13_time response.pdf
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-
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
Next Time:Root Locus Analysis
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