first so I can standardize the output to that. Seems effective. I'm happy with the result. But I'm curious. If you want a totally smooth ADSR how would you do it instead? $\endgroup$
    – mike
    Commented Dec 30, 2020 at 6:12

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