Chapter 11, Problem 11.25P

To determine

(a)

The control gains for PID controller by ultimate cycle method of Ziegler-Nichols and its closed loop characteristic roots.

Answer to Problem 11.25P

For PID action $K_p=2.1,K_I=1.64,K_D=0.672$ and its characteristic roots are $s=-1.28\pm 1.48i,-0.22\pm 1.29i$.

Explanation of Solution

Given:

The equation is $G_p(s)=\frac{4p}{(s^2+4\zeta s+4)(s+p)}$.

$\zeta =0.5,p=1$.

Concept Used:

Ultimate cycle technique of root locus method.

Calculation:

$\begin{array}{l}{G}_p(s)=\frac{4p}{(s^2+4\zeta s+4)(s+p)}\\ \zeta =0.5,p=1\\ G_p(s)=\frac{4\times 1}{(s^2+4\times 0.5s+4)(s+1)}\\ G_p(s)=\frac{4}{(s^2+2s+4)(s+1)}\\ G_p(s)=\frac{4}{s^3+3s^2+6s+4}\end{array}$

The ultimate cycle method starts by using proportional action only. The transfer function for P action with this plant is

$\frac{C(s)}{R(s)}=\frac{4K_p}{s^3+3s^2+6s+4K_p+4}$

The characteristic equation is

$s^3+3s^2+6s+4K_p+4=0$

To apply the ultimate cycle method, we must find the ultimate period $P_u$ and the associated gain is $K_{P_u}$. To do this analytically, we note that the sustained oscillations occur when a pair of roots is purely imaginary, and the rest of the roots have negative real parts. To calculate this put $s=j{\omega }_u$ where ${\omega }_u$ is unknown frequency of oscillation.

Replacing the values in the above characteristic equation we get,

${(j{\omega }_u)}^3+3{(j{\omega }_u)}^2+6(j{\omega }_u)+4K_{pu}+4=0$

Separating the real and the imaginary parts we get,

$-j{\omega }_u{}^3-3{\omega }_u{}^2+6j{\omega }_u+4K_{pu}+4=0$

Real part is $-3{\omega }_u{}^2+4K_{pu}+4=0$

Imaginary part is $-j{\omega }_u{}^3+6j{\omega }_u=0$

Where $j{\omega }_u$ yields non-oscillatory solution

$\begin{array}{l}j(-{\omega }_u{}^3+6{\omega }_u)=0\\ -{\omega }_u{}^3+6{\omega }_u=0\\ {\omega }_u(-{\omega }_u{}^2+6)=0\\ {\omega }_u=\sqrt{6}=2.45\\ {\omega }_u=0\end{array}$

$\begin{array}{l}-3{\omega }_u{}^2+4K_{pu}+4=0\\ {\omega }_u{}^2=6\\ -3\times 6+4K_{pu}+4=0\\ K_{pu}=3.5\end{array}$

Whereas

$\begin{array}{l}{P}_u=\frac{2\pi }{{\omega }_u}\\ P_u=\frac{2\pi }{2.45}=2.56\end{array}$

For PID action

$\begin{array}{l}{K}_p=0.6K_{P_u}\\ K_{P_u}=3.5\\ K_p=2.1\\ T_1=0.5P_u\\ P_u=2.56\\ T_1=1.28\\ K_I=\frac{K_p}{T_1}=1.64\\ T_D=0.125P_u=0.32\\ K_D=T_D{K}_P\\ K_D=0.672\end{array}$

For PID action $K_p=2.1,K_I=1.64,K_D=0.672$.

Closed loop characteristic roots are defined as,

For PID action

$\frac{C(s)}{R(s)}=\frac{4\left(s^2{K}_D+s{K}_p+K_I\right)}{s^4+3s^3+(6+K_D)s^2+\left(4+K_p\right)s+4K_I}$

Replacing the values of $K_p=2.1,K_I=1.64,K_D=0.672$

$\begin{array}{l}\frac{C(s)}{R(s)}=\frac{4\left(s^2{K}_D+s{K}_p+K_I\right)}{s^4+3s^3+(6+K_D)s^2+\left(4+K_p\right)s+4K_I}\\ K_p=2.1,K_I=1.64,K_D=0.672\\ \frac{C(s)}{R(s)}=\frac{4\left(0.672s^2+2.1s+1.64\right)}{s^4+3s^3+(6+0.672)s^2+\left(4+2.1\right)s+4\times 1.64}\\ \frac{C(s)}{R(s)}=\frac{4\left(0.672s^2+2.1s+1.64\right)}{s^4+3s^3+6.672s^2+6.1s+6.56}\end{array}$

Its characteristic roots are $s=-1.28\pm 1.48i,-0.22\pm 1.29i$.

Conclusion:

For PID action $K_p=2.1,K_I=1.64,K_D=0.672$ and its characteristic roots are $s=-1.28\pm 1.48i,-0.22\pm 1.29i$.

To determine

(b)

The effect of a variation in the parameter $p$ over the range $0.4\le \zeta \le 0.6$ by the root locus technique.

Answer to Problem 11.25P

The effect of a variation in the parameter $p$ over the range $0.4\le \zeta \le 0.6$

$0.45\pm 1.95i\le p\le -0.37\pm 1.97i$.

Explanation of Solution

Given:

The equation is $G_p(s)=\frac{4p}{(s^2+4\zeta s+4)(s+p)}$.

$0.4\le \zeta \le 0.6$.

Concept Used:

Root locus method.

Calculation:

$\begin{array}{l}{G}_p(s)=\frac{4p}{(s^2+4\zeta s+4)(s+p)}\\ \zeta =0.6\\ G_p(s)=\frac{4p}{(s^2+4\times 0.6s+4)(s+p)}\\ G_p(s)=\frac{4p}{(s^2+2.4s+4)(s+p)}\end{array}$

The characteristic equation is $1+G(s)H(s)=0$

$\begin{array}{l}1+G(s)H(s)=0\\ H(s)=1\\ G_p(s)=\frac{4p}{(s^2+2.4s+4)(s+p)}\\ 1+\frac{4p}{(s^2+2.4s+4)(s+p)}=0\\ (s^2+2.4s+4)(s+p)+4p=0\\ s^3+(p+2.4)s^2+(2.4p+4)s+8p=0\end{array}$

Calculating the value of $p$ to determine the roots of the equation

$\begin{array}{l}{s}^3+(p+2.4)s^2+(2.4p+4)s+8p=0\\ s^{3}12.4p+4\\ s^{2}p+2.48p\\ s^1\frac{\left(p+2.4\right)\left(2.4p+4\right)-8p}{p+2.4}0\\ s^{0}8p0\end{array}$

For Stability $K>0$

$\begin{array}{l}\frac{\left(p+2.4\right)\left(2.4p+4\right)-8p}{p+2.4}=0\\ 2.4p^2+4p+5.76p+9.6-8p=0\\ 2.4p^2+1.76p+9.6=0\\ p=-0.37\pm 1.97i\\ 8p=0\\ p=0\end{array}$

$\begin{array}{l}{G}_p(s)=\frac{4p}{(s^2+4\zeta s+4)(s+p)}\\ \zeta =0.4\\ G_p(s)=\frac{4p}{(s^2+4\times 0.4s+4)(s+p)}\\ G_p(s)=\frac{4p}{(s^2+1.6s+4)(s+p)}\end{array}$

The characteristic equation is $1+G(s)H(s)=0$

$\begin{array}{l}1+G(s)H(s)=0\\ H(s)=1\\ G_p(s)=\frac{4p}{(s^2+1.6s+4)(s+p)}\\ 1+\frac{4p}{(s^2+1.6s+4)(s+p)}=0\\ (s^2+1.6s+4)(s+p)+4p=0\\ s^3+(p+1.6)s^2+(1.6p+4)s+8p=0\end{array}$

Calculating the value of $p$ to determine the roots of the equation

$\begin{array}{l}{s}^3+(p+1.6)s^2+(1.6p+4)s+8p=0\\ s^{3}11.6p+4\\ s^{2}p+1.68p\\ s^1\frac{(p+1.6)(1.6p+4)-8p}{p+1.6}0\\ s^{0}8p0\end{array}$

For Stability $K>0$

$\begin{array}{l}\frac{(p+1.6)(1.6p+4)-8p}{p+1.6}=0\\ (p+1.6)(1.6p+4)-8p=0\\ p=0.45\pm 1.95i\\ 8p=0\\ p=0\end{array}$

The effect of a variation in the parameter $p$ over the range $0.4\le \zeta \le 0.6$

$0.45\pm 1.95i\le p\le -0.37\pm 1.97i$.

Conclusion:

The effect of a variation in the parameter $p$ over the range $0.4\le \zeta \le 0.6$

$0.45\pm 1.95i\le p\le -0.37\pm 1.97i$.

To determine

(c)

The effect of a variation in the parameter $p$ over the range $0.4\le \zeta \le 0.6$ by the root locus technique.

Answer to Problem 11.25P

The effect of a variation in the parameter $p$ over the range $0.5\le p\le 1.5$ is $\pm 2i\le p\le 0$.

Explanation of Solution

Given:

The equation is $G_p(s)=\frac{4p}{(s^2+4\zeta s+4)(s+p)}$.

$\zeta =0.5$.

Concept Used:

Root locus method.

Calculation:

$\begin{array}{l}{G}_p(s)=\frac{4p}{(s^2+4\zeta s+4)(s+p)}\\ \zeta =0.5\\ G_p(s)=\frac{4p}{(s^2+4\times 0.5s+4)(s+p)}\\ G_p(s)=\frac{4p}{(s^2+2s+4)(s+p)}\end{array}$

The general root locus equation is defined as $1+K\frac{N(s)}{D(s)}$, $1+KP(s)$

So, $K=p$, as the order of denominator is greater than the order of the numerator, then then the order of equation as well the number of roots is determined by $D(s)$. Order of $D(s)$ is equal to the number of poles of $P(s)$.

The characteristic equation is $1+G(s)H(s)=0$

$\begin{array}{l}1+G(s)H(s)=0\\ H(s)=1\\ G_p(s)=\frac{4p}{(s^2+2s+4)(s+p)}\\ 1+\frac{4p}{(s^2+2s+4)(s+p)}=0\end{array}$

$1+\frac{4p}{(s^2+2s+4)(s+p)}=0$

Poles are

$\begin{array}{l}(s^2+2s+4)(s+p)=0\\ p=0.5\\ (s^2+2s+4)(s+0.5)=0\\ s^3+2.5s^2+5s+2=0\\ s=-0.5,-1\pm 1.732i\end{array}$

$\begin{array}{l}(s^2+2s+4)(s+p)=0\\ p=1.5\\ (s^2+2s+4)(s+1.5)=0\\ s^3+3.5s^2+7s+6=0\\ s=-1.5,-1\pm 1.732i\end{array}$

The path starts at the poles of $P(s)$ with $K=0$ ,when $K=0$ then $1+K\frac{N(s)}{D(s)}$ shows that roots are defined by the denominator equal to zero or by the poles.

The path terminates with $K=\infty$ either at the zeros of $P(s)$ or by leaving the plot.

When $K\to \infty$ there are two conditions to satisfy the equation

It requires that $N(s)\to 0$ i.e. $s$ approaches to zero of $P(s)$

As $K\to \infty$ then $D(s)$ must approaches to infinity which can occur only if $s\to \infty$.

Calculating the value of $p$ to determine the roots of the equation

$\begin{array}{l}(s^2+2s+4)(s+p)+4p=0\\ s^3+(p+2)s^2+(2p+4)s+8p=0\\ s^{3}12p+4\\ s^{2}p+28p\\ s^1\frac{\left(p+2\right)\left(2p+4\right)-8p}{p+2}0\\ s^{0}8p0\end{array}$

For Stability $K>0$

$\begin{array}{l}\frac{\left(p+2\right)\left(2p+4\right)-8p}{p+2}=0\\ 2p^2+4p+4p+8-8p=0\\ 2p^2+8=0\\ p=\pm 2i\\ 8p=0\\ p=0\end{array}$

The effect of a variation in the parameter $p$ over the range $0.5\le p\le 1.5$ is $\pm 2i\le p\le 0$.

Conclusion:

The effect of a variation in the parameter $p$ over the range $0.5\le p\le 1.5$ is $\pm 2i\le p\le 0$.