Time Response of Linear, Time-Invariant (LTI) Systems Robert Stengel, Aircraft Flight Dynamics MAE 331, 2018

Transcription

1 Time Response of Linear, Time-Invariant LTI Systems Robert Stengel, Aircraft Flight Dynamics MAE 331, 2018 Learning Objectives Methods of time-domain analysis Continuous- and discrete-time models Transient response to initial conditions and inputs Steady-state equilibrium response Phase-plane plots Response to sinusoidal input Reading: Flight Dynamics , Airplane Stability and Control Sections Copyright 2018 by Robert Stengel. All rights reserved. For educational use only Linear, Time-Invariant LTI System Model Dynamic equation ordinary differential equation Δ!xt = FΔxt + GΔut + LΔwt, Δxt o given Output equation algebraic transformation Δyt = H x Δxt + H u Δut + H w Δwt State and output dimensions need not be the same [ ] = n 1 [ ] = r 1 2 dim Δxt dim Δyt 1

2 System Response to Inputs and Initial Conditions Solution of a linear dynamic model Δ!xt = FtΔxt + GtΔut + LtΔwt, t Δxt o given Δxt = Δxt o + [ Fτ Δxτ + Gτ Δuτ + Lτ Δwτ ]dτ t o... has two parts Unforced homogeneous response to initial conditions Forced response to control and disturbance inputs Initial condition response Step input response 3 Response to Initial Conditions 4 2

3 Unforced Response to Initial Conditions Neglecting forcing functions t Δxt = Δxt o + [ FΔxτ ]dτ = e F t t o Δxto = Φ t t o Δxt o t o The state transition matrix, Φ, propagates the state from t o to t by a single multiplication e F t t o = Matrix Exponential = I + F t t o = Φ t t o + 1 2! F t t o ! F t t 3 o +... = State Transition Matrix 5 Φ = I + F δt Initial-Condition Response via State Transition Δxt 1 = Φ t 1 t o Δxt o Δxt 2 = Φ t 2 t 1 Δxt 1 Δxt 3 = Φ t 3 t 2 Δxt ! F δt Incremental propagation of Δx If t k+1 t k = δt = constant, state transition matrix is constant ! F δt Δxt 1 = Φ δtδxt o = ΦΔxt o Δxt 2 = ΦΔxt 1 = Φ 2 Δxt o Δxt 3 = ΦΔxt 2 = Φ 3 Δxt o Propagation is exact 6 3

4 Discrete-Time Dynamic Model Response to continuous controls and disturbances t k+1 Δxt k+1 = Δxt k + [ FΔxτ + GΔuτ + LΔwτ ]dτ t k Response to piecewise-constant controls and disturbances x Δxt k+1 = Φ δtδxt k + Φ δt t k+1 e F τ t k = ΦΔxt k + ΓΔut k + ΛΔwt k dτ GΔut k + LΔwt k t k [ ] With piecewise-constant inputs, control and disturbance effects taken outside the integral Discrete-time model of continuous system = Sampled-data model 7 Sampled-Data Control- and Disturbance-Effect Matrices Δxt k = ΦΔxt k 1 + ΓΔut k 1 + ΛΔwt k 1 Γ = e Fδt IF 1 G = I 1 2! Fδt + 1 3! F2 δt 2 1 4! F3 δt Gδt Λ = e Fδt IF 1 L = I 1 2! Fδt + 1 3! F2 δt 2 1 4! F3 δt Lδt As δt becomes very small Φ I + Fδt δt 0 Γ Gδt δt 0 Λ Lδt 8 δt 0 4

5 Discrete-Time Response to Inputs Propagation of Δx, with constant Φ, Γ, and Λ Δxt 1 = ΦΔxt o + ΓΔut o + ΛΔwt o Δxt 2 = ΦΔxt 1 + ΓΔut 1 + ΛΔwt 1 Δxt 3 = ΦΔxt 2 + ΓΔut 2 + ΛΔwt 2 δt = t k +1 t k 9 Continuous- and Discrete-Time Short-Period System Matrices Continuous-time analog system " F = " G = " L = Sampled-data digital system δt = 0.01 s δt = t k +1 t k δt = 0.1 s Φ = Γ = Λ = Φ = Γ = Λ = δt = 0.5 s δt has a large effect on the digital model Φ = Γ = Λ =

6 Learjet 23 M N = 0.3, h N = 3,050 m V N = 98.4 m/s Δ!q t Δ!α t Continuous- and Discrete- Time Short-Period Models Differential Equations Produce State Rates of Change = Δq t Δα t Difference Equations Produce State Increments Δδ E t δt = 0.1sec Δq k +1 Δα k +1 = Δq k Δα k ΔδE k Note individual acceleration and difference sensitivities to state and control perturbations 11 Initial-Condition Response Δ!x 1 Δ!x 2 = FΔx t + ΔxGΔu t Δy 1 Δx = H x Δx t 1 + H u Δu Δδ E t = Δ!x t Δy t Δx 1 Δy 2 = Δx Δδ E Angle of Attack Initial Condition Short-Period Linear Model - Initial Condition F = [ ; ]; G = [-9.069;0]; Hx = [1 0;0 1]; sys = ssf,g,hx,0; xo = [1;0]; [y1,t1,x1] = initialsys, xo; xo = [2;0]; [y2,t2,x2] = initialsys, xo; plott1,y1,t2,y2, grid Pitch Rate Initial Condition figure xo = [0;1]; initialsys, xo, grid Doubling the initial condition doubles the output 12 6

7 Historical Factoids Commercial Aircraft of the 1940s Pre-WWII designs, reciprocating engines Development enhanced by military transport and bomber versions Douglas DC-4 adopted as C-54 Boeing Stratoliner 377 from B-29, C-97 Lockheed Constellation 749 from C Commercial Propeller-Driven Aircraft of the 1950s Reciprocating and turboprop engines Douglas DC-6, DC-7, Lockheed Starliner 1649, Vickers Viscount, Bristol Britannia, Lockheed Electra 188 Bristol Brabazon Jumbo Turboprop 14 7

8 Brabazon Committee study for a post-wwii jet-powered mailplane with small passenger compartment dehavilland Vampire, 1943 dehavilland Swallow, Commercial Jets of the 1950s Low-bypass ratio turbojet engines dehavilland DH 106 Comet st commercial jet transport engines buried in wings early takeoff accidents Boeing derived from prototype 1954 engines on pylons below wings largest aircraft of its time Sud-Aviation Caravelle st aircraft with twin aftmounted engines dehavilland Comet Boeing 707 Sud-Aviation Caravelle 16 8

9 Superposition of Linear Responses 17 Step Response Short-Period Linear Model - Step F = [ ; ]; G = [-9.069;0]; Hx = [1 0;0 1]; sys = ssf, -G, Hx,0; -1*Step sys2 = ssf, -2*G, Hx,0; -1*Step Step response stepsys, sys2, grid " Δ x 1 Δ x 2 " Δy 1 Δy 2 Δδ E t = Step Input 0, t < 0 1, t 0 = " " Δx Δx 2 + " ΔδE 0 = " 1 0 " Δx Δx 2 + " 0 ΔδE 0 Stability, speed of response, and damping are independent of the initial condition or input Doubling the input doubles the output 18 9

10 Superposition of Linear Step Responses Short-Period Linear Model - Superposition F = [ ; ]; G = [-9.069;0]; Hx = [1 0;0 1]; sys = ssf, -G, Hx,0; -1*Step xo = [1; 0]; t = [0:0.2:20]; u = ones1,lengtht; [y1,t1,x1] = [y2,t2,x2] = lsimsys,u,t,xo; lsimsys,u,t; u = zeros1,lengtht; [y3,t3,x3] = lsimsys,u,t,xo; plott1,y1,t2,y2,t3,y3, grid " Δ x 1 Δ x 2 = " " Δx Δx 2 + " ΔδE 0 " Δy 1 Δy 2 = " 1 0 " Δx Δx 2 + " 0 ΔδE 0 Stability, speed of response, and damping are independent of the initial condition or input 19 2 nd -Order Comparison: Continuous- and Discrete-Time LTI Longitudinal Models Differential Equations Produce State Rates of Change Phugoid Δ!V t Δ γ! t ΔV t Δγ t ΔδT t Short Period Δ!q t Δ!α t = Δq t Δα t Δδ E t 0 δt = 0.1sec Difference Equations Produce State Increments Phugoid ΔV k +1 Δγ k +1 = ΔV k Δγ k ΔδT k Short Period Δq k +1 Δα k +1 = Δq k Δα k ΔδE k 20 10

11 Equilibrium Response 21 Equilibrium Response Dynamic equation Δ xt = FΔxt + GΔut + LΔwt At equilibrium, the state is unchanging 0 = FΔxt + GΔut + LΔwt Constant values denoted by.* Δx* = F 1 GΔu * +LΔw * 22 11

12 Steady-State Condition If the system is also stable, an equilibrium point is a steady-state point, i.e., Small disturbances decay to the equilibrium condition 2 nd -order example System Matrices! F = " f 11 f 12 f 21 f 22 ; G =! g 1 g " 2 ; L =! l 1 l " 2 Equilibrium Response with Constant Inputs " Δx 1 * Δx 2 * " f 22 f 12 = f 21 f 11 " f 11 f 22 f 12 f 21 + * g 1 g 2,. Δu *+ l * l 2,. Δw * - Requirement for Stability si F = Δ s = s 2 + f 11 + f 22 s + f 11 f 22 f 12 f 21 = s λ 1 s λ 2 = 0 < 0 Re λ i 23 Equilibrium Response of Approximate Phugoid Model Equilibrium state with constant thrust and wind perturbations Δx P * = F 1 P G P Δu P * +L P Δw P * ΔV * Δγ * = 0 1 g V N L V V N D V gl V + -, -.- T δt L δt V N ΔδT * + D V L V V N ΔV W * /

13 Equilibrium Response of Approximate Phugoid Model ΔV * = L δt L V ΔδT * * + ΔV W Δγ * = 1 g T D δt + L V δt *ΔδT * L V Steady horizontal wind affects velocity but not flight path angle With L δt ~ 0, steady-state velocity perturbation depends only on the horizontal wind Constant thrust perturbation produces steady climb rate Corresponding dynamic response to thrust step, with L δt = 0 25 Equilibrium Response of Approximate Short-Period Model Equilibrium state with constant elevator and wind perturbations Δx SP * = F 1 SP G SP Δu SP * +L SP Δw SP * Δq * Δα * L α M V α N 1 M 1 = q 3 2 * L α - M V q + M 3 α +, N. / 43 M δ E L δ E V N ΔδE * M α L α V N Δα W *

14 Equilibrium Response of Approximate Short-Period Model L α M V δ E Δq * = N * ΔδE * L α M q + M α V N * M δ E Δα * * = ΔδE + Δα W L α M V q + M α N * with L δe = 0 Steady pitch rate and angle of attack response to elevator perturbation are not zero Steady vertical wind affects steady-state angle of attack but not pitch rate Dynamic response to elevator step with L δe = 0 27 Phase Plane Plots 28 14

15 A 2 nd -Order Dynamic Model Δ!x 1 Δ!x 2 = ω n 2ζω n Δx 1 Δx Δu 1 Δu 2 Δx 1 t : Displacement or Position Δx 2 t : Rate of change of Position ω n : Natural frequency, rad/s ζ : Damping ratio, - 29 State Phase Plane Plots " Δ x 1 Δ x 2 " ω n 2ζω n " Δx 1 Δx 2 + " 1 1 " 0 2 Δu 1 Δu 2 2nd-Order Model - Initial Condition Response clear z = 0.1; Damping ratio wn = 6.28; Natural frequency, rad/s F = [0 1;-wn^2-2*z*wn]; G = [1-1;0 2]; Hx = [1 0;0 1]; sys = ssf, G, Hx,0; t = [0:0.01:10]; xo = [1;0]; [y1,t1,x1] = initialsys, xo, t; plott1,y1 grid on figure ploty1:,1,y1:,2 grid on Cross-plot of one component against another Time is not shown explicitly 30 15

16 Dynamic Stability Changes the State-Plane Spiral Damping ratio = 0.1 Damping ratio = 0.3 Damping ratio = Scalar Frequency Response 32 16

17 Speed Control of Direct-Current Motor Angular Rate Control Law C = Control Gain ut = C et where et = y c t yt 33 Characteristics of the Motor Simplified Dynamic Model Rotary inertia, J, is the sum of motor and load inertias Internal damping neglected Output speed, yt, rad/s, is an integral of the control input, ut Motor control torque is proportional to ut Desired speed, y c t, rad/s, is constant Control gain, C, scales command-following error to motor input voltage 34 17

18 dyt dt = ut J = Cet J Model of Dynamics and Speed Control Dynamic equation = C [ J y c t yt ], y 0 given Integral of the equation, with y0 = 0 yt = 1 J t 0 utdt = C J t 0 etdt = C J t 0 [ y c t yt ]dt Direct integration of y c t Negative feedback of yt 35 Solution of the integral, with step command yt = y c * 1 e * " C t J Step Response of Speed Controller y c " 0, t < 0 t = 1, t 0 + -,- = y c 1 e λt +, = y c 1 e t τ + *,- where λ = C/J = eigenvalue or root of the system rad/s τ = J/C = time constant of the response sec 36 18

19 Angle Control of a DC Motor Control law with angle and angular rate feedback ut = c 1 [ y c t y 1 t] c 2 y 2 t Closed-loop dynamic equation, with yt = I 2 xt! " x 1 t x 2 t =! 0 1 c 1 / J c 2 / J "! " x 1 t x 2 t +! 0 c 1 / J " y c ω n = c 1 J ; ζ = c 2 J 2ω n 37 Step Response of Angle Controller, with Angle and Rate Feedback Single natural frequency, three damping ratios Step Response of Damped Angle Control ω n = c 1 J ; ζ = c 2 J 2ω n c 1 /J = 1 c 2 /J = 0, 1.414, F1 = [0 1;-1 0]; G1 = [0;1]; F1a = [0 1; ]; F1b = [0 1; ]; Hx = [1 0;0 1]; Sys1 Sys2 Sys3 = ssf1,g1,hx,0; = ssf1a,g1,hx,0; = ssf1b,g1,hx,0; stepsys1,sys2,sys

20 Angle Response to a Sinusoidal Angle Command y C t = y Cpeak sinωt Output wave lags behind the input wave Input and output amplitudes different Amplitude Ratio AR = y peak y Cpeak Phase Angle φ = 360 Δt peak Period, deg 39 Effect of Input Frequency on Output Amplitude and Phase Angle y c t = sin t / 6.28, deg ω n = 1 rad / s ζ = With low input frequency, input and output amplitudes are about the same Rate oscillation leads angle oscillation by ~90 Lag of angle output oscillation, compared to input, is small 40 20

21 At Higher Input Frequency, Phase Angle Lag Increases y c t = sin t, deg 41 At Even Higher Frequency, Amplitude Ratio Decreases and Phase Lag Increases y c t = sin 6.28t, deg 42 21

22 Angle and Rate Response of a DC Motor over Wide Input- Frequency Range Input Frequencies previous slides Very low damping Moderate damping High damping Long-term response of a dynamic system to sinusoidal inputs over a range of frequencies Determine experimentally from time response or Compute the Bode plot of the system s transfer functions TBD 43 Next Time: Transfer Functions and Frequency Response Reading: Flight Dynamics Learning Objectives Frequency domain view of initial condition response Response of dynamic systems to sinusoidal inputs Transfer functions Bode plots 44 22

23 Supplemental Material 45 Example: Aerodynamic Angle, Linear Velocity, and Angular Rate Perturbations Learjet 23 M N = 0.3, h N = 3,050 m V N = 98.4 m/s Aerodynamic angle and linear velocity perturbations Δα! Δw V N Δα = 1 Δw = = 1.7 m s Δβ! Δv V N Δβ = 1 Δv = = 1.7 m s Angular rate and linear velocity perturbations Δw wingtip = Δp b 2 Δw nose = Δq x nose x cm Δp = 1 / s = = 0.09 m s Δq = 1 / s [ ] = = 0.11m s Δr = 1 / s [ ] = = 0.11m s Δv nose = Δr x nose x cm 46 23

24 Continuous- and Discrete-Time Dutch-Roll Models Differential Equations Produce State Rates of Change Δ!r t Δ! β t Δr t Δβ t Difference Equations Produce State Increments Δδ R t 0 δt = 0.1sec Δr k +1 Δβ k Δr k Δβ k Δδ R k 47 Continuous- and Discrete-Time Roll-Spiral Models Δ!p t Δ!φ t Differential Equations Produce State Rates of Change Δp t Δφ t Difference Equations Produce State Increments Δδ A t δt = 0.1sec Δp k +1 Δφ k Δp k Δφ k Δδ A k 48 24

25 4 th - Order Comparison: Continuousand Discrete-Time Longitudinal Models Phugoid and Short Period Differential Equations Produce State Rates of Change Δ!V t Δ γ! t Δ!q t Δ!α t = ΔV t Δγ t Δq t Δα t ΔδT t Δδ E t Difference Equations Produce State Increments δt = 0.1sec ΔV k+1 Δγ k+1 Δq k+1 Δα k+1 = ΔV k Δγ k Δq k Δα k ΔδT k Δδ E k 49 25