Question 10.6.2: Suppose the plant shown in Figure 10.6.1 has the parameter v......

a. The error equation is

E(s)=\Omega_r(s)-\Omega(s)

=\left(1-\frac{K_P}{5 s  +  2  +  K_P}\right) \Omega_r(s)+\frac{1}{5 s  +  2  +  K_P} T_d(s)

=\left(\frac{5 s  +  2}{5 s  +  2  +  K_P}\right) \Omega_r(s)+\frac{1}{5 s  +  2  +  K_P} T_d(s)

The steady-state command error for a unit-step command with no disturbance is

e_{s s}= \underset{s \rightarrow 0}{lim}  s\left(\frac{5 s  +  2}{5 s  +  2  +  K_P}\right) \frac{1}{s}=\frac{2}{2  +  K_P}

The error will be 0.05 rad/s if K_P = 38. The time constant for this value of K_P is

\tau=\frac{I}{c  +  K_P}=\frac{5}{2  +  K_P}=\frac{1}{8} s

The steady-state error due to a unit-step disturbance is

e_{s s}=\underset{s \rightarrow 0}{lim}  s\left(\frac{1}{5 s  +  2  +  K_P}\right) \frac{1}{s}=\frac{1}{2  +  K_P}=\frac{1}{40}=0.025  rad / s

If both inputs are applied, the total steady-state error will be

e_{s s} = 0.05 + 0.025 = 0.075 rad/s

and the steady-state speed will be

ω_{s s} = 1 − 0.075 = 0.925 rad/s

Note that we can make the command error, the disturbance error, and the time constant smaller only by making K_P larger than 38, but this increases the maximum required torque and probably the cost of the system.
b. Using Ω_r (s) = 1/s² with the final value theorem, we find the steady-state command error is

e_{s s}=\underset{s \rightarrow 0}{lim}  s\left(\frac{5 s  +  2}{5 s  +  40}\right) \frac{1}{s^2}=\infty

The speed ω(t) never catches up with the command input ω_r (t), and the steady-state error is infinite.