REGULATING THE REFERENCE PATH MAE 461: DYNAMICS AND CONTROLS 7 REGULATING THE REFERENCE PATH (CONTINUOUSLY-ACTING ACTUATORS) This Chapter considers the regulation problem restricted to the case in which the control force is produced by a continuously-acting actuator. The control force is a continuous function of time. We’ll consider in this chapter control forces that are linearly proportional to displacements, integrals of displacements, velocities, and linear combinations of them. In each case, we’ll see how the dynamic performance is regulated by the control. 1. Displacement Feedback Consider an undamped single degree of freedom system. In the absence of a control force, the equation governing the motion and the uncontrolled system response are (7 – 1) )sin()cos( ,000000tvtxxkxxmuuuωωω+==+&&REGULATING THE REFERENCE PATH MAE 461: DYNAMICS AND CONTROLS In the presence of displacement feedback, we get (7 – 2) )sin()cos( , ,000tvtxxgxfkxxmβββ+=−==+&& where g is called the displacement feedback control gain. In Eq. (7 – 1) and Eq. (7 – 2), we have mgkmk+==βω 0 in which β is called the closed-loop frequency of the system (See Fig. 7 – 1). )3,1,0().( and )( :17 Fig00=====− gkmvxtxtxu Since displacements are continuous functions of time, so too are displacement feedback control forces. Therefore, displacement feedback can be relatively easy to produce in devices that can generate forces that are continuous functions of time. 0 10 20 30 40-1 -0.50 0.51REGULATING THE REFERENCE PATH MAE 461: DYNAMICS AND CONTROLS Now let’s see how displacement feedback changes system performance. First, consider peak-overshoot. From Eq. (7 – 1), the peak-overshoot of the uncontrolled system is (7 – 3) 2002)/(ωvxPOou+= and from Eq. (7 – 2), the peak-overshoot of the controlled system is (7 – 4) uoooPOvxvxvxPO2002202202)/()/()/(ωββ++=+= In Eq. (7 – 4) we find that peak-overshoot can not be reduced significantly if .000ωvx >> Indeed, if a system is initially at rest when it’s initially displaced, its peak-overshoot will be equal to its initial displacement regardless of the stiffness in the system. On the other hand, if ,00=x Eq. (7 – 4) reduces to (7 – 5) uPOPOβω0= This case also arises when the system is intermittently subjected to impulsive forces. From Eq. (7 – 5), the desired closed-loop frequency is (See Fig. 7 – 2) (7 – 6) 0ωβ⎟⎠⎞⎜⎝⎛=POPOuREGULATING THE REFERENCE PATH MAE 461: DYNAMICS AND CONTROLS Equation (7-6) is applied to lightly damped systems as an approximation. (Recall that Eq.(7 – 1) is undamped, so Eq. (7 – 6) is exact only for undamped systems.) Next, let’s examine the effect of displacement feedback on settling time. From Eq. (7 – 1) and (7 – 2) displacement feedback clearly has no effect of settling time. Finally, turning to steady-state error, let’s subject the system to a unit step function in order to create a steady-state error. The steady-state response in the absence of displacement feedback and the steady-state response in the presence of displacement feedback are (7 – 7) 211 ,1⎟⎟⎠⎞⎜⎜⎝⎛=+==uuPOPOkgkxkx Notice that displacement feedback can reduce steady-state error but not eliminate it. To practically eliminate it, a very large control force would be needed. The displacement feedback control force is realized using Eq. (7 – 2) and (7 – 5) in which the control gain is (7 – 8) 2220() 1uPOgm kPOβω⎡⎤⎛⎞=−= −⎢⎥⎜⎟⎝⎠⎢⎥⎣⎦REGULATING THE REFERENCE PATH MAE 461: DYNAMICS AND CONTROLS In summary, displacement feedback is predominantly used to control peak-overshoot, but can also reduce steady-state error. It has no effect on settling time. The displacement feedback control force depends on the static properties of the uncontrolled system, i.e., its stiffness. 2. Velocity Feedback Consider an undamped single degree-of-freedom system subjected to a velocity feedback control force. The uncontrolled response was given in Eq. (7 – 1). The equation governing the controlled response is (7 – 9) xhffkxxm&&&−==+ , Velocity feedback is realized using forcing functions that imitate velocities. Velocities are piece-wise continuous functions. The functions are continuous except during instants when jump discontinuities occur. Jump discontinuities occur when the system is subjected to instantaneous impulses caused by impact loads. It follows that velocity feedback forces can be realized in devices that can generate continuous functions of time but they’ll have difficulty when jump discontinuities occur. Dividing Eq. (7 – 9) by m, we get (7 – 10) mhxxx2 ,0220==++αωα&&& Equation (7 – 10) describes the closed-loop system, and admits the homogeneous solutionREGULATING THE REFERENCE PATH MAE 461: DYNAMICS AND CONTROLS (7 – 11) )]sin()cos([000txvtxextββαβα++=− 220αωβ−= where we can see in Eq. (7 – 11) that α represents the closed-loop decay rate of the system. Now, let’s see how velocity feedback changes system performance. First, consider peak-overshoot. When the damping rate is considerably smaller than the natural frequency of the system, the peak-overshoot of the closed-loop system can be approximated by Eq. (7 – 3) (7 – 12) 02/120020 ,])([ωαβα<<≈++=uPOxvxPO Indeed, velocity feedback does not significantly reduce peak-overshoot when the closed-loop damping rate is small compared to the system’s natural frequency. However, as the damping rate increases, the peak-overshoot decreases (See Fig. 7 – 2). 0 10 20 30 40-1 -0.50 0.51REGULATING THE REFERENCE PATH MAE 461: DYNAMICS AND CONTROLS )5.0,1.0,1,0().( and )( :27 Fig2100======− hhkmvxtxtxu Next consider the effect of velocity feedback on settling time. From Eq. (7 – 9) and Eq. (7 – 10) velocity feedback causes the system motion to dampen at the rate α depending on the control gain h that is applied. On the other hand, velocity feedback has no effect on steady-state error, since steady-state error is a static phenomenon. From Eq. (7 – 10), the velocity feedback gain used to achieve this performance is )137( − mTmhs)10ln(22 ==α In summary, velocity feedback is predominantly used to control settling time, but also can reduce peak-overshoot. It has no effect on