Is this a leaky integrator?

Question

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), 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 similar to the cited sites?
2. So is this indeed a leaky integrator?
3. What can I say about the values A and can be for stability of the leaky integrator? Should indeed hold:

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 $H(s)$.