2-DOF Pole Placement Controller Design and Implementation using SBHS

$\displaystyle A^g$	$\displaystyle =1-0.9723086z^{-1}$
$\displaystyle A^b$	$\displaystyle =1$
$\displaystyle B^{g}$	$\displaystyle =0.0116304$
$\displaystyle B^{b}$	$\displaystyle =1$
Let us now define the transient specifications. We choose Rise Time =100 s and Overshoot ( $\epsilon$ )= 5%. Number of samples in one rise time (), [1], is calculated as
$\displaystyle N_r$	$\displaystyle \le\frac{\text{Rise time}}{\text{Sampling time}}$
	$\displaystyle =100$
$% latex2html id marker 12347 $\displaystyle \therefore, \omega$$	$\displaystyle =\frac{\pi}{2N_r}$
	$\displaystyle =0.015708$
and
$\displaystyle \rho$	$\displaystyle \le \epsilon ^{\omega / \pi}$
	$\displaystyle =0.98513$
The closed loop characteristic polynomial is given by,
$\displaystyle \phi _{cl}$	$\displaystyle = 1-2\rho cos\omega z^{-1} + \rho ^2z^{-2}$
	$\displaystyle =1-1.9700229z^{-1}+0.9704870z^{-2}$
Refer to the Scilab code `desired.sci` to calculate , $\omega$ , $\rho$ and $\phi_{cl}$ . The code is available in the same location as `myc2d.sci`. But according to equation 6.26,
$\displaystyle A^bR_1+z^{-k}B^bS_1$	$\displaystyle =\phi_{cl}$
Recall that we have not considered external disturbance in the block diagram shown in figure 6.3. However, we can still up to some extent take care of the disturbances. This is achieved by using the Internal Model Principle. If a model of step is present inside the loop, step disturbances can be rejected [1]. We can apply this by forcing to have this term. A step model is given by
$\displaystyle 1(z)$	$\displaystyle =\frac{1}{1-z^{-1}} = \frac 1{\Delta}$
Therefore,
$\displaystyle R_c$	$\displaystyle =B^g\Delta R_1$
$\Delta$ has a root which lies on the unit circle. Hence it has to be treated as a bad part and should not be canceled out. Hence, we should make sure that all of the occurrences of have this term.

All the above calculations are incorporated into a single Scilab code named twodof_para.sce ^6.2.

$\displaystyle G(s)$	$\displaystyle =\frac{0.42}{35.61s+1}$	(6.29)
with a time constant of $\tau = 35.6 s$ and gain . After discretization with sampling time = 1 s, we obtain
$\displaystyle G(z)$	$\displaystyle =\frac{0.0116304z^{-1}}{1-0.9723086z^{-1}}$	(6.30)

$\displaystyle \phi _{cl}$	$\displaystyle =A^b\Delta R_1+z^{-k}B^bS_1$	(6.31)
Hence,
$\displaystyle A^b\Delta R_1+z^{-k}B^bS_1$	$\displaystyle =1-1.9700229z^{-1}+0.9704870z^{-2}$	(6.32)

$\displaystyle R_c$	$\displaystyle =0.0116304-0.0229175z^{-1}+0.0112871z^{-2}$
$\displaystyle S_c$	$\displaystyle =0.0004641-0.0004512z^{-1}$
$\displaystyle T_c$	$\displaystyle =1-0.9723z^{-1}$
$\displaystyle \gamma$	$\displaystyle =0.0004641$