Chapter 10, Problem 10.29P

To determine

(a)

The values of gain $K_2\text{and}{K}_I$ of the I controller with an internal feedback loop of first order plant for the following collection of roots:

Case 1. For $s=-10,-8$ (Root separation factor is 1.25).

Case 2. For $s=-10,-20$ (Root separation factor is 2).

Case 3. For $s=-10,-50$ (Root separation factor is 5).

Answer to Problem 10.29P

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

Case 1. For $s=-10,-8$, $K_I=160$ and $K_2=34$.

Case 2. For $s=-10,-20$, $K_I=400$ and $K_2=58$.

Case 3. For $s=-10,-50$, $K_I=1000$ and $K_2=118$.

Explanation of Solution

Given:

The I controller with an internal feedback loop of first order plant is as shown below:

Where, the parameter values are as given:

$I=c=2$

Also, the performance specifications require the time constant of the system to be $\tau =0.1$.

Concept Used:

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

$\begin{array}{l}\frac{\Omega \left(s\right)}{{\Omega }_r\left(s\right)}=\frac{K_I}{I{s}^2+s\left(c+K_2\right)+K_I}\\ \frac{\Omega \left(s\right)}{T_d\left(s\right)}=\frac{-s}{I{s}^2+s\left(c+K_2\right)+K_I}\end{array}$

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:

$\begin{array}{l}\frac{\Omega \left(s\right)}{{\Omega }_r\left(s\right)}=\frac{K_I}{I{s}^2+s\left(c+K_2\right)+K_I}\\ \frac{\Omega \left(s\right)}{T_d\left(s\right)}=\frac{-s}{I{s}^2+s\left(c+K_2\right)+K_I}\end{array}$

Therefore, the characteristic equation for the system is:

$I{s}^2+s\left(c+K_2\right)+K_I=0$

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

$\begin{array}{l}I{s}^2+s\left(c+K_2\right)+K_I=0\\ \Rightarrow 2s^2+s\left(2+K_2\right)+K_I=0\because I=c=2\\ \Rightarrow s^2+s\frac{\left(2+K_2\right)}{2}+\frac{K_I}{2}=0\end{array}$

Case 1. When the root separation factor is 1.25.

The given set of roots for this root separation are:

$s=-10,s=-8$

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

$\begin{array}{l}s=-10,s=-8\\ \Rightarrow \left(s+10\right)\left(s+8\right)=0\\ \Rightarrow s^2+18s+80=0\end{array}$

On comparing this equation with $s^2+s\frac{\left(2+K_2\right)}{2}+\frac{K_I}{2}=0$, we get

$\begin{array}{l}\frac{K_I}{2}=80\\ \Rightarrow K_I=160\end{array}$

And

$\begin{array}{l}\frac{\left(2+K_2\right)}{2}=18\\ \Rightarrow 2+K_2=36\\ \Rightarrow K_2=34\end{array}$

Case 2. When the root separation factor is 2.

The given set of roots for this root separation are:

$s=-10,s=-20$

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

$\begin{array}{l}s=-10,s=-20\\ \Rightarrow \left(s+10\right)\left(s+20\right)=0\\ \Rightarrow s^2+30s+200=0\end{array}$

On comparing this equation with $s^2+s\frac{\left(2+K_2\right)}{2}+\frac{K_I}{2}=0$, we get

$\begin{array}{l}\frac{K_I}{2}=200\\ \Rightarrow K_I=400\end{array}$

And

$\begin{array}{l}\frac{\left(2+K_2\right)}{2}=30\\ \Rightarrow 2+K_2=60\\ \Rightarrow K_2=58\end{array}$

Case 3. When the root separation factor is 5.

The given set of roots for this root separation are:

$s=-10,s=-50$

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

$\begin{array}{l}s=-10,s=-50\\ \Rightarrow \left(s+10\right)\left(s+50\right)=0\\ \Rightarrow s^2+60s+500=0\end{array}$

On comparing this equation with $s^2+s\frac{\left(2+K_2\right)}{2}+\frac{K_I}{2}=0$, we get

$\begin{array}{l}\frac{K_I}{2}=500\\ \Rightarrow K_I=1000\end{array}$

And

$\frac{\left(2+K_2\right)}{2}=60$

$\begin{array}{l}\Rightarrow 2+K_2=120\\ \Rightarrow K_2=118\end{array}$.

Conclusion:

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

Case 1. For $s=-10,-8$, $K_I=160$ and $K_2=34$.

Case 2. For $s=-10,-20$, $K_I=400$ and $K_2=58$.

Case 3. For $s=-10,-50$, $K_I=1000$ and $K_2=118$.

To determine

(b)

To Plot:The response $\omega \left(t\right)$ for the unit-step command response ${\omega }_r\left(t\right)$ for all the cases in sub-part (a)

Also, discuss the impact of root separation factor on the responses obtained in all these cases through the perspective of rise time, the overshoot and the maximum required torque.

Answer to Problem 10.29P

The impact of root separation factor are as follows:

Case 1. For $s=-10,-8$ (Root separation factor is 1.25), $t_r\approx 0.42\text{seconds}$ and $T_m\approx 7.12\text{N}\cdot \text{m}$.

Case 2. For $s=-10,-20$ (Root separation factor is 2), $t_r\approx 0.075\text{seconds}$ and $T_m\approx 10.52\text{N}\cdot \text{m}$.

Case 3. For $s=-10,-50$ (Root separation factor is 5), $t_r\approx 0.04\text{seconds}$ and $T_m\approx 13.78\text{N}\cdot \text{m}$.

Thus, we see that with an increase in the root separation factor, the rise time decreases, whereas there is no overshoot in the response due to the overdamped nature of the responses as shown in figure 1. While for the required torque, the value of maximum torque increased with an increase in the value of root separation factor as shown in figure 2.

Explanation of Solution

Given:

The I controller with an internal feedback loop of first order plant is as shown below:

Where, the parameter values are as given:

$I=c=2$

Also, the performance specifications require the time constant of the system to be $\tau =0.1$.

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

Case 1. For $s=-10,-8$, $K_I=160$ and $K_2=34$.

Case 2. For $s=-10,-20$, $K_I=400$ and $K_2=58$.

Case 3. For $s=-10,-50$, $K_I=1000$ and $K_2=118$.

Concept Used:

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

$\begin{array}{l}\frac{\Omega \left(s\right)}{{\Omega }_r\left(s\right)}=\frac{K_I}{I{s}^2+s\left(c+K_2\right)+K_I}\\ \frac{\Omega \left(s\right)}{T_d\left(s\right)}=\frac{-s}{I{s}^2+s\left(c+K_2\right)+K_I}\end{array}$.

Calculation:

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

$\begin{array}{l}\frac{\Omega \left(s\right)}{{\Omega }_r\left(s\right)}=\frac{K_I}{I{s}^2+s\left(c+K_2\right)+K_I}\\ \frac{\Omega \left(s\right)}{T_d\left(s\right)}=\frac{-s}{I{s}^2+s\left(c+K_2\right)+K_I}\end{array}$

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

$\Omega \left(s\right)=\frac{K_I}{I{s}^2+s\left(c+K_2\right)+K_I}{\Omega }_r\left(s\right)-\frac{s}{I{s}^2+s\left(c+K_2\right)+K_I}{T}_d\left(s\right)$

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

$\begin{array}{l}\Omega \left(s\right)=\frac{K_I}{I{s}^2+s\left(c+K_2\right)+K_I}{\Omega }_r\left(s\right)-\frac{s}{I{s}^2+s\left(c+K_2\right)+K_I}{T}_d\left(s\right)\\ \Rightarrow \Omega \left(s\right)=\frac{K_I}{2s^2+s\left(2+K_2\right)+K_I}{\Omega }_r\left(s\right)-\frac{s}{2s^2+s\left(2+K_2\right)+K_I}{T}_d\left(s\right)\because I=c=2\end{array}$

And from the block diagram shown in figure, we have

$\begin{array}{l}T\left(s\right)=\frac{K_I}{s}\cdot E\left(s\right)-K_2\cdot \Omega \left(s\right)\\ \Rightarrow T\left(s\right)=\frac{K_I}{s}\cdot \left({\Omega }_r\left(s\right)-\Omega \left(s\right)\right)-K_2\cdot \Omega \left(s\right)\because E\left(s\right)={\Omega }_r\left(s\right)-\Omega \left(s\right)\\ \Rightarrow T\left(s\right)=\frac{K_I}{s}\cdot {\Omega }_r\left(s\right)-\left(\frac{K_I}{s}+K_2\right)\cdot \Omega \left(s\right)\\ \Rightarrow T\left(s\right)=\left(\frac{K_I}{s}-\left(\frac{K_I}{s}+K_2\right)\cdot \frac{K_I}{I{s}^2+s\left(c+K_2\right)+K_I}\right){\Omega }_r\left(s\right)+\left(\frac{K_I}{s}+K_2\right)\cdot \frac{s}{I{s}^2+s\left(c+K_2\right)+K_I}{T}_d\left(s\right)\\ \because \Omega \left(s\right)=\frac{K_I}{I{s}^2+s\left(c+K_2\right)+K_I}{\Omega }_r\left(s\right)-\frac{s}{I{s}^2+s\left(c+K_2\right)+K_I}{T}_d\left(s\right)\\ \Rightarrow T\left(s\right)=\frac{K_I}{s}\left(1-\frac{s{K}_2+K_I}{I{s}^2+s\left(c+K_2\right)+K_I}\right){\Omega }_r\left(s\right)+\frac{s{K}_2+K_I}{I{s}^2+s\left(c+K_2\right)+K_I}{T}_d\left(s\right)\end{array}$

On keeping the values of the parameters such that $I=c=2$

$\begin{array}{l}T\left(s\right)=\frac{K_I}{s}\left(1-\frac{s{K}_2+K_I}{I{s}^2+s\left(c+K_2\right)+K_I}\right){\Omega }_r\left(s\right)+\frac{s{K}_2+K_I}{I{s}^2+s\left(c+K_2\right)+K_I}{T}_d\left(s\right)\\ \Rightarrow T\left(s\right)=\left(\frac{K_I\left(Is+c\right)}{I{s}^2+s\left(c+K_2\right)+K_I}\right){\Omega }_r\left(s\right)+\frac{s{K}_2+K_I}{I{s}^2+s\left(c+K_2\right)+K_I}{T}_d\left(s\right)\\ \Rightarrow T\left(s\right)=\frac{K_I\left(2s+2\right)}{2s^2+s\left(2+K_2\right)+K_I}{\Omega }_r\left(s\right)+\frac{s{K}_2+K_I}{2s^2+s\left(2+K_2\right)+K_I}{T}_d\left(s\right)\end{array}$

Case 1. For $s=-10,-8$, $K_I=160$ and $K_2=34$:

Since, $\Omega \left(s\right)=\frac{K_I}{2s^2+s\left(2+K_2\right)+K_I}{\Omega }_r\left(s\right)-\frac{s}{2s^2+s\left(2+K_2\right)+K_I}{T}_d\left(s\right)$

$\begin{array}{l}\Rightarrow \Omega \left(s\right)=\frac{160}{2s^2+s\left(2+34\right)+160}{\Omega }_r\left(s\right)-\frac{s}{2s^2+s\left(2+34\right)+160}{T}_d\left(s\right)\\ \because K_2=34,K_I=160\\ \Rightarrow \Omega \left(s\right)=\frac{160}{2s^2+36s+160}{\Omega }_r\left(s\right)-\frac{s}{2s^2+36s+160}{T}_d\left(s\right)\end{array}$

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

$\begin{array}{l}\Omega \left(s\right)=\frac{160}{2s^2+36s+160}{\Omega }_r\left(s\right)-\frac{s}{2s^2+36s+160}{T}_d\left(s\right)\\ \Rightarrow \Omega \left(s\right)=\frac{160}{2s^2+36s+160}\cdot \frac{1}{s}-\frac{s}{2s^2+36s+160}\cdot 0\\ \Rightarrow \Omega \left(s\right)=\frac{160}{2s^2+36s+160}\cdot \frac{1}{s}\end{array}$

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

$\begin{array}{l}\Omega \left(s\right)=\frac{160}{2s^2+36s+160}\cdot \frac{1}{s}=\frac{80}{s^2+18s+80}\cdot \frac{1}{s}\\ \Rightarrow \Omega \left(s\right)=\frac{1}{s}-\frac{5}{s+8}+\frac{4}{s+10}\end{array}$

On taking inverse Laplace transform of this, we have

$\begin{array}{l}\Omega \left(s\right)=\frac{1}{s}-\frac{5}{s+8}+\frac{4}{s+10}\\ \Rightarrow \omega \left(t\right)=1-5e^{-8t}+4e^{-10t}\end{array}$

Also, on keeping the gain values in the expression for torque:

$\begin{array}{l}T\left(s\right)=\frac{K_I\left(2s+2\right)}{2s^2+s\left(2+K_2\right)+K_I}{\Omega }_r\left(s\right)+\frac{s{K}_2+K_I}{2s^2+s\left(2+K_2\right)+K_I}{T}_d\left(s\right)\\ \Rightarrow T\left(s\right)=\frac{160\left(2s+2\right)}{2s^2+s\left(2+34\right)+160}{\Omega }_r\left(s\right)+\frac{34s+160}{2s^2+s\left(2+34\right)+160}{T}_d\left(s\right)\\ \Rightarrow T\left(s\right)=\frac{160\left(2s+2\right)}{2s^2+36s+160}{\Omega }_r\left(s\right)+\frac{34s+160}{2s^2+36s+160}{T}_d\left(s\right)\end{array}$

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

$\begin{array}{l}T\left(s\right)=\frac{160\left(2s+2\right)}{2s^2+36s+160}{\Omega }_r\left(s\right)+\frac{34s+160}{2s^2+36s+160}{T}_d\left(s\right)\\ \Rightarrow T\left(s\right)=\frac{160\left(2s+2\right)}{2s^2+36s+160}\cdot \frac{1}{s}+\frac{34s+160}{2s^2+36s+160}\cdot 0\\ \Rightarrow T\left(s\right)=\frac{160\left(s+1\right)}{s^2+18s+80}\cdot \frac{1}{s}\end{array}$

On simplifying the expression for this torqueusing partial fraction expansion, we get:

$\begin{array}{l}T\left(s\right)=\frac{160\left(s+1\right)}{s^2+18s+80}\cdot \frac{1}{s}\\ \Rightarrow T\left(s\right)=\frac{2}{s}+\frac{70}{s+8}-\frac{72}{s+10}\end{array}$

On taking inverse Laplace transform, we have

$\begin{array}{l}T\left(s\right)=\frac{2}{s}+\frac{70}{s+8}-\frac{72}{s+10}\\ \Rightarrow T\left(t\right)=2+70e^{-8t}-72e^{-10t}\end{array}$

Case 2. For $s=-10,-20$, $K_I=400$ and $K_2=58$:

Since, $\Omega \left(s\right)=\frac{K_I}{2s^2+s\left(2+K_2\right)+K_I}{\Omega }_r\left(s\right)-\frac{s}{2s^2+s\left(2+K_2\right)+K_I}{T}_d\left(s\right)$

$\begin{array}{l}\Rightarrow \Omega \left(s\right)=\frac{400}{2s^2+s\left(2+58\right)+400}{\Omega }_r\left(s\right)-\frac{s}{2s^2+s\left(2+58\right)+400}{T}_d\left(s\right)\\ \because K_2=58,K_I=400\end{array}$

$\Rightarrow \Omega \left(s\right)=\frac{400}{2s^2+60s+400}{\Omega }_r\left(s\right)-\frac{s}{2s^2+60s+400}{T}_d\left(s\right)$

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

$\begin{array}{l}\Omega \left(s\right)=\frac{400}{2s^2+60s+400}{\Omega }_r\left(s\right)-\frac{s}{2s^2+60s+400}{T}_d\left(s\right)\\ \Rightarrow \Omega \left(s\right)=\frac{400}{2s^2+60s+400}\cdot \frac{1}{s}-\frac{s}{2s^2+60s+400}\cdot 0\\ \Rightarrow \Omega \left(s\right)=\frac{400}{2s^2+60s+400}\cdot \frac{1}{s}\end{array}$

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

$\begin{array}{l}\Omega \left(s\right)=\frac{400}{2s^2+60s+400}\cdot \frac{1}{s}=\frac{200}{s^2+30s+200}\cdot \frac{1}{s}\\ \Rightarrow \Omega \left(s\right)=\frac{1}{s}-\frac{2}{s+10}+\frac{1}{s+20}\end{array}$

On taking inverse Laplace transform of this, we have

$\begin{array}{l}\Omega \left(s\right)=\frac{1}{s}-\frac{2}{s+10}+\frac{1}{s+20}\\ \Rightarrow \omega \left(t\right)=1-2e^{-10t}+e^{-20t}\end{array}$

Also, on keeping the gain values in the expression for torque:

$\begin{array}{l}T\left(s\right)=\frac{K_I\left(2s+2\right)}{2s^2+s\left(2+K_2\right)+K_I}{\Omega }_r\left(s\right)+\frac{s{K}_2+K_I}{2s^2+s\left(2+K_2\right)+K_I}{T}_d\left(s\right)\\ \Rightarrow T\left(s\right)=\frac{400\left(2s+2\right)}{2s^2+s\left(2+58\right)+400}{\Omega }_r\left(s\right)+\frac{58s+400}{2s^2+s\left(2+58\right)+400}{T}_d\left(s\right)\\ \Rightarrow T\left(s\right)=\frac{400\left(2s+2\right)}{2s^2+60s+400}{\Omega }_r\left(s\right)+\frac{58s+400}{2s^2+60s+400}{T}_d\left(s\right)\end{array}$

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

$\begin{array}{l}T\left(s\right)=\frac{400\left(2s+2\right)}{2s^2+60s+400}{\Omega }_r\left(s\right)+\frac{58s+400}{2s^2+60s+400}{T}_d\left(s\right)\\ \Rightarrow T\left(s\right)=\frac{400\left(2s+2\right)}{2s^2+60s+400}\cdot \frac{1}{s}+\frac{58s+400}{2s^2+60s+400}\cdot 0\end{array}$

$\Rightarrow T\left(s\right)=\frac{400\left(2s+2\right)}{2s^2+60s+400}\cdot \frac{1}{s}$

On simplifying the expression for this torqueusing partial fraction expansion, we get:

$\begin{array}{l}T\left(s\right)=\frac{400\left(2s+2\right)}{2s^2+60s+400}\cdot \frac{1}{s}=\frac{400\left(s+1\right)}{s^2+30s+200}\cdot \frac{1}{s}\\ \Rightarrow T\left(s\right)=\frac{2}{s}+\frac{36}{s+10}-\frac{38}{s+20}\end{array}$

On taking inverse Laplace transform, we have

$\begin{array}{l}T\left(s\right)=\frac{2}{s}+\frac{36}{s+10}-\frac{38}{s+20}\\ \Rightarrow T\left(t\right)=2+36e^{-10t}-38e^{-20t}\end{array}$

Case 3. For $s=-10,-50$, $K_I=1000$ and $K_P=118$:

Since, $\Omega \left(s\right)=\frac{K_I}{2s^2+s\left(2+K_2\right)+K_I}{\Omega }_r\left(s\right)-\frac{s}{2s^2+s\left(2+K_2\right)+K_I}{T}_d\left(s\right)$

$\begin{array}{l}\Rightarrow \Omega \left(s\right)=\frac{1000}{2s^2+s\left(2+118\right)+1000}{\Omega }_r\left(s\right)-\frac{s}{2s^2+s\left(2+118\right)+1000}{T}_d\left(s\right)\\ \because K_2=118,K_I=1000\\ \Rightarrow \Omega \left(s\right)=\frac{1000}{2s^2+120s+1000}{\Omega }_r\left(s\right)-\frac{s}{2s^2+120s+1000}{T}_d\left(s\right)\end{array}$

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

$\begin{array}{l}\Omega \left(s\right)=\frac{1000}{2s^2+120s+1000}{\Omega }_r\left(s\right)-\frac{s}{2s^2+120s+1000}{T}_d\left(s\right)\\ \Rightarrow \Omega \left(s\right)=\frac{1000}{2s^2+120s+1000}\cdot \frac{1}{s}-\frac{s}{2s^2+120s+1000}\cdot 0\\ \Rightarrow \Omega \left(s\right)=\frac{1000}{2s^2+120s+1000}\cdot \frac{1}{s}\end{array}$

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

$\Omega \left(s\right)=\frac{1000}{2s^2+120s+1000}\cdot \frac{1}{s}=\frac{500}{s^2+60s+500}\cdot \frac{1}{s}$

$\Rightarrow \Omega \left(s\right)=\frac{1}{s}-\frac{5}{4\left(s+10\right)}+\frac{1}{4\left(s+50\right)}$

On taking inverse Laplace transform of this, we have

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

Also, on keeping the gain values in the expression for torque:

$\begin{array}{l}T\left(s\right)=\frac{K_I\left(2s+2\right)}{2s^2+s\left(2+K_2\right)+K_I}{\Omega }_r\left(s\right)+\frac{s{K}_2+K_I}{2s^2+s\left(2+K_2\right)+K_I}{T}_d\left(s\right)\\ \Rightarrow T\left(s\right)=\frac{1000\left(2s+2\right)}{2s^2+s\left(2+118\right)+1000}{\Omega }_r\left(s\right)+\frac{118s+1000}{2s^2+s\left(2+118\right)+1000}{T}_d\left(s\right)\\ \Rightarrow T\left(s\right)=\frac{1000\left(2s+2\right)}{2s^2+120s+1000}{\Omega }_r\left(s\right)+\frac{118s+1000}{2s^2+120s+1000}{T}_d\left(s\right)\end{array}$

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

$\begin{array}{l}T\left(s\right)=\frac{1000\left(2s+2\right)}{2s^2+120s+1000}{\Omega }_r\left(s\right)+\frac{118s+1000}{2s^2+120s+1000}{T}_d\left(s\right)\\ \Rightarrow T\left(s\right)=\frac{1000\left(2s+2\right)}{2s^2+120s+1000}\cdot \frac{1}{s}+\frac{118s+1000}{2s^2+120s+1000}\cdot 0\\ \Rightarrow T\left(s\right)=\frac{1000\left(2s+2\right)}{2s^2+120s+1000}\cdot \frac{1}{s}\end{array}$

On simplifying the expression for this torqueusing partial fraction expansion, we get:

$\begin{array}{l}T\left(s\right)=\frac{1000\left(2s+2\right)}{2s^2+120s+1000}\cdot \frac{1}{s}=\frac{1000\left(s+1\right)}{s^2+60s+500}\cdot \frac{1}{s}\\ \Rightarrow T\left(s\right)=\frac{2}{s}+\frac{45}{2\left(s+10\right)}-\frac{49}{2\left(s+50\right)}\end{array}$

On taking inverse Laplace transform, we have

$\begin{array}{l}T\left(s\right)=\frac{2}{s}+\frac{45}{2\left(s+10\right)}-\frac{49}{2\left(s+50\right)}\\ \Rightarrow T\left(t\right)=2+\frac{45}{2}{e}^{-10t}-\frac{49}{2}{e}^{-50t}\end{array}$

Therefore, on plotting these speed responses and torque responses in figure 1 and 2 respectively, we have:

As can be seen in the figure above, the responses in all the three cases are overdamped in nature thus the responses doesn’t have overshoots in transient period. However, the responses are overdamped thus, their rise times are affected by the choice of the root separation factor. Therefore, with an increase in the root separation factor in the given cases, the rise time for the system response improves or say decreases.

Here, in the case of required torque the maximum value of torque increases with an increase in the root separation factor as shown in figure 2.

Conclusion:

The impact of root separation factor are as follows:

Case 1. For $s=-10,-8$ (Root separation factor is 1.25), $t_r\approx 0.42\text{seconds}$ and $T_m\approx 7.12\text{N}\cdot \text{m}$.

Case 2. For $s=-10,-20$ (Root separation factor is 2), $t_r\approx 0.075\text{seconds}$ and $T_m\approx 10.52\text{N}\cdot \text{m}$.

Case 3. For $s=-10,-50$ (Root separation factor is 5), $t_r\approx 0.04\text{seconds}$ and $T_m\approx 13.78\text{N}\cdot \text{m}$.

Thus, we see that with an increase in the root separation factor, the rise time decreases, whereas there is no overshoot in the response as the responses are overdamped in nature. While for the required torque, the value of maximum torque increased with an increase in the value of root separation factor.