《航空器的稳定与控制》（英文版）Lecture 9 Basic Longitudinal Control

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 be￾havior 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

Fa2004 16.33385 Root locus basics U Gc(s o Assume that the plant transfer function is of the form S-2 G=K K S-p and the controller transfer function is Ge(s)=k=kIl(s-zci Signals are u control commands y output/measurements r reference input e response error This is the unity feedback form. We could add the controller gc in the feedback path without changing the pole locations Will add disturbances later

� � Fall 2004 16.333 8–5 Root Locus Basics r e u y /OO / Gc(s) / Gp(s) / − • Assume that the plant transfer function is of the form Gp = Kp Np = Kp� i (s − zpi) Dp i (s − ppi) and the controller transfer function is Nc Gc(s) = Kc = Kc� i (s − zci) Dc i (s − pci) • Signals are: u control commands y output/measurements r reference input e response error • This is the unity feedback form. We could add the controller Gc in the feedback path without changing the pole locations. • Will add disturbances later

Fa2004 16.33386 Basic questions Analysis: Given N and De, where do the closed loop poles go as a function of k Synthesis: Given Kp, Np and Dp, how should we chose Kc, to put the closed loop poles in the desired locations? Block diagram analysis: Since y=G Gce and e=r-g, then easy to show that S r 1+gg vhere c(s KcKpNNp DDp+ KcKpNCNp is the closed loop transfer function The denominator is called the characteristic equation oc(s)and the roots of s)=0 are called the closed-loop poles(CLp) The CLp are clearly functions of Kc for a given Kp, Np, Dp, Nc, Dc a" locus of roots"[Evans, 1948

Fall 2004 16.333 8–6 • Basic questions: – Analysis: Given Nc and Dc, where do the closed loop poles go as a function of Kc? – Synthesis: Given Kp, Np and Dp, how should we chose Kc, Nc, Dc to put the closed loop poles in the desired locations? • Block diagram analysis: Since y = GpGce and e = r − y, then easy to show that y GcGp =r 1 + GcGp ≡ Gcl(s) where KcKpNcNp Gcl(s) = DcDp + KcKpNcNp is the closed loop transfer function • The denominator is called the characteristic equation φc(s) and the roots of φc(s) = 0 are called the closed­loop poles (CLP) . • The CLP are clearly functions of Kc for a given Kp, Np, Dp, Nc, Dc ⇒ a “locus of roots” [Evans, 1948]

Fa2004 16.3338-7 Root Locus Analysis General root locus is hard to determine by hand and requires matlab tools such as locus (num, den) to obtain full result but we can get some important insights by developing a short set of plotting rules Full rules in FPE, page 260 e basic questions 1. What points are on the root locus? 2. Where does the root locus start? 3. Where does the root locus end? 4. When/ where is the locus on the real line? 5. Given that so is found to be on the locus, what gain is need for that to become the closed-loop pole location? Question #1: is point so on the root locus? Assume that Ne and De are known let DD and K=Kkn then (s)=1+KLd(s)=0 or equivalently for values of s for which La(s)=-1/K, with K real For K positive, So is on the root locus if ∠L(50)=180°±l·360°,l=0.,1 If K negative, so is on the root locus if [0 locus] ∠L(s0)=0°±·360°,7=0,1 These are known as the phase conditions

Fall 2004 16.333 8–7 Root Locus Analysis • General root locus is hard to determine by hand and requires Matlab tools such as rlocus(num,den) to obtain full result, but we can get some important insights by developing a short set of plotting rules. – Full rules in FPE, page 260. • Basic questions: 1. What points are on the root locus? 2. Where does the root locus start? 3. Where does the root locus end? 4. When/where is the locus on the real line? 5. Given that s0 is found to be on the locus, what gain is need for that to become the closed­loop pole location? • Question #1: is point s0 on the root locus? Assume that Nc and Dc are known, let Ld = Nc Dc Np Dp and K = KcKp then φc(s) = 1 + KLd(s) = 0 or equivalently for values of s for which Ld(s) = −1/K, with K real. – For K positive, s0 is on the root locus if ∠Ld(s0) = 180◦ ± l · 360◦ , l = 0, 1, . . . – If K negative, s0 is on the root locus if [0◦ locus] ∠Ld(s0) = 0◦ ± l · 360◦ , l = 0, 1, . . . These are known as the phase conditions

Fa2004 16.33388 Question#2: Where does the root locus start? Q=1+K D. →DD+ KN.N=0 So if K-0. then locus starts at solutions of dd,=0 which are the poles of the plant and compensator Question #3: Where does the root locus end? Already shown that for so to be on the locus, must have La(so) K So if K→∞, the poles must satisfy NN Ld DD There are several possibilities 1. Poles are located at values of s for which NN=0, which are the zeros of the plant and the compensator 2. If Loop la( s) has more poles than zeros As|s→∞,|La(s)→0, but we must ensure that the phase condition is still satisfied

Fall 2004 16.333 8–8 • Question #2: Where does the root locus start? NcNp φc = 1 + K = 0 DcDp ⇒ DcDp + KNcNp = 0 So if K → 0, then locus starts at solutions of DcDp = 0 which are the poles of the plant and compensator. • Question #3: Where does the root locus end? Already shown that for s0 to be on the locus, must have 1 Ld(s0) = −K So if K → ∞, the poles must satisfy: NcNp Ld = = 0 DcDp • There are several possibilities: 1. Poles are located at values of s for which NcNp = 0, which are the zeros of the plant and the compensator 2. If Loop Ld(s) has more poles than zeros – As | | s → ∞ | , Ld(s)| → 0, but we must ensure that the phase condition is still satisfied

Fa2004 16.3338-9 More details as k→: Assume there are n zeros and p poles of la(s) Then for large s d(s So the root locus degenerates to 1+ (s-a)-n0 So n poles head to the zeros of la(s) Remaining p-n poles head to s= oo along asymptotes defined by the radial lines 1809+3600·(-1) P-7 so that the number of asymptotes is governed by the number of poles compared to the number of zeros(relative degree) If zi are the zeros if ld and p; are the poles, then the centroid of the asymptotes is given by ∑n-∑ Exampl pIe: L(S)

� � Fall 2004 16.333 8–9 • More details as K → ∞: – Assume there are n zeros and p poles of Ld(s) – Then for large |s|, 1 Ld(s) ≈ (s − α)p−n – So the root locus degenerates to: 1 1 + = 0 n (s − α)p− – So n poles head to the zeros of Ld(s) – Remaining p − n poles head to | | s = ∞ along asymptotes defined by the radial lines 180◦ + 360◦ · (l − 1) φl = l = 1, 2, . . . p − n so that the number of asymptotes is governed by the number of poles compared to the number of zeros (relative degree). – If zi are the zeros if Ld and pj are the poles, then the centroid of the asymptotes is given by: p n pj − zi α = p − n • Example: L(s) = s−4 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 Root Locus Real Axis Imaginary Axis