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raccourcis clavier

Continuous Control System to Digital Control System

Assume the transfer function is given by

D(s) = \frac{U(s)}{E(s)} = K_{0} \frac{s+a}{s+b}

difference equation

$u(k) = (1-bT)u(k-1) + K_{0}(aT-1)E(k-1) + K_{0}e(k)$

The corresponding z-transform

\begin{aligned} \frac{U(z)}{E(z)} &= \frac{K_{0}(aT-1)z^{-1} + K_{0}}{1+(bT-1)z^{-1}} = \frac{K_{0}z + K_{0}(aT-1)}{z+(bT-1)} \\ &=[K_{0}(aT-1) + zK_{0}]/[z+(bT-1)] \end{aligned}

z-transform of difference equation

example: Given $D(s) = \frac{a}{s+a}$ , $u(kT) = u(k)$

$U(s)(s+a) = aE(s)$ (Laplace transform) gives $\frac{u(k+1)-u(k)}{T} + au(k) = a e(k)$

difference equation is $u(k+1) = (1-aT)u(k) + aTe(k)$

z-transform is $\frac{U(z)}{E(z)}=\frac{aT z^{-1}}{1+(aT -1)z^{-1}}=\frac{aT}{z+(aT-1)}$

discrete equivalent

Consider the example

\begin{aligned} D(s) &= \frac{U(s)}{E(s)} = \frac{a}{s+a} \to U(s)s = aE(s) - aU(s) \\ &\to u^{'}(t) = -au(t) + ae(t) \end{aligned}

u(t) = \int_0^t [-au(\tau) + ae(\tau)] d \tau

for discrete system

$u(kT) = u(kT - T) + \int_{kT-T}^{kT} [-au(\tau) + ae(\tau)] d \tau$

We can use the following approximation methods for the second term from $D(z)$ to $D(s)$

$D(s)$	rule	z-transfer function $D(z)$	approximation	z-plane to s-plane	stability
$\frac{a}{s+a}$	forward	$\frac{a}{(z-1)/T+a}$	$s \gets \frac{z-1}{T}$	$z \gets sT + 1$	discrete $\to$ continuous
$\frac{a}{s+a}$	backward	$\frac{a}{(z-1)/(Tz)+a}$	$s \gets \frac{z-1}{Tz}$	$z \gets \frac{1}{1-Ts}$	discrete $\gets$ continuous
$\frac{a}{s+a}$	trapzoid	$\frac{a}{(2/T)[(z-1)/(z+1)]+a}$	$s \gets \frac{2}{T} \frac{z-1}{z+1}$	$z \gets \frac{1+Ts/2}{1-Ts/2}$	discrete $\leftrightarrow$ continuous

Assume the transfer function is given by

D(s) = \frac{U(s)}{E(s)} = K_{0} \frac{s+a}{s+b}

difference equation

$u(k) = (1-bT)u(k-1) + K_{0}(aT-1)E(k-1) + K_{0}e(k)$

The corresponding z-transform

\begin{aligned} \frac{U(z)}{E(z)} &= \frac{K_{0}(aT-1)z^{-1} + K_{0}}{1+(bT-1)z^{-1}} = \frac{K_{0}z + K_{0}(aT-1)}{z+(bT-1)} \\ &=[K_{0}(aT-1) + zK_{0}]/[z+(bT-1)] \end{aligned}

z-transform of difference equation

example: Given $D(s) = \frac{a}{s+a}$ , $u(kT) = u(k)$

$U(s)(s+a) = aE(s)$ (Laplace transform) gives $\frac{u(k+1)-u(k)}{T} + au(k) = a e(k)$

difference equation is $u(k+1) = (1-aT)u(k) + aTe(k)$

z-transform is $\frac{U(z)}{E(z)}=\frac{aT z^{-1}}{1+(aT -1)z^{-1}}=\frac{aT}{z+(aT-1)}$

discrete equivalent

Consider the example

\begin{aligned} D(s) &= \frac{U(s)}{E(s)} = \frac{a}{s+a} \to U(s)s = aE(s) - aU(s) \\ &\to u^{'}(t) = -au(t) + ae(t) \end{aligned}

u(t) = \int_0^t [-au(\tau) + ae(\tau)] d \tau

for discrete system

$u(kT) = u(kT - T) + \int_{kT-T}^{kT} [-au(\tau) + ae(\tau)] d \tau$

We can use the following approximation methods for the second term from $D(z)$ to $D(s)$

$D(s)$	rule	z-transfer function $D(z)$	approximation	z-plane to s-plane	stability
$\frac{a}{s+a}$	forward	$\frac{a}{(z-1)/T+a}$	$s \gets \frac{z-1}{T}$	$z \gets sT + 1$	discrete $\to$ continuous
$\frac{a}{s+a}$	backward	$\frac{a}{(z-1)/(Tz)+a}$	$s \gets \frac{z-1}{Tz}$	$z \gets \frac{1}{1-Ts}$	discrete $\gets$ continuous
$\frac{a}{s+a}$	trapzoid	$\frac{a}{(2/T)[(z-1)/(z+1)]+a}$	$s \gets \frac{2}{T} \frac{z-1}{z+1}$	$z \gets \frac{1+Ts/2}{1-Ts/2}$	discrete $\leftrightarrow$ continuous