2.1. INTRODUCTORY EXAMPLES: PENDULUM PROBLEMS 17
In this example, we concentrate on answering thefirst question. When we
attempt to keep the pendulum upright, our operating range is a small range
around the unstable equilibrium position. As such, we have alocalcontrol
problem, and we can simplify the mathematical model in Equation (2.1) by
linearizing it around the equilibrium point. Forφ=θ−πsmall, we keep only
thefirst-order term in the Taylor expansion ofsinθ,thatis,−(θ−π),sothat
the linearization of Equation (2.1) is the linear model (thelinear differential
equation)
φ®(t)−φ(t)=u(t) (2.2)
and the control objective is manipulatingu(t)to bringφ(t)andφ ̇(t)to zero
from any small nonzero initialφ(0),φ ̇(0).
Note that Equation (2.2) is a second-order differential equation. It is conve-
nient to replace Equation (2.2) by a system offirst-order differential equations
in terms ofφ(t)andφ ̇(t).Here,letx(t)be the vector
x(t)=μ
x 1 (t)
x 2 (t)∂
=
μ
φ(t)
φ ̇(t)∂
so that
x ̇(t)=μ
φ ̇(t)
φ®(t)∂
=
μ
x 2 (t)
x ̇ 2 (t)∂
With this notation we see that the original model, Equation (2.1), is written as
x ̇=f(x,u) (2.3)wherefis nonlinear, and
f=μ
f 1
f 2∂
wheref 1 (x,u)=x 2 andf 2 (x,u)=−sin (x 1 +π)+u.Sincefis continuously
differentiable andf(0,0) = 0, we can linearizefaround(x,u)=(0,0)as
x ̇=Ax+Bu (2.4)where the matricesAandBare
A =
√ ∂f 1
∂x 1∂f 1
∂x 2
∂f 2
∂x 1∂f 2
∂x 2!
=
μ
01
10∂
B =
√ ∂f
1
∂u
∂f 2
∂u!
=
μ
0
1∂
with both Jacobian matricesAandBevaluated at(x,u)=(0,0).
Thus, in the state-space representation, Equation (2.2) is replaced by Equa-
tion (2.4). Note that, in general, systems of the form (2.4) are calledlinear
systems,andwhenAandBdo not depend on time, they are calledtime-
invariantsystems.