Time ResponseRobert Stengel, Aircraft Flight Dynamics

MAE 331, 2010

Time-domain analysis

Transient response to initial conditions and inputs

Steady-state (equilibrium) response

Continuous- and discrete-time models

Phase-plane plots

Response to sinusoidal input

Copyright 2010 by Robert Stengel. All rights reserved. For educational use only.http://www.princeton.edu/~stengel/MAE331.html

http://www.princeton.edu/~stengel/FlightDynamics.html

Linear, Time-Invariant (LTI)Longitudinal Model

! !V (t)

! !" (t)

! !q(t)

! !#(t)

$

%

&&&&&

'

(

)))))

=

*DV

*g *Dq

*D#

LV

VN

0Lq

VN

L#VN

MV

0 Mq

M#

* LVVN

0 1 * L#VN

$

%

&&&&&&&&

'

(

))))))))

!V (t)

!" (t)

!q(t)

!#(t)

$

%

&&&&&

'

(

)))))

+

0 T+T 0

0 0 L+F /VN

M+E 0 0

0 0 *L+F /VN

$

%

&&&&&

'

(

)))))

!+E(t)

!+T (t)

!+F(t)

$

%

&&&

'

(

)))

Steady, level flight

Simplified control effects

Neglect disturbance effects

What can we do with it? Integrate equations to obtain time histories of initial condition, control, and

disturbance effects

Determine modes of motion

Examine steady-state conditions

Identify effects of parameter variations

Define frequency response

Gain insights aboutsystem dynamics

Linear, Time-Invariant

System Model

General model contains Dynamic equation (ordinary differential equation)

Output equation (algebraic transformation)

! x (t) = F!x(t) + G!u(t) + L!w(t), !x(to) given

!y(t) = Hx!x(t) + Hu!u(t) + Hw!w(t)

State and output dimensions need not be the same

dim !x(t)[ ] = n "1( )

dim !y(t)[ ] = r "1( )

System Response to Inputsand Initial Conditions

Solution of the linear, time-invariant (LTI) dynamic model

!!x(t) = F!x(t) +G!u(t) + L!w(t), !x(to ) given

!x(t) = !x(to ) + F!x(" ) +G!u(" ) + L!w(" )[ ]to

t

# d"

... has two parts

Unforced (homogeneous) response to initial conditions

Forced response to control and disturbance inputs

Response toInitial Conditions

Unforced Response

to Initial Conditions

The state transition matrix, !, propagates thestate from to to t by a single multiplication

!x(t) = !x(to) + F!x(" )[ ]

to

t

# d" = eF t$ to( )!x(t

o) = % t $ t

o( )!x(to )

eF t! to( ) = Matrix Exponential

= I + F t ! to( ) +1

2!F t ! to( )"# $%

2

+1

3!F t ! to( )"# $%

3

+ ...

= & t ! to( ) = State Transition Matrix

Neglecting forcing functions

Initial-Condition Response

via State Transition

! = I + F "t( ) +1

2!F "t( )#$ %&

2

+1

3!F "t( )#$ %&

3

+ ...

!x(t1) = " t

1# t

o( )!x(to )

!x(t2) = " t

2# t

1( )!x(t1)

!x(t3) = " t

3# t

2( )!x(t2 )

If (tk+1 tk) = !t = constant,state transition matrix isconstant

!x(t1) = " #t( )!x(t

o) = "!x(t

o)

!x(t2) = "!x(t

1) = "

2!x(t

o)

!x(t3) = "!x(t

2) = "

3!x(t

o)

Propagation of "x

Discrete-Time Dynamic Model

!x(tk+1) = !x(t

k) + F!x(" ) +G!u(" ) + L!w(" )[ ]

tk

tk+1

# d"

!x(tk+1) = " #t( )!x(tk ) +" #t( ) e

$F % $ tk( )&'

()

tk

tk+1

* d% G!u(tk ) + L!w(tk )[ ]

= "!x(tk) + +!u(t

k) + ,!w(t

k)

Response to continuous controls and disturbances

Response to piecewise-constant controls and disturbances

! = eF" t

# = eF" t$ I( )F$1G

% = eF" t$ I( )F$1L

Ordinary Difference Equation

Control- and Disturbance-Effect

Matrices

! = eF" t # I( )F#1G

= I #1

2!F"t +

1

3!F2"t 2 #

1

4!F3"t 3 + ...$

%&'()G"t

* = eF" t # I( )F#1L

= I #1

2!F"t +

1

3!F2"t 2 #

1

4!F3"t 3 + ...$

%&'()L"t

!x(tk) = "!x(t

k#1) + $!u(t

k#1) + %!w(t

k#1)

As !t becomes very small

!" t#0

$ #$$ I + F"t( )

%" t#0

$ #$$ G"t

&" t#0

$ #$$ L"t

Discrete-Time Response to Inputs

!x(t1) = "!x(t

o) + #!u(t

o) + $!w(t

o)

!x(t2) = "!x(t

1) + #!u(t

1) + $!w(t

1)

!x(t3) = "!x(t

2) + #!u(t

2) + $!w(t

2)

!

Propagation of "x, with constant !, #, and $

!t = tk+1

" tk

Continuous- and Discrete-Time

Short-Period System Matrices

!t = 0.1 s

!t = 0.5 s

F =!1.2794 !7.9856

1 !1.2709

"

#$

%

&'

G =!9.069

0

"

#$

%

&'

L =!7.9856

!1.2709

"

#$

%

&'

! =0.845 "0.694

0.0869 0.846

#

$%

&

'(

) ="0.84

"0.0414

#

$%

&

'(

* ="0.694

"0.154

#

$%

&

'(

! =0.0823 "1.475

0.185 0.0839

#

$%

&

'(

) ="2.492

"0.643

#

$%

&

'(

* ="1.475

"0.916

#

$%

&

'(

Continuous-time(analog) system

Discrete-time (digital) system

!t has a large effecton the digital model

!t = tk+1

" tk

! =0.987 "0.079

0.01 0.987

#

$%

&

'(

) ="0.09

"0.0004

#

$%

&

'(

* ="0.079

"0.013

#

$%

&

'(

!t = 0.01 s

Example: Aerodynamic

Angle, Linear Velocity, and

Angular Rate Perturbations

Learjet 23

MN = 0.3, hN = 3,050 m

VN = 98.4 m/s

!" ! !wVN

; !" = 1# !w = 0.01745 $ 98.4 = 1.7m s

!% ! !vVN

; !% = 1# !v = 0.01745 $ 98.4 = 1.7m s

!p = 1 / s; !wwingtip = !pb2

"#

$% = 0.01745 & 5.25 = 0.09m s

!q = 1 / s; !wnose = !q xnose ' xcm[ ] = 0.01745 & 6.4 = 0.11m s

!r = 1 / s; !vnose = !r xnose ' xcm[ ] = 0.01745 & 6.4 = 0.11m s

Aerodynamic angle and linear velocity perturbations

Angular rate and linear velocity perturbations

Example: Continuous- andDiscrete-Time Models

! !q

! !"

#

$%%

&

'((=

)1.3 )8

1 )1.3

#

$%

&

'(

!q

!"

#

$%%

&

'((+

)9.1

0

#

$%

&

'(!*E

!!p

! !"

#

$%%

&

'(() *1.2 0

1 0

#

$%

&

'(

!p

!"

#

$%%

&

'((+

2.3

0

#

$%

&

'(!+A

!!r

! !"

#

$%%

&

'(() *0.11 1.9

*1 *0.16

#

$%

&

'(

!r!"

#

$%%

&

'((+

*1.10

#

$%

&

'(!+R

Note individual acceleration and difference sensitivities to state and control perturbations

Short Period

Roll-Spiral

DutchRoll

!qk+1

!" k+1

#

$

%%

&

'

((=

0.85 )0.7

0.09 0.85

#

$%

&

'(

!qk

!" k

#

$

%%

&

'

((+

)0.84

)0.04

#

$%

&

'(!*Ek

!pk+1!"k+1

#

$%%

&

'(() 0.89 0

0.09 1

#

$%

&

'(

!pk!"k

#

$%%

&

'((+

0.24

*0.01

#

$%

&

'(!+Ak

!rk+1

!"k+1

#

$%%

&

'(() 0.98 0.19

*0.1 0.97

#

$%

&

'(

!rk

!"k

#

$%%

&

'((+

*0.110.01

#

$%

&

'(!+Rk

Differential Equations Produce

State Rates of Change

Difference Equations

Produce State Increments

Learjet 23

MN = 0.3, hN = 3,050 m

VN = 98.4 m/s !t = 0.1sec

Initial-Condition Response

Doubling the initial condition doubles the output

!!x1

!!x2

"

#

$$

%

&

''=

(1.2794 (7.9856

1 (1.2709

"

#$

%

&'

!x1

!x2

"

#

$$

%

&

''+

(9.069

0

"

#$

%

&'!)E

!y1

!y2

"

#

$$

%

&

''=

1 0

0 1

"

#$

%

&'

!x1

!x2

"

#

$$

%

&

''+

0

0

"

#$

%

&'!)E

% Short-Period Linear Model - Initial Condition

F = [-1.2794 -7.9856;1. -1.2709];

G = [-9.069;0];

Hx = [1 0;0 1];

sys = ss(F, G, Hx,0);

xo = [1;0];

[y1,t1,x1] = initial(sys, xo);

xo = [2;0];

[y2,t2,x2] = initial(sys, xo);

plot(t1,y1,t2,y2), grid

figure

xo = [0;1];

initial(sys, xo), grid

Angle of AttackInitial Condition

Pitch RateInitial Condition

Phase Plane Plots

State (Phase)-Plane Plots

Cross-plot of one component againstanother

Time or frequency not shown explicitly

% 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 = ss(F, G, Hx,0);

t = [0:0.01:10];

xo = [1;0];

[y1,t1,x1] = initial(sys, xo, t);

plot(t1,y1)

grid on

figure

plot(y1(:,1),y1(:,2))

grid on

!!x1

!!x2

"

#$$

%

&''(

0 1

)*n

2 )2+*n

"

#$$

%

&''

!x1

!x2

"

#$$

%

&''+

1 )10 2

"

#$

%

&'

!u1

!u2

"

#$$

%

&''

Dynamic Stability Changesthe State-Plane Spiral

Damping ratio = 0.1 Damping ratio = 0.3 Damping ratio = 0.1

Superposition ofLinear Responses

Step Response

Stability, speed of response,and damping areindependent of the initialcondition or input

Doubling the inputdoubles the output

!!x1

!!x2

"

#

$$

%

&

''=

(1.2794 (7.9856

1 (1.2709

"

#$

%

&'

!x1

!x2

"

#

$$

%

&

''+

(9.069

0

"

#$

%

&'!)E

!y1

!y2

"

#

$$

%

&

''=

1 0

0 1

"

#$

%

&'

!x1

!x2

"

#

$$

%

&

''+

0

0

"

#$

%

&'!)E

% Short-Period Linear Model - Step

F = [-1.2794 -7.9856;1. -1.2709];

G = [-9.069;0];

Hx = [1 0;0 1];

sys = ss(F, -G, Hx,0); % (-1)*Step

sys2 = ss(F, -2*G, Hx,0); % (-1)*Step

% Step response

step(sys, sys2), grid

!"E t( ) =0, t < 0

#1, t $ 0

%&'

('

Superposition of Linear Responses

Stability, speed of response, and damping areindependent of the initial condition or input

% Short-Period Linear Model - Superposition

F = [-1.2794 -7.9856;1. -1.2709];

G = [-9.069;0];

Hx = [1 0;0 1];

sys = ss(F, -G, Hx,0); % (-1)*Step

xo = [1; 0];

t = [0:0.2:20];

u = ones(1,length(t));

[y1,t1,x1] = lsim(sys,u,t,xo);

[y2,t2,x2] = lsim(sys,u,t);

u = zeros(1,length(t));

[y3,t3,x3] = lsim(sys,u,t,xo);

plot(t1,y1,t2,y2,t3,y3), grid

!!x1

!!x2

"

#

$$

%

&

''=

(1.2794 (7.9856

1 (1.2709

"

#$

%

&'

!x1

!x2

"

#

$$

%

&

''+

(9.069

0

"

#$

%

&'!)E

!y1

!y2

"

#

$$

%

&

''=

1 0

0 1

"

#$

%

&'

!x1

!x2

"

#

$$

%

&

''+

0

0

"

#$

%

&'!)E

Example: Continuous- andDiscrete-Time LTILongitudinal Models

Short Period

Phugoid

! !V! !"

#

$%%

&

'(() *0.02 *9.8

0.02 0

#

$%

&

'(

!V!"

#

$%%

&

'((+

4.7

0

#

$%

&

'(!+T

! !q

! !"

#

$%%

&

'((=

)1.3 )8

1 )1.3

#

$%

&

'(

!q

!"

#

$%%

&

'((+

)9.1

0

#

$%

&

'(!*E

!qk+1

!" k+1

#

$

%%

&

'

((=

0.85 )0.7

0.09 0.85

#

$%

&

'(

!qk

!" k

#

$

%%

&

'

((+

)0.84

)0.04

#

$%

&

'(!*Ek

!Vk+1

!"k+1

#

$%%

&

'((=

1 )0.980.002 1

#

$%

&

'(

!Vk

!"k

#

$%%

&

'((+

0.47

0.0005

#

$%

&

'(!*Tk

Learjet 23MN = 0.3, hN = 3,050 mVN = 98.4 m/s

Differential Equations Produce

State Rates of Change

Difference Equations

Produce State Increments

!t = 0.1sec

Example: Superposition ofContinuous- and Discrete-Time

Longitudinal Models

Phugoid and Short Period

! !V! !"

! !q

! !#

$

%

&&&&&

'

(

)))))

=

*0.02 *9.8 0 00.02 0 0 1.3

0 0 *1.3 *8*0.002 0 1 *1.3

$

%

&&&&

'

(

))))

!V!"

!q

!#

$

%

&&&&&

'

(

)))))

+

4.7 0

0 0

0 *9.10 0

$

%

&&&&

'

(

))))

!+T!+E

$

%&

'

()

!Vk+1!" k+1!qk+1!# k+1

$

%

&&&&&

'

(

)))))

=

1 *0.98 *0.002 *0.060.002 1 0.006 0.12

0.0001 0 0.84 *0.69*0.0002 0.0001 0.09 0.84

$

%

&&&&

'

(

))))

!Vk!" k!qk!# k

$

%

&&&&&

'

(

)))))

+

0.47 0.0005

0.0005 *0.0020 *0.840 *0.04

$

%

&&&&

'

(

))))

!+Tk!+Ek

$

%&&

'

())

Learjet 23MN = 0.3, hN = 3,050 mVN = 98.4 m/s

Differential EquationsProduce State Rates ofChange

DifferenceEquations ProduceState Increments

!t = 0.1sec

Equilibrium Response

Equilibrium Response

!!x(t) = F!x(t) +G!u(t) + L!w(t)

0 = F!x(t) +G!u(t) + L!w(t)

!x* = "F"1G!u *+L!w *( )

Dynamic equation

At equilibrium, the state is unchanging

Denoting constant values by (.)*

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

F =f11

f12

f21

f22

!

"

##

$

%

&&; G =

g1

g2

!

"

##

$

%

&&; L =

l1

l2

!

"

##

$

%

&&

!x1*

!x2*

"

#$$

%

&''= (

f22

( f12

( f21

f11

"

#$$

%

&''

f11f22

( f12f21( )

g1

g2

)

*++

,

-..!u *+

l1

l2

)

*++

,

-..!w *

"

#$$

%

&''

2nd-order example

sI ! F = " s( ) = s2 + f11+ f

22( )s + f11 f22 ! f12 f21( )

= s ! #1( ) s ! #2( ) = 0

Re #i( ) < 0

System Matrices

Equilibrium Response

Requirement forStability

Equilibrium Response of

Approximate Phugoid Model

!xP* = "F

P

"1G

P!u

P*+L

P!w

P*( )

!V *

!" *#

$%%

&

'((= )

0VN

LV

)1g

VNDV

gLV

#

$

%%%%%

&

'

(((((

T*T

L*T

VN

#

$

%%%

&

'

(((!*T * +

DV

)LV

VN

#

$

%%%

&

'

(((!V

W

*

+

,--

.--

/

0--

1--

Equilibrium state with constant thrust and wind perturbations

Steady-State Response of

Approximate Phugoid Model

!V * = "L#T

LV

!#T * + !VW

*

!$ * =1

gT#T + L#T

DV

LV

%

&'(

)*!#T *

With L!T ~ 0, steady-state velocity depends only on thehorizontal wind

Constant thrust produces steady climb rate

Corresponding step response, with L!T = 0

Equilibrium Response ofApproximate Short-Period Model

!xSP* = "F

SP

"1G

SP!u

SP*+L

SP!w

SP*( )

!q*

!" *#

$%%

&

'((= )

L"

VN

M"

1 )Mq

#

$

%%%

&

'

(((

L"

VN

Mq+ M"

*+,

-./

M0E

)L0E

VN

#

$

%%%

&

'

(((!0E* )

M"

)L"VN

#

$

%%%

&

'

(((!"

W

*

1

233

433

5

633

733

Equilibrium state with constant elevator and wind perturbations

Steady-State Response ofApproximate Short-Period Model

Steady pitch rate and angle of attack are not zero

Vertical wind reorients angle of attack

!q* = "

L#

VN

M$E%&'

()*

L#

VN

Mq+ M#

%&'

()*

!$E*

!# * = "M$E( )

L#

VN

Mq+ M#

%&'

()*

!$E + !#W

*

with L!E = 0

Scalar Frequency Response

Speed Control ofDirect-Current Motor

u(t) = Ce(t)

where

e(t) = yc (t) ! y(t)

Control Law (C = Control Gain)

Angular Rate

Characteristics

of the Motor

Simplified Dynamic Model

Rotary inertia, J, is the sum of motor and load

inertias

Internal damping neglected

Output speed, y(t), rad/s, is an integral of thecontrol input, u(t)

Motor control torque is proportional to u(t)

Desired speed, yc(t), rad/s, is constant

Model of Dynamics

and Speed Control

Dynamic equation

y(t) =1

Ju(t)dt

0

t

! =C

Je(t)dt

0

t

! =C

Jyc (t) " y(t)[ ]dt

0

t

!

dy(t)

dt=u(t)

J=Ce(t)

J=C

Jyc (t) ! y(t)[ ], y 0( ) given

Integral of the equation, with y(0) = 0

Direct integration of yc(t)

Negative feedback of y(t)

Step Response of

Speed Controller

y(t) = yc 1! e!

C

J

"#$

%&'t(

)**

+

,--= yc 1! e

.t() +, = yc 1! e! t /(

)*+,-

where " = C/J = eigenvalue or

root of the system (rad/s)

# = J/C = time constant ofthe response (sec)

Step input :

yC (t) =0, t < 0

1, t ! 0

"#$

%$

Solution of the integral,

with step commandyct( ) =

0, t < 0

1, t ! 0

"#$

%$

Angle Control of a DC Motor

Closed-loop dynamic equation, with y(t) = I2 x(t)

u(t) = c1yc (t) ! y1(t)[ ]! c2y2(t)

!x1(t)

!x2(t)

!

"

##

$

%

&&=

0 1

'c1/ J 'c

2/ J

!

"##

$

%&&

x1(t)

x2(t)

!

"

##

$

%

&&+

0

c1/ J

!

"##

$

%&&yc

Control law with angle and angular rate feedback

!n= c

1J ; " = c

2J( ) 2!n

c1 /J = 1

c2 /J = 0, 1.414, 2.828% Step Response of Damped

Angle Control

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 = ss(F1,G1,Hx,0); Sys2 = ss(F1a,G1,Hx,0);

Sys3 = ss(F1b,G1,Hx,0);

step(Sys1,Sys2,Sys3)

Step Response of Angle Controller,

with Angle and Rate Feedback

Single natural frequency, three damping ratios

Angle Response to a Sinusoidal

Angle Command

Amplitude Ratio (AR) =ypeak

yC peak

Phase Angle = !360"tpeak

Period, deg

Output wavelags behind theinput wave

Input and outputamplitudesdifferent

Effect of Input Frequency on Output

Amplitude and Phase Angle

With low inputfrequency, inputand outputamplitudes areabout the same

Lag of angleoutput oscillation,compared to input,is small

Rate oscillationleads angleoscillation by ~90deg

yc(t) = sin t / 6.28( ), deg !

n= 1 rad / s

" = 0.707

At Higher Input Frequency, Phase

Angle Lag Increasesyc(t) = sin t( ), deg

At Even Higher Frequency,

Amplitude Ratio Decreases and

Phase Lag Increasesyc(t) = sin 6.28t( ), deg

Angle and RateResponse of a

DC Motor overWide Input-FrequencyRange

! Long-term responseof a dynamic systemto sinusoidal inputsover a range offrequencies

! Determineexperimentally fromtime response or

! Compute the Bodeplot of the system!stransfer functions(TBD)

Very low damping

Moderate damping

High damping

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