Response of a Proportional Control System Suppose the plant shown in Figure 10.6.1 has the parameter values I = 5 and c = 2 in a consistent set of units in which time is measured in seconds. (a) Determine the smallest value of the gain KP required so that the steady-state command error will be no

Question

Response of a Proportional Control System

Suppose the plant shown in Figure 10.6.1 has the parameter values I = 5 and c = 2 in a consistent set of units in which time is measured in seconds. (a) Determine the smallest value of the gain [latex]K_{P}[/latex] required so that the steady-state command error will be no greater than 0.05 rad/s if [latex]ω_{r}[/latex] is a unit-step input. Evaluate the resulting time constant and the steady-state disturbance error for a unit-step disturbance. (b) Obtain the steady-state error due to a unit-ramp command.

Accepted Answer

a. The error equation is
[latex]E(s) = \Omega_{r}(s) − \Omega(s) [/latex]

[latex]= \left( 1 − \frac{K_{P}}{5s + 2 + K_{P}} \right) \Omega_{r}(s) + \frac{1}{5s + 2 + K_{P}} T_{d} (s) [/latex]

[latex]= \left(\frac{ 5s + 2}{5s + 2 + K_{P}} \right) \Omega_{r}(s) + \frac{1}{5s + 2 + K_{P}} T_{d} (s) [/latex]
The steady-state command error for a unit-step command with no disturbance is
[latex]e_{ss} = \lim_{s→0} s \left(\frac{ 5s + 2 }{5s + 2 + K_{P}} \right) \frac{1}{s} = \frac{2}{2 + K_{P}} [/latex]

The error will be 0.05 rad/s if [latex]K_{P} = 38[/latex]. The time constant for this value of [latex]K_{P} [/latex] is
[latex]τ = \frac{I}{c + K_{P}} = \frac{5}{2 + K_{P}} = \frac{1}{8} s [/latex]
The steady-state error due to a unit-step disturbance is
[latex]e_{ss} = \lim_{s→0} s \left(\frac{1}{5s + 2 + K_{P}} \right) \frac{1}{s} = \frac{1}{2 + K_{P}} = \frac{1}{40} = 0.025 rad/s [/latex]
If both inputs are applied, the total steady-state error will be
[latex]e_{ss} = 0.05 + 0.025 = 0.075 rad/s [/latex]
and the steady-state speed will be
[latex]ω_{ss} = 1 − 0.075 = 0.925 rad/s [/latex]
Note that we can make the command error, the disturbance error, and the time constant smaller only by making [latex]K_{P}[/latex] larger than 38, but this increases the maximum required torque and probably the cost of the system.
b. Using [latex]\Omega_{r}(s) = 1/s^{2}[/latex] with the final value theorem, we find the steady-state command error is
[latex]e_{ss} = \lim_{s→0} s \left(\frac{5s + 2}{5s + 40} \right) \frac{1}{s^{2}} = \infty [/latex]
The speed ω(t) never catches up with the command input [latex]ω_{r}(t)[/latex], and the steady-state error is infinite.

Question 10.6.2: Response of a Proportional Control System Suppose the plant ...

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