⌘ '

raccourcis clavier

closed loop system

For determine discrete systems and vice versa.

$G(z)$ with a Zero-Order hold

G(z) = (1-z^{-1}) Z[\frac{G_p(s)}{s}] = G^{*}(z) - z^{-1}G^{*}(z)

example: Find $G(z)$ if $G_p(s) = \frac{s+2}{s+1}$

\begin{aligned} G^{*}(s) &= \frac{G_p(s)}{s} = \frac{2}{s} - \frac{1}{s+1} \\[8pt] g^{*}(t) &= 2 - e^{-t} \quad (\text{inverse Laplace transform}) \\ g^{*}(kT) &= 2 - e^{-kT} \\ G^{*}(z) = \frac{2z}{z-1} - \frac{z}{z-e^{-T}} \end{aligned}

a. $C(z) = G_1(z) G_2(z) E(z)$ b. $C(z) = Z[G_1(s) G_2(s)]E(z)$

Note

The product of $G_1(s)G_2(s)$ must be evaluated first

The output of open-loop system is

C(z) = G(z)D(z)E(z)

\begin{aligned} E(z) &= R(z) - Z[G(s)H(s)]E(z) \\[12pt] &\because \frac{C(z)}{R(z)} = \frac{G(z)}{1+Z[G(s)H(s)]} \end{aligned}

C(z) = \frac{Z[G(s)R(s)]}{1+Z(G(s)H(s))}

\frac{C(z)}{R(z)} = \frac{G_{1}(z)G_{1}(z)}{1+G_{1}(z)Z(G_{2}(s)H(s))}

T(z) = \frac{G(z)}{1+G(z)}