2.2. STATE VARIABLES AND LINEAR SYSTEMS 31
Example 2.1 (Motion of an automobile)A classical example of a simpli-
fied control system is the motion of a car subject to acceleration and braking
controls. A simplified mathematical model of such a system is
d^2 s
dt^2+a
ds
dt+bs=f(t)wheres(t)represents position at timet,sothatdsdt(t)represents velocity and
d^2 s(t)
dt^2 represents acceleration. The basic idea of the state variable approach
is to select variables that represent the state of the system. Certainly, the
positions(t)and the velocitydsdt(t) both represent states of the system. If we
letx 1 (t)=s(t), then we are assigning a state variablex 1 (t)to represent the
position of the system. The velocityx ̇ 1 (t)=s ̇(t)can be assigned another state
variable
x 2 (t)=x ̇ 1 (t)This is one of the state equations. Here, we have expressed one state in terms
of the other. Proceeding further, note thatx® 1 (t)=x ̇ 2 (t)=®s(t)yields the
acceleration term. From the second-order differential equation®s(t)=−as ̇(t)−
bs(t)+f(t),we have
x ̇ 2 (t)=−ax 2 (t)−bx 1 (t)+f(t)This is the second of the state equations. For annthorder differential equation
there must benfirst-order state equations. In this case, for a second-order
differential equation, we have twofirst-order differential equations. Casting
these two equations in vector-matrix form, we can write the set of state equations
as ∑
x ̇ 1 (t)
x ̇ 2 (t)
∏
=
∑
01
−b −a∏∑
x 1 (t)
x 2 (t)∏
+
∑
0
1
∏
f(t)that is of the form
x ̇(t)=Ax(t)+Bu(t)whereu(t)=f(t).
To obtain both the position and the velocity of the system as outputs, we
can selecty 1 (t)andy 2 (t)to represent the statesx 1 (t)andx 2 (t), respectively.
Placing these quantities in vector-matrix form, we obtain the output equation
∑
y 1 (t)
y 2 (t)∏
=
∑
10
01
∏∑
x 1 (t)
x 2 (t)∏
+
∑
0
0
∏
f(t)that is of the form
y(t)=Cx(t)+Du(t)Note again that the outputs are expressed in terms of the system states.