0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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

$$G(z)=\frac{1-p}{z-p}\tag{1}$$

which is

$$g[n]=(1-p)p^{(n-1)}u[n-1]\tag{2}$$

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.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.