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I am trying to understand this equation:

enter image description here

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?

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Have a look at these notes.

In particular, this slide:

Indirect design approach

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.

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  • $\begingroup$ 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? $\endgroup$ – mike Dec 20 '19 at 19:24
  • $\begingroup$ @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. $\endgroup$ – Peter K. Dec 20 '19 at 19:28
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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.

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