2.9. LINEARIZATION 79
higher thanfirst-order terms gives
x ̇i(t)=fi[x 0 (t),u 0 (t)] +Xnj=1∂fi
∂xjØ Ø Ø Ø Ø Ø
x 0 ,u 0(xj−x 0 j)+Xnj=1∂fi
∂ujØ Ø Ø Ø Ø Ø
x 0 ,u 0(uj−u 0 j) (2.42)where,i=1, 2 ,...,n.
Now, if we let
4 xi = xi−x 0 i
4 ui = ui−u 0 ithen
4 x ̇i=x ̇i−x ̇ 0 i
Sincex ̇ 0 i=fi[x 0 (t),u 0 (t)], Equation (2.42) can be written as
4 x ̇i=Xnj=1∂fi
∂xjØ Ø Ø Ø Ø Ø
x 0 ,u 0( 4 xj)+Xnj=1∂fi
∂ujØ Ø Ø Ø Ø Ø
x 0 ,u 0( 4 uj) (2.43)Equation (2.43) can be expressed in vector-matrix form as
4 x ̇=A∗ 4 x+B∗ 4 u (2.44)where
A∗=
∂f 1
∂x 1∂f 1
∂x 2 ∑∑∑∂f 1
∂xn
∂f 2
∂x 1∂f 2
∂x 2 ∑∑∑∂f 2
∂xn
..
...
.
... ..
.
∂fn
∂x 1∂fn
∂x 2 ∑∑∑∂fn
∂xn
B∗=
∂f 1
∂u 1∂f 1
∂u 2 ∑∑∑∂f 1
∂un
∂f 2
∂u 1∂f 2
∂u 2 ∑∑∑∂f 2
∂un
..
...
.
... ..
.
∂fn
∂u 1∂fn
∂u 2 ∑∑∑∂fn
∂un
(2.45)
BothA∗ andB∗ are evaluated at the nominal operating point. In general,
however, Equation (2.44), although linear, may contain time varying elements.
Example 2.8 (Linearization)Suppose we wish to linearize the following state
equations of a nonlinear system:
x ̇ 1 (t)=− 1
x^22 (t)(2.46)
x ̇ 2 (t)=x 1 (t)u(t) (2.47)These equations are to be linearized about the nominal trajectory[x 01 (t),x 02 (t)]
whichis the solution to the equations with initial conditionsx 01 (0) =x 02 (0) = 1
and inputu(t)=0.
Integrating Equation (2.47) with respect to timetunder the specified initial
conditions, we get
x 2 (t)=x 2 (0) = 1 (2.48)