Question 10.6.4: Use of Internal Feedback Suppose the plant to be controlled ...
Use of Internal Feedback
Suppose the plant to be controlled is
G_{p}(s) = \frac{1}{I s + c}where I = 5 and c = 2. Figure 10.6.6 shows a proposed alternative to PI control for this plant. It uses an internal feedback loop to adjust the output of the controller. The dominant time constant is specified to be 0.5 s.
a. Compute the required values for K_{2} and K_{I} for each of the following three cases:
(1) ζ = 0.707, (2) ζ = 1, and (3) the secondary time constant must be 0.05 s.
Evaluate the steady-state command error and the steady-state disturbance error for each case given that both the command input ω_{r}(t) and the disturbance T_{d} (t) are unit-step functions.
b. Plot the output response ω(t) and the actuator response T (t) for each case, given that the command input ω_{r}(t) is a unit-step function and the disturbance T_{d} (t) is zero.
c. Evaluate the steady-state command error for each case given that the command input ω_{r}(t) is a unit-ramp function and the disturbance T_{d} (t) is zero.
d. Evaluate the frequency response characteristics of the disturbance transfer function
a. From the figure, we obtain the output equation as follows:
\Omega(s) = \frac{1}{5s + 2} [T (s) − T_{d} (s)] (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
(5s + 2) \Omega(s) = T (s) − T_{d} (s) = \frac{K_{I}}{s} [\Omega_{r}(s) − \Omega(s)] − K_{2} \Omega(s) − T_{d} (s)
Solve for Ω(s):
\Omega(s) = \frac{K_{I}}{5s^{2} + (2 + K_{2})s + K_{I}} \Omega_{r}(s) − \frac{s}{5s^{2} + (2 + K_{P} )s + K_{I}} T_{d} (s) (4)
For the error equation,
E(s) = \Omega_{r}(s) − \Omega(s)
Solve for E(s):
E(s) = \frac{5s^{2} + (2 + K_{2})s}{5s^{2} + (2 + K_{2})s + K_{I}} \Omega_{r}(s) + \frac{s}{5s^{2} + (2 + K_{2} )s + K_{I}} T_{d} (s) (5)
For the actuator equation,
T (s) = \frac{K_{I}}{s} E(s) − K_{2} \Omega(s) = \frac{K_{I}}{s} [\Omega_{r}(s) − \Omega(s)] − K_{2} \Omega(s)
Solve for T (s):
T (s) = \frac{(5s + 2)K_{I}}{5s^{2} + (2 + K_{2})s + K_{I}} \Omega_{r}(s) + \frac{K_{2}s + K_{I}}{5s^{2} + (2 + K_{2})s + K_{I}} T_{d} (s) (6)
The characteristic equation is
5s^{2} + (2 + K_{2})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.
| Case | K_{2} | K_{I} | ζ | Maximum torque | Ramp command error |
| 1 | 18 | 40 | 0.707 | 7.23 | 0.5 |
| 2 | 18 | 20 | 1 | 4.29 | 1 |
| 3 | 108 | 200 | 1.74 | 8.06 | 5.5 |
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 \Omega_{r}(s) = 1/s and T_{d} (s) = 0 gives
e_{ss} = \lim_{s→0} s E(s) = \lim_{s→0} s \frac{5s^{2} + (2 + K_{2})s}{5s^{2} + (2 + K_{2})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 \Omega_{r}(s) = 0 and T_{d} (s) = 1/s gives
e_{ss} = \lim_{s→0} s E(s) = \lim_{s→0} s \frac{s}{5s^{2} + (2 + K_{2})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 tf and 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 \Omega_{r}(s) = 1/s^{2} and T_{d} (s) = 0 gives
e_{ss} = \lim_{s→0} s E(s) = \lim_{s→0} s \frac{5s^{2} + (2 + K_{2})s}{5s^{2} + (2 + K_{2})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.
