Introducing Eqs. (4.64b,c) into Eq. (4.64a) the transfer function model may be rewritten as:

y(s) = x_{1}(s)  +  x_{2}(s)         (4.65a)

where

x_{1}(s) = \frac{K}{\tau s  +  1}  u(s)        (4.65b)

x_{2}(s) = \frac{K_{d}}{\tau _{d} s  +  1}  d(s)        (4.65c)

If we now rearrange Eqs. (4.65b,c) as was done in Example (4.1) we will have:

(\tau s  +  1) x_{1}(s) = K u(s)        (4.66a)

(\tau _{d} s  +  1) x_{2}(s) = K_{d} d(s)        (4.66b)

from where we may now deduce one possible set of differential equations which upon Laplace transformation would give rise to Eq. (4.66a,b) and which are, respectively:

\tau \frac{dx_{1}}{dt}  +  x_{1}(t) = K u(t)        (4.67a)

for the process, and

\tau _{d} \frac{dx_{2}}{dt}  +  x_{2}(t) = K_{d}  d(t)        (4.67b)

for the disturbance. The process output is obtained as:

y(t) = x_{1}(t)  +  x_{2}(t)         (4.68)

Along with the initial conditions with x_{1}(0) = 0, and x_{2}(0) = 0, Eqs. (4.67a,b) and (4.68) constitute one possible state-space equivalent of the model given in Eq. (4.64.)

Remarks

In the special case for which \tau = \tau _{d}, we may add Eq. (4.66a) to Eq. (4.66b) to obtain (by means of Eq. (4.65a)):

(\tau s  +  1) y(s) = K u(s)  +  K_{d}  d(s)

and the differential equation extracted from here will simply be:

\tau \frac{dy}{dt}  +  y(t) = K u(t)  +  K_{d}  d(t)

We may then set y(t) = x(t) and obtain an equivalent state-space representation:

\tau \frac{dx}{dt}  +  x(t) = K u(t)  +  K_{d}  d(t)

y(t) = x(t)