16.333 Lecture #9 Basic Longitudinal Control BasIc aircraft control concepts Basic control approaches
16.333 Lecture # 9 Basic Longitudinal Control • Basic aircraft control concepts • Basic control approaches
Fa2004 16.3338-1 Basic Longitudinal Control Goal: analyze aircraft longitudinal dynamics to determine if the be- havior is acceptable, and if not, then modify it using feed back control Note that we could (and will)work with the full dynamics model but for now, let's focus on the short period approximate model from lecture 7-5 An asn+B where de is the elevator input, and Zu/m Fw (Mw+ Mizu m)Iw(Ma+ Mai Uo B (M6e +Mi zse/ Add that 6=9, so se= g Take the output as 0, input is de, then form the transfer function 6(s)1q 5(s)S6(s) [01](s-Ay)-B As shown in the code, for the 747(40Kft, M=0.8)this reduces to 6(s) 1.1569s+0.3435 6(s)s(s2+0.7410s+0.92 x66 so that the dominant roots have a frequency of approximately 1 rad/ sec and damping of about 0.4
� � � � � � Fall 2004 16.333 8–1 Basic Longitudinal Control • Goal: analyze aircraft longitudinal dynamics to determine if the behavior is acceptable, and if not, then modify it using feedback control. • Note that we could (and will) work with the full dynamics model, but for now, let’s focus on the short period approximate model from lecture 7–5. x˙ sp = Aspxsp + Bspδe where δe is the elevator input, and w Zw/m U0 xsp = q , Asp = I−1 (Mw + Mw˙Zw/m) I−1 (Mq + Mw˙U0) yy yy Zδe/m Bsp = I−1 (Mδe + Mw˙Zδe/m) yy Add that θ • ˙ = q, so sθ = q • Take the output as θ, input is δe, then form the transfer function θ(s) 1 q(s) � � = = 0 1 (sI − Asp) −1 Bsp δe(s) s δe(s) • As shown in the code, for the 747 (40Kft, M = 0.8) this reduces to: θ(s) 1.1569s + 0.3435 = δe(s) −s(s2 + 0.7410s + 0.9272) ≡ Gθδe(s) so that the dominant roots have a frequency of approximately 1 rad/sec and damping of about 0.4
Fa2004 16.3338-2 Figure 1: Pole-zero map for Basic problem is that there are vast quantities of empirical data to show that pilots do not like the flying qualities of an aircraft with this combination of frequency and damping What do they prefer? Acceptable Satisfacto 0.40.60.81 Figure2:“ Thumb print” criterion . This criterion has been around since the 1950s but it is still valid Good target: frequency a 3 rad / sec and damping of about 0.6 Problem is that the short period dynamics are no where near these numbers, so we must modify them
Fall 2004 16.333 8–2 −1 −0.9 −0.8 −0.7 −0.6 −0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 0.09 0.95 0.68 0.54 0.42 0.3 0.2 0.09 0.84 1 0.95 0.68 0.54 0.42 0.3 0.2 0.84 1 0.2 0.4 0.6 0.8 0.8 0.2 0.4 0.6 Pole−Zero Map Real Axis Imaginary Axis Figure 1: Polezero map for Gqδe • Basic problem is that there are vast quantities of empirical data to show that pilots do not like the flying qualities of an aircraft with this combination of frequency and damping – What do they prefer? Acceptable 0.1 1 0 2 3 4 5 6 7 0.2 0.4 2 4 s � �s Unacceptable Poor Satisfactory 0.6 0.8 1 Undamped natural frequency rad/sec Damping ratio Figure 2: “Thumb Print” criterion • This criterion has been around since the 1950’s, but it is still valid. • Good target: frequency ≈ 3 rad/sec and damping of about ≈ 0.6 • Problem is that the short period dynamics are no where near these numbers, so we must modify them
Fa2004 16.3338-3 Could do it by redesigning the aircraft but it is a bit late for that Pole-Zero Map 095: × ○ × Figure 3: Pole-zero map and target pole locations Of course there are plenty of other things that we will consider when we design the controllers Small steady state error to commands S within limits No oscillations Speed control
Fall 2004 16.333 8–3 – Could do it by redesigning the aircraft, but it is a bit late for that... −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 −3 −2 −1 0 1 2 3 0.84 3.5 0.7 0.56 0.44 0.32 0.2 0.1 0.7 0.44 0.32 0.95 0.2 0.1 2.5 0.84 0.56 3 0.95 2 1.5 1 0.5 Pole−Zero Map Real Axis Imaginary Axis Figure 3: Polezero map and target pole locations • Of course there are plenty of other things that we will consider when we design the controllers – Small steady state error to commands – δe within limits – No oscillations – Speed control
Fa2004 16.3338-4 First Short Period Autopilot First attempt to control the vehicle response: measure g and fe back to the elevator command s Unfortunately the actuator is slow, so there is an apparent lag in the response that we must model 4 S+4 66(S Dynamics: de is the actual elevator deflection, de is the actuator command created by our controller 4 8=Gose(s)8 8=H(S6H()=s+4 The control is just basic proportional feed back 6e=-ke(6-6 Which gives that 8=-Gese(S)H(s ke(8-8c or that a(s Gose s)H(ske 8c(s) 1+ Gese s)H(s ke Looks good, but how do we analyze what is going on? Need to be able to predict where the poles are going as a function ofke→ Root locus
Fall 2004 16.333 8–4 First Short Period Autopilot • First attempt to control the vehicle response: measure θ and feed it back to the elevator command δe. – Unfortunately the actuator is slow, so there is an apparent lag in the response that we must model δc e / 4 s + 4 δa e / Gθδe(s) θ / o kθ − OO −θc o • Dynamics: δa is the actual elevator deflection, δe c is the actuator e command created by our controller 4 θ = Gθδe(s)δe a ; δa = H(s)δe c e ; H(s) = s + 4 The control is just basic proportional feedback δc = −kθ(θ − θc) e Which gives that θ = −Gθδe(s)H(s)kθ(θ − θc) or that θ(s) Gθδe(s)H(s)kθ = θc(s) 1 + Gθδe(s)H(s)kθ • Looks good, but how do we analyze what is going on? – Need to be able to predict where the poles are going as a function of kθ ⇒ Root Locus