# What type of z transform is this?

I am trying to understand this equation:

It comes from:

$$Force = mass * acceleration$$

$$F(t) = m * a(t)$$

$$F(s) = m * (s^2 * y(s) - s*y0 - y0)$$, where $$y0=0$$

$$F(s) = m * s^2 * y(s)$$

$$y(s) = F(s)/(m*s^2)$$

Then I don't understand what type of substitution they are using for s. It appears they substitute:

$$s = (1-z^{-1})/T$$

And then the last $$z^{-1}$$ is just to indicate they are delaying the whole thing one sample for another reason.

What type of s substitution is this called?

Have a look at these notes.

In particular, this slide:

shows that using a backward difference model sets up the approximation: $$\frac{d}{dt} \approx \frac{z - 1}{Tz} = \frac{1 - z^{-1}}{T}$$

There are many ways to approximate the continuous-time derivative with discrete-time difference operations. That's all that substitution is doing: choosing one way of making the approximation.

• Got it. So it's the backward difference model. Is it easiest for someone without a lot of knowledge to just do trial and error to see which method will work in a given situation? I was trying something before and either the backward or forward difference didn't work but Tustin's did. Is Tustin's usually the safest bet? – mike Dec 20 '19 at 19:24
• @mike The closer you approximate the "real" derivative operation, the better (in general). You could go even more complex, but Tustin's (trapezoidal) method is usually accurate enough for most operations. – Peter K. Dec 20 '19 at 19:28

What type of s substitution is this called?

This isn't a z transform per se. It's a method (called backwards difference) to approximate the transfer function of a system sampled time given a transfer function in continuous time.