chapter 1

1. Plan of the Lecture
I Review: control, feedback, etc.
I Today’s topic: state-space models of systems; linearization

2. Plan of the Lecture
I Review: control, feedback, etc.
I Today’s topic: state-space models of systems; linearization
Goal: a general framework that encompasses all examples of
interest. Once we have mastered this framework, we can
proceed to analysis and then to design.

3. Plan of the Lecture
I Review: control, feedback, etc.
I Today’s topic: state-space models of systems; linearization
Goal: a general framework that encompasses all examples of
interest. Once we have mastered this framework, we can
proceed to analysis and then to design.
Reading: FPE, Sections 1.1, 1.2, 2.1–2.4, 7.2, 9.2.1.
Chapter 2 has lots of cool examples of system models!!
Week 3 data
2-4 to 2-7 for Reading Purpose. vv imp. articles

4. Notation Reminder
We will be looking at dynamic systems whose evolution in
time is described by differential equations with external
inputs.
We will not write the time variable t explicitly, so we use
x instead of x(t)
ẋ instead of x0
(t) or
dx
dt
ẍ instead of x00
(t) or
d2
x
dt2
etc.

5. Example 1: Mass-Spring System
u
m
x

6. Example 1: Mass-Spring System
u
m
x
Newton’s second law (translational motion):
F
|{z}
total force
= ma

7. Example 1: Mass-Spring System
u
m
x
Newton’s second law (translational motion):
F
|{z}
total force
= ma = spring force + friction + external force

8. Example 1: Mass-Spring System
u
m
x
Newton’s second law (translational motion):
F
|{z}
total force
= ma = spring force + friction + external force
spring force = −kx (Hooke’s law)
friction force = −ρẋ (Stokes’ law — linear drag, only an approximation!!)

9. Example 1: Mass-Spring System
u
m
x
Newton’s second law (translational motion):
F
|{z}
total force
= ma = spring force + friction + external force
spring force = −kx (Hooke’s law)
friction force = −ρẋ (Stokes’ law — linear drag, only an approximation!!)
F = −kx − ρẋ + u

10. Example 1: Mass-Spring System
u
m
x
Newton’s second law (translational motion):
F
|{z}
total force
= ma = spring force + friction + external force
spring force = −kx (Hooke’s law)
friction force = −ρẋ (Stokes’ law — linear drag, only an approximation!!)
mẍ = −kx − ρẋ + u

11. Example 1: Mass-Spring System
u
m
x
Newton’s second law (translational motion):
F
|{z}
total force
= ma = spring force + friction + external force
spring force = −kx (Hooke’s law)
friction force = −ρẋ (Stokes’ law — linear drag, only an approximation!!)
mẍ = −kx − ρẋ + u
Move x, ẋ, ẍ to the LHS, u to the RHS:
mẍ + ρẋ + kx = u

12. Example 1: Mass-Spring System
u
m
x
Newton’s second law (translational motion):
F
|{z}
total force
= ma = spring force + friction + external force
spring force = −kx (Hooke’s law)
friction force = −ρẋ (Stokes’ law — linear drag, only an approximation!!)
mẍ = −kx − ρẋ + u
Move x, ẋ, ẍ to the LHS, u to the RHS:
ẍ +
ρ
m
ẋ +
k
m
x =
u
m

13. Example 1: Mass-Spring System
u
m
x
Newton’s second law (translational motion):
F
|{z}
total force
= ma = spring force + friction + external force
spring force = −kx (Hooke’s law)
friction force = −ρẋ (Stokes’ law — linear drag, only an approximation!!)
mẍ = −kx − ρẋ + u
Move x, ẋ, ẍ to the LHS, u to the RHS:
ẍ +
ρ
m
ẋ +
k
m
x =
u
m
2nd-order linear ODE

14. Example 1: Mass-Spring System
u
m
x
ẍ +
ρ
m
ẋ +
k
m
x =
u
m
2nd-order linear ODE

15. Example 1: Mass-Spring System
u
m
x
ẍ +
ρ
m
ẋ +
k
m
x =
u
m
2nd-order linear ODE
Canonical form: convert to a system of 1st-order ODEs

16. Example 1: Mass-Spring System
u
m
x
ẍ +
ρ
m
ẋ +
k
m
x =
u
m
2nd-order linear ODE
Canonical form: convert to a system of 1st-order ODEs
ẋ = v (definition of velocity)
v̇ = −
ρ
m
v −
k
m
x +
1
m
u

17. Example 1: Mass-Spring System
u
m
x
State-space model: express in matrix form
ẋ
v̇
=
0 1
− k
m − ρ
m
x
v
+
0
1
m
u

18. Example 1: Mass-Spring System
u
m
x
State-space model: express in matrix form
ẋ
v̇
=
0 1
− k
m − ρ
m
x
v
+
0
1
m
u
Important: start reviewing your linear algebra now!!
I matrix-vector multiplication; eigenvalues and eigenvectors; etc.

19. General n-Dimensional State-Space Model
state x =



x1
.
.
.
xn


 ∈ Rn
input u =



u1
.
.
.
um


 ∈ Rm

20. General n-Dimensional State-Space Model
state x =



x1
.
.
.
xn


 ∈ Rn
input u =



u1
.
.
.
um


 ∈ Rm


ẋ1
.
.
.
ẋn

 =


A
n × n
matrix




x1
.
.
.
xn

 +


B
n × m
matrix




u1
.
.
.
um



21. General n-Dimensional State-Space Model
state x =



x1
.
.
.
xn


 ∈ Rn
input u =



u1
.
.
.
um


 ∈ Rm


ẋ1
.
.
.
ẋn

 =


A
n × n
matrix




x1
.
.
.
xn

 +


B
n × m
matrix




u1
.
.
.
um


ẋ = Ax + Bu

22. Partial Measurements
state x =



x1
.
.
.
xn


 ∈ Rn
input u =



u1
.
.
.
um


 ∈ Rm

23. Partial Measurements
state x =



x1
.
.
.
xn


 ∈ Rn
input u =



u1
.
.
.
um


 ∈ Rm
output y =



y1
.
.
.
yp


 ∈ Rp
y = Cx C– p × n matrix

24. Partial Measurements
state x =



x1
.
.
.
xn


 ∈ Rn
input u =



u1
.
.
.
um


 ∈ Rm
output y =



y1
.
.
.
yp


 ∈ Rp
y = Cx C– p × n matrix
ẋ = Ax + Bu
y = Cx

25. Partial Measurements
state x =



x1
.
.
.
xn


 ∈ Rn
input u =



u1
.
.
.
um


 ∈ Rm
output y =



y1
.
.
.
yp


 ∈ Rp
y = Cx C– p × n matrix
ẋ = Ax + Bu
y = Cx
Example: if we only care about (or can only measure) x1, then
y = x1 = 1 0 . . . 0





x1
x2
.
.
.
xn






26. State-Space Models: Bottom Line
ẋ = Ax + Bu
y = Cx

27. State-Space Models: Bottom Line
ẋ = Ax + Bu
y = Cx
State-space models are useful and convenient for
writing down system models for different types
of systems, in a unified manner.

28. State-Space Models: Bottom Line
ẋ = Ax + Bu
y = Cx
State-space models are useful and convenient for
writing down system models for different types
of systems, in a unified manner.
When working with state-space models, what are
states and what are inputs?

29. State-Space Models: Bottom Line
ẋ = Ax + Bu
y = Cx
State-space models are useful and convenient for
writing down system models for different types
of systems, in a unified manner.
When working with state-space models, what are
states and what are inputs?
— match against ẋ = Ax + Bu

30. Example 2: RL Circuit
I
VS
VR
VL
+
+
+

31. Example 2: RL Circuit
I
VS
VR
VL
+
+
+
−VS + VR + VL = 0 Kirchhoff’s voltage law
VR = RI Ohm’s law
VL = L ˙
I Faraday’s law
−VS + RI + L ˙
I = 0

32. Example 2: RL Circuit
I
VS
VR
VL
+
+
+
−VS + VR + VL = 0 Kirchhoff’s voltage law
VR = RI Ohm’s law
VL = L ˙
I Faraday’s law
−VS + RI + L ˙
I = 0
˙
I = −
R
L
I +
1
L
VS (1st-order system)

33. Example 2: RL Circuit
I
VS
VR
VL
+
+
+
−VS + VR + VL = 0 Kirchhoff’s voltage law
VR = RI Ohm’s law
VL = L ˙
I Faraday’s law
−VS + RI + L ˙
I = 0
˙
I = −
R
L
I +
1
L
VS (1st-order system)
I – state, VS – input

34. Example 2: RL Circuit
I
VS
VR
VL
+
+
+
−VS + VR + VL = 0 Kirchhoff’s voltage law
VR = RI Ohm’s law
VL = L ˙
I Faraday’s law
−VS + RI + L ˙
I = 0
˙
I = −
R
L
I +
1
L
VS (1st-order system)
I – state, VS – input
Q: How should we change the circuit in order to implement a
2nd-order system?

35. Example 2: RL Circuit
I
VS
VR
VL
+
+
+
−VS + VR + VL = 0 Kirchhoff’s voltage law
VR = RI Ohm’s law
VL = L ˙
I Faraday’s law
−VS + RI + L ˙
I = 0
˙
I = −
R
L
I +
1
L
VS (1st-order system)
I – state, VS – input
Q: How should we change the circuit in order to implement a
2nd-order system? A: Add a capacitor.

36. Example 3: Pendulum
`
mg
mg sin ✓
✓
✓
Te
external
torque

37. Example 3: Pendulum
`
mg
mg sin ✓
✓
✓
Te
external
torque
Newton’s 2nd law (rotational motion):

38. Example 3: Pendulum
`
mg
mg sin ✓
✓
✓
Te
external
torque
Newton’s 2nd law (rotational motion):
T
|{z}
total
torque
= J
|{z}
moment
of inertia
α
|{z}
angular
acceleration

39. Example 3: Pendulum
`
mg
mg sin ✓
✓
✓
Te
external
torque
Newton’s 2nd law (rotational motion):
T
|{z}
total
torque
= J
|{z}
moment
of inertia
α
|{z}
angular
acceleration
= pendulum torque + external torque

40. Example 3: Pendulum
`
mg
mg sin ✓
✓
✓
Te
external
torque
Newton’s 2nd law (rotational motion):
T
|{z}
total
torque
= J
|{z}
moment
of inertia
α
|{z}
angular
acceleration
= pendulum torque + external torque
pendulum torque = −mg sin θ
| {z }
force
· `
|{z}
lever arm

41. Example 3: Pendulum
`
mg
mg sin ✓
✓
✓
Te
external
torque
Newton’s 2nd law (rotational motion):
T
|{z}
total
torque
= J
|{z}
moment
of inertia
α
|{z}
angular
acceleration
= pendulum torque + external torque
pendulum torque = −mg sin θ
| {z }
force
· `
|{z}
lever arm
moment of inertia J = m`2

42. Example 3: Pendulum
`
mg
mg sin ✓
✓
✓
Te
external
torque
Newton’s 2nd law (rotational motion):
T
|{z}
total
torque
= J
|{z}
moment
of inertia
α
|{z}
angular
acceleration
= pendulum torque + external torque
pendulum torque = −mg sin θ
| {z }
force
· `
|{z}
lever arm
moment of inertia J = m`2
−mg` sin θ + Te = m`2
θ̈

43. Example 3: Pendulum
`
mg
mg sin ✓
✓
✓
Te
external
torque
Newton’s 2nd law (rotational motion):
T
|{z}
total
torque
= J
|{z}
moment
of inertia
α
|{z}
angular
acceleration
= pendulum torque + external torque
pendulum torque = −mg sin θ
| {z }
force
· `
|{z}
lever arm
moment of inertia J = m`2
−mg` sin θ + Te = m`2
θ̈
θ̈ = −
g
`
sin θ +
1
m`2
Te (nonlinear equation)

44. Example 3: Pendulum
θ̈ = −
g
`
sin θ +
1
m`2
Te (nonlinear equation)

45. Example 3: Pendulum
θ̈ = −
g
`
sin θ +
1
m`2
Te (nonlinear equation)
For small θ, use the approximation sin θ ≈ θ

46. Example 3: Pendulum
θ̈ = −
g
`
sin θ +
1
m`2
Te (nonlinear equation)
For small θ, use the approximation sin θ ≈ θ
sinHΘL
Θ
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
θ̈ = −
g
`
θ +
1
m`2
Te

47. Example 3: Pendulum
θ̈ = −
g
`
sin θ +
1
m`2
Te (nonlinear equation)
For small θ, use the approximation sin θ ≈ θ
sinHΘL
Θ
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
θ̈ = −
g
`
θ +
1
m`2
Te
State-space form: θ1 = θ, θ2 = θ̇
θ̇2 = −
g
`
θ +
1
m`2
Te = −
g
`
θ1 +
1
m`2
Te

48. Example 3: Pendulum
θ̈ = −
g
`
sin θ +
1
m`2
Te (nonlinear equation)
For small θ, use the approximation sin θ ≈ θ
sinHΘL
Θ
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
θ̈ = −
g
`
θ +
1
m`2
Te
State-space form: θ1 = θ, θ2 = θ̇
θ̇2 = −
g
`
θ +
1
m`2
Te = −
g
`
θ1 +
1
m`2
Te
θ̇1
θ̇2
=
0 1
−g
` 0
θ1
θ2
+
0
1
m`2
Te

49. Linearization
Taylor series expansion:
f(x) = f(x0) + f0
(x0)(x − x0) +
1
2
f00
(x0)(x − x0)2
+ . . .
≈ f(x0) + f0
(x0)(x − x0) linear approximation around x = x0

50. Linearization
Taylor series expansion:
f(x) = f(x0) + f0
(x0)(x − x0) +
1
2
f00
(x0)(x − x0)2
+ . . .
≈ f(x0) + f0
(x0)(x − x0) linear approximation around x = x0
Control systems are generally nonlinear:
ẋ = f(x, u) nonlinear state-space model
x =



x1
.
.
.
xn


 u =



u1
.
.
.
um


 f =



f1
.
.
.
fn




51. Linearization
Taylor series expansion:
f(x) = f(x0) + f0
(x0)(x − x0) +
1
2
f00
(x0)(x − x0)2
+ . . .
≈ f(x0) + f0
(x0)(x − x0) linear approximation around x = x0
Control systems are generally nonlinear:
ẋ = f(x, u) nonlinear state-space model
x =



x1
.
.
.
xn


 u =



u1
.
.
.
um


 f =



f1
.
.
.
fn



Assume x = 0, u = 0 is an equilibrium point: f(0, 0) = 0
This means that, when the system is at rest and no control is
applied, the system does not move.

52. Linearization
Linear approx. around (x, u) = (0, 0) to all components of f:
ẋ1 = f1(x, u), . . . , ẋn = fn(x, u)

53. Linearization
Linear approx. around (x, u) = (0, 0) to all components of f:
ẋ1 = f1(x, u), . . . , ẋn = fn(x, u)
For each i = 1, . . . , n,
fi(x, u) = fi(0, 0)
| {z }
=0
+
∂fi
∂x1
(0, 0)x1 + . . . +
∂fi
∂xn
(0, 0)xn
+
∂fi
∂u1
(0, 0)u1 + . . . +
∂fi
∂um
(0, 0)um

54. Linearization
Linear approx. around (x, u) = (0, 0) to all components of f:
ẋ1 = f1(x, u), . . . , ẋn = fn(x, u)
For each i = 1, . . . , n,
fi(x, u) = fi(0, 0)
| {z }
=0
+
∂fi
∂x1
(0, 0)x1 + . . . +
∂fi
∂xn
(0, 0)xn
+
∂fi
∂u1
(0, 0)u1 + . . . +
∂fi
∂um
(0, 0)um
Linearized state-space model:
ẋ = Ax + Bu, where Aij =
∂fi
∂xj x=0
u=0
, Bik =
∂fi
∂uk x=0
u=0

55. Linearization
Linear approx. around (x, u) = (0, 0) to all components of f:
ẋ1 = f1(x, u), . . . , ẋn = fn(x, u)
For each i = 1, . . . , n,
fi(x, u) = fi(0, 0)
| {z }
=0
+
∂fi
∂x1
(0, 0)x1 + . . . +
∂fi
∂xn
(0, 0)xn
+
∂fi
∂u1
(0, 0)u1 + . . . +
∂fi
∂um
(0, 0)um
Linearized state-space model:
ẋ = Ax + Bu, where Aij =
∂fi
∂xj x=0
u=0
, Bik =
∂fi
∂uk x=0
u=0
Important: since we have ignored the higher-order terms, this
linear system is only an approximation that holds only for small
deviations from equilibrium.

56. Example 3: Pendulum, Revisited
Original nonlinear state-space model:
θ̇1 = f1(θ1, θ2, Te) = θ2 — already linear
θ̇2 = f2(θ1, θ2, Te) = −
g
`
sin θ1 +
1
m`2
Te

57. Example 3: Pendulum, Revisited
Original nonlinear state-space model:
θ̇1 = f1(θ1, θ2, Te) = θ2 — already linear
θ̇2 = f2(θ1, θ2, Te) = −
g
`
sin θ1 +
1
m`2
Te
Linear approx. of f2 around equilibrium (θ1, θ2, Te) = (0, 0, 0):
∂f2
∂θ1
= −
g
`
cos θ1
∂f2
∂θ2
= 0
∂f2
∂Te
=
1
m`2
∂f2
∂θ1
0
= −
g
`
∂f2
∂θ2
0
= 0
∂f2
∂Te
0
=
1
m`2

58. Example 3: Pendulum, Revisited
Original nonlinear state-space model:
θ̇1 = f1(θ1, θ2, Te) = θ2 — already linear
θ̇2 = f2(θ1, θ2, Te) = −
g
`
sin θ1 +
1
m`2
Te
Linear approx. of f2 around equilibrium (θ1, θ2, Te) = (0, 0, 0):
∂f2
∂θ1
= −
g
`
cos θ1
∂f2
∂θ2
= 0
∂f2
∂Te
=
1
m`2
∂f2
∂θ1
0
= −
g
`
∂f2
∂θ2
0
= 0
∂f2
∂Te
0
=
1
m`2
Linearized state-space model of the pendulum:
θ̇1 = θ2
θ̇2 = −
g
`
θ1 +
1
m`2
Te valid for small deviations from equ.

59. General Linearization Procedure
I Start from nonlinear state-space model
ẋ = f(x, u)

60. General Linearization Procedure
I Start from nonlinear state-space model
ẋ = f(x, u)
I Find equilibrium point (x0, u0) such that f(x0, u0) = 0

61. General Linearization Procedure
I Start from nonlinear state-space model
ẋ = f(x, u)
I Find equilibrium point (x0, u0) such that f(x0, u0) = 0
Note: different systems may have different equilibria,
not necessarily (0, 0), so we need to shift variables:
x = x − x0 u = u − u0
f(x, u) = f(x + x0, u + u0) = f(x, u)

62. General Linearization Procedure
I Start from nonlinear state-space model
ẋ = f(x, u)
I Find equilibrium point (x0, u0) such that f(x0, u0) = 0
Note: different systems may have different equilibria,
not necessarily (0, 0), so we need to shift variables:
x = x − x0 u = u − u0
f(x, u) = f(x + x0, u + u0) = f(x, u)
Note that the transformation is invertible:
x = x + x0, u = u + u0

63. General Linearization Procedure

64. General Linearization Procedure
I Pass to shifted variables x = x − x0, u = u − u0
ẋ = ẋ (x0 does not depend on t)
= f(x, u)
= f(x, u)
— equivalent to original system

65. General Linearization Procedure
I Pass to shifted variables x = x − x0, u = u − u0
ẋ = ẋ (x0 does not depend on t)
= f(x, u)
= f(x, u)
— equivalent to original system
I The transformed system is in equilibrium at (0, 0):
f(0, 0) = f(x0, u0) = 0

66. General Linearization Procedure
I Pass to shifted variables x = x − x0, u = u − u0
ẋ = ẋ (x0 does not depend on t)
= f(x, u)
= f(x, u)
— equivalent to original system
I The transformed system is in equilibrium at (0, 0):
f(0, 0) = f(x0, u0) = 0
I Now linearize:
ẋ = Ax + Bu, where Aij =
∂fi
∂xj x=x0
u=u0
, Bik =
∂fi
∂uk x=x0
u=u0

67. General Linearization Procedure
I Why do we require that f(x0, u0) = 0 in equilibrium?

68. General Linearization Procedure
I Why do we require that f(x0, u0) = 0 in equilibrium?
I This requires some thought. Indeed, we may talk about
a linear approximation of any smooth function f at
any point x0:
f(x) ≈ f(x0)+f0
(x0)(x−x0) — f(x0) does not have to be 0

69. General Linearization Procedure
I Why do we require that f(x0, u0) = 0 in equilibrium?
I This requires some thought. Indeed, we may talk about
a linear approximation of any smooth function f at
any point x0:
f(x) ≈ f(x0)+f0
(x0)(x−x0) — f(x0) does not have to be 0
I The key is that we want to approximate a given
nonlinear system ẋ = f(x, u) by a linear system
ẋ = Ax + Bu (may have to shift coordinates:
x 7→ x − x0, u 7→ u − u0)

70. General Linearization Procedure
I Why do we require that f(x0, u0) = 0 in equilibrium?
I This requires some thought. Indeed, we may talk about
a linear approximation of any smooth function f at
any point x0:
f(x) ≈ f(x0)+f0
(x0)(x−x0) — f(x0) does not have to be 0
I The key is that we want to approximate a given
nonlinear system ẋ = f(x, u) by a linear system
ẋ = Ax + Bu (may have to shift coordinates:
x 7→ x − x0, u 7→ u − u0)
Any linear system must have an equilibrium point at
(x, u) = (0, 0):
f(x, u) = Ax + Bu f(0, 0) = A0 + B0 = 0.
Example 2-5 is solved in Lec 2-1