1. No category

y(t) = Cx(t)
(7.1)
which represents the open-loop system or plant to be controlled. Our
focus is on the application of state feedback control laws of the form
u(t) = −Kx(t) + r(t)
(7.2)
with the goal of achieving desired performance characteristics for the
closed-loop state equation
x(t)
˙ = (A − BK )x(t) + Br(t)
Introduction
y(t) = Cx(t)
D
x0
u(t)
150
B
x(t)
∫
+
+
OBSERVABILITY
x(t)
C
+
y(t)
+
A
We begin the chapter with an analysis of observability patterned after
our introduction to controllability. This suggests a duality that exists
between controllability and observability that we develop in detail. This
pays immediate dividends in that various observability-related results can
be established with modest effort
by referring to the corresponding result
Observability
for controllability and invoking duality. In particular, we investigate relationships between observability and state coordinate transformations as
well as formulate Popov-Belevich-Hautus tests for observability. We conInstructor:
Dr.
Vardy
for observability
clude the chapter
by illustrating
theAndrew
use of MATLAB
analysis and revisit the MATLAB Continuing Example as well as Continuing
Examples 1 and 2.
Adapted from the notes of
Gabriel Oliver Codina
4.1 FUNDAMENTAL RESULTS
ENGI 7825: Control Systems II
For the n–dimensional linear time-invariant state equation
x(t)
˙ = Ax(t) + Bu(t)
y(t) = Cx(t) + Du(t)
x(t0 ) = x0
►  If we
have
a model
of ouronsystem,
and can
apply
an input
The
effect
of state
feedback
the open-loop
block
diagram
of and
Figure 1.1
output.
is measure
shown in the
Figure
7.1. Thus the following quantities are known:
■  A,state
B, C, feedback
and D
The
control law (7.2) features a constant state feedback
u(t)
gain■  matrix
K of dimension m × n and a new external reference input r(t)
■  y(t)
necessarily
having the same dimension m × 1 as the open-loop input u(t),
►  However,
state
vector x(t)
may
notin
bethis
observable.
be
as
well as thethe
same
physical
units.
Later
chapter weThis
will can
modify
problematic when trying to design a compensator (K below):
D
x0
r(t) +
we assume that the input signal u(t) and the output signal y(t) can be
measured over a finite time interval and seek to deduce the initial state
x(t0 ) = x0 by processing this information in some way. As noted earlier, if the initial state can be uniquely determined, then this, along with
knowledge of the input signal, yields the entire state trajectory via
►  Recall that the state vector’s current value is given by:
! t
x(t) = eA(t−t0 ) x0 +
eA(t−τ ) Bu(τ ) dτ for t ≥ t0
How to Recover x(t)
u(t)
−
B
IfSince
the system
(i.e.toA,beB,known,
C, andthe
D), zero-state
t0, the input,
and output
u(t) is model
assumed
response
can beare
known then
only unknown
on y(t),
the RHS
is x0. Ifinwe
knew
x0 then
extracted
from the
the complete
response
also known,
order
to isolate
wezero-input
could recover
anycomponent
value of x(t).
the
response
via
"! provides
►  The output, y(t),
a window on the #system
t
)
CeA(t−τ
Bu(τ )dτ
+ Du(t) = CeA(t−t0 ) x0
y(t) −
t
A(t−t )
t0 = Ce
y(t)
x 0 + ∫ Ce A(t−τ )Bu(τ )d τ + Du(t)
t
0
0
which depends directly on the unknown initial state. Consequently, we can
assume without loss of generality that u(t) ≡ 0 for all t ≥ t0 and instead
zero-input output: y (t)
y (t): zero-state output
consider the homogeneouszi state equation zs
►  We can set u(t) = 0 and try and solve for x0:
x(t)
˙
= Ax(t)
x(tto
x0whether it is even
(4.2)
0) =
►  In general, we would like a simple test
say
y(t) = Cx(t)
possible to solve for x0.
which directly produces the zero-input response component of Equation (4.1).
x(t)
x(t)
C
+
+ y(t)
+6.3 Observability
153
K
FIGURE 7.1 Closed-loop system block diagram.
The concept of observability is dual to that of controllability. Roughly speaking, controllability
studies the possibility of steering the state from the input; observability studies the possibility
of estimating the state from the output. These two concepts are defined under the assumption
that the state equation or, equivalently, all A, B, C, and D are known. Thus the problem of
observability is different from the problem of realization or identification, which is to determine
or estimate A, B, C, and D from the information collected at the input and output terminals.
Consider the n-dimensional p-input q-output state equation
Observability
x˙ = Ax + Bu
(6.22)
y = Cx + Du
t0
► 
•
+
state equation describes. The physical significance of the controllability index is not transparent
here; but it becomes obvious in the discrete-timeAcase. As we will discuss in later chapters, the
controllability index can also be computed from transfer matrices and dictates the minimum
degree required to achieve pole placement and model matching.
6.3 Observability
(4.1)
(7.3)
where A, B, C, and D are, respectively, n × n, n × p, q × n, and q × p constant matrices.
Definition 6.O1 The state equation (6.22) is said to be observable if for any unknown
initial state x(0), there exists a finite t1 > 0 such that the knowledge of the input u and
the output y over [0, t1 ] suffices to determine uniquely the initial state x(0). Otherwise,
the equation is said to be unobservable.
Example 6.6 Consider the network shown in Fig. 6.5. If the input is zero, no matter what the
initial
the capacitor
the output is
is identically
zero because
of thecan
symmetry
In voltage
otheracross
words,
the is,system
observable
if we
of the four resistors. We know the input and output (both are identically zero), but we cannot
solveuniquely
for the
initial
x(0).
determine
the initial
state.state
Thus the
network or, more precisely, the state equation
that describes the network is not observable.
Example 6.7 Consider the network shown in Fig. 6.6(a). The network has two state variables:
the current x1 through the inductor and the voltage x2 across the capacitor. The input u is a
Figure 6.5 Unobservable network.
+
+
1!
u
−
1F 1!
+
1!
x
−
1
y
1!
−
OBSERVABILITY EXAMPLES
158
Observability matrix
tf
t0
⎢ CA ⎥
T
) ⎢ T A⎥ (t0 −τ )
eA(t0 −τ
dτ W −1 (t0 , tf )x
⎢ CAe2 ⎥
Q =BB
⎢
⎥
$
%

= eA(tf −t0 ) x − W (t0 , tf )W −1⎢⎢(t0 ,nt−1f⎥⎥)x
⎣CA
#
⎦
It can be proved
is OBSERVABILITY
observable if and only if: rank (Q) = n
= 0 that a system
158
that the arbitrarily
selected
► thereby
For theverifying
general multiple-output
(p) case,
A−1
is
state xisn controllable
matrix
and Ctoisthe
p x n.
= M (tan, tfn)M(t
0 , tf ) x0
n-1
Then,which
Q consists
of n matrix
blocks C, CA, CA02 …CA
, each with
!
origin,
concludes
the proof.
dimension
p
x
n,
stacked
one
on
top
of
another.
Thus,
Q
has
dimension
= x0
Returning to the claim made at the beginning
of this section, given anp x
n, having more rows than
columns.
which
uniquely recovers
the initial left
state.to the reader
controllable state equation,
a straightforward
calculation
►  For the single-output case, C consists of a single row, yielding a square n x n
shows
that the matrix
input signal
defined abysingle-output linear system is observable
observability
Q. Therefore,
if and only ifT the
associated observability
matrix Q is nonsingular. (|Q|≠0)
AT (t −t) −1
A(t −t )
u(t) = B e
W 4.2
(t0 , OBSERVABILITY
tf )(e 0 f xf − xEXAMPLES
0 ) for t0 ≤ t ≤ tf
0
As it happens for controllability, observability is invariant with respect to
steers
the state
trajectory from the initial state x(t0 ) = x0 to the final state
coordinate
transformation.
xf lying 4.1
anywhere
inthe
thetwo–dimensional
state space. For
the
x(tf ) = xf , with x0 and Example
Consider
single-output
state equation
► 
x(t"f!) = x"f is!often
special158
case OBSERVABILITY
in which x(t0 ) = 0 ∈ Rn , !the final
" state
!
"
1 5a controllable
x1 (t)
−2
x˙1 (t)
referred to as reachable from the origin. Consequently,
state
=
+
u(t)
−1
8 4statex2equation.
(t)
2
= M (treferred
, tfas
)x˙x2a0(t)reachable
equation is also sometimes
0 , tf )M(t0to
3.2
= x0
which uniquely recovers the
initial state. y(t) = [ 2
CONTROLLABILITY
EXAMPLES
2]
!
"
x1 (t)
+ [0]u(t)
x2 (t)
which thesingle-input
associated (A,
B) pair is the same
as in Example 3.1. The
two–dimensional
linear
Example 3.1 Given theforfollowing
4.2 OBSERVABILITY
EXAMPLES
observability
matrix Q is found as follows:
state equation,
we now assess
its controllability.
!
# !
#!
# !
#
x˙1 (t)
1 5
x1 (t)
−2 C = [ 2 2 ]
=
+
u(t)
Example 4.1
single-output
state equation
x˙2 (t)Consider 8the two–dimensional
x2 (t)
4
2
Example 2
► 
Consider the following system (same one we sawCA
for=
controllability):
[ 18 18 ]
!
"
!
"!
"
!
"
so
!
"
The controllability matrix
follows:−2
5 as
x1 (t)
x˙1 (t) P is 1found
2
u(t) 2
! x˙2#(t) = 8 !4 x#2 (t) + !2Q =
#
−2
8
−2 18
8 18
B=
AB = ! x1 (t) "P =
Clearly
= 0 so the
state 2equation
2 y(t)
−8 is not observable. Because rank
2 ]−8
+ [0]u(t)
= [ 2 |Q|
Q < 2 but Qx2is(t)not the 2 × 2 zero matrix, we have rankQ = 1 and
|P | = 0, somatrix
the state
To see why variables.
this
nullityQ
= 1.ofis1,not
► Clearly,
The observability
hasequation
a rank
so controllable.
there are unobservable
which the
associatedTo
(A,see
B)
pairthis
is the
same
as in Example
3.1. The we again use the state
is true,for
consider
a different
state
definition
why
state
equation
is not observable,
►  Try using
different
stateQvariables:
observability
matrix
is found
follows:
# !as
#
! coordinate
transformation
given
by:
x1 (t)
z1 (t)
=
!
" !
"!
"
2+"] x2 (t)
z2 (t) C =x![1z2(t)
x1 (t)
1 0
x1 (t)
1 (t)
=
CA = [ z18(t) 18 ]
x (t) + x (t)
=
The corresponding transform is
2
1
2
The
associated
coordinate transformation
xis(t ) ⎤
⎡ z!1 (t ) ⎤ (see
⎡1 " 0Section
⎤ ⎡ x1 (t ) ⎤ 2.5)
so
−1 ⎡ 1
► 
1 1
x2 (t)
⎢ z (the
⎥ = T equation
⎢ x (t )⎥
2 ⎥ = 2⎢1 1⎥ ⎢ x (t )state
which
yields
⎣
⎦ ⎣ 2 ⎦
2 t )⎦ transformed
⎣ 2 ⎦
x(t) =Q
T=⎣z(t)
116
CONTROLLABILITY
18
!
# ! 18
#
!
#
!
" !
"!
" !
"
►  This yields
system:
(t) state
1 z˙0(t) z1 (t)
z1 (t) rank−2
−4 5 Because
Clearlythe
|Q|transformed
= 0 sox1the
observable.
= equation 1is not=
+
u(t)
x2 (t)
z2 (t)
Applying this coordinate
transformation
state
Q < 2 but
Q is not
the 2 yields
× 2−1
zerothe
we 0have9 rankQ
= 1 and 0
(t) transformed
z˙12matrix,
z2 (t)
nullityQ
=
1.
"
!
equation
► 
is decoupled
from !
" To see
!y(t),
! equation
"
"
! The output,
z1 (t)
why this"state
is not observable,
we again
use the state
is−4
further
from −2 y(t) = [ 0 2 ] z (t) + [0]u(t)
z1 (t)
z˙11(t)
(t)which
5 decoupled
coordinate
transformation
given+by:
2
=
u(t)
to determine.
z˙ 2 (t) so z1(t) is0 impossible
z2 (t)
9
0
!
" !
" !
"!
"
z1 (t)
x1 (t)
1 0
x1 (t)
=
=
z
x
x
(t)
(t)
+
x
(t)
1
1
(t)
2
1
2
2
depend on the input u(t), so this state variable
We see that z˙ 2 (t) does not
is not controllable. which yields the transformed state equation
!
"
!
"!
"

x˙1 (t)
 x˙2 (t)  =
'
0
0
!
"
4
4
►  First
0 0 0 1 0
5 (t)
calculate ex˙5Ât(t)(t0 = 0):
COVERED xON
BOARD
►  Now
solve for z0: COVERED ON BOARD
►  We
cannot solve for z1(0) because it has no effect on y(t)
that a lack of observability is the least of this
system’s problems---it is also unstable! (the e9t terms)
►  Note
!
!
"
z1 (t) single-input
−4 5
−2
z˙ 1 (t) three-dimensional
state
Example 3.2 Given the following
=
+
u(t)
0 9
0
z˙ 2 (t)
z2 (t)
equation, that is,

Example 4.3 We investigate the observability of the three-dimensional
state equation
= M −1 (t0 , tf )M(t0 , tf ) x0

 
  

= x0x˙1 (t)
x1 (t)
0 0 −a0
b0
which uniquely recovers
the
initial
state.
 x˙2 (t)  =  1 0 −a1   x2 (t)  +  b1  u(t)
x˙2 (t)
x2 (t)
0 1 −a2
b2
EXAMPLES
161
 OBSERVABILITY

(t)
x
4.2 OBSERVABILITY EXAMPLES
1
Example 4.3 We investigate
ofthe three-dimensional
0 observability
0 1 ]  x2 (t)
y(t) = [ the
state equation
x2 (t)
  the two–dimensional
   state equation

Example 
4.1 Consider
single-output
x˙1 (t) reader 0will0 recall
(t)
−a0 is ax1state-space
b0
which
of the transfer
"
!
"
!
"
!
" b1 realization
!
 x˙the




 u(t)
(t) = 1 0 −a1
x (t)  + 
function2 x˙1 (t)
1 5
x1 (t) 2
−2
2
x˙2 (t)
x2s(t)
b2b
=0 1 −a2
+
u(t)
b
+
b
s
+
2
1
0
x˙2 (t)
8 4  x2 (t) 
2
H (s)!= 3 "
x s(t)+ a s 2 + a1 s + a0
x11(t)  2

0 [ 20 21 ]] x2 (t) + [0]u(t)
y(t)y(t)
= [=
(t)
The observability matrix Q xis
found as follows:
x22 (t)
 as in Example 3.1. The
is the same
for
which
the
associated
(A,
B)
pair
Determine
matrix:
C
whichthe
theobservability
reader will recall
is a state-space
realization of the transfer
observability matrix Q is found as follows:

function
CA
Q=
2
b2 s + b12s + b0
H (s) = C3= [ 2 CA
22 ]
s + a
2 s + a1 s + a0

1
CA = [ 180 18 ]0
The
observability
matrix
Q
is
found
as
follows:
−a2 
=
so
!  0 "1
2
2 2

Q =C  1 −a2 a2 − a1
18 18
 CA 
=independent
ThisClearly
matrix
hasobservability
ofQ3,state
the
characteristic
polynomial
This
matrix
is
to
the controllability
matrix P from
|Q|
=a 0rank
so the
equation
isofnot
observable.
Because
rank
2 identical
CA
coefficients
a2 , Q
a2,is3.3.
a0.notSo
the
system
fully
observable.
We could
solve
for
The
observability
matrix
Qhave
is independent
the transfer
Q < 2Example
but
the
2
× 2 zeroismatrix,
we
rankQ
= 1 of
and

x0 through:
function-numerator
coefficients
b0 ,1b1 , and b2 . The determinant of the
0
0
nullityQ
= 1.
 0|Q|is =
1 −1
−a0,2 sothe
To see
why this state
equation
not
observable,
we state
again equation
use the state
= is
observability
matrix
"=
is observable.
coordinate
given1 by:
a22 − a1 of the characteristic polynomial
Notetransformation
that this outcome
is−a
independent
2
coefficients
a0 , a!1 , and a2 , so" a state-space
in this form is
!
"
!
" ! realization
"
This observability
identical
to the
P from
(t) matrix
x1 (t)
1for0 anyx1system
1 (t)matrix isThis
always zobservable.
is also
truecontrollability
order
n, as we will
=
=
Example
3.3. zThe
x1 (t) +matrix
x2 (t) of the transfer
x2 (t) Q is 1independent
1
2 (t) observability
demonstrate
shortly.
!
function-numerator coefficients b0 , b1 , and b2 . The determinant of the
observability
matrix
is |Q| = −1
0, so the state equation is observable.
whichExample
yields the
transformed
state" =
equation
4.4 Consider
the
five-dimensional, two-output homogeneous
Note that this outcome is independent of the characteristic polynomial
!
" !
"!
" !
"
state equation
coefficients a0 , az˙ 11(t)
, and a2 ,−4
so a5 state-space
in this form is
z1 (t) realization
−2
u(t)
always observable.
any +system
order
= is also
 true
 will
for
 n, as we
0
9
(t)
0
z˙ 2 (t)This
z
2
(t)
x
˙
0
0
0
0
0
x1 (t)
1
demonstrate shortly.
!
"
!
1 0 0 0 0   x2 (t) 
 x˙2 (t)  z
1 (t)




0
2
+
[0]u(t)
y(t)
=
[
]
x3 (t) 
(t)  =z(t)
0 0 0 two-output
0 0  homogeneous
 x˙3the
Example 4.4 Consider
2
 x˙ (t)five-dimensional,
 
0 0 1 0 0   x4 (t) 
4
state equation
x˙5 (t)observable?
x (t)
0 0 0 1If observable
0
►  Why is this system
we

not
 5


x˙1 (t)
0 0 0 0 0
x1 (t)
should be able
to
solve
for
z
given
the
output
y(t)
0
 x˙2 (t)   1 0 0 0 0   x2 (t) 

 


 x˙3 (t)  =  0 0 0 0 0   x3 (t) 
 x˙ (t)   0 0 1 0 0   x (t) 
Example 1
OBSERVABILITY TEST
Given a system defined by its linear state equation,
the observability
CONTROLLABILITY
EXAMPLES matrix
115 is
defined as:
⎡ C ⎤
!
"
= eA(tf −t0 ) x −
161
OBSERVABILITY
z1 (t)
'+ [0]u(t)
(
y(t) (
=
[0 2] 
x1 (t) z2 (t)
1
0
0


x2 (t) +
0
1
1 u(t)
2
Determining Observability through Matlab
Matlab function:
obsv() calculates the obsevability matrix for state-space systems.
Syntax
Q=obsv(sys)!
! !Q=obsv(A, C) or Q=obsv(sys.A, sys.C)
■ 
3