Lecture #12 Aircraft Lateral Autopilots Multi-loop closure Heading Control: linear Heading control: nonlinear
Lecture # 12 Aircraft Lateral Autopilots • Multiloop closure • • Heading Control: linear Heading Control: nonlinear
Fa2004 16.33310-1 Lateral Autopilots We can stabilize/ modify the lateral dynamics using a variety of dif- ferent feedback architectures o Look for good sensor/ actuator pairings to achieve desired behavior e Example: Yaw damper Can improve the damping on the dutch-roll mode by adding a feedback on r to the rudder df=kr(rc-r) Servo dynamics Hr=3. 32added to rudder 8a=H. c 1618s3+0.7761s2+0.03007s+0.1883 +3.967s4+3.06s3+3.642s2+1.71s+0.01223 Lateral autopilot: r to rudder 050340.16 × Per Ave Figure 2: Lateral pole-zero map gser
Fall 2004 16.333 10–1 Lateral Autopilots • We can stabilize/modify the lateral dynamics using a variety of different feedback architectures. δa - -p 1 - φ s δr - Glat(s) -r 1 - ψ s • Look for good sensor/actuator pairings to achieve desired behavior. • Example: Yaw damper – Can improve the damping on the Dutchroll mode by adding a feedback on r to the rudder: δc = kr(rc − r) r 3.33 – Servo dynamics Hr = s+3.33 added to rudder δa = Hrδc r r – System: 1.618s3 + 0.7761s2 + 0.03007s + 0.1883 Gδr cr = −s5 + 3.967s4 + 3.06s3 + 3.642s2 + 1.71s + 0.01223 −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 0.94 0.86 0.76 0.64 0.5 0.34 0.16 3.5 0.76 0.985 0.64 0.5 0.34 0.16 1 0.86 0.94 0.985 3 2.5 2 1.5 0.5 Lateral autopilot: r to rudder Real Axis Imaginary Axis c r Figure 2: Lateral polezero map Gδ r
Fa2004 16.33310-2 Note that the gain of the plant is negative(Kplant0, so must draw a 180 locus(neg feedback ateral autopilot: r to8 with k>0 Lateral autopilot: r to 8 with k<0 Figure 3: Lateral pole-zero map. Definitely need kr <0 Root locus with hr <0 looks pretty good as we have authority over the four poles kr=-1.6 results in a large increase in the dutch-roll damping and spiral/roll modes have combined into a damped oscillation Yaw damper looks great, but this implementation has a problem There are various flight modes that require a steady yaw rate (ssf). For example, steady turning flight Our current yaw damper would not allow this to happen -it would create the rudder inputs necessary to cancel out the motion Exact opposite of what we want to have happen, which is to damp out any oscillations about the steady turn
� Fall 2004 16.333 10–2 • Note that the gain of the plant is negative (Kplant 0, so must draw a 180◦ locus (neg feedback) −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 1 0.5 0.25 0.25 0.25 0.75 0.5 0.5 1 0.75 0.25 0.5 Lateral autopilot: r to δ r with k>0 Real Axis Imaginary Axis −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 1 0.25 0.5 0.25 0.5 0.75 0.25 0.5 0.75 Lateral autopilot: r to δ r with k<0 Real Axis Imaginary Axis Figure 3: Lateral polezero map. Definitely need kr < 0 • Root locus with kr < 0 looks pretty good as we have authority over the four poles. – kr = −1.6 results in a large increase in the Dutchroll damping and spiral/roll modes have combined into a damped oscillation. • Yaw damper looks great, but this implementation has a problem. – There are various flight modes that require a steady yaw rate (rss = 0). For example, steady turning flight. – Our current yaw damper would not allow this to happen – it would create the rudder inputs necessary to cancel out the motion !! – Exact opposite of what we want to have happen, which is to damp out any oscillations about the steady turn
Fa2004 16.33310-3 Yaw Damper Part 2 Can avoid this problem to some extent by filtering the feed back signal Feedback only a high pass version of the r signal High pass cuts out the low frequency content in the signal steady state value of r would not be fed back to the controller New yaw damper: 8c=kr(rc- Hu(sr)where Hu(s)=Tsi is the washout filt Washout filter with a4_2 Figure 4: Washout filter with T= 4. 2 New control picture k H2(s)
Fall 2004 16.333 10–3 Yaw Damper: Part 2 • Can avoid this problem to some extent by filtering the feedback signal. – Feedback only a high pass version of the r signal. – High pass cuts out the low frequency content in the signal ⇒ steady state value of r would not be fed back to the controller. • New yaw damper: δc = kr(rc − Hw(s)r) where Hw(s) = τs is the r τs+1 “washout” filter. 10−2 10−1 100 101 10−2 10−1 100 Washout filter with τ=4.2 |H w(s)| Freq (rad/sec) Figure 4: Washout filter with τ = 4.2 • New control picture δa p 1 φ - - - s Glat(s) δc rc r r 1 ψ - - H - - - kr r(s) – 6 s H w(s)
Fa2004 16.33310-4 Lateral autopilot: r to rudder WITH washout filter 025 0.75 O}0 0.5 Figure 5: Root Locus with the washout filter included Zero in Hu(s) traps a pole near the origin, but it is slow enough that it can be controlled by the pilot · Obviously has changed the closed loop pole locations(◆→√),but kr=-1.6 still seems to give a well damped response
Fall 2004 16.333 10–4 −2 −1.5 −1 −0.5 0 0.5 1 −1.5 −1 −0.5 0 0.5 1 1.5 0.5 0.25 0.5 0.75 0.75 0.25 1 0.5 0.25 Lateral autopilot: r to rudder WITH washout filter Real Axis Imaginary Axis Figure 5: Root Locus with the washout filter included. • Zero in Hw(s) traps a pole near the origin, but it is slow enough that it can be controlled by the pilot. • Obviously has changed the closed loop pole locations (� ⇒ �), but kr = −1.6 still seems to give a well damped response
Fa2004 16.333105 E Without Washout With Washout 0.5 15 Time Without Washout 15 Time Figure 6: Impulse response of closed loop system with and without the Washout filter(T= 4.2). Commanded Tc=0, and both have(8)ss =0, but without the filter, Tss=0, whereas with it, Tss #0 For direct comparison with and without the filter, applied impulse as re to both closed-loop systems and then calculated r and dr Bottom plot shows that control signal quickly converges to zero in both cases, i.e., no more control effort is being applied to correct the motion But only the one with the washout filter produces a zero control input even though the there is a steady turn=, the controller will not try to fight a commanded steady turn
� Fall 2004 16.333 10–5 0 5 10 15 20 25 30 −0.5 0 0.5 1 Time Response r Without Washout With Washout 0 5 10 15 20 25 30 −0.5 0 0.5 1 Time Control δ r Without Washout With Washout Figure 6: Impulse response of closed loop system with and without the Washout filter (τ = 4.2). Commanded rc = 0, and both have (δr)ss = 0, but without the filter, rss = 0, whereas with it, rss = 0. • For direct comparison with and without the filter, applied impulse as rc to both closedloop systems and then calculated r and δr. • Bottom plot shows that control signal quickly converges to zero in both cases, i.e., no more control effort is being applied to correct the motion. • But only the one with the washout filter produces a zero control input even though the there is a steady turn ⇒ the controller will not try to fight a commanded steady turn
Fa2004 16.33310-6 Heading Autopilot Design So now have the yaw damper added correctly and want to control the heading yl Need to bank the aircraft to accomplish this Result is a "coordinated turn"with angular rate y FRoA◆HW R PATH oF U AlC R Aircraft banked to angle so that vector sum of mg and mvo l is along the body z-aXIs ng in the body y-axis direction gives muo? cos o= mg sin g tan Since typically 1, we have gives the desired bank angle for a specified turn rate
Fall 2004 16.333 10–6 Heading Autopilot Design • So now have the yaw damper added correctly and want to control the heading ψ. – Need to bank the aircraft to accomplish this. – Result is a “coordinated turn” with angular rate ψ˙ Aircraft banked to angle φ so that vector sum of mg and mU0ψ˙ • is along the body zaxis – Summing in the body yaxis direction gives mu0ψ˙ cos φ = mg sin φ U0ψ˙ tan φ = g • Since typically φ � 1, we have U0ψ˙ φ ≈ g gives the desired bank angle for a specified turn rate
Fa2004 16.33310-7 Problem now is that yb tends to be a noisy signal to base out bank angle on, so we generate a smoother signal by filtering it Assume that the desired heading is known vd and we want y to follow v,d relatively slowly Choose dynamics T1 +yb=yd yd T1s+ with T1=15-20sec depending on the vehicle and the goals A low pass filter that eliminates the higher frequency noise Filtered heading angle satisfies which we can use to create the desired bank angle 719
Fall 2004 16.333 10–7 Problem now is that ψ˙ • tends to be a noisy signal to base out bank angle on, so we generate a smoother signal by filtering it. – Assume that the desired heading is known ψd and we want ψ to follow ψd relatively slowly – Choose dynamics τ1ψ˙ + ψ = ψd ψ 1 ⇒ = ψd τ1s + 1 with τ1=1520sec depending on the vehicle and the goals. – A low pass filter that eliminates the higher frequency noise. • Filtered heading angle satisfies 1 ψ˙ = (ψd − ψ) τ1 which we can use to create the desired bank angle: U0 φd = ψ˙ = U0 (ψd − ψ) g τ1g
Fa2004 16.333108 Roll Control e Given this desired bank angle we need a roll controller to ensure that the vehicle tracks it accurately Aileron is best actuator to use: Ba= ho(od -o)-kpp To design hio and kp, can just use the approximation of the roll mode cp= Lpp o=p which gives L For the design, add the aileron servo dynamics Ha(s) 0.15s+1 8a= ha(s8 ● PD controller ko(sy +1)+kood, adds zero at s=-1/ I-Pick y=2/3 0.120.06 s Axe igure 7: Root Locus for roll loop -closed Loop poles for Kp=-20, Ko=-30
−7 −6 −5 −4 −3 −2 −1 0 −10 −8 −6 −4 −2 0 2 4 6 8 10 0.88 0.68 0.52 0.38 0.28 0.2 0.12 0.06 0.68 0.38 0.28 0.2 0.12 0.06 8 0.88 10 0.52 6 2 4 6 8 10 2 4 Root Locus Real Axis Imaginary Axis � Fall 2004 16.333 10–8 Roll Control • Given this desired bank angle, we need a roll controller to ensure that the vehicle tracks it accurately. – Aileron is best actuator to use: δa = kφ(φd − φ) − kpp • To design kφ and kp, can just use the approximation of the roll mode I� I� ¨ ˙ xxp˙ = Lpp + Lδaδa xxφ − Lpφ˙ = Lδaδa φ = p which gives φ Lδa = δa s(I� xxs − Lp) • For the design, add the aileron servo dynamics 1 Ha(s) = , δa = Ha(s)δc a 0.15s + 1 a • PD controller δc a = −kφ(sγ + 1) + kφφd, adds zero at s = −1/γ – Pick γ = 2/3 Figure 7: Root Locus for roll loop – closed Loop poles for Kp = −20, Kφ = −30
Fa2004 16.33310-9 Reponse to initial roll of 15 degs Reponse to initial roll of 15 degs Figure 8: Roll response to an initial roll offset
Fall 2004 16.333 10–9 0 1 2 3 4 5 6 7 8 9 10 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 Reponse to initial roll of 15 degs φ p 0 1 2 3 4 5 6 7 8 9 10 −2 0 2 4 6 8 10 x 10−3 Reponse to initial roll of 15 degs ψ r β Figure 8: Roll response to an initial roll offset
