# What is the decay rate (time constant) of two One-Pole LPF's cascaded in series?

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 $$t_2$$?

Thanks.

• pick which LPF has the dominant pole (the slowest decay). asymptotically that is the decay rate of the whole thing. Commented Dec 29, 2020 at 19:19
• 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.
– mike
Commented Dec 29, 2020 at 23:06

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

$$h_{tot}(t)=\frac{e^{-\alpha_1t}-e^{-\alpha_2t}}{\alpha_2-\alpha_1}u(t)\tag{4}$$

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.

• 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.
– mike
Commented Dec 29, 2020 at 23:12
• 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? Commented Dec 29, 2020 at 23:42
• 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?
– mike
Commented Dec 30, 2020 at 6:12