EE 422G Notes: Chapter 7
Instructor: Zhang
Chapter 7 State-Variable Technique
7-1 Introduction
What is it about? Another method for system description and analysis.
What we already have?
(Why introduced Laplace Transform? to avoid the complexity of solving
high-order
differential equations!)
Is Laplace Transform method good enough or do we need more or other
techniques?
(1) How effective Laplace Transform is in solving high-order differential
equations with non-zero initial conditions?
(2) How effective Laplace Transform is in handing multivariable (multi-input
multi-output) system?
(3) Classical control theory: Laplace Transform based
Modern control theory: state equation based
Many modern control system design methods require state equation
description
of system to apply!
Characteristics of state-variable technique
(state equation description based):
(1) Uniform structure (form) for all linear-systems:
despite the order, the numbers of inputs and outputs,
and forms of the input functions.
Always : x  Ax  Bu
state equation
y  Cx  Du output equation
x : state vector, x = (x1, … xn)T
xjs : state variable
u = (u1, …, um)T : input vector
ujs : inputs
y = (y1,…yp)T : output vector
yjs : outputs
Page 7-1
EE 422G Notes: Chapter 7
Instructor: Zhang
A : nxn B: nxm
C: pxn D: pxm
Difference: dimensions of the vectors and matrices.
(2) Uniform method of solution:
Laplace: different equations, different input function different solution!
State: Different solution method for different inputs? No difference !
Different solution method for different systems (with different
orders)?
No! The same!
 uniform form, uniform methods for analysis and design!
Fundamental Characteristic
Laplace Transform, differential equation, …, convolution:
external inputs  external outputs!
State equation (state-variable technique):
External inputs  internal state variables (as a bridge)
 external outputs
help to understand the system better because of use of “internal state”!
What’s the state of a system?
7-2 State-Variable Concepts
(1) Example: University’s rank (simplified and idealized)
What’s the fundamental measurement of university system (dynamic system)?
Money? Very important, not a direct measurement of
university with great inertia!
What are the direct measurement which need long time to build
and change?
Faculty quality
Student quality
Good faculty => Research , Teaching 
Reputation , Good student 
Good students => student Reputation 
Graduate quality 
Good Faculty 
Faculty quality and student quality:
Page 7-2
EE 422G Notes: Chapter 7
Instructor: Zhang
State of university (dynamic system)
State equations:
UK’s ranking: output  c1 x f  c 2 x s ------- output equation
(xf Faculty quality , xs: student quality)
State equation:
Change of faculty quality:
x f  a11 x f  a12 xs  a13  salary
Change of student quality:
x s  b21 x f  b22 xs  b23  Scholorship  b24  Bashetball
At 2000: x f (2000), xs (2000) : initial state
At and after 2000: (state appreciation, private donations)
=> (salary, scholarship)
Administration, basketball: future inputs
=> UK’s ranking
(determined by initial state and future inputs)
(2) State, state-variables, state-space, trajectory:
State of a system at t 0 : includes the minimum information necessary to specify
completely the condition of the system at t0 and allow determination of all
system outputs at t>t0 when inputs up to time t are specified.
State : a set of state variables
State:
x 
x f
 xs 
(state vector)
State Space:
x f 
x

Set of all possible
 x  (All possible paired values
 s
( x f (1), x s (1)); ( x f (2), x s (2));...) )
Trajectory of the state
 x (2000)
x f

 x s (2000) 
 x (2001)
x f

 x s (2001) 
……
Page 7-3
EE 422G Notes: Chapter 7
Instructor: Zhang
a curve in two-dimensional space.
 x1 
x    
 x n 
In general
7-3 Form of the state equations
1. Form
Example :
 x1  a11 x1  a12 x 2  b11u1  b12u 2

 x 2  a 21 x1  a 22 x 2  b21u1  b22 u 2
 x  Ax  Bu
 y1  c11 x1  c12 x2  d11u1  d12 u 2

 y2  c21 x1  c22 x2  d 21u1  d 22 u 2
 y  Cx  Du
x 
x   1 ,
 x2 
a
A   11
a21
a12 
,
a22 
b
B   11
b21
b12 
b22 
y 
y   1,
 y2 
c
C   11
c21
c12 
,
c22 
d
D   11
d 21
d12 
d 22 
u 
u   1
u 2 
Page 7-4
EE 422G Notes: Chapter 7
Instructor: Zhang
In general :
 x1 
x     ,
 x n 
x  Ax  Bu

y  cx  du
A:nn
C : pn
( x : n 1 ,
 u1 
u     ,
u m 
 y1 
 
y    ,
yp 
 
state equation
output equation
B :nm
D:cm
u : m 1 ,
y : p 1 )
(2) Simulation example
 x1  a11 x1  a12 x2  b11u1  b12 u2

 x2  a21 x1  a22 x2  b21u1  b22 u2
y  c1 x1  c2 x2  d1u1  d 2u2
Page 7-5
EE 422G Notes: Chapter 7
Instructor: Zhang
(3) Block Diagram of state equation
 x  Ax  Bu

 y  cx  du
Homework
1. Given x (t )  Ax (t )  Bu (t ) with initial condition x(t0 )  x0 where x0 is the
given initial state. Construct an algorithm (recursive equation) to calculate
x(t0  (k  1)t ), (k  0,1,...) based on x(t0  kt ) and u(t0  kt ) where
t  0 is a small time interval.
2. Given the initial condition problem
 x  Ax  Bu

 x(t0 )  x0
It is known that the solution of this problem is unique. Prove that the unique
solution of this problem is
x(t )  e
A( t  t 0 )
t
x0   e A(t   ) Bu ( )d
t0
d
f ( , t )
f ( , t )d  f ( , t )  t  
d )
(Note:

dt a

t
a
t
t
3. Use Laplace transform to prove
Y (s)  [c(sI  A) 1 B  D]U (s)
 x (t )  Ax(t )  Bu (t ),
when 
 y (t )  Cx(t )  Du (t )
x(0)  0
Page 7-6
EE 422G Notes: Chapter 7
Instructor: Zhang
7-4 Time-Domain solution of the state equations
Focus: Find x(t) (t  t0) which satisfies
 x  Ax  Bu

 x(t0 )  x0
x0 given
1. Mathematical Preparation
(1) Matrix Exponential eAt
Scalar Exponential
e at  1  at 
a2 2 a3 3
t  t 
2!
3!
Introducing Notation
e At  I  At 
where :
1 2 2 1 33
A t  A t 
2!
3!
A : n  n matrix
0 
1
   n
I : n  n Identity matrix


0
1 
              
              
2



A : AA  

 
 



 

            

  
n n
Aj : nn
n n
n n
j = 0,1,2,…
t: scalar
e At : n  n
Properties of eAt

de At
 Ae At
dt
e At  I  At 
1 2 2 1 33
A t  A t 
2!
3!
Page 7-7
EE 422G Notes: Chapter 7
Instructor: Zhang
de At
2
3
 0  A  A 2 t  A 3t 2  
dt
2!
3!
1
 A( I  At  A2t 2  )
2!
 Ae At
e A0  I

1 2 2
A 0   I
2!
*
(Inverse of n  n matrix eAt is e-At)
(e At ) 1  e At
1
1
 I  ( A)t  ( A) 2 t 2  ( A)3 t 3  
2!
3!
1
1
 I  At  A2t 2  A3t 3  
2!
3!
1
1
 I  A(t )  A2 (t ) 2  A3 (t )3  
2!
3!
t
f ( , t )
f ( , t )d  f ( , t )   t  
d )

t
a
e A0  I  A  0 
d t
(2)
dt a
A( t  )
Bu ( )
If f ( , t )  e
t
d t
f ( , t )
f
(

,
t
)
d


f
(

,
t
)

Then
 t d
 t
dt t
t
0
But
0
f (, t )  t  e A(t t ) Bu (t )  IBu (t )  Bu (t ) ,
f ( , t )
 Ae A(t   ) Bu ( )
t
t
t
d
A( t  )
Bu ( )d  Bu (t )  A e A(t  ) Bu ( )d
Therefore dt  e
t0
t0
Page 7-8
EE 422G Notes: Chapter 7
Instructor: Zhang
t
Z (t )   e A(t   ) Bu ( )d
Denote
t0
=>
t
d
Z (t )   e A(t   ) Bu ( )d
dt t
0
 AZ (t )  Bu (t )
Z (t )  AZ (t )  Bu (t )
i. e.
2. Numerical solution of state equations
(Eq. (7-9)----Eq. (7-15))
3. Analytic solution of homogeneous equaiton
 x  Ax

 x(t 0 )  x0
zero-input response
question : what do we mean that x(t) = f(t) is a solution of a differential
equation with a
given initial condition ?
(1) x 
df (t )
=Af(t)
dt
(2) f (t ) t  0  x0
for example, if we assume x(t)=At
then x  A  A( At )  Ax (t )
 x(t)=At is not a solution of x (t )  Ax (t )
 x  Ax
question : Is x(t )  e x0 a solution of 
?
 x(t 0 )  x0
At
x (t ) 
d At
(e x0 )  Ae At x0  Ax (t )
dt
Page 7-9
EE 422G Notes: Chapter 7
=> x  Ax
Instructor: Zhang
 x  Ax
satisfies the first of 
 x(t 0 )  x0
However,
x(t0 )  e At 0 x0
If t0  0 and A  0 , e At0  I
=> x(t0 )  x0
Therefore,
 x  Ax
 x(t 0 )  x0
x(t )  e At x0 is not a solution of 
 x  Ax
Question: If we find a solution for 
 x(t 0 )  x0
Can we find a second different solution for it?
 x  Ax
No! The solution for 
is unique
 x(t 0 )  x0
when A = constant matrix!
 x  Ax
Let’s find “any” solution for 
. It will be the unique solution!
 x(t 0 )  x0
Suggested solution
x(t )  e A(t t 0 ) x0
verification:
de A(t  t 0 ) x0
(1) x 
 Ae A(t  t 0 ) x0  Ax
dt
A(t  t )
(2) x(t0 )  e 0 0 x0  Ix0  x0
=> x(t )  e A(t t 0 ) x0 : unique solution
Page 7-10
EE 422G Notes: Chapter 7
4. Analytic solution of
Instructor: Zhang
 x  Ax  Bu

 x(t 0 )  0
(a)
(b)
.
(zero-state response)
t
Suggested solution : x(t )   e A(t   ) Bu ( )d
t0
Verification:
From mathematical preparation:
t
z (t )   e A(t   ) Bu ( ) d  our suggested x(t)
t0
=>
z (t )  Az (t )  Bu (t )
Hence, suggested solution x(t) satisfies
x (t )  Ax (t )  Bu (t ) (a)
t0
Further, x(t0 )   e A(t 0   ) Bu ( )d  0 (b)
t0
 x  Ax  Bu
5. Solution of  x(t )  x
0
 0
(Given problem)
Solution: zero-input response + zero-state response
x(t )  e
A( t t 0 )
t
x0   e A(t  ) Bu ( )d
t0
6. State Transition Matrix (t )  e At
No input, x(t) is a transition of x0 at t0 to t>t0:
x(t )  e A(t t 0 ) x0  (t  t0 ) x0
7. Output
t
y (t )  C (t  t0 ) x0   C (t   ) Bu ( )d  Du (t )
t0
Page 7-11
EE 422G Notes: Chapter 7
Instructor: Zhang
7-5 Frequency-Domain solution of the State Equation
x(0)  x0
x  Ax  Bu ,
(t0  0)
1. Solution
sX ( s )  x0  AX ( s )  BU ( s )
( sI  A) X ( s )  x0  BU ( s )
X ( s )  ( sI  A) 1 x0  ( sI  A) 1 BU ( s )
x(t )  L1 [( sI  A) 1 x0 ]  L1 [( sI  A) 1 BU ( s )]
2. What is (sI  A) 1 ?
L1[(sI  A) 1 ]  e At  (t )
 (sI  A) 1 : Laplace transform of the state transition matrix or of eAt.
Denote: (s)  (sI  A) 1
What is (sI  A) 1 BU (s) ?
( s )(BU ( s)) ---- product of
 (s ) and BU (s ) !
 L1[(sI  A) 1 BU (s)] : convolution of
L1[( sI  A) 1 ]  (t ) and L1[ BU (s)]  Bu (t ) !
Hence
t
L [( sI  A) BU ( s)]  (t ) * Bu (t )   (t   ) Bu ( )d
1
1
0
3. Output Laplace Transform and Transfer Function Matrix
Y ( s)  CX ( s)  DU ( s)
 C ( sI  A) 1 x0  c( sI  A) 1 BU ( s)  DU ( s)
When x0  0
Y ( s)  C ( sI  A) 1 BU ( s)  DU ( s)
 [C ( sI  A) 1 B  D]U ( s)
Page 7-12
EE 422G Notes: Chapter 7
Instructor: Zhang
( p  m)
Denote H (s)
 C ( sI  A) 1 B  D
=> Y ( s )  H ( s )U ( s )
H(s): Transfer function matrix
 H11 ( s) H12 ( s)  H1m ( s) 


H (s)   

 H p1 ( s) H p 2 ( s)  H pm ( s)


Hij(s) : transfer function between the jth input and ith output.
4. Impulse Response Matrix H(t)

H (t )  L1 [ H ( s)]
 H (t )  L1 [C ( sI  A) 1 B  D]
 Ce At B  D (t )
 (t ) is m  1 vector impulse
5. Is that all for state equation technique?
No! We have not found effective way for eAt yet!
7-6
Finding the state Transition Matrix
1. Based on Definition
 (t )  e At  I  At 
(Note:
1 2 2 1 33
A t  A t 
2!
3!
1 k
A must go to 0nn as k   ; otherwise, eAt does not exist)
k!
Why numerical calculation possible? Convergence of
1 k
A
k!
Example 7-1 Easy example
Example 7-2 Not easy example. I can not do it!
2. Based on (s)  (sI  A) 1
(t )  L1[(sI  A) 1 ]
Page 7-13
EE 422G Notes: Chapter 7
Instructor: Zhang
Key : (1) Find (sI  A) 1 ;
(2) For each element of (sI  A) 1 , obtain partial-fraction expansion
1
0

 6  5
Example 7-3 A  
1  s  1 
 s 0  0
( sI  A)  

 


0 s   6  5 6 s  5
 a11
a
 21
a12 
a22 
1
 a22  a12 
 a
a11 

21

a11 a22  a12 a21
 s  5 1
  6 s
s 1 

1
( sI  A)  
 

s ( s  5)  6
 6 s  5
 s  5 1
s5

  6 s
   ( s  2)( s  3)
 

6
( s  2)( s  3) 
 ( s  2)( s  3)
1
1

( s  2)( s  3) 

s

( s  2)( s  3) 
s5
3
2


 3e  2t  2e  3t
( s  2)(s  3) s  2 s  3
1
1
1


 e  2t  e  3t
( s  2)(s  3) s  2 s  3
6
1 
 1
 6 

 6[e  2t  e  3t ]

( s  2)(s  3)
 s  2 s  3
s
2
3


 2e  2t  3e  3t
( s  2)(s  3) s  2 s  3
eAt = …
(sI-A)-1 : Systematic way exists! But very complex operation!
Page 7-14
EE 422G Notes: Chapter 7
Instructor: Zhang
3. Diagonalization
A
nn
A  PP
P : nn
 : nn
1
t
diagonal
e : easy to
matrix
find
A k  ( PP 1 )( PP 1 )  ( PP 1 )  Pk P 1
A2 2
t 
2!
1
 PP 1  PP 1t  P2 P 1t 2  
2!
1
 P[ I  t  2 t 2  ]P 1
2!
t 1
 Pe P
e At  I  t 
4. Cayley – Hamilton Theorem Based Method
(t )  e At   0 (t ) I  1 (t ) A   2 (t ) A2     n 1 (t ) An 1
nXn
This method will be simple if the eigenvalues of A( j ) are distinct:
Replacing A by  j and I by 1 in the above equation.
Replacing by 1 : e 1t   0 (t )  1 (t )1   2 (t )1     n 1 (t )1
……
n 1
2
 t
n 1
Replacing by n : e n   0 (t )  1 (t )n   2 (t )n     n 1 (t )n
(mention (1) 0 (t ),1 (t ), 2 (t ) … …are the same in each of the n equations.)
(2)  j ' s are given numbers! )
2
Example 7-4 The same A in examples 7-2, 7-3.
1
0
A

 6  5
(1) Find eigenvalues : easy by using Matlab
Page 7-15
EE 422G Notes: Chapter 7
Instructor: Zhang
a = [0 1 ; -6 –5];
eiga = eig(a);
eiga
eiga =
-2
-3
Analytic Method:
| I  A |

1
6  5
1  2,
  (  5)  6  2  5  6  0
2  3,
(2)
1  2  e 2t   0 (t )  1 (t )(2)
(1)
2  3  e  3t   0 (t )  1 (t )(3)
(2)
(1)  (2)  e  2t  e  3t  1 (t )
  0 (t )  e  3t  31 (t )  e  3t  3e  2t  3e  3t
 3e  2t  2e  3t
Page 7-16
EE 422G Notes: Chapter 7
Instructor: Zhang
(3)
e At   0 (t ) I   1 (t ) A
3e  2t  2e 3t
 
0
0


 
0
3e  2t  2e 3t   6e 2t  6e 3t

 3e 2t  2e 3t
e  2 t  e  3t 


 2t
 3t
 2e  2t  3e 3t 
 6e  6e
e  2 t  e  3t 

 5e 2t  5e 3t 
The same as obtained in Example 7-2 and 7-3.
Question: What about some repeated eigenvalues?
Method can be modified! EE611
7-7 State Equation of Electrical Networks
Obtain State Equations for Systems
Two examples: From electrical networks
From transfer functions (system realization)
For electrical network: select
iL
and
vC
as state variables.
Algorithm
Step 1: Select each iL and vc as state variables
diL
Step 2: For each iL , write a KVL ( iL 
will be included)
dt
For each vc , write a KCL ( vC will be included)
Step 3: Other KCL and KVL, and element relation to eliminate “other”
variables (other than states (iL, vc ) and sources (input).
=>state equation.
Step 4 : Output equation
Page 7-17
EE 422G Notes: Chapter 7
Instructor: Zhang
Example 7-7
Step 1: Label vc as x1
iL as x2
Step 2: For vc , write KCL :
iR1  Cx1  x2
(1)
For iL , write KVL
x1  Lx2  iR2 R2
(2)
Step 3 : Other variables (undesired): iR1 , iR2
iR1
v  x1
 s
R1
Must be expressed in
terms of state
variables and sources!
i R 2  i L  x2
 vs  x1
 Cx1  x2
(1)'

  R1
 x  Lx  R x
 1
2
2 2
1
1
1


x


x

x

vs
1
2
 1
CR1
C
CR1

 x  1 x  R2 x
 2 L 1 L 2
1
 1


  x1   1 
 x1   CR1
C
 
     CR1  vs
R x


 x2   1
 2  2   0 
 L
L 
Step 4 : Output equation y = v0
Page 7-18
EE 422G Notes: Chapter 7
Instructor: Zhang
x 
y  R2 x2  0 1  1 
 x2 
7-8 State Equation from Transfer Functions
State Equation => Tell how to realize and simulate (system
realization) the systems
Problem in this section :
Y ( s) bm s m  bm 1s m 1    b1s  b0
 n
Given
U (s)
s  an 1s n 1    a1s  a0
Find
x  Ax  Bu
y  Cx  Du
( m  n)
(u --- scalar, y --- scalar)
1
Such that C ( sI  A) B  D equals the given transfer function.
1. Basic solution
Example :
Y ( s)
4
4
4
4U (S ) 4U (s)



 Y ( s) 

 4 X 1 ( s)  4 X 2 ( s)
U (s) (s  1)( s  2) s  1 s  2
s 1 s  2
U ( s)
 x1   x1  u
s 1
U (s)
X 2 ( s) 
 x2  2 x2  u
s2
 x1   1 0   x1  1
 x    0  2  x   1u
 2   
 2 
X 1 (s) 
x 
y  4  4 1 
 x2 
General Case
Y ( s) bm s m  bm 1s m 1    b1s  b0

U ( s)
( s  p1 )(s  p2 )  ( s  pn )
Page 7-19
EE 422G Notes: Chapter 7
Instructor: Zhang
Assumption: No repeated poles
No zero-pole cancellation
Partial-Fraction Expansion:
Bn
B1
B2
Y (s)



U ( s ) s  p1 s  p 2
s  pn
Y (s) 
B U (s)
B1U ( s ) B2U ( s )

 n
s  p1
s  p2
s  pn
 B1 X 1 ( s )  B2 X 2 ( s )    Bn X n ( s )
 x1 
x 
y (t )  B1 , B2 ,  Bn  2  ---- output equation

 
 xn 
(
U ( s)
 x j  p j x j  u(t )
)
s  pj
0   x1  1
     u (t ) ---- state equation

   
pn   xn  1
X j ( s) 
 x1   p1
 
  
 x n   0
Drawback (not serious): Complex pole
 Complex coefficients!
There are ways to fix it!
2. Repeated pole : No zero-pole cancellation
Y ( s)
A
B
  


U ( s)
( s  pi ) 2 s  pi
Denote:
Yi ( s ) 
AU ( s )
BU ( s ) 

 AX i1 ( s )  BX i2 ( s )
( s  pi ) 2 s  pi
Page 7-20
EE 422G Notes: Chapter 7

Instructor: Zhang
X i2 ( s ) 
U ( s)
s  pi
x i2  pi  xi2  u (t )
X i1 ( s) 
X i2 ( s )
s  pi
x i1  pi  xi1  xi2 ( s)
Sub-block :
 xi1   pi
 x   
 i2   0
1   xi1  0
u
 
pi   xi2  1
 xi 
yi   A B  1 
 xi2 
How to incorporate this method into system realization, see examples.
Example 7-9
Y (s)
s 2  3s  9

U ( s ) 5s 5  8s 4  24 s 3  34 s 2  23s  6
3.5
 4.75 5.875
7
1.125
Y (s) 




3
2
s 1 s  2 s  3
( s  1)
( s  1)
0
0
0   x1  0
 x1   1 1
 x   0  1 1
0
0   x 2  0 
2
  
   
 x3    0
0 1 0
0   x3   1u (t )
  
   

x
0
0
0

2
0
4
  
  x4  1
 x5   0
0
0
0  3  x5  1
Page 7-21
EE 422G Notes: Chapter 7
Instructor: Zhang
 x1 
x 
 2
y (t )  3.5  4.75 5.875  7 1.125 x3 
 
 x4 
 x5 
(3) Matlab Use
From state-equation => Transfer Function
[num, den] = ss2tf [A, B, C, D]
to
From transfer Function => state-equation
[A, B, C, D] = tf2ss [num, den]
Page 7-22