In this paper on page 5 equation (10) is supposed to be the reverse z-transform of equation (5) on page 4.

$$\frac{U(z)}{\bar{U}(z)} = G(z) = \frac{1-p}{z-p} \quad \leftrightarrow \quad u(k) = \bar{u} \cdot (1-p^k)$$

Hint: This is related to settling time somehow...

Questions: Why is suddenly the dependence of $\bar u$ on k neglected? Second how is this transform calculated? I would have done it like this:

$$G(z) = \frac{1-p}{z-p} \quad \leftrightarrow \quad u(k) = \bar u(k)*(1-p) \cdot p^{k-1} \cdot \epsilon [k-1] $$ with $\epsilon [k]$ being the step-function and $*$ being convolution. What am I missing right here?


1 Answer 1


As we've already seen in your previous question about that paper, the authors of that paper are not very much into details when it comes to math and signal processing. What Eq. $(10)$ shows is definitely not the inverse $\mathcal{Z}$ transform of


which is


as you've figured out by yourself. What Eq. $(10)$ in the paper does show is the step response, given by

$$s[n]=\sum_{k=-\infty}^{n}g[k]=(1-p)\sum_{k=1}^np^{(k-1)}=1-p^n,\quad n\ge 0\tag{3}$$

So if you apply a step $\tilde{\mu}\cdot u[n]$ at the input, the system's response is

$$\mu[n]=\tilde{\mu}\left(1-p^n\right),\quad n\ge 0\tag{4}$$

as given by Eq. $(10)$ in the paper.


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