State Variables and Linear Time-Invariant Systems
CEE 541. Structural Dynamics
Department of Civil and Environmental Engineering
Duke University
Henri P. Gavin
Fall 2014
1 Linearity and Time Invariance
A system G that maps an input u(t) to an ouput y(t) is a linear system if and
only if
(α1 y1 (t) + α2 y2 (t)) = G[α1 u1 (t) + α2 u2 (t)]
(1)
where y1 = G[u1 ], y2 = G[u2 ] and α1 and α2 are scalar constants.
A system G that maps an input u(t) to an ouput y(t) is a time-invariant
system if and only if
y(t − to ) = G[u(t − to )] .
(2)
Systems described by
ẋ(t) = Ax(t) + Bu(t) ,
x(0) = xo
y(t) = Cx(t) + Du(t)
(4)
are linear and time-invariant.
variable
x
u
y
A
B
C
D
description
state vector
input vector
output vector
dynamics matrix
input matrix
output matrix
feedthrough matrix
(3)
dimension
n by 1
r by 1
m by 1
n by n
n by r
m by n
m by r
2
CEE 541. – Structural Dynamics – Duke University – Fall 2014 – H.P. Gavin
2 Example: a spring-mass-damper oscillator
An externally-forced spring-mass-damper oscillator is described by
mr̈(t) + cṙ(t) + kr(t) = f (t) ,
r(0) = do , ṙ(0) = vo .
(5)
Assuming a solution of the form r(t) = r̄eλt ,
(mλ2 + cλ + k)r̄ = 0 .
This equation is valid for r̄ = 0, the trivial solution, and
v
u
u c2
c
k
λ=−
±t
−
2m
4m m
Note that λ can be complex, λ = σ ± ıω. Now, defining the natural frequency
√
ωn2 = k/m, and the damping ratio ζ = c/(2 mk), we find c/(2m) = ζωn , so,
q
λ = −ζωn ± ζ 2 ωn2 − ωn2
q
= −ζωn ± ωn ζ 2 − 1
(6)
and if ζ < 1, the root may be written
q
λ = −ζωn ± ıωn 1 − ζ 2 .
(ı =
√
−1) Note that if ζ > 0 then σ < 0 and the system is stable.
The second-order ordinary differential equation (5) may be written as two
first-order ordinary differential equations, by defining a state vector of the
position and velocity, x = [r ṙ]T .

d r
dt ṙ



=
0
1
−k/m −c/m


r
ṙ



+

0 
f (t) ,
1/m


r(0)
ṙ(0)



=
do
vo


(7)
In terms of a desired response from this system, we may be interested in the
force on the foundation, fF , and the acceleration of the mass, both of which
can be computed directly through a linear combination of the states and the
input.







f
k
c
r
0
F

=

+
 f (t)
(8)
r̈
−k/m −c/m
ṙ
1/m
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3
Linear Time Invariant Systems
A single degree of freedom oscillator as well as many other types of dynamical
systems may be described in a general sense using state variable descriptions,
ẋ(t) = Ax(t) + Bu(t) ,
x(0) = xo
y(t) = Cx(t) + Du(t) .
3 Free Response
If u(t) = 0, then the free response y(t) of the system to initial conditions xo
is
y(t) = CeAt xo ,
(9)
where eAt is called a matrix exponential.
In Matlab, y(:,p)=C*expm(A*t(p))*xo;
4 Unit Impulse Response Function
If xo = 0 and the input is a unit impulse, u(t) = Im δ(t), the responses y(t) of
the system are called the unit impulse response function, H(t),
H(t) = CeAt B .
(10)
The set of free responses is richer than the set of unit impulse responses.
In Matlab, H(p,:,:)=C*expm(A*t(p))*B;
5 Forced Response
The forced response of a linear time invariant system is
y(t) = CeAt xo + C
Z t
eA(t−τ )
0
Bu(τ ) dτ + Du(t).
(11)
An efficient method for computing y(t) for a arbitrary inputs u(t) is provided
in the last sections of this document.
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CEE 541. – Structural Dynamics – Duke University – Fall 2014 – H.P. Gavin
6 The Matrix Exponential
The matrix exponential is defined for real square matrices, An×n , as
∞
∆ X
eA =
Ak /k! = I + A + AA/2 + AAA/6 + AAAA/24 + · · ·
(12)
k=0
Properties of the matrix exponential:
• If A = AT then eA > 0.
• If A = −AT then [eA ][eA ]T = I
• eAt eBt = e(A+B)t
• [eA ]T = eA
T
• [eA ]−1 = e−A
• eSAS
−1
eSAS
−1
eSAS
−1
•
d At
dt e
= SeA S −1 for any square matrix Sn×n .
1
1
= I + SAS −1 + SAS −1 SAS −1 + SAS −1 SAS −1 SAS −1 + · · ·
2
6
1
1
= SS −1 + SAS −1 + SAAS −1 + SAAAS −1 + · · ·
2
6
= S[I + A + AA/2 + AAA/6 + · · · ]S −1
= SeA S −1 .
(13)
= AeAt
1
1
1
eAt = I + At + AAt2 + AAAt3 + AAAAt4 + · · ·
2
6
24
d At
1
1
e
= A + AA + AAAt2 + AAAAt3 + · · ·
dt
2
6
1
1
1
= A[I + At + AAt2 + AAAt3 + AAAAt4 + · · · ]
2
6
24
d At
e
= AeAt
(14)
dt
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5
Linear Time Invariant Systems
• A
R t Aτ
0 e dτ
= eAt − I
Z t
A eAτ dτ
0
Z t
=
d Aτ
e dτ
dτ
0
t
eAτ 0
At
=
= e
− eA0
= eAt − I
Z t
A eAτ dτ
0
Z t
−Aτ
−A
•
R t A(t−τ )
dτ
0e
0
= eAt − I
(15)
dτ = e−At − I
e
(16)
= A−1 (eAt − I)
Z t
eA(t−τ ) dτ
0
=
Z t
eAt e−Aτ dτ
0
Z
At t −Aτ
= e
0
e
dτ
= eAt (−A−1 ) e−At − I
= −eAt A−1 e−At + eAt A−1
= −A−1 AeAt A−1 e−At + A−1 AeAt A−1
= −A−1 eAAA
−1
t −At
e
+ A−1 eAAA
−1
t
= −A−1 eAt e−At + A−1 eAt
= −A−1 + A−1 eAt
= A
=
−1
At
e
−I
(17)
Z t
eAτ dτ
0
(18)
• The solution to ẋ(t) = Ax(t) + Bu(t), x(0) = xo , for t ≥ 0 is
At
x(t) = e xo +
Z t
eA(t−τ )
0
Bu(τ ) dτ .
(19)
As a special case of the above, if the input to the system is vδ(t) where
v is a m by 1 vector and δ(t) is the Dirac delta function, and if xo = 0,
then the output y(t) is called the unit impulse response function, h(t).
h(t) = CeAt Bv .
(20)
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CEE 541. – Structural Dynamics – Duke University – Fall 2014 – H.P. Gavin
∆
q
With definitions of the natural frequency, ωn = k/m, the damping ratio,
q
√
∆
∆
ζ = c/(2 mk), and the damped natural frequency, ωd = ωn |ζ 2 − 1|, let

A=
0
1
−k/m −c/m



=
0
1
2
−ωn −2ζωn


(21)
.
For this dynamics matrix, the matrix exponential depends on the damping
ratio, ζ, as follows:
damping
undamped
underdamped
damping ratio
ζ=0
0<ζ<1
eAt
e
At
"
cos ωn t
−ωn sin ωn t

cos ωd t + √ ζ
=
eAt = e−ζωn t
1
ωn
sin ωn t
cos ωn t
1−ζ 2

− √ ωd
1−ζ 2
critically damped
overdamped
ζ=1
ζ>1
e
At
=e
−ωn t
eAt = e−ζωn t
#
1
ωd
sin ωd t
sin ωd t
cos ωd t − √ ζ
1 + ωn t
t
−ωn2 t 1 − ωn t

cosh ωd t + √ ζ
sinh ωd t
1−ζ 2

− √ ω2d
ζ −1
sinh ωd t

sin ωd t
1−ζ 2
1
ωd
sin ωd t

sinh ωd t
cosh ωd t − √ ζ2
ζ −1
sinh ωd t
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

7
Linear Time Invariant Systems
7 Eigenvalues and Diagonalization
Consider the dynamics matrix A of a linear time invariant, (LTI) system.
If the input, u(t) is zero, then ẋ = Ax. Assuming a solution of the form
x(t) = x̄eλt , and substituting this solution into ẋ = Ax, results in:
x̄λeλt = Ax̄eλt
or
x̄λ = Ax̄ ,
(22)
which is a standard eigenvalue problem, in which x̄ is the eigen-vector and λ is
the eigen-value. If A is a n×n matrix, then there are n eigen-values, λ1 , · · · , λn
and n associated eigen-vectors, x̄1 , · · · , x̄n . For the dynamics matrix given in
equation (7), there are two eigenvalues.
v
u
u
t
c2
k
c
±
−
2m
4m
m
q
= −ζωn ± ωn ζ 2 − 1
λ1,2 = −
(23)
(24)
The dynamics matrix contains all the information required to determine the
natural frequencies and damping ratios of the system.
The n eigen-vectors can be assembled, column-by-column into a matrix,
X̄ = [x̄1 x̄2 · · · x̄n ] .
Pre-multiplying the eigen-problem by X̄ −1 ,

X̄
−1
AX̄ = X̄
−1
X̄Λ = diag(λi ) =

λ1





...
λn





This is called a diagonalization of the dynamics matrix A.
Now, consider the linear transformation of coordinates, x(t) = X̄q(t),
q(t) = X̄ −1 x(t). Substituting this change of coordinates into equation (3),
X̄ q̇(t) = AX̄q(t) + Bu(t),
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CEE 541. – Structural Dynamics – Duke University – Fall 2014 – H.P. Gavin
and pre-multiplying by X̄ −1
q̇ = X̄ −1 AX̄q + X̄ −1 Bu,
or

q̇ =






λ1
...
λn
q̇ = Āq + B̄u ,


q


+ X̄ −1 Bu ,
q(0) = X̄ −1 xo
q(0) = X̄ −1 xo
(25)
y = C X̄q + Du
y = C̄q + Du
(26)
The n differential equations q̇i = λi qi + B̄i u are uncoupled. The state qi (t) is
independent of all the other states qj (t), j 6= i. Note that if Λ is complex, so
are X̄, Ā, B̄, and C̄.
Consider one of the un-coupled equations from equation (25), for the unforced
case u = 0
q̇i (t) = λi qi (t) ,
qi (0) = 1 .
This equation has a solution
qi (t) = eλi t ,
where λi is, in general, a complex value,
λi = σi ± ıωi ,
and qi + qi∗ is real-valued.
1 λi t 1 λ∗i t
e + e
2
2
1
1 σi t
= e (cos ωi t + ı sin ωi t) + eσi t (cos ωi t − ı sin ωi t)
2
2
σi t
= e cos ωi t ,
qi (t) + qi∗ (t) =
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9
Linear Time Invariant Systems
8 Transfer Function and Frequency Response Function
Taking the Laplace transform of equation (3), (and ignoring the effects of
initial conditions) gives
sX(s) = AX(s) + BU (s) ,
(27)
Y (s) = CX(s) + DU (s) ,
(28)
which can be written as Y (s) in terms of U (s) as follows,
Y (s)
= H(s) = C[sI − A]−1 B + D.
(29)
U (s)
This is the transfer function of the linear dynamic system. The same may be
carried out on equations (25) and (26), resulting in
Y (s)
= H(s) = C̄[sI − Ā]−1 B̄ + D.
U (s)
Recall that the matrix Ā = diag(λi ), so, H(λi ) = ∞. The eigen-values λi of
A are called the poles of the system.
Equation (29) may be used to determine the complex-valued frequency response function of any dynamic system, by evaluating the transfer function
along the line s = ıω.
Y (ω)
= H(ω) = C[ıωI − A]−1 B + D.
(30)
U (ω)
Assuming that the inputs u(t) are sinusoidal with frequency ω and unit amplitude, the magnitude of the frequency response function,
q
|H(ω)| = <[H(ω)]2 + =[H(ω)]2
gives the amplitude of the responsess y(t). The phase of H(ω),


=[H(ω)] 
∠H(ω) = arctan −
<[H(ω)]
gives the phase angle between the input u(t) and the output y(t). So if
u(t) = cos(ωt), then y(t) = |H(ω)| cos(ωt + ∠H(ω)).
In Matlab, [mag,pha] = bode(A,B,C,D); generates plots of |H(ω)| and
∠H(ω) for a system defined by matrices A, B, C, and D.
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CEE 541. – Structural Dynamics – Duke University – Fall 2014 – H.P. Gavin
9 Transforming Continuous Time Systems to Discrete Time Systems (I)
Making a substitution of t = ∆t into equation (19), and specifying that
u(τ ) = u(ti ) = ui for 0 ≤ τ ≤ ∆t,
x(∆t) = eA∆t xi +
= e
A∆t
Z ∆t
eA(∆t−τ )
0
−1
A∆t
xi + A
dτ Bui
− I Bui ,
e
and defining xi+1 = x(∆t),
xi+1 =
h
eA∆t xi + A−1 eA∆t − I B ui ,
i
h
i
= [Ad ] xi + [Bd ] ui
(31)
where Ad and Bd are the discrete-time dynamics matrix and the discrete-time
input matrix, Ad = eA∆t and Bd = A−1 (Ad − I)B.
Note that Bd exists even though A may not be invertible. Consider the
diagonalization A = X̄ −1 ΛX̄. The inverse of A may be expressed A−1 =
X̄ −1 Λ−1 X̄, where X̄ X̄ −1 = I. So,
A−1 (eAt − I) = X̄ −1 Λ−1 X̄ X̄ −1 (eΛt − I)X̄ = X̄ −1 Λ−1 (eΛt − I)X̄,
which contains diagonal terms (eλi − 1)/λi , and
eλi − 1
=1.
lim
λi →0
λi
To compute Bd without inverting A, note the following:
∆t2
∆t3
∆t4
AA +
AAA +
AAAA + · · ·
2
6
24
∆t2
∆t3
∆t4
A−1 (eA∆t − I) = ∆t +
A+
AA +
AAA + · · ·
2
6
24
∆t2
∆t3
∆t4
−1 A∆t
Bd = A (e
− I)B = ∆tB +
AB +
AAB +
AAAB + · · · ,
2
6
24
eA∆t − I = A∆t +
and, if

M =
An×n Bn×r
0r×n 0r×r


,
(32)
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11
Linear Time Invariant Systems
then,
M ∆t
e
∆t3
∆t4
∆t2
MM +
MMM +
MMMM + · · ·
= I + ∆tM +
2
6
24






2
3
∆t
∆t
A
B
AA
AB
AAA
AAB
+

+

 + ···
= I + ∆t 
2
6
0 0
0
0
0
0

=

I + ∆tA +
∆t2
2 AA
+ · · · ∆tB +
∆t2
2 AB
0
+
0
∆t3
6 AAB
+ ···


,
so

eM ∆t = 
Ad Bd
0 0


,
(33)
and the discrete time dynamics matrix Ad and input matrix Bd may be computed using a single matrix exponential computation.
In Matlab:
[n,r] = size(B);
M = [ A B ; zeros(r,n+r) ];
eMdt = expm(M*dt);
Ad = eMdt(1:n,1:n);
Bd = eMdt(1:n,n+1:n+r);
With the matrices Ad and Bd , the transient response may be computed digitally using equation (31).
x(:,1) = x0;
for p=1:points-1
x(:,p+1) = Ad * x(:,p) + Bd * u(:,p);
end
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CEE 541. – Structural Dynamics – Duke University – Fall 2014 – H.P. Gavin
10 Transforming Continuous Time Systems to Discrete Time Systems (II)
Making a substitution of t = ∆t into equation (19), and specifying that the
input changes linearly over a time increment, ∆t,
1
u(τ ) = u(ti ) +
(u(ti+1 ) − u(ti )) τ = ui + u0i τ for 0 ≤ τ ≤ ∆t , (34)
∆t
gives
x(∆t) = eA∆t xi +
A∆t
= e
A∆t
= e
xi +
Z ∆t
eA(∆t−τ )
0
Z ∆t
A(∆t−τ )
0
e
B u(τ ) dτ
dτ B ui +
Z ∆t
eA(∆t−τ )
0
−2
A∆t
xi + A−1 eA∆t − I B ui + A
e
B u0i τ dτ
− I − A∆t B u0i .
Defining xi+1 = x(∆t),
xi+1 =
h
eA∆t xi + A−1 eA∆t − I B ui + A−2 eA∆t − I − A∆t B u0i ,
i
h
= [Ad ] xi + [Bd ] ui + [Bd0 ] u0i
i
h
i
(35)
where Ad , Bd , and Bd0 are the discrete-time dynamics matrix and the discretetime input matrices. If u(τ ) is constant within the time increment, then u0i = 0
and Bd0 need not be computed.
The input matrices Bd and Bd0 exist even though A may not be invertible. To
compute Bd without inverting A, note the following:
∆t2
∆t3
∆t4
A∆t
e
− I = A∆t +
AA +
AAA +
AAAA + · · ·
2
6
24
∆t2
∆t3
∆t4
A−1 (eA∆t − I) = ∆t +
A+
AA +
AAA + · · ·
2
6
24
∆t2
∆t3
∆t4
−1 A∆t
Bd = A (e
− I)B = ∆tB +
AB +
AAB +
AAAB + · · · ,
2
6
24
and,
∆t2
∆t3
∆t4
e
− I − A∆t =
AA +
AAA +
AAAA + · · ·
2
6
24
∆t2 ∆t3
∆t4
−2 A∆t
A (e
− I − A∆t) =
+
A+
AA + · · ·
2
6
24
∆t2
∆t3
∆t4
0
−2 A∆t
Bd = A (e
− I − A∆t)B =
B+
AB +
AAB + · · · ,
2
6
24
A∆t
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13
Linear Time Invariant Systems
If


∆ 
M(n+2m)×(n+2r) = 

An×n Bn×r 0n×r
0r×n 0r×r Ir×r
0r×n 0r×r 0r×r






(36)
,
then,
eM ∆t = I + ∆tM +
∆t2
∆t3
∆t4
MM +
MMM +
MMMM + · · ·
2
6
24






AA AB B
AAA AAB AB
A B 0
3
2
 ∆t 


 ∆t 
+
= I + ∆t  0 0 I  +
0
0
0
0
0  + ···


 0
2
6
0
0
0
0
0
0
0 0 0

I + ∆tA +
= 

∆t2
2 AA
+ ···
∆tB +
0
0
∆t2
2 AB
+
0
0
∆t3
6 AAB
+ ···
∆t2
2 B
+
∆t3
6 AB
0
0
+ ···


 ,
so,

M ∆t
e
=





Ad Bd
0 0
0 0

0
Bd 

0 

,
(37)
0
and the discrete time dynamics matrix Ad and input matrices Bd and Bd0 may
be computed using a single matrix exponential computation.
In Matlab:
[n,m]
M
eMdt
Ad
Bd
Bd_
Bd0
Bd1
=
=
=
=
=
=
=
=
size(B);
[ A B zeros(n,r) ; zeros(r,n+r) eye(r) ; zeros(r,n+2*r) ];
expm(M*dt);
eMdt(1:n,1:n);
eMdt(1:n,n+1:n+r);
eMdt(1:n,n+r+1:n+2*r);
Bd - Bd_ / dt;
Bd_ / dt;
With the matrices Ad , Bd , and Bd0 , the transient response may be computed
digitally using equation (35).
x(:,1) = x0;
for p=1:points-1
x(:,p+1) = Ad * x(:,p) + Bd0 * u(:,p) + Bd1 * u(:,p+1);
end
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CEE 541. – Structural Dynamics – Duke University – Fall 2014 – H.P. Gavin
lsym.m http://www.duke.edu/∼hpgavin/lsym.m
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function y = lsym (A ,B ,C ,D ,u ,t , x0 )
% LSYM S i m u l a t i o n o f a c o n t i n u o u s −time l i n e a r s ys t e m t o a r b i t r a r y i n p u t s .
% y = LSYM(A, B, C,D, u , t , x0 ) computes t h e time r e s p o n s e o f t h e l i n e a r s y st e m :
%
.
%
x = Ax + Bu
%
y = Cx + Du
%
%
A
:
r e a l n by n dynamics
matrix
%
B
:
r e a l n by r i n p u t
matrix
%
C
:
r e a l m by n o u t p u t
matrix
%
D
:
r e a l m by r
feedthrough matrix
%
u
:
real
r by p m a t r i x o f sampled i n p u t s
%
t
:
r e a l 1 by p v e c t o r o f u n i f o r m l y s p a c e d p o i n t s i n time
%
x0
:
r e a l n by 1 v e c t o r o f i n i t i a l s t a t e s , d e f a u l t s t o z e r o
%
y
:
r e a l m by p m a t r i x o f t h e sy s te m o u t p u t s
%
%
LSYM r e t u r n s a semi−a n a l y t i c e x a c t s o l u t i o n u s i n g a
%
c o n t i n u o u s −to−d i s c r e t e c o n v e r s i o n b a s e d on t h e m a t r i x e x p o n e n t i a l ,
%
and assuming u i s a l i n e a r ramp b e t w e e n sample p o i n t s .
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[n ,r , m ] = abcddim (A ,B ,C , D ); % d e t e r m i n e and c h e c k c o m p a t a b i l i t y o f m a t r i x dim ’ s
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points = s i z e (u ,2);
% number o f d a t a p o i n t s
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= zeros (m , points );
y
% memory a l l o c a t i o n f o r t h e o u t p u t
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i f ( nargin == 7 )
x
= x0 ;
else
x
= zeros (n ,1);
end
% i n i t i a l conditions for the state
% i n i t i a l c o n d i t i o n s are zero
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dt
=
% uniform time−s t e p v a l u e
t (2) - t (1);
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M
eMdt
Ad
Bd
Bd_
Bd0
Bd1
=
=
=
=
=
=
=
[ A B zeros (n , r ) ; zeros (r , n + r ) eye( r ) ; zeros (r , n +2* r ) ];
expm( M * dt );
% matrix e x p o n e n t i a l
eMdt (1: n ,1: n );
% d i s c r e t e −time dynamics m a t r i x
eMdt (1: n , n +1: n + r );
% d i s c r e t e −time i n p u t m a t r i x
eMdt (1: n , n + r +1: n +2* r );
% d i s c r e t e −time i n p u t m a t r i x
Bd - Bd_ / dt ;
% d i s c r e t e −time i n p u t m a t r i x f o r time p
Bd_ / dt ;
% d i s c r e t e −time i n p u t m a t r i x f o r time p+1
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y (: ,1) = C * x
+
D * u (: ,1);
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f o r p = 1: points -1
46
x
47
= Ad * x
+
Bd0 * u (: , p )
+
Bd1 * u (: , p +1);
48
y (: , p +1) = C
49
* x
+
D * u (: , p +1);
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end
CC BY-NC-ND H.P. Gavin
Linear Time Invariant Systems
15
abcddim.m http://www.duke.edu/∼hpgavin/abcddim.m
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function [n , r , m ] = abcddim (a , b , c , d )
% Usage : [ n , r , m] = abcddim ( a , b , c , d )
%
% Check f o r c o m p a t i b i l i t y o f t h e d i m e n s i o n s o f t h e m a t r i c e s d e f i n i n g
% t h e l i n e a r sy s t em ( a , b , c , d ) .
%
% Returns n = number o f s ys t e m s t a t e s ,
%
r = number o f sy s te m i n p u t s ,
%
m = number o f s y st e m o u t p u t s .
%
% Returns n = r = m = −1 i f t h e s y s te m i s n o t c o m p a t i b l e .
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% A. S . Hodel <s c o t t e @ e n g . auburn . edu>
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i f ( nargin ˜= 4)
error ( ’ usage : abcddim (a , b , c , d ) ’ );
end
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n = -1; r = -1; m = -1;
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[ am , an ] = s i z e ( a );
i f ( am ˜= an ) , error ( ’ abcddim : a is not square ’ ); end
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[ bm , bn ] = s i z e ( b );
i f ( bm ˜= am ) , error ( ’ abcddim : a and b are not compatible ’ ); end
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[ cm , cn ] = s i z e ( c );
i f ( cn ˜= an ) , error ( ’ abcddim : a and c are not compatible ’ ); end
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[ dm , dn ] = s i z e ( d );
i f ( cm ˜= dm ) , error ( ’ abcddim : c and d are not compatible ’ ); end
i f ( bn ˜= dn ) , error ( ’ abcddim : b and d are not compatible ’ ); end
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n = an ;
r = bn ;
m = cm ;
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% −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− abcddim .m
CC BY-NC-ND H.P. Gavin
16
CEE 541. – Structural Dynamics – Duke University – Fall 2014 – H.P. Gavin
damp.m http://www.duke.edu/∼hpgavin/damp.m
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function [ wnout , z ] = damp (a , delta_t )
% DAMP N a t u r a l f r e q u e n c y and damping f a c t o r f o r c o n t i n u o u s s y s t e m s .
%
[Wn, Z ] = DAMP(A) r e t u r n s v e c t o r s Wn and Z c o n t a i n i n g t h e
%
n a t u r a l f r e q u e n c i e s and damping f a c t o r s o f A.
The v a r i a b l e A
%
can be i n one o f s e v e r a l f o r m a t s :
%
%
1) I f A i s s q u a r e , i t i s assumed t o be t h e s t a t e −s p a c e
%
’A’ m a t r i x .
%
2) I f A i s a row v e c t o r , i t i s assumed t o be a v e c t o r o f
%
t h e p o l y n o m i a l c o e f f i c i e n t s from a t r a n s f e r f u n c t i o n .
%
3) I f A i s a column v e c t o r , i t i s assumed t o c o n t a i n
%
root locations .
%
%
When i n v o k e d w i t h o u t l e f t hand arguments DAMP p r i n t s t h e
%
e i g e n v a l u e s w i t h t h e i r n a t u r a l f r e q u e n c y and damping f a c t o r i n a
%
t a b u l a r f o r m a t on t h e s c r e e n .
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[m , n ] = s i z e ( a );
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i f (n <1 || m <1) , wnout =0; z =0; return ; end
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i f ( m == n )
r = e ig ( a );
e l s e i f ( m == 1)
r = ( roots ( a ));
e l s e i f ( n == 1)
r = a;
else
error ( ’ Must be a vector or a square matrix . ’ );
end
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i f ( nargin == 2 ) , r = log ( r )/ delta_t ; end
% d i s c r e t e time s ys t em
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f o r k = 1: n
wn ( k ) = abs( r ( k ));
z(k)
= - ( r e a l ( r ( k )) - 2* eps) / ( wn ( k ) + 2* eps );
end
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[ wns , idx ] = sort (abs( wn ));
% s o r t by i n c r e a s i n g n a t u r a l f r e q u e n c y
wn = wn ( idx );
z = z ( idx );
r = r ( idx );
wd = wn .* sqrt ( abs ( z .ˆ2 - 1 ) );
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i f nargout == 0
% D i s p l a y r e s u l t s on t h e s c r e e n .
f p r i n t f ( ’ \ n ’ );
fprintf (’
Natural
Damped \ n ’ );
fprintf (’
Frequency
Frequency
Eigenvalue \ n ’ );
fprintf (’
( cyc / sec )
Damping ( cyc / sec )
real
imag \ n ’ );
fprintf (’
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -\ n ’ );
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f o r idx = 1:1: n
fprintf (’
end
%10.5 f
%10.5 f
%10.5 f
%10.5 f
%10.5 f
\ n ’ , wn ( idx )/(2* pi ) , z ( idx ) , wd ( idx )/(2* p
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return % S u p p r e s s o u t p u t
end
wnout = wn ;
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% −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− DAMP
CC BY-NC-ND H.P. Gavin
17
Linear Time Invariant Systems
mybode.m http://www.duke.edu/∼hpgavin/mybode.m
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function [ mag , pha ] = mybode (a ,b ,c ,d , iu ,w ,axes , figno , leg , tol )
% [ mag , pha ] = mybode ( a , b , c , d , iu , w , axes , f i g n o , l e g , t o l )
% p l o t s t h e magnitude and p h a s e o f t h e
% s t e a d y −s t a t e harmonic r e p o n s e o f a c o n t i n u o u s time l i n e a r dynamic sy s te m .
% where :
% a , b , c , d a r e t h e s t a n d a r d s t a t e −s p a c e m a t r i c e s o f a dynamic sy s te m
% i u i s t h e s e l e c t e d i n p u t ( column o f B)
( d e f a u l t : i u = 1)
% w is a vector of frequencies
( d e f a u l t : w = l o g s p a c e ( −2 ,2 ,200)∗2∗ p i
% a x i s i s e i t h e r x , y , n , or b t o i n d i d a t e wich a x e s s h o u l d be l o g −s c a l e d . . .
%
x , y , n e i t h e r , or b o t h . The d e f a u l t i s b o t h . The y−a x i s f o r t h e p h a s e p l o t
% i s a l w a y s l i n e a r l y −s c a l e d .
% mag and pha a r e t h e magnitude and p h a s e o f t h e f r e q u e n c y r e s p o n s e f c t n
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%
Henri Gavin , Dept . C i v i l E n g i n e e r i n g , Duke U n i v e r s i t y , h e n r i . gavin@duke . edu
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% Krajnik , Eduard , ’A s i m p l e and r e l i a b l e p h a s e unwrapping a l g o r i t h m , ’
% h t t p : / /www. mathnet . or . k r / mathnet / p a p e r f i l e / Czech / Eduard / p h a s e . ps
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[n ,m , p ] = abcddim (a ,b ,c , d );
% check f o r c o m p a t i b i l e dimensions
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if
if
if
if
( nargin
( nargin
( nargin
( nargin
<
<
<
<
10) tol = 1e -16; end
7) axes = ’b ’;
end
6) w = logspace ( -2 ,2 ,200)*2* pi ;
5) iu = 1;
end
% default
% default
end
%
% default
rcond l e v e l
plot formatting
d e f a u l t frequency axis
input
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nw = length ( w );
biu = b (: , iu );
diu = d (: , iu );
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I = sqrt ( -1.0);
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%warning o f f
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g = zeros ( nw , p ); % a l l o c a t e memory f o r t h e f r e q u e n c y r e s p o n s e f u n c t i o n , g
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% compute t h e f r e q u e n c y r e s p o n s e f u n c t i o n , g
f o r j =1: nw
%
g ( j , : ) = [ c ∗ ( (w( j ) ∗ I ∗ e y e ( n)−a ) \ b i u ) + d i u ] ’ ;
[u ,s , v ] = svd( w ( j )* I *eye( n ) - a );
% s i n g u l a r v a l u e decomp . o f A
idx = max( find ( diag ( s ) > s (1 ,1)* tol ) );
char_eq = v (: ,1: idx ) * inv ( s (1: idx ,1: idx )) * u (: ,1: idx ) ’;
g (j ,:) = [ c * char_eq * biu + diu ] ’;
end
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mag = abs( g );
pha = zeros ( nw , p );
pha (1 ,:) = -atan2(imag( g (1 ,:)) , r e a l ( g (1 ,:))); % −ve s i g n !
pha (2: nw ,:) = ones( nw -1 ,1)* pha (1 ,:) - cumtrapz ( angle ( g (2: nw ,:)./ g (1: nw -1 ,:)));
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%pha
= unwrap(−atan2 ( imag ( g ) , r e a l ( g ) ) ) ;
% −ve s i g n !
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i f (nargout < 2 )
% s e t up t h e p l o t s
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i f ( nargin >= 8) , f i g u r e ( figno ); e l s e f i g u r e (13); end
clf
subplot (211)
i f ( axes == ’x ’)
semilogx ( w /2/ pi , mag );
% p l o t t h e magnitude r e s p o n s e
e l s e i f ( axes == ’y ’)
semilogy ( w /2/ pi , mag );
% p l o t t h e magnitude r e s p o n s e
CC BY-NC-ND H.P. Gavin
18
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CEE 541. – Structural Dynamics – Duke University – Fall 2014 – H.P. Gavin
e l s e i f ( axes == ’n ’)
plot ( w /2/ pi , mag );
else
log log ( w /2/ pi , mag );
end
i f ( nargin >= 9) , legend ( leg ); end
% p l o t t h e magnitude r e s p o n s e
% p l o t t h e magnitude r e s p o n s e
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axis ([ min( w )/2/ pi , max( w )/2/ pi , min(min( mag )) , max(max( mag )) ]);
ylabel ( ’ magnitude ’ );
grid on
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subplot (212)
i f ( axes == ’n ’ || axes == ’y ’)
plot ( w /(2* pi ) , pha *180/ pi );
% p l o t the phase
else
semilogx ( w /(2* pi ) , pha *180/ pi );
% p l o t the phase
end
i f ( nargin >= 9) , legend ( leg ); end
pha_min = f l o o r (min(min( pha *180/ pi ))/90)*90;
pha_max = c e i l (max(max( pha *180/ pi ))/90)*90;
set (gca , ’ ytick ’ , [ pha_min : 90 : pha_max ])
axis ([ min( w )/2/ pi max( w )/2/ pi pha_min pha_max ]);
xlabel ( ’ frequency ( Hertz ) ’)
ylabel ( ’ phase ( degrees ) ’)
grid on
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end
end
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% −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− MYBODE
CC BY-NC-ND H.P. Gavin
19
Linear Time Invariant Systems
11 Numerical Example: a spring-mass-damper oscillator
For the single degree of freedom oscillator of Section 2, let’s say m = 2 ton,
c = 1.4 N/mm/s, k = 6.8 N/mm, do = 5.5 mm and vo = 2.1 mm/s. Note
that these units are consisent. (1 N)=(1 kg)(1 m/s2 )=(1 ton)(1 mm/s2 )
For these values, the linear time invariant system description of equations (7)
and (8) become

ẋ(t) =


y(t) =




0
1 
0 
x(t) + 
u(t) ,
−3.4 −0.7
0.5



x(0) = 
5.5
2.1


(38)

6.8 1.4 
0 
x(t) + 
u(t)
−3.4 −0.7
0.5
(39)
1. What are the natural frequencies and damping ratios of this system?
>> A = [ 0 1 ; -3.4 -0.7 ]
% the dynamics matrix
A =
0.00000
-3.40000
1.00000
-0.70000
>> eig(A)
% eigenvalues of the dynamics matrix
ans =
-0.3500 + 1.8104i
-0.3500 - 1.8104i
>> wn = abs(eig(A))
% absolute values of the eig’s of A are omega_n
wn =
1.8439
1.8439
>> z = -real(eig(A)) ./ wn
% ratio of real eig(A) to omega_n is damping ratio
z =
0.18981
0.18981
CC BY-NC-ND H.P. Gavin
20
CEE 541. – Structural Dynamics – Duke University – Fall 2014 – H.P. Gavin
>> wd = imag(eig(A))
% imaginary parts of the eig’s of A are omega_d
wd =
1.8104
-1.8104
q
So, the natural frequency is 1.84 rad/s (the same as k/m), the damping
√
ratios is 19% (the same as c/ q 4mk), and the damped natural frequency
is 1.81 rad/s (the same as ωn |ζ 2 − 1|).
These calculations can be done in one step with the m-file damp.m
>> damp(A)
Natural
Damped
Frequency
Frequency
Eigenvalue
(cyc/sec)
Damping
(cyc/sec)
real
imag
----------------------------------------------------0.29347
0.18981
0.28813
-0.35000
1.81039
0.29347
0.18981
0.28813
-0.35000
-1.81039
2. What is the free response of this system to the specified initial conditions
xo ?
>>
>>
>>
>>
>>
>>
>>
>
>
>>
>>
>>
>>
dt = 0.01;
%
n = 1000;
%
t = [0:n-1]*dt;
%
xo = [ 5.5 ; 2.1];
%
C = [ 6.8 1.4 ; -3.4 -0.7 ];
%
y = zeros(2,n);
%
for k=1:n
y(:,k) = C*expm(A*t(k))*xo;
end
plot(t,y)
legend(’foundation force, N’, ’mass
xlabel(’time, s’)
ylabel(’outputs, y_1 and y_2’)
time step value, sec
number of time steps
time values, starting at t=0
initial state (mm, mm/s)
output matrix
initialize outputs
acceleration, mm/sˆ2’)
The free response is plotted in figure 1.
CC BY-NC-ND H.P. Gavin
21
Linear Time Invariant Systems
foundation force, N
mass acceleration, mm/s2
40
outputs, y1 and y2
30
20
10
0
-10
-20
0
2
4
6
8
10
time, s
Figure 1. Free response of the linear time invariant system given in equations (38) and (39).
3. What is the response of this system to a unit impuse, u(t) = δ(t)?
>>
>>
>>
>>
>
>
>>
>>
>>
>>
B =
v =
h =
for
[ 0 ; 0.5 ];
% input matrix
1;
zeros(2,n);
% initialize unit impulse responses
k=1:n
h(:,k) = C*expm(A*t(k))*B*v;
end
plot(t,h)
legend(’foundation force, N’, ’mass acceleration, mm/sˆ2’)
xlabel(’time, s’)
ylabel(’unit impuse responses, h_1(t) and y_2(t)’)
The unit impuse response is plotted in figure 2.
4. What is the response of this system to external forcing u(t) = 50 cos(πt)?
>> D = [ 0 ; 0.5 ];
% feedthrough matrix
>> u = 50 * cos(pi*t);
% external forcing
>> y = lsym(A,B,C,D,u,t,xo); % use the "lsym" command
>> plot(t,y)
>>
legend(’foundation force, N’, ’mass acceleration, mm/sˆ2’)
>>
xlabel(’time, s’)
>>
ylabel(’forced harmonic responses, y_1(t) and y_2(t)’)
CC BY-NC-ND H.P. Gavin
22
CEE 541. – Structural Dynamics – Duke University – Fall 2014 – H.P. Gavin
foundation force, N
mass acceleration, mm/s2
unit impuse responses, h1(t) and h2(t)
1.5
1
0.5
0
-0.5
-1
0
2
4
6
8
10
time, s
Figure 2. Unit impulse response of the linear time invariant system given in equations (38) and
(39).
The forced harmonic response is plotted in figure 3.
5. What is the frequency response function from u(t) to y(t) for this system?
>> w = 2*pi*logspace(-1,0,100);
>> bode(A,B,C,D,1,w);
% frequency axis data
The magnitude and phase of the steady-state forced harmonic response
is plotted in figure 4. Note how the magnitude and phase of the frequency response shown in figure 4 can be used to predict the steady
steate response of the system to a forcing of u = 10 cos(πt). (The forcing
frequency is 0.5 Hertz).
This is a fairly simple example. Nevertheless, by simply changing the definitions of the system matrices, A, B, C, and D, and of the input forcing
u(t), any linear time invariant system may be analyzed using the same sets
of Matlab commands.
CC BY-NC-ND H.P. Gavin
23
Linear Time Invariant Systems
60
input forcing
40
20
0
-20
-40
-60
0
2
4
6
8
10
responses, y1(t) and y2(t)
time, s
60
40
20
0
-20
-40
foundation force, N
mass acceleration, mm/s2
-60
-80
0
2
4
6
8
10
time, s
magnitude
Figure 3. Forced response of the linear time invariant system given in equations (38) and (39).
100
10
-1
100
phase (degrees)
180
90
0
-90
-180
100
frequency (Hertz)
Figure 4. Frequency response of the linear time invariant system given in equations (38) and
(39).
CC BY-NC-ND H.P. Gavin