Chapter 12, Problem 12.1P

To determine

The design of PD compensator of the given plant.

Answer to Problem 12.1P

The closed loop transfer function of the plant with $\text{PD}$ compensator is $\frac{7.25+3.535s}{5s^2+3.535s+1.25}$with$K_P=7.25$and$K_D=3.535$.

Explanation of Solution

Given information:

The transfer function of the plant is $G_p\left(s\right)=\frac{\Theta \left(s\right)}{\Phi \left(s\right)}=\frac{1}{5s^2-6}$. For a closed loop system, the damping ratio ($\zeta$) is $0.707$ and the angular frequency (${\omega }_n$) is $0.5$.

The structure of the system with $\text{PD}$ compensation is shown below.

Figure-(1)

Write the closed loop transfer function of the system.

$\frac{C\left(s\right)}{R\left(s\right)}=\frac{\left(K_P+K_{D}s\right)G_P\left(s\right)}{1+\left(K_P+K_{D}s\right)G_P\left(s\right)}$........ (I)

Here, the plant function is $G_P\left(s\right)$, the constants of $\text{PD}$ controller are $K_P$ and $K_D$.

Substitute $\frac{1}{5s^2-6}$ for $G_P\left(s\right)$ in Equation (I).

$\begin{array}{c}\frac{C\left(s\right)}{R\left(s\right)}=\frac{\left(K_P+K_{D}s\right)\times \frac{1}{5s^2-6}}{1+\left(K_P+K_{D}s\right)\times \frac{1}{5s^2-6}}\\ =\frac{\left(K_P+K_{D}s\right)}{5s^2-6+\left(K_P+K_{D}s\right)}\\ =\frac{K_P+K_{D}s}{5\left(s^2+\frac{K_D}{5}s+\frac{K_P-6}{5}\right)}\end{array}$........ (II)

Write the characteristic equation of the above system.

$s^2+\frac{K_D}{5}s+\frac{K_P-6}{5}$........ (III)

Compare with the standard characteristic equation of the second degree closed loop system.

$s^2+2\zeta {\omega }_{n}s+{\omega }_n^2=0$........ (IV)

Compare the equation (III) and (IV).

${\omega }_n^2=\frac{K_P-6}{5}$........ (V)

$2\zeta {\omega }_n=\frac{K_D}{5}$........ (VI)

Substitute $0.5$ for ${\omega }_n$ in the Equation (V).

$\begin{array}{c}{\left(0.5\right)}^2=\frac{K_P-6}{5}\\ 0.25=\frac{K_P-6}{5}\\ 1.25=K_P-6\\ K_P=7.25\end{array}$

Substitute $0.5$ for ${\omega }_n$ and $0.707$ for $\zeta$ in the Equation (VI).

$\begin{array}{c}2\times 0.707\times 0.5=\frac{K_D}{5}\\ K_D=3.535\end{array}$

Substitute $7.25$ for $K_P$ and $3.535$ for $K_D$ in the Equation (II).

$\begin{array}{c}\frac{C\left(s\right)}{R\left(s\right)}=\frac{7.25+3.535s}{5\left(s^2+\frac{3.535}{5}s+\frac{7.25-6}{5}\right)}\\ =\frac{7.25+3.535s}{5s^2+3.535s+1.25}\end{array}$.

Conclusion:

The closed loop transfer function of the plant with $\text{PD}$ compensator is $\frac{7.25+3.535s}{5s^2+3.535s+1.25}$with$K_P=7.25$and$K_D=3.535$.