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I am currently preparing for an exam where on part is control theory. This is one task from a former exam where I don't know how to solve it (translated from German):

Calculate the z-transformed transfer function $G(z)$ of the time-discrete plant ("Ersatzregelstrecke"). The time-continous plant $G_s(s)$ is given. The sampling rate $T_A$ should be viewed as given.

$$G_s(s) = \frac{1}{s^2 + 2s + 1}$$

Note: $$\mathcal{Z} \left \{\frac{a^2}{s \cdot (s+a)^2} \right\} = \frac{z}{z-1} - \frac{z}{z-e^{-aT}} - \frac{aTz \cdot e^{-aT}}{(z-e^{-aT})^2}$$

I know that I can simplify $G_s(s) = \frac{1}{(s+1)^2}$, but this is obviously not of the form $\frac{a^2}{s \cdot (s+a)^2}$...

How can I calculate this?

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  • $\begingroup$ According to Wikipedia, the correct solution is $\frac {e^{-a \cdot T_A} \cdot T_A \cdot z} {(z-e^{-a \cdot T_A})^2}$. $\endgroup$ Sep 20 '15 at 10:06
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In order to solve it, you have to divide $ G_s(s)$ by s first and then you have it in the form on which you can perform the Z-transform.

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Observe that $G(s)$ is formed by $$\frac{1}{s^2} \to \frac{1}{(s+1)^2}$$ with the corresponding Z-transform/Laplace pairs $$\frac{1}{s^2} \to \frac{Tz}{(z-1)^2}$$ and $$\frac{1}{s+a} \to \frac{z}{z-e^{-aT}}$$ With this information you should be able to make the conversion adhering to all the rules of linearity, e.t.c, thats if you wanna find the representation from scratch. Found this: A more elegant way is to go back to the time domain and compute the Z-transform sum. Or just use the tables that have become very popular Table of Laplace and Z Transforms.

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