1.1.1
Linearization via Taylor Series
In order to linearize general nonlinear systems, we will use the Taylor Series expansion of functions.
Consider a function f (x) of a single variable x, and suppose that x̄ is a point such that f (x̄) = 0. In this
case, the point x̄ is called an equilibrium point of the system ẋ = f (x), since we have ẋ = 0 when x = x̄
(i.e., the system reaches an equilibrium at x̄). Recall that the Taylor Series expansion of f (x) around the
point x̄ is given by
df 1 d2 f 1 d3 f 2
f (x) = f (x̄) +
(x − x̄) +
(x − x̄) +
(x − x̄)3 + · · · .
dx x=x̄
2 dx2 x=x̄
6 dx3 x=x̄
This can be written as
df f (x) = f (x̄) +
(x − x̄) + higher order terms.
dx | {zx=x̄}
a
For x sufficiently close to x̄, these higher order terms will be very close to zero, and so we can drop them
to obtain the approximation
f (x) ≈ f (x̄) + a(x − x̄) .
Since f (x̄) = 0, the nonlinear differential equation ẋ = f (x) can be approximated near the equilibrium
point by
ẋ = a(x − x̄) .
To complete the linearization, we define the perturbation state (also known as delta state) δx = x − x̄,
and using the fact that δẋ = ẋ, we obtain the linearized model
δẋ = aδx .
Note that this linear model is valid only near the equilibrium point (how “near” depends on how nonlinear
the function is).
Extension To Functions of Multiple States and Inputs
The extension to functions of multiple states and inputs is very similar to the above procedure. Suppose
the evolution of state xi is given by
ẋi = fi (x1 , x2 , . . . , xn , u1 , u2 , . . . , um ) ,
for some general function fi . Suppose that the equilibrium points are given by x̄1 , x̄2 , . . . , x̄n , ū1 , ū2 , . . . , ūm ,
so that
fi (x̄1 , x̄2 , . . . , x̄n , ū1 , ū2 , . . . , ūm ) = 0 ∀i ∈ {1, 2, . . . , n} .
Note that the equilibrium point should make all of the functions fi equal to zero, so that all states in the
system stop moving when they reach equilibrium. The linearization of fi about the equilibrium point is
then given by
n
m
X
X
∂fi ∂fi fi (x1 , . . . , xn , u1 , . . . , um ) ≈
(xj − x̄j ) +
(uj − ūj ) .
∂xj xj =x̄j
∂uj uj =ūj
j=1
j=1
If we define the delta states and inputs δxj = xj − x̄j (for 1 ≤ j ≤ n) and δuj = uj − ūj (for 1 ≤ j ≤ m),
the linearized dynamics of state xi are given by
n
m
X
X
∂fi ∂fi δẋi =
δxj +
δuj .
∂xj xj =x̄j
∂uj uj =ūj
j=1
j=1
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Note: Sometimes the “δ” notation is dropped in the linearized equation, with the implicit understanding
that we are working with a linearized system.
Example. Linearize the nonlinear state-space model
ẋ1 = x21 + sin x2 − 1
ẋ2 = −x32 + u
y = x1 + x2
around the equilibrium point x̄1 = 1, x̄2 = 0, ū = 0.
Solution.
6
2
Obtaining The Transfer Function From A Linear State-Space Model
Since we can generally convert nonlinear models to a linear model (in a small region around the equilibrium
point), we will focus on linear state-space models of the form
ẋ = Ax + Bu,
x ∈ Rn , u ∈ Rm , y ∈ Rp
y = Cx .
for the rest of the course. Since this model represents a linear system, we can ask how the matrices A, B
and C relate to the transfer function of the system. To see this, take the Laplace Transform of the above
state space equations:
sX(s) − x(0) = AX(s) + BU(s)
Y(s) = CX(s) .
Note that this includes the initial conditions of all the states. The first equation can be rearranged to solve
for X(s) as follows:
(sI − A)X(s) = x(0) + BU(s) ⇔ X(s) = (sI − A)−1 x(0) + (sI − A)−1 BU(s) .
The term I represents the n × n identity matrix. Substituting this into the equation for Y(s), we obtain
Y(s) = C(sI − A)−1 x(0) + C(sI − A)−1 BU(s) .
The transfer function of the state-space model ẋ = Ax + Bu, y = Cx (when x(0) = 0) is
H(s) = C(sI − A)−1 B .
Note that H(s) is a p × m matrix, and thus it is a generalization of the transfer function for standard
single-input single-output systems. In fact, it is a matrix where entry i, j is a transfer function describing
how the j–th input affects the i–th output. When p = 1 and m = 1, we get the transfer function that we
studied in the first part of the course.
Example. Calculate the transfer function for the state space model
0
1
0
ẋ =
x+
u, y = 1 0 x .
−2 −3
4
| {z }
|
{z
}
|{z}
C
A
B
Solution.
7
Note that the above solution agrees with the transfer function at the beginning of the section.
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