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## Q. 4.1

REALIZATION OF A FIRST-ORDER TRANSFORM-DOMAIN TRANSFER FUNCTION.

Given the following transfer function model:

$y(s) = g(s) u(s)$        (4.60a)

with

$g(s) = \frac{K}{\tau s + 1}$       (4.60b)

find the equivalent state-space representation.

## Verified Solution

Introducing Eq. (4.60b) into Eq. (4.60a) and rearranging, the transform-domain model may be rewritten as:

$(\tau s + 1)y(s) = Ku(s)$        (4.61a)

From here it is now clear that the most obvious differential equation that would give rise to Eq. (4.61a) upon Laplace transform is:

$\tau \frac{dy}{dt} + y(t) = Ku(t)$        (4.61b)

with the initial condition, y(0) = 0. This may be rewritten as:

$\tau \frac{dx}{dt} + x(t) = Ku(t)$        (4.62a)

$y(t) = x(t)$        (4.62b)

This is therefore a state-space equivalent of the model given in Eq. (4.60). Observe, however, that for any arbitrary constant c, we may define:

$y(t) = cx(t)$

and by this device, rewrite Eq. (4.61b) as:

$\tau \frac{dx}{dt} + x(t) = \frac{K}{c}u(t)$

Thus, another acceptable state space equivalent of the model in Eq. (4.61a) will be:

$\tau \frac{dx}{dt} + x(t) = \frac{K}{c}u(t)$        (4.63a)

$y(t) = cx(t)$       (4.63b)

Thus for various values of the constant c, we are able to generate a corresponding number of equally acceptable state-space equivalents of Eq. (4.61a). Each of these equivalent forms will have the same input/output relation between y(t) and u(t), but will have different definitions of the state x(t).