Is there an equation that will graph the shape or predict the maximum amplitude of a single sample impulse fed through a One-Pole LPF?

If you feed a single sample impulse of a given amplitude through a One-Pole LPF you get an envelope that looks like this:

The output of the LPF from its peak will settle into an exponential decay where time to reach $$1/e$$ amplitude is given by $$1/(2πf)$$ where $$f$$ is the cutoff frequency of the LPF.

I am wondering if there's any way to predict the maximum amplitude of the envelope from the sample rate, LPF cutoff freq, and amplitude of the single sample impulse. In general, is there a non-recursive equation that can describe or approximate the curve that results from these three factors?

If such an equation can be developed or exists, then one can take the derivative and solve where it equals zero to find the max.

Is this possible?

Alternatively, I can solve it numerically by just iterating through a temporary setup of a single sample impulse into an LPF of the given parameters until (output_1 < output). This will find the max as well but this is tedious and computationally consuming.

I would hope some equation can do this faster.

Thanks for any thoughts.

• you should know that the impulse response that you have graphed there is not from a One-pole filter. It is at least two poles.. – robert bristow-johnson Dec 28 '20 at 22:38
• and "single sample Dirac impulse" is a misnomer. you might mean "single-sample Kronecker impulse" – robert bristow-johnson Dec 28 '20 at 22:42
• Okay thanks. I took out the "Dirac" if that was a misnomer. But I'm not sure what you mean about the filter type. Are you saying the shape is wrong for a One-Pole and that's how you know it's not? Either way, I think the principle is the same. I am referring to a filter like earlevel.com/main/2012/12/15/a-one-pole-filter where as the author there says: "Note that if you feed the one-pole lowpass an impulse, it will yield a perfect exponential decay. To look at it another way, the feedback path is an iterative solution to calculating an exponential curve." – mike Dec 28 '20 at 22:44
• all this depends, mike, if your problem is about continuous-time (like an RC LPF) or is a discrete time. either way, if it's one-pole, the maximum occurs immediately and it's about scaling. if it's two-pole, you can have the peak occur at some later time. but how big the peak is about scaling. – robert bristow-johnson Dec 28 '20 at 22:52
• //Are you saying the shape is wrong for a One-Pole and that's how you know it's not?// Yes. – robert bristow-johnson Dec 28 '20 at 22:53

As mentioned in the comments, an impulse response with the shape shown in your question can only be obtained by a system with two real-valued poles. In continuous time, with two distinct poles, the total impulse response is

$$h(t)=\frac{e^{-\alpha_1t}-e^{-\alpha_2t}}{\alpha_2-\alpha_1}u(t)\tag{1}$$

It's a basic exercise to determine the location of the maximum of $$(1)$$, and the result is

$$t_{max}=\frac{\log\left(\frac{\alpha_1}{\alpha_2}\right)}{\alpha_1-\alpha_2},\qquad\alpha_1\neq\alpha_2\tag{2}$$

For $$\alpha_1=\alpha_2=\alpha$$, the total impulse response is

$$h(t)=te^{-\alpha t}u(t)\tag{3}$$

and the location of the maximum is

$$t_{max}=\frac{1}{\alpha}\tag{4}$$

The figure below shows an example of an impulse response $$(1)$$ with $$\alpha_1=0.4$$ and $$\alpha_2=0.6$$, and with the location of the maximum predicted by $$(2)$$.