Linear Systems and Control
Lecture # 1
State models
– p. 1/3
We will deal with dynamical systems modeled by n coupled
first-order ordinary differential equations:
ẋ1 = f1 (t, x1 , . . . , xn , u1 , . . . , um )
ẋ2 = f2 (t, x1 , . . . , xn , u1 , . . . , um )
..
..
.
.
ẋn = fn (t, x1 , . . . , xn , u1 , . . . , um )
ẋi =
dxi
dt
,
t is the time variable
u1 , . . . , um are the input variables
x1 , . . . , xn are the state variables
– p. 2/3
Vector Notation:



x1


u1
 x2 



 u2
 .. 
x =  . , u = 
 ..
 . 
 .
 .. 


um
xn







 , f (t, x, u) = 






f1 (t, x, u)
f2 (t, x, u)
..
.
..
.
fn (t, x, u)









ẋ = f (t, x, u)
– p. 3/3
Output equation:



y=


y1
y2
..
.
yp

y = h(t, x, u)





 , h(t, x, u) = 




h1 (t, x, u)
h2 (t, x, u)
..
.
hp (t, x, u)






State Model:
ẋ = f (t, x, u) State equation
y = h(t, x, u) Output equation
– p. 4/3
Example: Consider a single-input–single-output (SISO)
system represented by the nth-order differential equation
y (n) = g t, y, y (1) , . . . , y (n−1) , u
where u is the input, y is the output, and
y (i) =
di y
dti
State variables:
x1 = y, x2 = y (1) , · · · , xn = y (n−1)
– p. 5/3
y (n) = g t, y, y (1) , . . . , y (n−1) , u
x1 = y, x2 = y (1) , · · · , xn = y (n−1)
ẋ1 = x2
ẋ2 = x3
..
..
.
.
ẋn−1 = xn
ẋn = g(t, x1 , x2 , . . . , xn , u)
– p. 6/3




f =



x2
x3
..
.
xn
g(t, x1 , x2 , . . . , xn , u)




 , h = x1



ẋ = f (t, x, u)
y = h(x)
– p. 7/3
Example: Consider a SISO system represented by the
input-output relation
ÿ + a1 ẏ + a2 y = b0 ü + b1 u̇ + b2 u
where u is the input, and y is the output
Rewrite equation as follows:
ÿ = b0 ü + (b1 u̇ − a1 ẏ) + (b2 u − a2 y)
Z
Z Z
y = b0 u + (b1 u − a1 y) +
(b2 u − a2 y)
{z
}
|
x2
|
{z
}
x1
– p. 8/3
Define:
x1 =
x2 =
Z
Z
(b1 u − a1 y)dt +
Z Z
(b2 u − a2 y)dt dt
(b2 u − a2 y)dt
Then,
y = b0 u + x1
ẋ1 = b1 u − a1 y + x2
= (b1 − b0 a1 )u − a1 x1 + x2
ẋ2 = b2 u − a2 y
= (b2 − b0 a2 )u − a2 x1
– p. 9/3
"
#"
#
"
−a1 1
b 1 − b 0 a1
x1
f =
+
−a2 0
x2
b 2 − b 0 a2
"
#
h
i x
1
+ b0 u
h =
1 0
x2
#
u
ẋ = f (t, x, u)
y = h(x)
– p. 10/3
Example: Pendulum Equation
l
θ
•
mg
mℓ2 θ̈ = −mgℓ sin θ − kℓ2 θ̇ + u
m
ℓ
u
k
Mass of the bob
Length of the rod
Applied torque
Friction coefficient
– p. 11/3
mℓ2 θ̈ = −mgℓ sin θ − kℓ2 θ̇ + u
θ̈ = −(g/ℓ) sin θ − (k/m)θ̇ + (1/mℓ2 )u
State variables : x1 = θ, x2 = θ̇
Output variable : y = θ
f =
"
x2
−(g/ℓ) sin x1 − (k/m)x2 + (1/mℓ2 )u
#
, h = x1
ẋ = f (x, u)
y = h(x)
– p. 12/3
Existence and Uniqueness of Solutions
Consider
ẋ = f (t, x, u)
y = h(t, x, u)
Assume that f and h are continuously differentiable in
(x, u) and continuous in t
Then there is a unique solution x(t) of
ẋ = f (t, x, u),
x(t0 ) = x0
over some time interval [t0 , t1 ].
Consequently, the output y(t) is unique
– p. 13/3
Classification of State models
Time-varying system:
ẋ = f (t, x, u)
y = h(t, x, u)
Time-invariant system:
ẋ = f (x, u)
y = h(x, u)
The response of a time-invariant system is invariant to
shifts of the time origin. Therefore, without loss of
generality, we can take the initial time to be t0 = 0
Why?
– p. 14/3
Let x(t) be the solution of
ẋ = f (x(t), u(t)),
x(t0 ) = x0
and x̃(t) be the solution of
x̃˙ = f (x̃(t), ũ(t)),
x̃(t0 + a) = x̃0
Take
x̃0 = x0 and ũ(t) = u(t − a), for t ≥ t0 + a
– p. 15/3
τ =t−a
ψ(τ ) = x̃(τ + a) = x̃(t)
dψ
dτ
dψ
=
dx̃
dt
= f (x̃(τ + a), ũ(τ + a))
= f (ψ(τ ), u(τ )),
dτ
By the uniqueness of solution
ψ(τ ) = x(t)
ψ(t0 ) = x0
t=τ
x̃(t) = ψ(τ ) = ψ(t − a) = x(t − a)
– p. 16/3
Classification of State models
Nonlinear system:
ẋ = f (t, x, u)
y = h(t, x, u)
Linear system:
ẋ = A(t)x + B(t)u
y = C(t)x + D(t)u
A is n × n, B is n × m, C is p × n, D is p × m
The response of a linear system satisfies the superposition
principle: Let
(xa (t), ya (t)) be the response to xa (t0 ) = xa0 & ua (t)
(xb (t), yb (t)) be the response to xb (t0 ) = xb0 & ub (t)
then, for any real numbers α and β , αxa (t) + βxb (t) is the
response to x(t0 ) = αxa0 + βxb0 & u(t) = αua (t) + βub (t)
– p. 17/3
Let x(t) be the solution of
ẋ = A(t)x + B(t)u
x(t0 ) = αxa0 + βxb0 , and u(t) = αua (t) + βub (t)
Let ψ(t) = αxa (t) + βxb (t)
ψ̇ =
=
=
=
αẋa + β ẋb
α[A(t)xa + B(t)ua ] + β[A(t)xb + B(t)ub ]
A(t)(αxa + βxb ) + B(t)(αua + βub )
A(t)ψ + B(t)u
By the uniqueness of solutions
x(t) = ψ(t) = αxa (t) + βxb (t)
– p. 18/3
Classification of Linear models
Time-varying:
ẋ = A(t)x + B(t)u
y = C(t)x + D(t)u
Time-invariant:
ẋ = Ax + Bu
y = Cx + Du
The response of a linear time-invariant system is invariant
to shifts of the time origin. Therefore, without loss of
generality, we can take the initial time to be t0 = 0
– p. 19/3
Linearization
Consider the nonlinear state equation
ẋ = f (t, x, u)
Let x̃(t) be the solution when u(t) = ũ(t) and x̃(t0 ) = x̃0
Suppose the input and initial state are slightly perturbed
u(t) = ũ(t) + uδ (t),
x0 = x̃0 + x0δ
where kuδ (t)k and kx0δ k are small
q
kxk = x21 + · · · + x2n
– p. 20/3
Let x(t) be the solution of
ẋ = f (t, x, u),
x(t0 ) = x0
and write
x(t) = x̃(t) + xδ (t)
It is reasonable to expect kxδ (t)k to be small
ẋ = f (t, x, u),
x(t0 ) = x0
x̃˙ + ẋδ = f (t, x̃ + xδ , ũ + uδ ),
x̃(t0 ) + xδ (t0 ) = x̃0 + x0δ
Expand the R.H.S. in a Taylor series about the nominal
solution
– p. 21/3
fi (t, x̃ + xδ , ũ + uδ ) = fi (t, x̃, ũ)
∂fi
n
+ Σj=1
(t, x̃, ũ)xδj
∂xj
m ∂fi
(t, x̃, ũ)uδj
+ Σj=1
∂uj
+ Higher order terms
– p. 22/3
fi (t, x̃ + xδ , ũ + uδ ) ≈ fi (t, x̃, ũ)
+
+
∂fi
∂x1
∂fi
∂u1
,...,
,...,
∂fi
∂xn
∂fi
∂um


xδ1
 .. 
 . 
xδn


uδ1
 .. 
 . 
uδm
and ∂f
∂u such that
∂f
∂fi
the (i, j) element of
is
∂u
∂uj
Define the Jacobian matrices
∂f
∂x
– p. 23/3
˙x̃ + ẋδ ≈ f (t, x̃, ũ) + ∂f (t, x̃, ũ)xδ
∂x
∂f
+
(t, x̃, ũ)uδ ,
∂u
x̃(t0 ) + xδ (t0 ) = x̃0 + x0δ
x̃˙ = f (t, x̃, ũ),
ẋδ ≈
∂f
∂x
x̃(t0 ) = x̃0
(t, x̃, ũ)xδ +
∂f
∂u
(t, x̃, ũ)uδ ,
xδ (t0 ) = x0δ
– p. 24/3
Let
A(t) =
∂f
∂x
(t, x̃, ũ),
B(t) =
ẋδ ≈ A(t)xδ + B(t)uδ ,
∂f
∂u
(t, x̃, ũ)
xδ (t0 ) = x0δ
Similarly
yδ ≈ C(t)xδ + D(t)uδ
C(t) =
∂h
(t, x̃, ũ),
∂x
Approximate Linear Model:
D(t) =
∂h
∂u
(t, x̃, ũ)
ẋδ = A(t)xδ + B(t)uδ
yδ = C(t)xδ + D(t)uδ
– p. 25/3
Example
ẋ1 = x2
ẋ2 = −x22 + u
y = sin x1 + x2 u
Let
t0 = 0, ũ(t) ≡ 0, ∀ t ≥ 0, x̃0 =
Nominal solution :
x̃(t) =
"
"
1
1
1 + ln(1 + t)
1
1+t
#
#
Linearize the system about the nominal solution
– p. 26/3
f (x, u) =
∂f
"
∂f1
∂x1
∂f2
∂x1
"
∂f1
∂x2
∂f2
∂x2
x2
−x22 + u
#
"
#
0
1
=
=
0 −2x2
∂x
#
"
∂f
0
1
A(t) =
(x̃, ũ) =
2
0 − 1+t
∂x
"
#
" #
∂f1
∂f
0
∂u
= ∂f2 =
=B
1
∂u
∂u
#
– p. 27/3
h(x, u) = sin x1 + x2 u
∂h
∂x
C(t) =
=
∂h
∂x
h
∂h
∂x1
(x̃, ũ) =
∂h
∂u
∂h
∂x2
h
i
=
h
cos x1 u
i
cos(1 + ln(1 + t))
= x2 ⇒ D(t) =
1
0
i
1+t
– p. 28/3
Calculation of the nominal solution
dx2
x22
ẋ1 = x2 ,
x1 (0) = 1
ẋ2 = −x22 ,
x2 (0) = 1
= −dt ⇒
Z
x2
1
dz
z2
= −
Z
t
dτ
0
x 2
1 1
1
− = −t ⇒ −
+ 1 = −t ⇒ x2 =
z 1
x2
1+t
Z t
1
x1 (t) = 1 +
dτ = 1 + ln(1 + t)
0 1+τ
– p. 29/3
Special Case
Consider the state equation
ẋ = f (x)
and suppose that x̄ is an equilibrium point
f (x̄) = 0
x(0) = x̄ ⇒ x(t) ≡ x̄, ∀ t ≥ 0
Linearization about x̄:
ẋδ = Axδ ,
A=
∂f
∂x
(x̄)
– p. 30/3
Example: The Pendulum equation(with u = 0)
ẋ1 = x2
ẋ2 = −(g/ℓ) sin x1 − (k/m)x2
y = x1
f =
"
x2
−(g/ℓ) sin x1 − (k/m)x2
#
, h = x1
Equilibrium Points:
ẋ1 = ẋ2 = 0 ⇒ x2 = 0,
x̄ =
"
kπ
0
#
,
sin x1 = 0
k = 0, ±1, ±2, . . .
– p. 31/3
"
#
x2
f =
−(g/ℓ) sin x1 − (k/m)x2
"
#
∂f
0
1
=
−(g/ℓ) cos x1 −(k/m)
∂x
"
#
0
1
Linearization at (0, 0) :
A=
−(g/ℓ) −(k/m)
"
#
0
1
Linearization at (π, 0) :
A=
(g/ℓ) −(k/m)
– p. 32/3