Chapter 10, Problem 10.22P

To determine

(a)

The values of gain $K_P\text{and}{K}_I$ of the PI controller of first-order plant for the following cases:

Case 1: For $\zeta =0.707$.

Case 2: For $\zeta =1$.

Case 3: For a root separation factor of 10.

Answer to Problem 10.22P

The values of gains for the PI controller are as follows:

Case 1: For $\zeta =0.707$, $K_I=200$ and $K_P=36$.

Case 2: For $\zeta =1$, $K_I=100$ and $K_P=36$.

Case 3: For a root separation factor of 10, $K_I=1000$ and $K_P=216$.

Explanation of Solution

Given:

The proportional integral controller of the first order plant is as shown below:

Where, the parameter values are as given:

$I=c=4$

Also, the performance specifications require the time constant of the system to be $\tau =0.2$.

Concept Used:

1. The transfer functions for the block diagram are as shown below:

$\begin{array}{l}\frac{\Omega \left(s\right)}{{\Omega }_r\left(s\right)}=\frac{s{K}_P+K_I}{I{s}^2+s\left(c+K_P\right)+K_I}\\ \frac{\Omega \left(s\right)}{T_d\left(s\right)}=\frac{-s}{I{s}^2+s\left(c+K_P\right)+K_I}\end{array}$

2. For a second order system having a characteristic equation of the form $s^2+2\zeta {\omega }_{n}s+{\omega }_n{}^2=0$, the characteristic roots are of the form $s=-\zeta {\omega }_{n}s\pm {\omega }_n\sqrt{1-{\zeta }^2}\forall \zeta \le 1$ .

Also, the corresponding time constant for the system would be $\tau =\frac{1}{\zeta {\omega }_n}$.

3. For a second order system, if the root separation factor is K then, we have following conclusions for the roots and their corresponding time constant, that is

   For roots $s_1\text{and}{s}_2$ ,

   $\frac{s_1}{s_2}=K$

The corresponding time constants of the system would be:

$\frac{{\tau }_1}{{\tau }_2}=\frac{\frac{1}{\mathrm{Re}\left(s_1\right)}}{\frac{1}{\mathrm{Re}\left(s_2\right)}}=\frac{\mathrm{Re}\left(s_2\right)}{\mathrm{Re}\left(s_1\right)}=\frac{1}{K}$

Calculation:

From the block diagram as shown, the transfer functions are as:

$\begin{array}{l}\frac{\Omega \left(s\right)}{{\Omega }_r\left(s\right)}=\frac{s{K}_P+K_I}{I{s}^2+s\left(c+K_P\right)+K_I}\\ \frac{\Omega \left(s\right)}{T_d\left(s\right)}=\frac{-s}{I{s}^2+s\left(c+K_P\right)+K_I}\end{array}$

Therefore, the characteristic equation for the system is:

$I{s}^2+s\left(c+K_P\right)+K_I=0$

On putting the values of parameters in this expression of characteristic equation:

$\begin{array}{l}I{s}^2+s\left(c+K_P\right)+K_I=0\\ \Rightarrow 4s^2+s\left(4+K_P\right)+K_I=0\because I=c=4\\ \Rightarrow s^2+s\frac{\left(4+K_P\right)}{4}+\frac{K_I}{4}=0\end{array}$

On comparing this equation with form $s^2+2\zeta {\omega }_{n}s+{\omega }_n{}^2=0$, we get

Undamped natural frequency:

$\begin{array}{l}{\omega }_n{}^2=\frac{K_I}{4}\\ \Rightarrow {\omega }_n=\sqrt{\frac{K_I}{4}}\\ \Rightarrow {\omega }_n=\frac{\sqrt{K_I}}{2}\text{rad/sec}\end{array}$

Damping ratio:

$\begin{array}{l}2\zeta {\omega }_n=\frac{\left(4+K_P\right)}{4}\\ \Rightarrow \zeta =\frac{\left(4+K_P\right)}{8{\omega }_n}\\ \Rightarrow \zeta =\frac{\left(4+K_P\right)}{8\frac{\sqrt{K_I}}{2}}\\ \Rightarrow \zeta =\frac{\left(4+K_P\right)}{4\cdot \sqrt{K_I}}\end{array}$

Thus, the time constant for the system is:

$\begin{array}{l}\tau =\frac{1}{\zeta {\omega }_n}\\ \Rightarrow \tau =\frac{1}{\frac{\left(4+K_P\right)}{4\cdot \sqrt{K_I}}\cdot \frac{\sqrt{K_I}}{2}}\because \zeta =\frac{\left(4+K_P\right)}{4\cdot \sqrt{K_I}},{\omega }_n=\frac{\sqrt{K_I}}{2}\text{rad/sec}\\ \Rightarrow \tau =\frac{8}{\left(4+K_P\right)}\end{array}$

As the performance specifications require the time constant of the system to be $\tau =0.2$.

Therefore, at $\tau =0.2$

$\begin{array}{l}\tau =\frac{8}{\left(4+K_P\right)}\\ \Rightarrow 0.2=\frac{8}{\left(4+K_P\right)}\\ \Rightarrow 4+K_P=40\\ \Rightarrow K_P=36\end{array}$

Case 1. When $\zeta =0.707$, the gain value of $K_I$ is:

Since, $\zeta =\frac{\left(4+K_P\right)}{4\cdot \sqrt{K_I}}$

$\begin{array}{l}\Rightarrow 0.707=\frac{\left(4+36\right)}{4\cdot \sqrt{K_I}}\because K_P=36\\ \Rightarrow 2.828\sqrt{K_I}=40\\ \Rightarrow \sqrt{K_I}=\frac{40}{2.828}=14.144\\ \Rightarrow K_I=200.05\approx 200\end{array}$

Case 2. When $\zeta =1$, the gain value of $K_I$ is:

Since, $\zeta =\frac{\left(4+K_P\right)}{4\cdot \sqrt{K_I}}$

$\begin{array}{l}\Rightarrow 1=\frac{\left(4+36\right)}{4\cdot \sqrt{K_I}}\because K_P=36\\ \Rightarrow \sqrt{K_I}=10\\ \Rightarrow K_I=100\end{array}$

Case 3. When the root separation factor is 10.

For this value of the root separation factor $\frac{s_1}{s_2}=10$, the ratio of time constants for the roots would be:

$\frac{{\tau }_1}{{\tau }_2}=\frac{1}{10}$

As given in the question system, performance requires time constant to be 0.2seconds or say the dominant time constant to be 0.2 seconds, then the secondary time constant would be 0.02 seconds. Therefore, the characteristic roots corresponding to these time constants would be:

$s=-\frac{1}{0.2}=-5,s=-\frac{1}{0.02}=-50$

Thus, the corresponding characteristic equation for the system will be:

$\begin{array}{l}s=-5,s=-50\\ \Rightarrow \left(s+5\right)\left(s+50\right)=0\\ \Rightarrow s^2+55s+250=0\end{array}$

On comparing this equation with $s^2+s\frac{\left(4+K_P\right)}{4}+\frac{K_I}{4}=0$, we get

$\begin{array}{l}\frac{K_I}{4}=250\\ \Rightarrow K_I=1000\end{array}$

And

$\begin{array}{l}\frac{\left(4+K_P\right)}{4}=55\\ \Rightarrow 4+K_P=220\\ \Rightarrow K_P=216\end{array}$.

Conclusion:

The values of gains for the PI controller areas follows:

Case 1. For $\zeta =0.707$, $K_I=200$ and $K_P=36$.

Case 2. For $\zeta =1$, $K_I=100$ and $K_P=36$.

Case 3. For a root separation factor of 10, $K_I=1000$ and $K_P=216$.

To determine

(b)

To plot:

The response $\Omega \left(s\right)$ for the unit-step command response ${\Omega }_r\left(s\right)$ for all the cases in sub-part (a). Also, compare the responses obtained in all these cases.

Answer to Problem 10.22P

The response $\Omega \left(s\right)$ for the unit-step command response ${\Omega }_r\left(s\right)$ for all the cases is shown in Figure 1. At the damping ratio $\zeta =0.707$, there is a high overshoot in the response in comparison to the other cases where damping ratios are 1 and 1.739 respectively.

Explanation of Solution

Given:

The proportional integral controller of first order plant is as shown below:

Where, the parameter values are as given:

$I=c=4$

Also, the performance specifications require the time constant of the system to be $\tau =0.2$.

The values of gains for the PI controller are as follows:

Case 1. For $\zeta =0.707$, $K_I=200$ and $K_P=36$.

Case 2. For $\zeta =1$, $K_I=100$ and $K_P=36$.

Case 3. For a root separation factor of 10, $K_I=1000$ and $K_P=216$.

Concept Used:

1. The transfer functions for the block diagram are as shown below:

$\begin{array}{l}\frac{\Omega \left(s\right)}{{\Omega }_r\left(s\right)}=\frac{s{K}_P+K_I}{I{s}^2+s\left(c+K_P\right)+K_I}\\ \frac{\Omega \left(s\right)}{T_d\left(s\right)}=\frac{-s}{I{s}^2+s\left(c+K_P\right)+K_I}\end{array}$.

Calculation:

From the block diagram as shown, the transfer functions are as:

$\begin{array}{l}\frac{\Omega \left(s\right)}{{\Omega }_r\left(s\right)}=\frac{s{K}_P+K_I}{I{s}^2+s\left(c+K_P\right)+K_I}\\ \frac{\Omega \left(s\right)}{T_d\left(s\right)}=\frac{-s}{I{s}^2+s\left(c+K_P\right)+K_I}\end{array}$

Therefore, the response $\Omega \left(s\right)$ for the system is:

$\Omega \left(s\right)=\frac{s{K}_P+K_I}{I{s}^2+s\left(c+K_P\right)+K_I}{\Omega }_r\left(s\right)-\frac{s}{I{s}^2+s\left(c+K_P\right)+K_I}{T}_d\left(s\right)$

On putting the values of parameters in this expression of characteristic equation:

$\begin{array}{l}\Omega \left(s\right)=\frac{s{K}_P+K_I}{I{s}^2+s\left(c+K_P\right)+K_I}{\Omega }_r\left(s\right)-\frac{s}{I{s}^2+s\left(c+K_P\right)+K_I}{T}_d\left(s\right)\\ \Rightarrow \Omega \left(s\right)=\frac{s{K}_P+K_I}{4s^2+s\left(4+K_P\right)+K_I}{\Omega }_r\left(s\right)-\frac{s}{4s^2+s\left(4+K_P\right)+K_I}{T}_d\left(s\right)\because I=c=4\end{array}$

Case 1. When $\zeta =0.707$, the gain values are $K_I=200$ and $K_P=36$:

Since, $\Omega \left(s\right)=\frac{s{K}_P+K_I}{4s^2+s\left(4+K_P\right)+K_I}{\Omega }_r\left(s\right)-\frac{s}{4s^2+s\left(4+K_P\right)+K_I}{T}_d\left(s\right)$

$\begin{array}{l}\Rightarrow \Omega \left(s\right)=\frac{36s+200}{4s^2+s\left(4+36\right)+200}{\Omega }_r\left(s\right)-\frac{s}{4s^2+s\left(4+36\right)+200}{T}_d\left(s\right)\\ \because K_P=36,K_I=200\\ \Rightarrow \Omega \left(s\right)=\frac{36s+200}{4s^2+40s+200}{\Omega }_r\left(s\right)-\frac{s}{4s^2+40s+200}{T}_d\left(s\right)\end{array}$

Therefore, for unit-step command response ${\Omega }_r\left(s\right)$ and zero disturbance torque $T_d\left(s\right)$, we have

$\begin{array}{l}\Omega \left(s\right)=\frac{36s+200}{4s^2+40s+200}{\Omega }_r\left(s\right)-\frac{s}{4s^2+40s+200}{T}_d\left(s\right)\\ \Rightarrow \Omega \left(s\right)=\frac{36s+200}{4s^2+40s+200}\cdot \frac{1}{s}-\frac{s}{4s^2+40s+200}\cdot 0\\ \Rightarrow \Omega \left(s\right)=\frac{36s+200}{4s^2+40s+200}\cdot \frac{1}{s}\end{array}$

On simplifying this response of $\Omega \left(s\right)$ using partial fraction expansion, we get

$\begin{array}{l}\Omega \left(s\right)=\frac{36s+200}{4s^2+40s+200}\cdot \frac{1}{s}\\ \Rightarrow \Omega \left(s\right)=\frac{1}{s}-\frac{\left(s+1\right)}{s^2+10s+50}\end{array}$

On taking inverse Laplace transform of this, we have

$\begin{array}{l}\Omega \left(s\right)=\frac{1}{s}-\frac{\left(s+1\right)}{s^2+10s+50}=\frac{1}{s}-\frac{\left(s+5\right)}{{\left(s+5\right)}^2+{\left(5\right)}^2}+\frac{4}{5}\cdot \frac{1}{{\left(s+5\right)}^2+{\left(5\right)}^2}\\ \Rightarrow \omega \left(t\right)=1-e^{-5t}\cos \left(5t\right)+0.8e^{-5t}\sin \left(5t\right)\end{array}$

Case 2. When $\zeta =1$, the gain values are $K_I=100$ and $K_P=36$:

Since, $\Omega \left(s\right)=\frac{s{K}_P+K_I}{4s^2+s\left(4+K_P\right)+K_I}{\Omega }_r\left(s\right)-\frac{s}{4s^2+s\left(4+K_P\right)+K_I}{T}_d\left(s\right)$

$\begin{array}{l}\Rightarrow \Omega \left(s\right)=\frac{36s+100}{4s^2+s\left(4+36\right)+100}{\Omega }_r\left(s\right)-\frac{s}{4s^2+s\left(4+36\right)+100}{T}_d\left(s\right)\\ \because K_P=36,K_I=100\\ \Rightarrow \Omega \left(s\right)=\frac{36s+100}{4s^2+40s+100}{\Omega }_r\left(s\right)-\frac{s}{4s^2+40s+100}{T}_d\left(s\right)\end{array}$

Therefore, for unit-step command response ${\Omega }_r\left(s\right)$ and zero disturbance torque $T_d\left(s\right)$, we have

$\begin{array}{l}\Omega \left(s\right)=\frac{36s+100}{4s^2+40s+100}{\Omega }_r\left(s\right)-\frac{s}{4s^2+40s+100}{T}_d\left(s\right)\\ \Rightarrow \Omega \left(s\right)=\frac{36s+100}{4s^2+40s+100}\cdot \frac{1}{s}-\frac{s}{4s^2+40s+100}\cdot 0\\ \Rightarrow \Omega \left(s\right)=\frac{36s+100}{4s^2+40s+100}\cdot \frac{1}{s}\end{array}$

On simplifying this response of $\Omega \left(s\right)$ using partial fraction expansion, we get

$\begin{array}{l}\Omega \left(s\right)=\frac{36s+100}{4s^2+40s+100}\cdot \frac{1}{s}\\ \Rightarrow \Omega \left(s\right)=\frac{1}{s}+\frac{4}{{\left(s+5\right)}^2}-\frac{1}{\left(s+5\right)}\end{array}$

On taking inverse Laplace transform of this, we have

$\begin{array}{l}\Omega \left(s\right)=\frac{1}{s}+\frac{4}{{\left(s+5\right)}^2}-\frac{1}{\left(s+5\right)}\\ \Rightarrow \omega \left(t\right)=1+e^{-5t}\cdot t-e^{-5t}\end{array}$

Case 3. When the root separation factor is 10.

That is, the gain values are $K_I=1000$ and $K_P=216$:

Since, $\Omega \left(s\right)=\frac{s{K}_P+K_I}{4s^2+s\left(4+K_P\right)+K_I}{\Omega }_r\left(s\right)-\frac{s}{4s^2+s\left(4+K_P\right)+K_I}{T}_d\left(s\right)$

$\begin{array}{l}\Rightarrow \Omega \left(s\right)=\frac{216s+1000}{4s^2+s\left(4+216\right)+1000}{\Omega }_r\left(s\right)-\frac{s}{4s^2+s\left(4+216\right)+1000}{T}_d\left(s\right)\\ \because K_P=216,K_I=1000\\ \Rightarrow \Omega \left(s\right)=\frac{216s+1000}{4s^2+220s+1000}{\Omega }_r\left(s\right)-\frac{s}{4s^2+220s+1000}{T}_d\left(s\right)\end{array}$

Therefore, for unit-step command response ${\Omega }_r\left(s\right)$ and zero disturbance torque $T_d\left(s\right)$, we have

$\begin{array}{l}\Omega \left(s\right)=\frac{216s+1000}{4s^2+220s+1000}{\Omega }_r\left(s\right)-\frac{s}{4s^2+220s+1000}{T}_d\left(s\right)\\ \Rightarrow \Omega \left(s\right)=\frac{216s+1000}{4s^2+220s+1000}\cdot \frac{1}{s}-\frac{s}{4s^2+220s+1000}\cdot 0\end{array}$

$\Rightarrow \Omega \left(s\right)=\frac{216s+1000}{4s^2+220s+1000}\cdot \frac{1}{s}$

On simplifying this response of $\Omega \left(s\right)$ using partial fraction expansion, we get

$\begin{array}{l}\Omega \left(s\right)=\frac{216s+1000}{4s^2+220s+1000}\cdot \frac{1}{s}\\ \Rightarrow \Omega \left(s\right)=\frac{1}{s}+\frac{4}{45\left(s+5\right)}-\frac{49}{45\left(s+50\right)}\end{array}$

On taking inverse Laplace transform of this, we have

$\begin{array}{l}\Omega \left(s\right)=\frac{1}{s}+\frac{4}{45\left(s+5\right)}-\frac{49}{45\left(s+50\right)}\\ \Rightarrow \omega \left(t\right)=1+\frac{4}{45}\cdot e^{-5t}-\frac{49}{45}\cdot e^{-50t}\end{array}$

On plotting the responses $\omega \left(t\right)$ for all the three cases, we have

Figure 1

From Figure 1,it is observed that there is a high overshoot in the responseat the damping ratio $\zeta =0.707$.

Conclusion:

Therefore, the response $\Omega \left(s\right)$ for the unit-step command response ${\Omega }_r\left(s\right)$ for all the cases is shown in Figure 1. At the damping ratio $\zeta =0.707$, there is a high overshoot in the response in comparison to the other cases where damping ratios are 1 and 1.739 respectively.