Download Discrete-Time State-Space Equations Outline Solution of State

Outline
• Discrete-time (DT) state equation from
solution of continuous-time state equation.
• Expressions in terms of constituent
matrices.
• Solution of DT state equation.
• Example.
Discrete-Time State-Space
Equations
M. Sami Fadali
Professor of Electrical Engineering
UNR
1
Solution of State Equation
• Analog systems with piecewise constant inputs over
a sampling period: relate state variables at the end
of each period by a difference equation.
• Obtain difference equation from the solution of the
analog state, over a sampling period .
• Solution of state equation for initial time
final time
2
Piece-wise Constant Input
1. Move input outside the integral.
2. Change the variable of integration
, and
Discrete-time state equation
= state vector at time
3
4
State & Input Matrices
Constituent Matrices
discrete state matrix
discrete input matrix
(same orders as their continuous counterparts).
• Discrete state matrix = state transition matrix of
the analog system evaluated at the sampling
period .
• Properties of the matrix exponential: integral of
the matrix exponential for invertible matrix
• Use expansion of the matrix exponential in
terms of the constituent matrices.
• Eigenvalues of discrete state matrix related
to those of the analog system.
5
Discrete-time State-space
Representation
Input Matrix
• Discrete state & output equation.
• Discrete-time state equation:
approximately valid for a general input
provided that the sampling
vector
period is sufficiently short.
• Scalar integrands: easily evaluate integral.
1
,
1
6
0
,
• Output equation evaluated at time
0
• Assume distinct eigenvalues (only one zero eigenvalue)
7
8
Example 7.15
Discrete state matrix
.
• Obtain the DT state equations for the system of
Example 7.7
.
.
• for a sampling period T=0.01 s.
• Solution: From Example 7.7, the state-transition
matrix is
10 11 1
0
0 0
10
0
0 0
0
0
0
10
10
10
1
1
1
9
0
0
0
1
10
100
1
10
100
• Simplifies to
90
9
Discrete-time Input matrix
.
.
10
MATLAB
.
MATLAB command to obtain
form
» pd = c2d(p,0.01)
Alternatively the matrices are obtained using
the MATLAB commands
» ad = expm(a * 0.05)
» bd = a\ (ad-eye(3) )* b
.
• Simplifies to
11
12
Solution of DT State-Space
Equation
Solution by Induction
2
• DT State Equation: state at time in terms
and the
of the initial condition vector
.
input sequence
• At
we have
0
0
• State-transition matrix for the DT system .
• State-transition matrix for time-varying DT system:
– not a matrix power
– dependent on both time k and initial time k0 .
• Solution=zero-input response+ zero-state response
13
14
Z-Transform Solution of DT State
Equation
Output Solution
• Substitute in output equation
• z-transform the discrete-time state equation
15
16
Matrix Inversion
Inverse z-transform
• Evaluate using the Leverrier algorithm.
• Partial fraction expansion then multiply by .
• Inverse z-transform
Z
• Analogous to the scalar transform pair
Z
17
18
DT State Matrix
Zero-state Response
• Parentheses: pertaining to the CT state matrix .
• Equality for any sampling period and any matrix
•
Known inverse transform for
•
Multiplication by
term.
: delay by .
Convolution theorem: inverse of product is the
convolution summation
• Same constituent matrices for DT state matrix & CT
state matrix A
• DT eigenvalues are exponential functions of the CT
eigenvalues times the sampling period.
19
20
Example 7.16
Alternative Expression
x ZS (t ) 
k 1  n
 
i 0
x ZS (t ) 
n
 j 1
Z je
 Z j Bd e
(a) Solve the state equation for a unit step input
and the initial condition vector x(0) = [1 0]T
(b) Use the solution to obtain the discrete-time state
equations for a sampling period of 0.1s.
(c) Solve the discrete-time state equations with the
same initial conditions and input as in (a) and
verify that the solution is the same as that of (a)
evaluated at multiples of the sampling period T.
 j  A  k i 1 T 
 j  A  k 1 T
j 1
 Bd u(i )

 k 1  j  A iT

u(i )
 e
 i 0


Useful when the summation over i can be
obtained in closed form.
21
1   x1  0
 x1   0

 x   2  3  x   1u
 2   
 2 
22
State-transition Matrix
Solution
(a)The resolvent matrix
• Partial fraction expansions
23
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Zero-input Response
Zero-state Response
1  t  1  1 2t 0
 2
x ZS (t )  
 e   2 2  e 1 1t 


2
1




 

 1
1
   e t 1t     e 2t 1t 
 1
2
1
 1 1  e 2t
t
x ZS (t )    1  e    
 1
2 2
1 2  1  t  1 e 2t
     e   
 0   1
2 2
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Total Response
26
(b) Discrete-time state equations
.
x(t )  x ZI (t )  x ZS (t )
 2  t  1 2t 1 2  1  t  1 e 2t
x(t )    e    e       e   
 2
2
 0   1
2 2
1 2  1  t  1 e 2t
     e   
 0   1
2 2
At the sampling points: t = multiples of 0.1s
27
.
.
.
.
.
• CT system response to a step input of duration one
sampling period, is the same as the response of a
system due to a piecewise constant input
28
z-transform of the zero-state
response
Zero-input Response
 2 1  0.1k 1 1 0.2 k
Adk   (01
. k)  

e
e
2 2
2 1
x ZI ( k )   (01
. k )x(0)
z
z
1 1
2 1
( z )  



0.1
0 . 2
2 2  ze
2 1 z  e
 2 1  0.1k 1 1 0.2 k 1

 
 
e
e
2 2
2 1
0
X ZS ( z )   ( z ) z 1BdU ( z )
1
2
  e 0.1k   e 0.2 k
2
 2 
Same as the zero-input response of the continuous-time
system at all sampling points k = 0, 1, 2, ...
2
 1
x ZI (t )    e t    e  2t
  2
2
29
Partial Fractions
z  e  z  1


30
 1
0.5  1 
x zs (k )       e 0.1k  0.5  e 0.2 k
2
 0   1
 1z 
 z
 10.5083


 z  1 z  0.9048 


 0 .5  z
1 z
 1
z
 
 0 .5  
X zs ( z )   
0.1
0.2
 0  z  1  1 z  e
 2  z e
 1z 
1  z


0.1 
0.1
 z  e  z  1 1  e  z  1 z  e 0.1 
z

Expand zero-state response
z
0.2
1
 2
z
z 0.0045 z 1 z
 1  1

 
0.1
 2 2  z  e 0.2  0.0861 z  1




 2  1 z  e

z
1
 9.5163  10 2  
0.1
 z  1
 1 z  e
 1
z
 9.0635  10 2  
0.2
 z  1
 2  z e
Identical to the zero-state response for the
continuous system at time t = 0.1 k, k = 0,1,2,...
 1z 
1  z

0.2 
1  e  z  1 z  e 0.2 
 1 e  2t
1 2  1 
x ZS (t )       e t   
2 2
 0   1
 1z 
 z
 5.5167 

 z  1 z  0.8187 
31
32
Zero-state response
n
x ZS (t )   Z j Bd e
 j  A  k 1 T
j 1
k 1
 ai 
i0
x ZS ( k ) 
n
Z
j
Bd e
 k 1  j  A iT

u(i )
 e
 i 0

1 ak
1 a
 j  A k  1 T
j 1
z-Transfer Function
• For zero initial conditions
1  e   j  A  kT 

  j  AT 
1 e

• Substitute
1  e  j  A  kT 
  Z j Bd 
 j  AT 
j 1
1 e

n
33
Impulse Response & Modes
Poles and Stability
• Inverse transform of the transfer function
Z
↔
,
1
,
0
• Substitute in terms of constituent matrices
1
Z
↔
,
,
34
1
0
35
• poles= eigenvalues of discrete-time state
matrix
= exponential functions of
(continuous-time state matrix ).
have negative real parts
• For stable ,
and have magnitude less than unity.
• Discretization yields a stable DT system
for a stable CT system.
36
Minimal Realizations
•
•
•
•
•
Decoupling Modes
Product
can vanish & eliminate eigenvalues
from the transfer function: if
,
, or both.
If cancellation occurs, the system is said to have an
–
: output-decoupling zero at
–
: an input-decoupling zero at
–
, &
: an input-output-decoupling zero
at
Poles of the reduced transfer function are a subset of
the eigenvalues of the state matrix .
A state-space realization that leads to pole-zero
cancellation is said to be reducible or nonminimal.
If no cancellation occurs, the realization is said to be
irreducible or minimal.
•
•
•
•
Output-decoupling zero at : the forced system
response does not include the mode .
Input-decoupling zero: the mode is decoupled
from or unaffected by the input.
Input-output-decoupling zero: the mode is
decoupled both from the input and the output.
These properties are related to the concepts of
controllability and observability discussed later
in this chapter.
37
38
Example 7.17
Solution
z
z
2 1
1 1
( z )  



0.1
0 . 2
2 2 ze
2 1 z  e
• Obtain the z-transfer function for the position
control system of Example 7.16
(a) With x1 as output. (b)
1 2  1 
1 e 0.2 0.0045
Bd      e 0.1   


 0  1
2 2
0.0861
With x1 + x2 as output.
The resolvent matrix and input matrix were
obtained in Example 7.16.
(a)
Output
:
1 1
 2
 1  1 1 0.0045

G ( z )  1 0
0.1
0.2  



 2 2  z  e  0.0861
 2  1 z  e
9.5163 10 2 9.0635 10 2


z  e 0.1
z  e 0.2
39
40
(c) Output x1+x2
Step Response
1 1
 2
 1  1 1 0.0045

G ( z )  1 1
0.1
0.2  



 2 2  z  e  0.0861
 2  1 z  e
0
9.0635 10 2


z  e 0.1
z  e 0.2
.
1. Output-decoupling zero at
since
2. System response to any input does not include
the decoupling term.
9.0635  10 2 z
Y ( z) 
z  e 0.2  z  1
 1z   0.5 z   1z 
9.0635 10 2  z


 0 .2
 0. 2

 z  1 z  0.8187 
1 e
 z  1 z  e 
Z-transform inverse
41
z-Transfer Function: MATLAB
• Let T =0.05s
» P = ss(Ad, Bd, C, D, 0.05)
» g = tf(P) % Obtain z-domain transfer function
» zpk(g) % Obtain transfer function poles and zeros
• The command reveals that the system has a zero
at 0.9048 and poles at (0.9048, 0.8187) with a
gain of 0.09035.
• With pole-zero cancellation, the transfer function
is the same as that of Example 7.17(b).
» minreal(g) % Cancel poles and zeros
43
.
42