B.2.3 Some preliminaries on longitudinal dynamic analysis
B.2.3.1 Simplify the equation variables
○Notations of the perturbed variables will be simplified as follows:
U u,   ,   ,  e  e and  T  T .
--- We have dropped all  signs on the perturbed variables. From now on, u ,
 ,  and  e are perturbations on their steady state quantities.
○Then, the equation set for the perturbed longitudinal motion becomes:
Luu  L   U 0 (   )  0
u  ( Du  Tu )u  ( D  g )  g  T  T
Muu  M   M   M      M  e
--- The left-hand-sides of these equations represent the internal interactions of the
system, the aircraft plus the air around it, whereas the right-hand-sides
represents the human input to the system.
○We attempt to find a set of time functions u (t ),  (t ) ,  ( t ) and  e ( t ) such
that the above equations are stratified; this set of u (t ),  (t ) ,  ( t ) and  e ( t )
are called the solution to the above equations
○A general approach to find the solution will uses the Laplace transformation
technique; a briefly outline of the technique follows.
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B.2.3.2 Laplace transformation of linear functions
○The Laplace transform of a real function f ( t ) is defined as follows:
F ( s )  L f (t )  


f (t )e j  st dt,s  C .
※ In order that this transformation may exist, it requires that
(a) f ( t )  , t   , and
(b) there exists a real   0 such that lim t  f ( t )  e t .
○Table of the Laplace transforms of some typical functions:
If  C , a complex conjugate
f ( t ): a ae t ae t sin  t
. pair of the characteristic function
a

a
a
F ( s ):
will appear simultaneously.
s s   ( s   )2   2
○Mathematical properties of the Laplace transform:
A. For functions f1( t ),, f n ( t ) all satisfying conditions (a) and (b), we have
La1 f1( t )an f n ( t )  a1L f1( t )an L f n ( t )
for an arbitrary set of real constants a1,, an .
dl f  l
d l 1 f

   f (t  0) .
B. If f ( t ) is differentiable, then L  l   s F ( s )  l 1
dt
dt


t 0
C. The mapping between f ( t ) and F ( s ) are one-to-one and reversible.
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B.2.3.3 Solving linear ODE using Laplace transform technique - An example
1. Equation of motion of the example system:
x0
x
mx(t )  f (t )  ks ( x(t )  x0 )   x (t )
Spring
Force
k
Mass
f
--- m, k s ,  : constants of the system
m

--- x0 : equilibrium state of the system
Friction Fixed base
2. Define, x (t )  x(t )  x0 , then x (t )  x (t )
and x(t )  x(t ) ; hence, mx(t )  f (t )  k s x (t )   x (t ) .
3. Assume that the system is initially at rest, i.e. x ( t  0)  x ( t  0)  0 .
 From property (B) of the Laplace transform, we have
Lx (t )  s Lx(t )  s X ( s ) and Lx(t )  s 2 Lx(t )  s 2 X ( s )
 This result and property (A) of the Laplace transform imply that:
Lmx(t )  ms2 X (s)  L f (t )  kx (t )   x (t )  F (s)  (k s   s) X (s) .
5. We thus obtain the Laplace transformed equation of motion:
k
1
1

X ( s)   2
F ( s),   , k s  s .
m s  s  k s
m
m
 We often call the denominator polynomial, Q(s)  s 2  s  k s , the
characteristic polynomial of the system. The roots of Q( s ) will
determine the characteristics of the motion, as is shown next.
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6. Q( s ) can be factorized into Q( s )  ( s  p1 )( s  p2 ) with p1, p2  C .
1
1
Hence, X ( s )  
F ( s) .
m ( s  p1 )( s  p2 )
7. Solution of x ( t ) for F ( s )  1, i.e. an unit impulse force:
-- - In this case, f (t )  0,t  0 ; hence, this is a solution with zero input.
 By means of partial fraction expansion, we have
X ( s) 
1
1

2
.

 1 
m ( s  p1 )( s  p2 ) s  p1 s  p2
 From property (c) of the Laplace transform, we thus have
x ( t )   1e p1t   2e p2t
pt
p t
--- This solution contains two characteristic functions: e 1 and e 2 , hence
two characteristic modes: p1 and p2 . If p1, p2  C , they will be
complex conjugate pair of each other, so are  1 and  2 .
--- Whether or not this solution will be
j
oscillatory, or diverging, or converging,
Stable
Unstable
depends on the locations of p1 and p2 on
region
region
the complex s -plane. (You may compare

p1, p2 with si given in B.2.3.2.)
s -plane
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8. Solution of x ( t ) for any nonzero f can also be derived:
◎ f (t )  1,t  0 , i.e. a constant input  F ( s )  1 s
 The partial fraction expansion should give us
X ( s) 
1
1
1

2


  1 
 3
m ( s  p1 )( s  p2 ) s s  p1 s  p2 s
 Accordingly, x ( t )   1e 1   2e 2   3 .
--- Here,  3 represents the particular solution of x due to input f .
pt
p t

◎ f (t )  sin t ,t  0 , i.e. a sinusoidal input  F ( s )  2
s  2
 This time, X ( s ) 
1


m ( s  p1 )( s  p2 )( s  j )( s  j )
3




s  p1 s  p2 s  j s  j

where 3 and  3 are complex conjugate pair.
pt
p t
 Accordingly, x (t )  1e 1   2e 2  2 sin( t   )
--- and  are the phase and the magnitude of 3 , or  3   e j .
--- The last term is the particular solution of x due to input f .
1
2
3
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【Some observations on the results of Section B.2.3.3】
○ Regardless of the function for the forcing input f (t ) , the solutions for x ( t )
always consist of terms that associate with the roots of Q(s ) .
f (t )  an impulse,  x (t )   1e p1t   2 e p2t
f (t )  1, t  0 
f (t )  sin t , t  0 
x (t )   1e p1t   2 e p2t   3
x (t )  1e p1t   2 e p2t  2 sin( t   )
The characteristic modes
○ These characteristic modes reflect the intrinsic properties of the system
● For the spring-mass system, these characteristic modes reflect the
ineraction among the spring, the friction, and the mass m .
● For an aircraft, the characteristic modes will reflect the interactions
between the aerodynamics and the inertial of the vehicle.
○ These characteristic modes determines the behavior of the system
● If any of these characteristic mode diverges, then the plant is unstable.
● Safe operation of the any dynamic system will require that all of the
characteristic modes are stable.
○ The characteristic solution of a system is the first subject to be analyzed.
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B.2.3.4 The Laplace-domain equations of the longitudinal motion
○For the sake of simplicity, let the Laplace transforms of u (t ),  (t ) ,  ( t ) and
 e ( t ) be denoted as u ( s),  ( s) ,  ( s) and  e ( s) , respectively.
○Then, the following Laplace domain equations of the perturbed longitudinal
motion is obtained by taking Laplace transform of the equations:
Luu( s)  ( sU 0  L ) ( s)  sU 0 ( s)  0
(s  Du  Tu )u(s)  ( D  g ) (s)  g (s)  T  T
Muu( s)  ( sM  M ) ( s)  ( sM  s2 ) ( s)   M  e ( s)
○The above equations can be lumped into as follows:
Lu
sU 0  L
 sU 0   u ( s )   0
0

 e ( s ) 
s  D  T





D  g
g
 ( s)  0
T 
u
u







(
s
)



Mu
sM   M  sM   s 2   ( s )   M  0  T
○We are interested on the following solutions of the above equations:
1. The solution that reflects the characteristics of an aircraft motion due to
the aerodynamics, the gravitation and the inertial of the vehicle.
2. The solution that reflects the aircraft response to input  e (s) and  T (s ) .
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B.2.3.5 Some comments on the characteristic solutions of a linear EOM
○ For any linear equation set A( s) x( s)  B( s) Z ( s) ,
 x1 ( s ) 
a11 ( s ) a12 ( s ) a13 ( s ) 
b11 ( s ) b12 ( s ) 
 z1 ( s ) 






x  x2 ( s ) , A  a 21 ( s ) a 22 ( s ) a 23 ( s ) , B  b 21 ( s ) b 22 ( s ) , Z  







z
(
s
)
 2 
 x3 ( s ) 
a 31 ( s ) a 32 ( s ) a 33 ( s ) 
b 31 ( s ) b 32 ( s ) 
where aij (s) and b ij (s ) are polynomials of s and xi (s) are the elements
of x(s ) , a Laplace domain solution for x(s ) will be x( s )  A( s) 1 B( s ) Z ( s)
○ Because A(s) is a 3  3 matrix, we will have
 g11 ( s ) g12 ( s ) g13 ( s ) 
1
A( s ) 1 
 g 21 ( s ) g 22 ( s ) g 23 ( s ), ( s )  det A( s ) ,

( s) 
g 31 ( s ) g 32 ( s ) g 33 ( s ) 
and therefore
 x1 ( s ) 
 g11 ( s ) g12 ( s ) g13 ( s )  b11 ( s ) b12 ( s ) 
 x ( s )  1  g ( s ) g ( s ) g ( s ) b ( s ) b ( s )  z1 ( s ) 
22
23
22
 2   ( s )  21
  21
  z 2 ( s )
 x3 ( s ) 
g 31 ( s ) g 32 ( s ) g 33 ( s )  b 31 ( s ) b 32 ( s ) 
where g ij (s) are also polynomials of s .
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○ For example, let
 s  2.25  1 0 
 1 0
A( s )   1.625
s  1, and B( s )  0.5 1




0
s 
 0.375
 0 0
Then, (s)  s 3  2.25s 2  1.6 2 s5  0.3 7 5 and
g11 ( s)  s 2
g12 ( s )  s
g13 ( s)  1
,
g 21 ( s)  1.625s  0.375 g 22 ( s)  s 2  2.25s
g 23 ( s )  s  2.25
g 31 ( s )  0.375s
g 32 ( s)  0.375 g 33 ( s)  s( s 2  2.25s  1.625)
Therefore,
( s 2  0.5s ) z1 ( s )  s z 2 ( s )
x1 ( s )  3
,
2
s  2.25s  1.625s  0.375
2
(0.5s  0.5s  0.375) z1 ( s)  ( s 2  2.25s) z 2 ( s)
x2 ( s ) 
,
3
2
s  2.25s  1.625s  0.375
and
 (0.375s  0.1875) z1 ( s)  0.375 z 2 ( s)
x1 ( s) 
s 3  2.25s 2  1.625s  0.375
==> No that all solutions have the same denominator (characteristics).
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○ It is seen that the determinant of A(z ) appears on all solutions.
--- Analogous to the example shown in section B.2.3.3, the roots of the
determinant of A(z ) will determine the characteristic functions of
the time domain solutions of x1 (t ) , x2 (t ) , and x3 (t ) .
○ In our problem, the EOM for the longitudinal aircraft motion is
Lu
sU 0  L
 sU 0   u ( s ) 
0

 0
 e ( s ) 
s  D  T





D  g
g
 ( s )   0 T 
u
u




  T ( s )
2

Mu
sM   M  sM   s   ( s ) 
 M  0 
 Therefore, the determinant of the matrix,
Lu
sU 0  L
 sU 0 

s  D  T
,
D

g
g
u
u



2

Mu
sM   M  sM   s 
will determine the characteristic of the longitudinal motion.
 However, a 3  3 matrix equation is difficult to solve.
 For most A/C, simplified and hence easier-to-solve equations of certain
low degree-of-freedom simplified motions may offer solutions that are
close approximate of the complete solutions.
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