I'm going through an exam question where I've been told that the samples $f(kT)$ of the following function \begin{equation}{F\left(z\right)=\frac{1}{1-0.819z^{-1}}} \end{equation} are applied to a discrete system with the pulse transfer function

$$\text{G}\left(z\right)=\frac{C(z)}{F(z)}=\frac{T}{2}{\left(\frac{1+z^{-1}}{1-z^{-1}}\right)}\quad\text{where}\quad T = 0.1\textrm{ s}$$

I'm then asked to determine the final value of the output of the system. I know that the final value theorem states

\begin{equation}{\lim_{k\to\infty}f\left(kt\right)=\lim_{z\to1}\left(1-z^{-1}\right)\cdot F\left(\text{z}\right)} \end{equation}

but how do I obtain a value of $F(z)$? I'm not sure how the two equations above are manipulated to obtain the $F(z)$ I should be using the final value theorem with.


1 Answer 1


Your final equation is incorrect: it should be

$${\lim_{k\to\infty}f\left(kt\right)=\lim_{z\to1}\left(\left(1-z^{-1}\right)H\left(\text{z}\right)\right)} $$

Where $H(z)= F(z)G(z)$ which is the $\mathcal Z$-transform of the output of the system that has a transfer function of $F(z)$ with $G(z)$ applied to the input.

From that you already have $F(z)$ from the first line, multiply it by $G(z)$ to get the $\mathcal Z$-transform of the function that would be at the output of the system, multiply that product by $(1-z^{-1})$, and then take the limit of the resulting equation as $z$ goes to $1$.

  • $\begingroup$ That's what I was thinking, I'm just not sure why they gave me the transfer function G(z) if I don't really need it. It threw me off a bit as I thought I might be missing something. $\endgroup$
    – Ca01an
    May 16, 2017 at 11:29
  • $\begingroup$ My mistake, the "F(z)" as I showed would be the resulting equation in z at the output of the system: (F(z)G(z))-- I will fix that $\endgroup$ May 16, 2017 at 11:41

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