Question 10.6.4: Suppose the plant to be controlled is Gp(s) = 1/I s + c wher......

a. From the figure, we obtain the output equation as follows:

\Omega(s)=\frac{1}{5 s  +  2}\left[T(s)-T_d(s)\right]        (1)

T(s)=\frac{K_I}{s} E(s)-K_2 \Omega(s)          (2)

E(s)=\Omega_r(s)-\Omega(s)              (3)

Substituting equations (2) and (3) into equation (1) and multiplying both sides by 5s + 2 gives

Solve for Ω(s):

\Omega(s)=\frac{K_I}{5 s^2  +  \left(2  +  K_2\right) s  +  K_I} \Omega_r(s)-\frac{s}{5 s^2  +  \left(2  +  K_P\right) s  +  K_I} T_d(s)          (4)

For the error equation,

Solve for E(s):

E(s)=\frac{5 s^2  +  \left(2  +  K_2\right) s}{5 s^2  +  \left(2  +  K_2\right) s  +  K_I} \Omega_r(s)+\frac{s}{5 s^2  +   \left(2  +  K_2\right) s  +  K_I} T_d(s)                (5)

For the actuator equation,

Solve for T (s):

T(s)=\frac{(5 s  +  2) K_I}{5 s^2  +  \left(2  +  K_2\right) s  +  K_I} \Omega_r(s)+\frac{K_2 s  +  K_I}{5 s^2  +  \left(2  +  K_2\right) s  +  K_I} T_d(s)                (6)

The characteristic equation is

5 s^2+\left(2+K_2\right) s+K_I=0                (4)

This has the same form as equation (4) of Example 10.6.3 with K_2 replacing K_P .

Therefore, the gain values obtained in that example can be used here. This gives the gain values shown in the following table.

Although the denominators in this example are the same as those in Example 10.6.3, the numerators are different. Applying the final value theorem to the error equation (5) with Ω_r (s) = 1/s and T_d (s) = 0 gives

e_{s s}=\underset{s \rightarrow 0}{lim}  s E(s)=\underset{s \rightarrow 0}{lim}  s \frac{5 s^2  +  \left(2  +  K_2\right) s}{5 s^2+\left(2  +  K_2\right) s  +  K_I} \frac{1}{s}=0

for all three cases. Thus the system has zero command error for a step command.
Applying the final value theorem to the error equation (5) with Ω_r (s) = 0 and T_d (s) = 1/s gives

e_{s s}=\underset{s \rightarrow 0}{lim}  s E(s)=\underset{s \rightarrow 0}{lim}  s \frac{s}{5 s^2  +  \left(2  +  K_2\right) s+K_I} \frac{1}{s}=0

for all three cases. Thus the system has zero disturbance error for a step disturbance.
b. The plots can be easily obtained with MATLAB using the \boxed{tf}  and  \boxed{step} functions applied to equations (1) and (3). The plots are shown in Figures 10.6.7 and 10.6.8. The rise times are much longer than for PI control, but the overshoots are much smaller. The peak torque required is 7.23, 4.29, and 8.06 for cases 1, 2, and 3, respectively. These are much less than the peak torque required with PI control.
c. Applying the final value theorem to the error equation (5) with Ω_r (s) = 1/s² and T_d (s) = 0 gives

e_{s s}=\underset{s \rightarrow 0}{lim}  s E(s)=\underset{s \rightarrow 0}{lim}  s \frac{5 s^2  +  \left(2  +  K_2\right) s}{5 s^2  +  \left(2  +  K_2\right) s  +  K_I} \frac{1}{s^2}=\frac{2  +  K_2}{K_I}

Thus the ramp command error is different for each of the three cases and much larger than for PI action.

d. The disturbance response of this system is identical to that of the PI controller because they both have the same disturbance transfer function.