Proportional Control of a First-Order Plant Consider proportional control of the first-order plant whose transfer function is 1/(Is +c). This can represent a rotational system whose model is Iω˙ + cω = T − Td , where the controlled variable is the speed ω, the actuator torque is T , and the

Question

Proportional Control of a First-Order Plant

Consider proportional control of the first-order plant whose transfer function is 1/(Is +c). This can represent a rotational system whose model is [latex]I \dot{ω} + cω = T − T_{d}[/latex] , where the controlled variable is the speed ω, the actuator torque is T , and the disturbance torque is [latex]T_{d}[/latex] . To illustrate the methods simply, we will assume that the actuator has a gain of 1 and has an instantaneous response. This implies that the actuator transfer function is [latex]G_{a}(s) = 1[/latex]. The block diagram is shown in Figure 10.6.1. Discuss the effect of the value of [latex]K_{P}[/latex] on the system performance when the inputs are step functions

Accepted Answer

The transfer functions are
[latex]\frac{\Omega(s)}{\Omega_{r}(s)} = \frac{K_{P}}{I s + c + K_{P}} [/latex] (1)
[latex]\frac{\Omega(s)}{T_{d} (s)} = \frac{− 1}{I s + c + K_{P}} [/latex] (2)
Steady-State Errors: For a unit-step command, [latex]\Omega_{r}(s) = 1/s [/latex], the speed approaches the steady-state value
[latex]ω_{ss} = \lim_{s→0} s \frac{K_{P}}{I s + c + K_{P}} \frac{1}{s} = \frac{K_{P}}{c + K_{P}} < 1 [/latex]
Thus, the steady-state speed is less than the desired value of 1, but it might be close enough if the damping c is small. The time required to reach this value is approximately four time constants, or [latex]4τ = 4I /(c + K_{P} )[/latex].
It is impossible for any control system to have a disturbance response that is zero for all time, because it takes time for any system to respond to the disturbance. In many applications, we are satisfied if the disturbance response is zero at steady state. In other applications, we might tolerate a small nonzero disturbance response. The performance specifications should indicate what is required for the design.
The response due to a unit-step disturbance, [latex]T_{d} (s) = 1/s[/latex], is found from equation (2).
[latex]\Omega(s) = \frac{−1}{I s + c + K_{P}} \frac{1}{s} [/latex]
The steady-state response to the disturbance is found with the final value theorem to be [latex]−1/(c + K_{P} )[/latex]. If [latex](c + K_{P} )[/latex]. is large, this response will be small. Note that this is the speed change caused by the disturbance, and is thus the disturbance error.
Frequency Response: The disturbance transfer function has the form of a low-pass filter whose low-frequency gain is [latex]1/(c + K_{P} )[/latex]., and whose bandwidth is [latex]1/τ = (c + K_{P} )/I[/latex]. Increasing the gain [latex]K_{P}[/latex] will increase the bandwidth, thus making the system sensitive to higher-frequency disturbances.
Actuator Requirements: It is important to predict what the actuator requirements will be. In the absence of a disturbance, the expression for the motor torque T can be found from the block diagram and equation (1). It is
[latex]T (s) = K_{P} [\Omega_{r}(s) − \Omega(s)] = K_{P} \left( 1 − \frac{K_{P}}{I s + c + K_{P}} ight) \Omega_{r}(s) [/latex]
or
[latex]T (s) = K_{P} \frac{I s + c}{I s + c + K_{P}} \Omega_{r}(s) [/latex]
If the commanded speed is a unit step, the torque response can be found from a partial fraction expansion to be
[latex]T (t) = \frac{K_{P}}{c + K_{P}} [c + K_{P} e^{−(c+K_{P} )t/I} ] [/latex]

Thus the torque decays exponentially to the value [latex]cK_{P}/(c + K_{P} )[/latex]. The maximum torque occurs at t = 0 and is [latex]T (0) = K_{P}[/latex] . Thus the larger the gain [latex]K_{P}[/latex] , the more torque is required. To speed up the system (by reducing [latex] au[/latex] ) or to reduce the steady-state error, we must increase [latex]K_{P}[/latex] and therefore must increase the available torque. This has implications for the size and cost of the motor that must be used.
Summary: The performance of proportional control action thus far can be summarized as follows. For the first-order plant 1/(I s + c), whose inputs are step functions,
1. The response to the command input never reaches the desired value if there is damping (c ≠ 0), although it can be made arbitrarily close to the desired value by choosing the gain [latex]K_{P}[/latex] large enough.
2. The response to the command input approaches its final value without oscillation. The time to reach this value is inversely proportional to [latex]K_{P}[/latex] .
3. The steady-state disturbance error is inversely proportional to the gain [latex]K_{P}[/latex] even if there is no damping.
4. For a step command input of magnitude M, the maximum required motor torque occurs at t = 0 and equals [latex]M K_{P}[/latex] . To reduce the time constant or the errors, we must increase [latex]K_{P}[/latex] and thus the maximum available torque.

Question 10.6.1: Proportional Control of a First-Order Plant Consider proport...

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