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(t

k) +" #t( ) e

$F % $ tk( )&'

()

tk

tk+1

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

k)[ ]

= "!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#1

G

= I #1

2!F"t +

1

3!F2"t 2 #

1

4!F3"t 3 + ...$

%&'()G"t

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

L

= 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 = !p b2

"#

$% = 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 0

0.02 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( f

12f21( )

g1

g2

)

*++

,

-..!u *+

l1

l2

)

*++

,

-..!w *

"

#$$

%

&''

• 2nd-order example

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

22( )s + f11f22! f

12f21( )

= 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 *+ !V

W

*

!$ * =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 oscillation“leads” 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

Next Time:Root Locus Analysis