Question 2 (Generalized Controller Design): Consider the open loop system GH (=)=- (=-0.75)(=-1)² a) Design a controlle...

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.

$\begin{array}{rcl}C\left(Z\right)& =& \frac{K_d.Z^2+K_{p}Z+K_i}{Z}\\ C\left(Z\right)& =& \frac{-1.50Z^2+0.25Z+0.125}{Z}\end{array}$

Put the value (3), (4), and (5) in the eq(1).

The transfer function with the controller is

$\begin{array}{rcl}T.F.& =& \frac{\left(-1.50\right).Z^2+0.25Z+0.125}{Z^4-2.75 Z^3+\left(2.5-1.50\right)Z^2+\left(0.25-0.75\right)Z+0.125}\\ T.F.& =& \frac{-1.50.Z^2+0.25Z+0.125}{Z^4-2.75 Z^3+Z^2+-0.5Z+0.125}\end{array}$