State-Space Formulation
Review of analysis techniques
 Continuous-time domain:
o Classical solution of differential equations
o Impulse response and convolution
o Fourier transform
o Laplace transform

Discrete-time domain:
o Classical solution of difference equations
o Impulse response and convolution
o Discrete-time Fourier transform
o Z-transform
State-variable characterization of systems
 It is another time-domain analysis technique for linear systems.

Advantages:
o Describes all signals in system – not just inputs and outputs
o Handles MIMO systems and time-varying systems readily
o Ideal for computer solutions

Disadvantage:
o Lots of work even for setting up a simple system
o Knowledge of advanced matrix mathematics required
Definitions
 State: A summary of the past information that affects the system’s future behavior.

State variables: A minimal set of variables needed to determine the future behavior of
the system from the system’s inputs.

State-space formulation: A mathematical description of the relationships of the input,
output, and the state of the system.
Continuous-time systems
 Differential equation:
y ( n) (t )  an 1 y ( n 1) (t )    a1 y ' (t )  a0 y (t )  b0u (t )
Solving for the highest order derivative:
y ( n) (t )  an 1 y ( n 1) (t )    a1 y ' (t )  a0 y (t )  b0u (t )
1

Block diagram:
u(t)
+
b0
y(n-1)(t)
y(n)(t)


y(n-2)(t)

y’(t)

y(t)
an-1
an-2
a0

Converting an nth order differential equation to a set of n first-order differential
equations: Let xi(t) represent the outputs of the integrators. As a result we can choose
them as the state variables of the system.
x1(t) = y(t)
x2(t) = y’(t) = x1‘(t)
x3(t) = y”(t) = x2‘(t)
…
xn(t) = y(n-1)(t) = xn-1‘(t)
Rewriting these equations:
x1‘(t) = x2(t)
x2‘(t) = x3(t)
…
xn-1‘(t) = xn(t)
xn‘(t) = y(n)(t) = b0 u(t) – a0 x1(t) – … – an-1xn(t)

Matrix formulation:
1
0
 x1 ' (t )  0
 x ' (t )  0
0
1
 2  

. 
.
.
.

 
.  
.
.
.


. 
.
.
.

 
0
0
 xn 1 ' (t )  0
 x ' (t )  a  a  a
1
2
 n   0
.
.
.
.
.
.
.
.
0
0  x1 (t )  0
.
0
0  x2 (t )  0
.
.
. 
.  .
 
  
.
.
.  
.   .  u (t )
.
.
. 
.  .
 
  
.
0
1  xn 1 (t )  0
.  an  2  an 1   xn (t ) b0 
2

 x1 (t )
 x (t )
 2 
y (t )  1 0  0 0  
  [0]  u (t )


 xn 1 (t )
 xn (t )
Vector notation:
x’(t) = A x(t) + B u(t)
y(t) = C x(t) + D u(t)
where
o x(t) is the state vector
o y(t) is the output
o u(t) is the input
o A is called the state (dynamic) matrix
o B is called the input matrix
o C is called the output (sensor) matrix
o D is called the feed through matrix
Now, an nth order differential equation has been converted to a set of n first-order
differential equations.

For a multi-input/multi-output system:
x’(t) = A x(t) + B u(t)
y(t) = C x(t) + D u(t)
For a system with n state variable, p inputs and q outputs:
A is n  n
B is n  p
C is q  n
D is q  p
x(t) is n  1
y(t) is q  1
u(t) is p  1

The transfer function matrix:
Taking the Laplace transform of the state-variable vector equation we obtain:
s X(s) – x(0) = A X(s) + B U(s)

X(s) = (s I – A)-1 {x(0) + B U(s)}
and
Y(s) = C X(s) + D U(s)
3

Y(s) = C (s I – A)-1 {x(0) + B U(s)} + D U(s)
Finally, setting x(0) = 0 gives us the transfer function matrix:
Y(s) = C (s I – A)-1 B U(s) + D U(s) = {C (s I – A)-1 B + D} U(s)
G(s) = C (s I – A)-1 B + D
or

Example 1:
y’’’(t) + y”(t) – 2y’(t) + 3y(t) = u(t)
y’’’(t)
u(t)
1

y”(t)
x3(t)

y’(t)
x2(t)

y(t)
x1(t)

-1
2
-3
 x1 ' (t )   0 1 0  x1 (t )  0
 x ' (t )   0 0
1   x2 (t )  0  u (t )
 2  
 x3 ' (t )   3 2  1  x3 (t )  1
 x1 (t ) 
y (t )  1 0 0   x2 (t )  0  u (t )
 x3 (t ) 

Example 2: Derivatives of u(t) are involved.
y”(t) + 2y’(t) + 3y(t) = 4u”(t) + 5u’(t) + 6u(t)
Step 1: Rearranging terms so that the derivatives are on the LHS
[y”(t) – 4u”(t)] + [2y’(t) – 5u’(t)] = -3y(t) + 6u(t)
Step 2: Construct the main path
Input
state variable
x2‘(t)
u(t)
K

x1‘(t)
x2(t)


4
state variable output
x1(t)

y(t)

Step 3: Start with the derivative terms and pass them through the first integrator
x2‘(t)= [y”(t) – 4u”(t)]
+ [2y’(t) – 5u’(t)]
x2(t) = [y’(t) – 4u’(t)]
+ [2y(t) – 5u(t)]

Step 4: Remove the non-derivative terms
x2(t) = [y’(t) – 4u’(t)]
+ [2y(t) – 5u(t)]
x1‘(t) = y’(t) – 4u’(t)

note: we will obtain
this term in step 9.
-[2y(t) – 5u(t)]
Step 5: Pass the result through the second integrator

x1‘(t)= y’(t) – 4u’(t)
x1(t) = y(t) – 4u(t)
Step 6: Remove the input term
x1(t) = y(t) – 4u(t)

y(t)
note: this is the feed through term.
4u(t)
Step 7: Use the input u(t) and feedback from x1(t) and x2(t) to construct x2‘(t)
u(t)
K

x2‘(t) = -3y(t) + 6u(t)
K1
x1(t) = y(t) – 4u(t)
K2
x2(t) = [y’(t) – 4u’(t)]
+ [2y(t) – 5u(t)]
Step 8: Compute the constants
o K2 = 0 since x2‘(t) has no derivative term.
o K1 = -3 in order to match the -3y(t) term in x2‘(t).
o -3y(t) + 6u(t) = -3 [y(t) – 4u(t)] + K u(t)  K = -6
5
Step 9: Use u(t) and feedback from x1(t) and x2(t)to construct x1’(t) = y’(t) – 4u’(t)
u(t)
K5
x2(t) = [y’(t) – 4u’(t)]
+ [2y(t) – 5u(t)]
y’(t) – 4u’(t)

K4
x1(t) = y(t) – 4u(t)
K3
Step 10: Compute the constants
o K4 = 1
o K3 = -2 in order to match the -2y(t) term.
o - [2y(t) – 5u(t)] = -2 [y(t) – 4u(t)] + K5 u(t)  K5 = -3
Step 11: Add feedback path to obtain x2‘(t) (from step 8)
x2‘(t)
u(t)
-6
x1‘(t)
x2(t)


x1(t)


y(t)

-3
Step 12: Add feedback/through paths to obtain x1‘(t) (from step 10)
-3
x2‘(t)
u(t)
-6


x1‘(t)
x2(t)
x1(t)


y(t)

-2
-3
Step 13: Add feed through path to obtain y(t) (from step 6)
4
-3
x2‘(t)
u(t)
-6

x1‘(t)
x2(t)



-2
-3
6
x1(t)
y(t)

Step 14: Formulate the state-space equations
x1‘(t) = -2 x1(t) + x2(t) – 3 u(t)
x2‘(t) = -3 x1(t) – 6 u(t)
y(t) = x1(t) + 4 u(t)
 x1 ' (t )   2 1  x1 (t )    3
 x ' (t )    3 0   x (t )   6  u (t )
  2   
 2  
 x (t ) 
y (t )  1 0   1   4  u (t )
 x2 (t )

Example 3: LRC circuit
State variables:
x1(t) – capacitor voltage
x2(t) – inductor current
Node voltage at 1: [x1(t) – u(t)] / 6 + x2(t) + x1‘(t) / 30 = 0, or
x1’(t) = -5x1(t) – 30x2(t) + 5u(t)
Inductor I-V equation: 5x2’(t) = x1(t) – u(t),
x2’(t) = [x1(t) – u(t)] / 5
Rewriting the equations we have:
x1’(t) = (-5)x1(t) + (-30)x2(t) + (5)u(t)
x2’(t) = (1/5)x1(t) + (0) x2(t) + (-1/5)u(t)
y(t) = (1)x1(t) + (0) x2(t) + (0) u(t)
7
or
Matrix notation:
 x1 ' (t )    5  30  x1 (t )   5 


 u (t )
 x ' (t )  0.2
0   x2 (t )  0.2
 2  
 x (t ) 
y (t )  1 0   1   0  u (t )
 x2 (t )
Block diagram:
5
x2‘(t)
u(t)
-1/5

x1‘(t)
x2(t)

-30


-5
1/5

Example 4: (text, p. 26)

Op Amp implementation:
o Inverter:
o Adder:
8
x1(t) = y(t)
o Integrator:
o Example:
  2 4
2
x'  
 x    u

 0 5
  4
y  3 10 x   2 u
Rearranging the terms we have:
 y  3x1  10 x 2  2u
x1 '  2 x1  4 x 2  2u

 x 2 '  5 x 2  4u
Constructing y:

Constructing x1:
9

Constructing x2:

Connecting the stages:
10

Linearization: Nonlinear systems can be linearized by ignoring the higher order
terms in the Taylor series (Text P. 18).
Discrete-time systems
 Difference equation:
y(k+n) + an-1 y(k+n-1) + … + a1 y(k+1) + a0 y(k) = b0 u(k)
Solving for the highest order delay:
y(k+n) = -an-1 y(k+n-1) – … – a0 y(k)+ b0 u(k)

Block diagram:
y(k+n)
u(k)
b0
+
z-1

y(k+n-1)
-1
z
y(k+n-2) y(k+1) z-1
y(k)
an-1
an-2
a0

Converting an nth order difference equation to a set of n first-order difference
equations: Let xi(k) represent the outputs of the unit delay. As a result we can choose
them as the state variables of the system.
x1(k) = y(k)
x2(k) = y(k+1) = x1(k+1)
x3(k) = y(k+2) = x2(k+1)
…
…
…
xn-1(k) = y(k+n-2) = xn-2(k+1)
xn(k) = y(k+n-1) = xn-1(k+1)
Rewriting these equations:
x1(k+1) = x2(k)
x2(k+1) = x3(k)
…
…
…
xn-1(k) = xn(k)
xn(k) = y(k+n) = b0 u(k) – a0 x1(k) – … – an-1xn(k)
11

Matrix formulation:
1
0
 x1 (k  1)  0
 x (k  1)  0
0
1
 2
 

. 
.
.
.

 
.  
.
.
.


. 
.
.
.

 
0
0
 xn 1 (k  1)  0
 x (k  1)  a  a  a
1
2
 n
  0
.
.
.
.
.
.
.
.
0
0  x1 (k )  0
.
0
0  x2 (k )  0
.
.
. 
.  .
 
  
.
.
.  
.   .  u (k )
.
.
. 
.  .
 
  
.
0
1  xn 1 (k )  0
.  an  2  an 1   xn (k ) b0 
 x1 (k )
 x (k )
 2 
y (k )  1 0  0 0  
  [0]  u (k )


 xn 1 (k )
 xn (k )

Vector notation:
x(k+1) = A x(k) + B u(k)
y(k) = C x(k) + D u(k)
where
o x(k) is the state vector
o y(k) is the output
o u(k) is the input
o A is called the state (dynamic) matrix
o B is called the input matrix
o C is called the output (sensor) matrix
o D is called the feed through matrix
Now, an nth order difference equation has been converted to a set of n first-order
difference equations.

For a multi-input/multi-output system:
x(k+1) = A x(k) + B u(k)
y(k) = C x(k) + D u(k)
For a system with n state variable, p inputs and q outputs:
A is n  n
x(k) is n  1
B is n  p
y(k) is q  1
C is q  n
u(k) is p  1
D is q  p
12

The transfer function matrix:
Taking the z-transform of the state-variable vector equation we obtain:
G(z) = C (z I – A)-1 B + D

Example:
y(k+2)+ 0.5 y(k+1) – 0.3 y(k) = 0.4 u(k)
y(k+1)
x2(k)
y(k+2)
u(k)
0.4

z-1
z-1
y(k)
x1(k)
-0.5
0.3
1   x1 (k )   0 
 x1 (k  1)   0
 x (k  1)  0.3  0.5   x (k )  0.4  u (k )
  2   
 2
 
 x (k ) 
y (k )  1 0  1   0 u (k )
 x2 ( k ) 
MIMO Example
For a multi-input/multi-output system:
x’(t) = A x(t) + B u(t)
y(t) = C x(t) + D u(t)
Example:
p’’’ + 2 p’ + 3 q – 4 r’ = 5 (t)
p + 8 q’ +9 q = 10 (t) – 11 (t)
p’’ – 6 r’’ + 7 r = 0
Let:
x1 = p
x4 = q
u1 = (t)
x2 = p’
x5 = r
u2 = (t)
x3 = p’’
x6 = r’
Substituting and rearranging the equations:
x3’ = -2 x2 –3 x4 + 4 x6 + 5 u1
x4’ = -(1/8) x1 – (9/8) x4 + (10/8) u1 – (11/8) u2
x6’ = (1/6) x3 + (7/6) x5
In matrix notation:
13

 0

 x1 '   0
 x ' 
 2 
 x3 '  0
  1
 x4 '  
 x5 '  8
  
 x6 '  0

 0

1
0
0
0
0
1
0
0
2
0
3
0
0
0

9
8
0
0
0
0
0
0
1
6
0
7
6


0 



0   x1  
 x  
 2
4  x  


3
  x   
0   4 
  x5  
 
1   x6  




0 



 x1 
x 
2
 p  1 0 0 0 0 0   0 0
 q   0 0 0 1 0 0   x3   0 0   u1 
  
 x  
 u 
 r  0 0 0 0 1 0  4  0 0  2 
 x5 
 
 x6 
14
0
0
5
10
8
0
0

0 

0 

0  u 

 1

11
  u 2 
8
0 


0 
