Page 213 - Modern Control Systems

Section 3.6 The Transfer Function from the State Equation 187

and
- = 0. (3.69)
x 2 + lx A gx 3
To obtain the necessary first-order differential equations, we solve for lx 4 in Equa-
tion (3.69) and substitute into Equation (3.68) to obtain
= u{t), (3.70)
Mx 2 + mgx 3
since M » m. Substituting x 2 from Equation (3.68) into Equation (3.69), we have

Mlx 4 - Mgx 2 + u(t) = 0. (3.71)
Therefore, the four first-order differential equations can be written as

X\ — *2i x 2

JC 3 = JC 4 , and X4 = X u{t) (3.72)
7 * ~ Wi -
Thus, the system matrices are
0 1 0 0 0
0 0 -mg/M 0 1/M
A = B = (3.73)
0 0 0 1 0
0 0 g/l 0 V(Ml)

3.6 THE TRANSFER FUNCTION FROM THE STATE EQUATION

Given a transfer function G(s), we can obtain the state variable equations using the
signal-flow graph model. Now we turn to the matter of determining the transfer
function G(s) of a single-input, single-output (SISO) system. Recalling Equations
(3.16) and (3.17), we have

x = Ax + Bw (3.74)
and

v = Cx + Du (3.75)
where v is the single output and u is the single input. The Laplace transforms of
Equations (3.74) and (3.75) are
sX(s) = AX(s) + BU(s) (3.76)

and
Y(s) = CX(s) + DU(s) (3.77)

where B is an n X 1 matrix, since u is a single input. Note that we do not include ini-
tial conditions, since we seek the transfer function. Rearranging Equation (3.76), we
obtain

(si - A)X(s) = BU(s).