Aircraft Control: Pitch Stabilization with PD and PID Feedback - Prof. Craig A. Woolsey, Study notes of Aerospace Engineering

Download Aircraft Control: Pitch Stabilization with PD and PID Feedback - Prof. Craig A. Woolsey and more Study notes Aerospace Engineering in PDF only on Docsity! Lecture 21: Introduction to Aircraft Control – Pitch Stabilization dc Figure 1: An airplane with canards. Consider a wind-tunnel model which is pinned to allow pitch rotation about the center of gravity. Rather than a conventional horizontal stabilizer with elevators, the model uses servo-actuated canards to provide longitudinal stability and control. For small angles of attack, the model dynamics are well-described by the second order ODE θ̈ − Mq θ̇ − Mαθ = Mδcδc (1) where δc represents the canard deflection. (Since the model is mounted in a wind tunnel, the angle of attack α is identical to the aircraft pitch angle θ.) Pitch damping due to a canard acts to oppose pitch rate just as it does for an aft tail. Thus, one finds that Mq < 0. The pitch stiffness, however, is diminished by a forward tail and, for the case shown in Figure 1, one finds that Mα > 0. The airplane is not statically stable. Given a desired pitch angle θd, define the error e = θd − θ. For simplicity, suppose that θd = 0 and define the following “proportional-derivative (PD)” feedback control law δc = kpe + kdė = −kpθ − kdθ̇. That is, apply a control deflection in which is the sum of terms directly proportional to the error and the rate of increase of the error. Substituting into the dynamic equation gives θ̈ + (Mδckd − Mq)θ̇ + (Mδckp − Mα)θ = 0. For stability, we require that Mδckd > Mq and Mδckp > Mα. (2) Since Mq < 0 and Mδc > 0, we could actually just choose kd = 0, which would give a simple “proportional feedback” control structure. On the other hand, because Mα > 0, we must choose kp large enough to dominate that destabilizing term. Moreover, in order to obtain an arbitrary closed-loop natural frequency and damping ratio, we should retain the derivative term and choose kp = 1 Mδc ( Mα + ω 2 n ) and kd = 1 Mδc (Mq + 2ζωn) . Now suppose that θd is some nonzero constant. Then we have θ̈ + (Mδckd − Mq)θ̇ + (Mδckp − Mα)θ = Mδckpθd. Assuming conditions (2) hold, one may use the final value theorem (FVT) to show that lim t→∞ θ = Mδckp Mδckp − Mα θd 1 If kp is chosen large enough, then θ will approach θd in time, however it will never really converge to the desired value. While letting kp → ∞ would make the error arbitrarily small, there are practical concerns associated with such “high gain” feedback, including actuator limits and destabilization of unmodeled dynamics. A better approach to eliminating the steady-state error is to incorporate an integral term in the controller. At this point, it may be easier to proceed in the s-domain rather than the t-domain Re-expressing the pitch dynamics (1) in the s-domain, we find that the plant transfer function is P (s) = θ(s) δc(s) = Mδc s2 − Mqs − Mα . Given the desired pitch angle history θd(t), the error signal is e(s) = θd(s) − θ(s). In order to stabilize the system, so that the angle of attack may be prescribed as desired, we will implement the PID compensator F (s) = δc(s) e(s) = kp + ki 1 s + kds = kps + ki + kds 2 s . (The fact that the degree of the numerator polynomial is higher than that of the denominator is a bit problematic, because it suggests that the compensator is acausal, i.e., that current outputs δc depend on future inputs e. In practice, there are simple ways around this problem.) µ d ±c µF + - P e Figure 2: The closed-loop control system. The feedback control structure is shown in Figure 2. To solve for the closed-loop transfer function (from θd to θ), we note that θ(s) = P (s)δc(s) = P (s)(F (s)e(s)) = P (s)F (s)(θd(s) − θ(s)). Solving for θ (and omitting the independent variable s) gives (1 + PF )θ = PFθd so that the closed-loop transfer function is H(s) = θ(s) θd(s) = PF 1 + PF . Substituting the definitions of P and F and manipulating a bit gives H(s) = ( kps+ki+kds 2 s ) ( Mδc s2−Mqs−Mα ) 1 + ( kps+ki+kds 2 s ) ( Mδc s2−Mqs−Mα ) = ( kps + ki + kds 2 ) (Mδc) s (s2 − Mqs − Mα) + Mδc (kps + ki + kds2) = Mδc ( kps + ki + kds 2 ) s3 + (Mδckd − Mq) s2 + (Mδckp − Mα) s + (Mδcki) . 2