Lecture 6 - Summary: EOM, State-space Model, Linearisation

and Stability

Last time

• Linearised dynamics trajectory solution

For the dynamics δ̇ = Aδ where δ(0) = δ0 then δ(t) = eAt δ0 where

1 2 1 3
eAt = I + At + (At) + (At) + . . .
2! 3!

• Qualitative behaviour about an equilibrium

The dynamics are stable if all of the eigenvalues of A have a negative real part.
The dynamics are unstable if any of the eigenvalues of A have a positive real part.

LHP: Left half plane, i.e., Re(λ) < 0. RHP: Right half plane, i.e., Re(λ) > 0.
(Courtesy of Franklin, Power and Emami-Naeini, Feedback control of Dynamic systems, 2006)

Today

• What we have achieved so far...

1
Newton’s laws Equations of State-space Equilibrium
Motion (EOM) Form
X =⇒ =⇒ =⇒  
Fx = max 0
g X1 =
X α̈ = − sin α ẋ1 = x2 0
Fy = may l g  
ẋ2 = − sin x1 0
X
l X2 =
Mz = Jα π

Linearisation Trajectory solution Qualitative Behaviour

about an
=⇒ equilibrium =⇒     =⇒
| |
Re(λi ) < 0 =⇒ stable
  δ(t) = c1 eλ1 t  v1  + c2 eλ2 t  v2 
0 1 | |
δ̇ = δ
− gl 0 Im(λi ) 6= 0 =⇒
oscillatory
Bottom row is applicable for any linear dynamic system.

• A little bit on linearisation of Ẋ = f (X, U ) about an equilibrium X, U

Flyball Governor

Developed in the 1780s, was an enabler of the successful Watt steam An approximation using a
engine. rotating rod (green) driving a
Objective: To achieve a constant rotation rate under load. solid rod (yellow) instead of
flyballs.

Equation of motion:
ml2 ml2 2 l
α̈ − ω sin α cos α + mg sin α + bα̇ = 0
3 3 2

5. Write EOM in state-space form

5.a. Group highest order derivatives on one side of the EOM

3g 3
α̈ = ω 2 sin α cos α − sin α − bα̇
2l ml2

dθ dn−1 θ
5.b. For a nth DE, define variables x1 , x2 , . . . , xn−1 as x1 := θ, x2 := dt , . . . , xn := dt etc.

2
    
x1 α
X= = with ẋ2 = α̈
x2 α̇

5.c. Substitute into EOM and write lower order ODE

ẋ1 = x2
3g 3
ẋ2 = ω 2 sin x1 cos x1 − sin x1 − bx2
2l ml2

Are these dynamics nonlinear? Next step? Yes, terms like sin x1 are nonlinear.

Finding the Equilibriums for the Autonomous System Ẋ = f (X)

1. For the state-space dynamics Ẋ = f (X), find all equilibriums by solving f (X) = 0.
Eqn 1) 0 = ẋ1 = x2 = 0
Eqn 2) Use Eqn 1...

3g 3
0 = ẋ2 = ω 2 sin x1 cos x1 − sin x1 − bx2
2l ml2
3g
= ω 2 sin x1 cos x1 − sin x1
 2l 
3g
= sin x1 ω 2 cos x1 −
2l

3g
Then either: Case (a) sin x1 = 0 or Case (b) ω 2 cos x1 − 2l =0
Case (a) sin x1 = 0 then (1) x1 = 0 or (2) x2 = π
3g 3g 3g
Case (b) cos x1 = 2lω 2 then (3) x1 = cos−1 2lω 2 , α∗ or (4) x1 = − cos−1 2lω 2 , −α∗

 
0
X1 = ,
0
 
π
X2 =
0
α∗
 
X3 =
0
−α∗
 
X4 =
0

3
Our Equilibriums Family

   
0 π
X1 = X2 =
0 0

α∗ −α∗
   
X3 = X4 =
0 0

2. Choose a specific equilibrium X about which you will obtain a linearisation

... if one exists
3. Define the incremental state variable as δ = X − X.
   
0 π
Unlike in class, the full notes contain the linear approximation about all of the X 1 = 0
, X2 =
0
equilibriums. ∗
−α∗
   
α
 X3 = , X4 =
x1 − α ∗ x1 + α ∗ 0 0
      
x1 x1 − π
δ1 = δ2 = δ3 = δ4 =
x2 x2 x2 x2
4. Calculate the linearised dynamics about X as δ̇ = Aδ, where A is the Jacobian evaluated at X .
" #
∂f1 ∂f1
∂x1 ∂x2
 ẋ1 = x2
A=

∂f2 ∂f2 
3g
∂x1 ∂x2
ẋ2 = ω 2 sin x1 cos x1 −

X=X sin x1

0
1

 2l
= 3
ω 2 cos2 x1 − sin2 x1 − 3g 3b

2l cos x1 − ml2 − bx2

X=X ml 2
 
0
1 
=
ω 2 2 cos2 x1 − 1 − 3g 3b

2l cos x 1 − ml 2

X=X
Then δ̇ = Aδ, with m = l = b = 1, g ≈ 10. We will leave ω as a variable to explore its impact on the
dynamics.
   
0 1 0 1
A1 = 3gl = ,
ω2 − 2
3b
− ml 2 ω 2 − 15 −3
   
0 1 0 1
A2 = 3gl =
ω2 + 2
3b
− ml 2 ω 2 + 15 −3
   
0 1 0 1
A3 = 9g 2 9g 2 = 225
2l2 ω 2 − ω2 − 4ω 2
3b
− ml 2 ω2 − ω2 −3
   
0 1 0 1
A4 = 2 2 = 675
9g
2l2 ω 2 − ω2 + 9g
4ω 2
3b
− ml 2 ω2 − ω2 −3

4
Qualitative Behavior Near Equilibrium Points

1. Find the eigenvalues of A where δ̇ = Aδ and A is the Jacobian evaluated as X .

2. Use the location of the eigenvalues in the complex plane to describe the dynamics δ̇ = Aδ.
The dynamics are exponential stability if all of the eigenvalues of A have a negative real part.
The dynamics are oscillatory if any of the eigenvalues of A have a nonzero imaginary part.

3. If an initial condition x(0) = X0 is of interest:

a) Find the right and left eigenvectors of A.
b) Examine the state trajectory solution δ0 = X0 − X for ci = uTi δ0

     
| | |
δ(t) = c1 eλ1 t  v1  + c2 eλ2 t  v2  + · · · + cn eλn t  vn  .
| | |

(We won’t have time for step 3 today!)

Eigenvalues and eigenvectors are going to be important so...

Eigenvalues and Eigenvectors Properties

For general A...

Eigenvalues: Find the roots of the polynomial

det (A − λI) = 0

Eigenvectors: For each eigenvalue λ solve for the right eigenvector v and left eigenvector u as

(A − λI) v = 0 (AT − λI)u = 0

For a 2 × 2 matrix A...

Eigenvalues:

1h p i
λ1 , λ2 = tr(A) ± tr(A)2 − 4 det A
2

1. Find the eigenvalues of A where δ̇ = Aδ and A is the Jacobian evaluated as X .

2. Use the location of the eigenvalues in the complex plane to describe the dynamics δ̇ = Aδ.
The dynamics are exponential stability if all of the eigenvalues of A have a negative real part.
The dynamics are oscillatory if any of the eigenvalues of A have a nonzero imaginary part.
 
0
For X 1 = , δ̇ = Aδ then
0
 
0 1
A=
ω 2 − 15 −3

Eigenvalues?

5
 1h p i 1h p i
λ1 , λ2 = −3 ± 9 + 4 (ω 2 − 15) λ1 , λ2 = tr(A) ± tr(A)2 − 4 det A
2 2
3 1p
=− ± −51 + 4ω 2
2 2

3 1p
λ1,2 = − ± −51 + 4ω 2
2 2

For what range of ω is the system oscillatory? Im(λi ) 6= 0

−51 + 4ω 2 < 0
√
51
|ω| < ≈ 3.5
2

3 1p
λ1,2 = − ± −51 + 4ω 2
2 2

For what range of ω is the equilibrium X 1 stable? Re(λi ) < 0

√
Case 1: −51 + 4ω 2 /2 is imaginary i.e.,−51 + 4ω 2 ≤ 0 (already done) and
√ √
Case 2: −51 + 4ω 2 /2 is real but smaller than 3/2, i.e., −51 + 4ω 2 < 3
p
−51 + 4ω 2 < 3
1
ω 2 < (9 + 51)
4√
60
|ω| < ≈ 3.9
2

So
|ω| < max {3.5, 3.9} = 3.9

Finding the Equilibriums for the Non-Autonomous System Ẋ = f (X, U )

1. For the state-space dynamics Ẋ = f (X, U ), find all equilibriums by solving f (X, U ) = 0.


2. Choose a specific equilibrium X, U about which you will obtain a linearisation ... if one
exists

3. Define the incremental state variable as δX = X − X and δU = U − U .

4. Calculate the linearised dynamics about X as δ̇X = AδX +BδU , where A and B is the Jacobian
evaluated at X and U with
∂f1 ∂f1  ∂f1 ∂f1 
··· ···
 
∂x1 ∂xn
 ∂u1 ∂um

 
A= 
 .. .. .. 
and B = 
 .. .. .. 
. . . 
 . . . 

∂fn ∂fn ∂fn ∂fn
∂x1 ··· ∂xn
 X=X
∂u1 ··· ∂um
 X=X
U =U U =U

6
Here, there are n states and m control inputs.
Matrix A is called the state matrix.
Matrix B is called the input matrix.

Revisit our Pendulum Problem

Assume we can control an applied force perpendicular to gravity to a pendulum. To emphasize, let the
force be u so

g u ẋ1 = x2
α̈ = − sin α + cos α
l ml g u
ẋ2 = − sin x1 + cos x1
l ml
1. For the state-space dynamics Ẋ = f (X, U ), find all equilibriums by solving f (X, U ) = 0.

0 = x2
g u
0 = − sin x1 + cos x1
l ml

Then x2 = 0 and u = mg tan x1 .

 
x1
X1 = , U 1 = mg tan x1
0

2. Choose a specific equilibrium X, U about which you will obtain a linearisation ... if one exists

π
 
X1 = 4 , U 1 = mg
0

3. Define the incremental state variable as δX = X − X and δU = U − U .

4. Calculate the linearised dynamics about X as δ̇X = AδX + BδU , where A and B is the Jacobian
evaluated at X and U with

∂f1 ∂f1  ∂f1 ∂f1 

··· ···
 
∂x1 ∂xn
 ∂u1 ∂um

 
A= 
 .. .. .. 
and B = 
 .. .. .. 
. . . 
 . . . 

∂fn ∂fn ∂fn ∂fn
∂x1 ··· ∂xn
 X=X
∂u1 ··· ∂um
 X=X
U =U U =U

7
 " #
∂f1 ∂f1 
∂x1 ∂x2
A=

∂f2 ∂f2 
∂x1 ∂x2
 X=X
U =U
 
0 1 
=
− gl cos x1 − u

sin x1 0  X=X
ml
U =U
 
0
√
1
= 2g
− l 0

" #
∂f1 
∂u1
B=

∂f2 
∂u1
 X=X
U =U
 
0 
= 1

 X=X
ml cos x 1
U =U
 
0
√
= 2
2ml

Then

δ̇X = AδX + BδU

   
√
0 1 0
√
= δX + δU
− l2g 0 2
2ml