Linearization of Nonlinear Models

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Linearization of Nonlinear Models

• Most chemical process models are nonlinear, but they are often linearized to perform a simulation and stability analysis.

• Linear models are easier to understand (than

nonlinear models) and are necessary for most control

system design methods.

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Single Variable Example

• A general single variable nonlinear model

• The function f(x) can be approximated by a Taylor series approximation around the steady-state operating point (x_s)

• Neglect the quadratic and higher order terms dx ( )

dt = f x

( )

² 2

( )

²

( ) ( ) 1 high order terms

s 2 s

s s s

x x

f f

f x f x x x x x

x x

∂ ∂

= + − + − +

∂ ∂

( )

( ) ( )

s

s s

x

f x f x f x x

x

≈ + ∂ −

∂

( )

s

s x

dx f

f x x x

dt x

= ≈ ∂ −

∂

At steady-state

( ) 0

s

dx f x

dt = =

The partial derivative of f(x) with respect to x, evaluated at the steady-state

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• Since the derivative of a constant (x_s) is zero

• We are often interested in deviations in a state from a steady- state operating point (deviation variable)

• Write in state-space form

( ) ( )

s

s x

d x x f

x x

dt x

− ≈ ∂ −

∂

(

^s

)

d x x dx

dt dt

= −

xs

dx f dt x x

≈ ∂

∂

x = − x x

s : the change or perturbation from a steady-state value

dx a x dt ≈

xs

a f

x

= ∂ where ∂

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One State Variable and One Input Variable

• Consider a function with one state variable and one input variable

• Using a Taylor Series Expansion for f(x,u)

• Truncating after the linear terms

( , ) x dx f x u

= dt = ɺ

( ) ( ) ( )

( )( ) ( )

2 2

2

, , ,

2 2

, ,

( , ) 1

2

1 high order terms

2

s s s s s s

s s s s

s s s s s

x u x u x u

s s s

x u x u

f f f

x f x u x x u u x x

x u x

f f

x x u u u u

x u u

∂ ∂ ∂

= + − + − + −

∂ ∂ ∂

∂ ∂

+ − − + − +

∂ ∂ ∂

ɺ

( ) ( )

, ,

( , )

s s s s

x u x u

f f

x f x u x x u u

x u

∂ ∂

≈ + − + −

∂ ∂

ɺ

( ) ( ) ( )

, ,

s s s s

s

s s

x u x u

d x x f f

x x u u

dt x u

− ≈ ∂ − + ∂ −

∂ ∂

zero

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• Using deviation variables, and

• Write in state-space form

• If there is a single output that is a function of the state and input

• Perform a Taylor series expansion and truncate high order terms

, ,

s s s s

x u x u

dx f f

x u

dt x u

∂ ∂

≈ +

∂ ∂

( ) ( )

, ,

( , ) ( , )

s s s s

x u x u

g g

g x u g x u x x u u

x u

∂ ∂

≈ + − + −

∂ ∂

s, s

x u

a f

x

= ∂

∂

x = −x xs u = −u u_s

dx a x b u

dt ≈ + ^where

s, s

x u

b f

u

= ∂

∂

( , ) y = g x u

( ) ( )

, ,

s s s s

s s s

x u x u

g g

y y x x u u

x u

∂ ∂

− = − + −

∂ ∂

y = c x +d u ^where

s, s

x u

c g

x

= ∂

∂ _,

s s

x u

d g

u

= ∂

∂

( ,_s _s) _s g x u = y

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Linearization of Multistate Models

• Two-state system

• Perform Taylor series expansion of the nonlinear functions and neglect high-order terms

1

1 dx 1( ,1 2, )

x f x x u

= dt = ɺ

( ) ( ) ( )

1 2

1 2 1 2

1 1 1

1 1 2 1 1 2 1 1 2 2

, ,

1 , , 2 , ,

( , , ) ( , , )

s s s

s s s s s s

x x u

x x u x x u

f f f

f x x u f x x u x x x x u u

x x u

∂ ∂ ∂

= + − + − + −

∂ ∂ ∂

2

2 dx 2( ,1 2, )

x f x x u

= dt = ɺ

1 2

( , , ) y = g x x u

( ) ( ) ( )

1 2

1 2 1 2

2 2 2

2 1 2 2 1 2 1 1 2 2

, ,

1 , , 2 , ,

( , , ) ( , , )

s s s

s s s s s s

x x u

x x u x x u

f f f

f x x u f x x u x x x x u u

x x u

∂ ∂ ∂

= + − + − + −

∂ ∂ ∂

( ) ( ) ( )

1 2

1 2 1 2

1 2 1 2 1 1 2 2

, ,

1 , , 2 , ,

( , , ) ( , , )

s s s

s s s s s s

x x u

x x u x x u

g g g

g x x u g x x u x x x x u u

x x u

∂ ∂ ∂

= + − + − + −

∂ ∂ ∂

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• For the linearization about the steady-state

• We can write the state-space model

1 2

( _s, _s, _s) _s g x x u = y

( )

¹ ² ¹ ² ¹ ²

[ ]

1 2

1 2 1 2

1 1 1

1 1

1 , , 2 , , 1 1 , ,

2 2

2 2 2 2 2

, ,

1 , , 2 , ,

s s s s s s s s s

s s s

s s s s s s

s

x x u x x u s x x u

s s s

x x u

x x u x x u

f f f

d x x

x x x x u

dt u u

x x

d x x f f f

dt x x u

∂ ∂  ∂ 

−

     

  ∂ ∂   −  ∂ 

 −  =    −  + ∂  −

  ∂ ∂    

    ∂

  ∂ ∂   

1( 1_s, 2_s, _s) 2( 1_s, 2_s, _s) 0 f x x u = f x x u =

(

¹ ¹

) (

² ²

)

1 d x x s 2 d x x s

dx dx

dt dt dt dt

− −

= =

[ ]

1 2

1 2 1 2

1 1

2 2 , ,

1 s, s, s 2 s, s, s s s s

s

s s

s x x u

x x u x x u

x x

g g g

y y u u

x x

x x u

 ∂ ∂  −  ∂ 

 

− = ∂ ∂  −  +∂  −

x = A x + B u y = C x + D u

ɺ

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Generalization

• Consider a general nonlinear model with n state variables, m input variables, and r output variables

• Elements of the linearization matrices

1 1 1 1

1 1

1 1 1 1

1 1

( , , , , , )

( , , , , , ) ( , , , , , )

( , , , , , )

n m

n n n m

n m

r r n m

x f x x u u

y g x x u u

=

ɺ ⋯ ⋯

⋮

ɺ ⋯ ⋯

⋯ ⋯

⋮

⋯ ⋯

x = A x + B u y = C x + D u

ɺ

x = f(x, u) y = g(x, u)

ɺ

Vector notation:

, i ij

j

A f

x

= ∂

∂

s s

x u ,

i ij

j

B f

u

= ∂

∂

s s

x u

, i ij

j

C g

x

= ∂

∂

s s

x u ,

i ij

j

D g

u

= ∂

∂

s s

x u

x = A x + B u y = C x + Du

ɺ

State-space form:

or

(The “overbar” is usually dropped)

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Example: Interacting Tanks

• Two interacting tank in series with outlet flowrate being function of the square root of tank height

• Modeling equations

1 1 1 2

F = R h −h ^F² ⁼ ^R² ^h²

F

F₁ F₂

( )

1 1

1 2 1 1 2

1 1

2 1 2

1 2 2 2 1 2

2 2

, ,

dh F R

h h f h h F

dt A A

dh R R

h h h f h h F

dt A A

= − − =

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• Assume only the second tank height is measured. The output, in deviation variable form is y = h₂- h_2s

• There are two state variables, one input variable, one one output variable

• The element of the A (Jacobian) and B matrices

1 1

11

1 , 1 1 2

1 1

12

2 , 1 1 2

2 1

21

1 , 2 1 2

2 1 2

22

2 , 2 1 2 2 2

2

2 2

s

s s

F

s s

F

s s

F

s s s

F

f R

A h A h h

f R

A h A h h

f R

A h A h h

f R R

A h A h h A h

= ∂ = −

∂ −

= ∂ =

∂ −

= ∂ =

∂ −

= ∂ = − −

∂ −

s

h

1

2 s

s

h h

 

=  

 

hs ¹ ¹ ¹

2 2

2

s

h h x

−

 

=     =  − 

x u = −F F_s

1 11

, 1

2 21

,

1

0

s

F

B f

F A

B f

F

= ∂ =

∂

= ∂ =

∂

s

h

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• Only the height of the second tank is measured

• The state-space model is

11

1 ,

2 12

2 ,

0

1

s

F

C g

h C g

h

= ∂ =

∂

= ∂ =

∂

s

h

1 2 2 2

( , , ) _s

y = g h h F = −h h

[ ]

1 1

1

1 1 2 1 1 2 1

1 2

2 1 1 2

2 1 2 2 1 2 2 2

2 2 1

2 2 2 0

s s s s

s s s s s

R R

dx

A h h A h h x

dt A u

x

dx R R R

dt A h h A h h A h

 

  −   

  =  − −    +

        

   − −   

 

   − − 

[ ]

¹

(

² ² ²

)

2

0 1 x _s

y y x h h

x

=    = = −

 

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Interpretation of Linearization

• Consider the single tank problem (assume F is constant)

• Linearization

(

^,

)

¹ ¹

5 dh F R

h f h F h

dt = −A A = = −

(

^,

)

⁰ ¹

( )

10 ^s

f h F ≈ − h h−

The linear approximation works well between 3.5 to 7 feet

The two functions are exactly equal at the steady-state value of 5 feet

0 1 2 3 4 5 6 7 8 9 10

-0.5 0 0.5 1

x (h)

f(x)

nonlinear linear

hs = 5

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Exercise: interacting tanks

• Two interacting tank in series with outlet flowrate being function of the square root of tank height

– Parameter values

– Input variable F = 5 ft³/min

– Steady-state height values : h_1s= 10, h_2s= 6

• Perform the following simulation using state-space model

– What are the responses of tank height if the initial heights are h₁(0)=12 ft and h₂(0)=7 ft ?

– Assume the system is at steady-state initially. What are the responses of tank height if

• F changes from 5 to 7 ft³/min at t = 0

• F has periodic oscillation of F = 5 + sin(0.2t)

• F changes from 5 to 4 ft³/min at t = 20

2.5 2.5

2 2

1 2 1 2

ft 5 ft

2.5 5ft 10 ft

min 6 min

R = R = A = A =

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Stability of State-Space Models

• A state-space model is said to be stable if the response x(t) is bounded for all u(t) that is bounded

• Stability criterion for state-space model

– The state-space model will exhibit a bounded response x(t) for all bounded u(t), if and only if all of the eigenvalues of A have negative real parts

(the stability is independent matrices B and C)

• Single variable equation has the solution

• The solution of is

– Stable if all of the eigenvalues of A are less than zero

– The response x(t) is oscillatory if the eigenvalues are complex

x ɺ = a x

( ) ^at (0) stable if 0

x t = e x ⇒ a <

x = Axɺ x( )t = e^A^tx(0)

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Exercise

• Consider the following system equations

– Find the responses of x(t) for and (slow subspace v.s. fast subspace)

• Consider the following system equations

– Find the responses of x(t) for and (stable subspace v.s. unstable subspace)

1 1 2

2 2

0.5

2 x x x

x x

= − +

= −

ɺ ɺ

(0) 1

0

=   

x   0.5547

(0) 0.8321

−

 

=  

 

x

1 1 2

2 1 2

2 2

x x x

= +

= −

ɺ ɺ

0.2703 (0) 0.9628

 

= − 

x 0.8719

(0) 0.4896

 

=  

 

x

Note: Find eigenvalue and eigenvector of A

>> [V, D] = eig(A)