Question 2 (Generalized Controller Design)
Consider the open loop system GH=
Closed loop system:
It is a system in which the output is dependent on the input of the system. The output has a feedback system that ia connected with the input. Examples are AC, voltage stabilizer, electric iron, etc.
Open loop system has $GH\left(S\right)=\frac{1}{\left(Z-0.75\right){\left(Z-1\right)}^{2}}$
The general controller block diagram is shown below.
the transfer function of the block diagram is
$\begin{array}{rcl}T.F.& =& \frac{Y\left(Z\right)}{R\left(Z\right)}=\frac{C\left(Z\right)GH\left(Z\right)}{1+C\left(Z\right)GH\left(Z\right)}\\ C\left(Z\right)& =& {K}_{p}+\frac{{K}_{i}}{Z}+{K}_{d}.Z\\ C\left(Z\right)& =& \frac{{K}_{d}.{Z}^{2}+{K}_{p}Z+{K}_{i}}{Z}\end{array}$
Put C(Z) in the transfer function equation.
$\begin{array}{rcl}T.F.& =& \frac{\left(\frac{{K}_{d}.{Z}^{2}+{K}_{p}Z+{K}_{i}}{Z}\right)\left({\displaystyle \frac{1}{\left(z-0.75\right){\left(z-1\right)}^{2}}}\right)}{1+\left(\frac{{K}_{d}.{Z}^{2}+{K}_{p}Z+{K}_{i}}{Z}\right)\left(\frac{1}{\left(z-0.75\right){\left(z-1\right)}^{2}}\right)}\\ & & \\ & =& \frac{{K}_{d}.{Z}^{2}+{K}_{p}Z+{K}_{i}}{Z\left(z-0.75\right){\left(z-1\right)}^{2}+{K}_{d}.{Z}^{2}+{K}_{p}Z+{K}_{i}}\\ & =& \frac{{K}_{d}.{Z}^{2}+{K}_{p}Z+{K}_{i}}{{Z}^{4}-2.75{Z}^{3}+\left(2.5+{K}_{d}\right){Z}^{2}+\left({K}_{p}-0.75\right)Z+{K}_{i}}.......\left(1\right)\end{array}$
Poles are at
$\begin{array}{rcl}{Z}_{1,2}& =& 0.25\pm j0.25\\ Pole,{P}_{1}& =& 0.25+j0.25\\ Pole,{P}_{2}& =& 0.25-j0.25\end{array}$
The polynomial is
$\begin{array}{rcl}Polynomial& =& \left[Z-\left(0.25+j0.25\right)\right]\left[Z-\left(0.25-j0.25\right)\right]\\ & =& {Z}^{2}-0.5Z+0.125.......\left(2\right)\end{array}$
Compare (1) with (2) and get
$\begin{array}{rcl}{K}_{i}& =& 0.125.......\left(3\right)\\ & & \\ 2.5+{K}_{d}& =& 1\\ {K}_{d}& =& -1.50......\left(4\right)\\ & & \\ {K}_{P}-0.75& =& -0.50\\ {K}_{P}& =& 0.25.........\left(5\right)\end{array}$
Put the above value in the C(Z) equation.
$\begin{array}{rcl}C\left(Z\right)& =& \frac{{K}_{d}.{Z}^{2}+{K}_{p}Z+{K}_{i}}{Z}\\ C\left(Z\right)& =& \frac{-1.50{Z}^{2}+0.25Z+0.125}{Z}\end{array}$
Put the value (3), (4), and (5) in the eq(1).
The transfer function with the controller is
$\begin{array}{rcl}T.F.& =& \frac{\left(-1.50\right).{Z}^{2}+0.25Z+0.125}{{Z}^{4}-2.75{Z}^{3}+\left(2.5-1.50\right){Z}^{2}+\left(0.25-0.75\right)Z+0.125}\\ T.F.& =& \frac{-1.50.{Z}^{2}+0.25Z+0.125}{{Z}^{4}-2.75{Z}^{3}+{Z}^{2}+-0.5Z+0.125}\end{array}$