We have implemented a way of approximating a data set with the following lines of code based on the dx/dt of wikipedia of leaky integrator.

dy = -A * y_previous + C   
y_current = y_previous + dy*delta_t 

However, I am uncertain if we are allowed to call the following truly a leaky integrator. I was a bit confused how to get the general solution with Laplace transforms from the wikipage with this because of the delta_t in our implementation. I have to add that I am slightly familiar with Laplace, but not with z-transformations and I was told I needed this since it's discrete.

With delta_t being the difference in time between t_current and t_previous, which is a constant value. I think we may have created a leaky integrator due to delta_t being a constant, if I would for instance call delta_t alpha (alpha), C to x_t, and rewrite it a bit, it seems more similar to sources I found online (1, 2, 3):


So then with the z-transform I have determined the transfer function:


My questions are:

  1. Is in this case indeed the delta_t kind of alpha similar to the cited sites?
  2. So is this indeed a leaky integrator?
  3. What can I say about the values A and alpha can be for stability of the leaky integrator? Should indeed hold:



1 Answer 1


Yes this is indeed what is referred to as a “leaky integrator”, as the discrete time approximation of such a device. The key element that makes it “Leaky” is having the positive real pole at $z < 1$ which is the equivalent of a continuous time real pole along the negative real axis.

With the pole at $z=1$, (and a zero at $z=0$) we get the discrete time approximation of a pure integrator, this is an accumulator with the transfer function given as:

$$H(z) = \frac{z}{z-1}$$

Any other constants added are gain scaling, but this represents an “integrator” equivalent as the impulse invariant mapping of $H(s) = 1/s$.

As we move the pole off of $z=1$ As given by:

$$H(z) = \frac{z}{z-\alpha}$$

With $\alpha$ as an arbitrary real constant between 0 and 1 (not using OP’s $\alpha$ but generally), it becomes a “leaky integrator” with $\alpha$ as the damping factor: the closer $\alpha$ is to 1, the less the “leak”. Any other constant as before is just a scaling factor.

As noted, this is the discrete time equivalent to moving the pole on the s-plane from $s=0$ that we get with $H(s) = 1/s$ into the left half plane to be:

$$H(s) = \frac{1}{s+ 1/\tau}$$

Where $\tau$ is the time constant indicating a decay given by $e^{-t/\tau}$ which comes directly from the inverse Laplace Transform of $H(s)$.

  • $\begingroup$ Thank you so much for your elaborate response! If I may ask another question, would it still be a leaky integrator if the delta_t was varying over t_1 to t_N, in other words not a constant? $\endgroup$ May 18, 2023 at 12:08
  • $\begingroup$ Is your delta_t varying such that you don't have a consistent sample rate, or is it more that the sampling rate varies over the longer term duration? In either case changing delta_t simply changes $\alpha$ and similarly the time constant $\tau$, which just means the bandwidth or leak rate is modified accordingly-- but it is still a leaky integrator as long as the $\alpha$ you end up with has a value between 0 and 1 not including either. $\endgroup$ May 18, 2023 at 12:23

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