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 298-313, 338-342 Airplane Stability and Control Sections 11.1-11.12 Copyright 2018 by Robert Stengel. All rights reserved. For educational use only. http://www.princeton.edu/~stengel/mae331.html http://www.princeton.edu/~stengel/flightdynamics.html 1 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
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
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 2 1 + 3! 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 2 + 1 2! F δt Incremental propagation of Δx If t k+1 t k = δt = constant, state transition matrix is constant 2 + 1 3! F δt 3 +... 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
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 3 +... Gδt Λ = e Fδt IF 1 L = I 1 2! Fδt + 1 3! F2 δt 2 1 4! F3 δt 3 +... Lδt As δt becomes very small Φ I + Fδt δt 0 Γ Gδt δt 0 Λ Lδt 8 δt 0 4
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 = 1.2794 7.9856 1 1.2709 9.069 0 7.9856 1.2709 Sampled-data digital system δt = 0.01 s δt = t k +1 t k δt = 0.1 s Φ = Γ = Λ = Φ = Γ = Λ = 0.987 0.079 0.01 0.987 0.09 0.0004 0.079 0.013 0.845 0.694 0.0869 0.846 0.84 0.0414 0.694 0.154 δt = 0.5 s δt has a large effect on the digital model Φ = Γ = Λ = 0.0823 1.475 0.185 0.0839 2.492 0.643 1.475 0.916 10 5
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 = 1.3 8 1 1.3 Δq t Δα t Difference Equations Produce State Increments + 9.1 0 Δδ E t δt = 0.1sec Δq k +1 Δα k +1 = 0.85 0.7 0.09 0.85 Δq k Δα k + 0.84 0.04 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 2 + 9.069 0 Δy 1 Δx = H x Δx t 1 + H u Δu Δδ E t = 1.2794 7.9856 Δ!x t1 1.2709 Δy t Δx 1 Δy 2 = 1 0 0 1 Δx 2 + 0 0 Δδ E Angle of Attack Initial Condition Short-Period Linear Model - Initial Condition F = [-1.2794-7.9856;1. -1.2709]; 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
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-69 13 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
Brabazon Committee study for a post-wwii jet-powered mailplane with small passenger compartment dehavilland Vampire, 1943 dehavilland Swallow, 1946 15 Commercial Jets of the 1950s Low-bypass ratio turbojet engines dehavilland DH 106 Comet 1951 1 st commercial jet transport engines buried in wings early takeoff accidents Boeing 707 1957 derived from 367-80 prototype 1954 engines on pylons below wings largest aircraft of its time Sud-Aviation Caravelle 1959 1 st aircraft with twin aftmounted engines https://www.youtube.com/watch?v=2bvhov0nxpq dehavilland Comet Boeing 707 Sud-Aviation Caravelle 16 8
Superposition of Linear Responses 17 Step Response Short-Period Linear Model - Step F = [-1.2794-7.9856;1. -1.2709]; 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 = " 1.2794 7.9856 " Δx 1 1 1.2709 Δx 2 + " 9.069 ΔδE 0 = " 1 0 " Δx 1 0 1 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
Superposition of Linear Step Responses Short-Period Linear Model - Superposition F = [-1.2794-7.9856;1. -1.2709]; 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 = " 1.2794 7.9856 " Δx 1 1 1.2709 Δx 2 + " 9.069 ΔδE 0 " Δy 1 Δy 2 = " 1 0 " Δx 1 0 1 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 0.02 9.8 0.02 0 ΔV t Δγ t + 4.7 0 ΔδT t Short Period Δ!q t Δ!α t = 1.3 8 Δq t 1 1.3 Δα t + 9.1 Δδ E t 0 δt = 0.1sec Difference Equations Produce State Increments Phugoid ΔV k +1 Δγ k +1 = 1 0.98 0.002 1 ΔV k Δγ k + 0.47 0.0005 ΔδT k Short Period Δq k +1 Δα k +1 = 0.85 0.7 0.09 0.85 Δq k Δα k + 0.84 0.04 ΔδE k 20 10
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
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 1 + - * 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 * / - 0-1- 24 12
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 * 5 3 6 3 73 26 13
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
A 2 nd -Order Dynamic Model Δ!x 1 Δ!x 2 = 0 1 2 ω n 2ζω n Δx 1 Δx 2 + 1 1 0 2 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 " 0 1 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
Dynamic Stability Changes the State-Plane Spiral Damping ratio = 0.1 Damping ratio = 0.3 Damping ratio = 0.1 31 Scalar Frequency Response 32 16
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
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
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, 2.828 F1 = [0 1;-1 0]; G1 = [0;1]; F1a = [0 1;-1-1.414]; F1b = [0 1;-1-2.828]; Hx = [1 0;0 1]; Sys1 Sys2 Sys3 = ssf1,g1,hx,0; = ssf1a,g1,hx,0; = ssf1b,g1,hx,0; stepsys1,sys2,sys3 38 19
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 = 0.707 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
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
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 342-357 Learning Objectives Frequency domain view of initial condition response Response of dynamic systems to sinusoidal inputs Transfer functions Bode plots 44 22
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 = 0.01745 98.4 = 1.7 m s Δβ! Δv V N Δβ = 1 Δv = 0.01745 98.4 = 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 = 0.01745 5.25 = 0.09 m s Δq = 1 / s [ ] = 0.01745 6.4 = 0.11m s Δr = 1 / s [ ] = 0.01745 6.4 = 0.11m s Δv nose = Δr x nose x cm 46 23
Continuous- and Discrete-Time Dutch-Roll Models Differential Equations Produce State Rates of Change Δ!r t Δ! β t 0.11 1.9 1 0.16 Δr t Δβ t Difference Equations Produce State Increments + 1.1 Δδ R t 0 δt = 0.1sec Δr k +1 Δβ k +1 0.98 0.19 0.1 0.97 Δr k Δβ k + 0.11 0.01 Δδ R k 47 Continuous- and Discrete-Time Roll-Spiral Models Δ!p t Δ!φ t Differential Equations Produce State Rates of Change 1.2 0 1 0 Δp t Δφ t Difference Equations Produce State Increments + 2.3 0 Δδ A t δt = 0.1sec Δp k +1 Δφ k +1 0.89 0 0.09 1 Δp k Δφ k + 0.24 0.01 Δδ A k 48 24
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 = 0.02 9.8 0 0 0.02 0 0 1.3 0 0 1.3 8 0.02 0 1 1.3 ΔV t Δγ t Δq t Δα t + 4.7 0 0 0 0 9.1 0 0 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 = 1 0.98 0.002 0.06 0.002 1 0.006 0.12 0.0001 0 0.84 0.69 0.002 0.0001 0.09 0.84 ΔV k Δγ k Δq k Δα k + 0.47 0.0005 0.0005 0.002 0 0.84 0 0.04 ΔδT k Δδ E k 49 25