# One more inverse z transform … a bit more complicated [closed]

Thank you for all the help teaching me. I am looking at one more inverse z transform and not understanding what to do with it.

$$F(t) = m \cdot a(t)$$ (i.e. Force = mass $$\times$$ acceleration)

$$F(s) = m \cdot d(s) \cdot s^2$$ (i.e. Where d(s) is displacement and initial position is 0)

Using $$s = \frac{1-z^{-1}}T$$:

$$F(z) = m \cdot d(z) \cdot \frac{1-z^{-1}}{T}^2$$

$$F(z) = m \cdot d(z) \cdot \frac{z^{-2} - 2z^{-1} + 1}{T^2}$$

I have displacement calculated per sample already from another equation, so I can just substitute that in for $$d(z)$$. But what on earth do I do with that to get $$f[n]$$?

Thanks and sorry for all the questions. I unfortunately have zero education on this and it's hard to find simple to understand resources explaining how to do these things.

• See my comments from the last post and think about what $d(z)z^{-2}$ would be.... – Dan Boschen Dec 21 '19 at 22:12
• Oh shoot Dan. Thanks for the prompt. I actually knew $d(z)z{^-2}$ is d[n-2]. My problem was I was being dumb about something else. I get $F[n] = m* \frac{d[n-2] - 2d[n-1] + d(n)}{T^2}$. Is that it? – mike Dec 21 '19 at 22:15
• Yes that look's right to me. – Dan Boschen Dec 21 '19 at 22:29
• Okay super easy. Thanks as always. – mike Dec 21 '19 at 22:34

$$F[n]=m \cdot \frac{d[n−2]−2d[n−1]+d[n]}{T^2}$$