Example 2:

Now let us consider that the system as described in the previous example is subject to a sampled data control system with sampling time T = 0.2 sec. Thus

The bi-linear transformation

$\displaystyle z = \frac{{1+ wT/2}}{{1-wT/2}}= \frac{{(1 + 0.1w)}}{{(1 - 0.1w)}}$

will transfer into w -plane, as

$\displaystyle G_w(w) = \frac{{\left( {1 + \frac{w}{{300}}} \right)\left( {1 - \frac{w}{{10}}} \right)}}{{w(w + 1)}}$ [please try the simplification]

We need first design a phase lead compensator so that PM of the compensated system is at least 50° with K_v = 2 . The compensator in w -plane is

$\displaystyle C(w) = K\frac{{1 + \tau w}}{{1 + \alpha \tau w}}\,\,\,\,\,\,\,\,\,\,\,\,0 < \alpha < 1$

Design steps are as follows.

• K has to be found out from the K_v requirement.

• Make ω_max = ω_{g_new}.

• Compute the gain crossover frequency ω_g and phase margin of the uncompensated system after introducing K in the system.

• At ω_g check the additional/required phase lead, add safety margin, find out $\phi _{\max }$ . Calculate α from the required $\phi _{\max }$

• Since the lead part of the compensator provides a gain of $20 \log_{10} \dfrac{1}{\sqrt \alpha}$ , find out the frequency where the logarithmic magnitude is $-20 \log_{10} \dfrac{1}{\sqrt \alpha}$ . This will be the new gain crossover frequency where the maximum phase lead should occur.

• Calculate τ from the relation $\displaystyle \omega_{g_{new}} = \omega_{\max}=\frac{1}{{\tau \sqrt \alpha }}$