## Chapter 4

## 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.

## Step-by-Step

## 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).

Some comments about this last example are in order here. Regarding the concept of realization, we note that:

- Several, equally acceptable, state-space realizations are possible for the transfer function model in Eq. (4.60). Thus while the transformation from the state-space to the transform domain is unique, the reverse transformation back to the state space is not.
- The nonuniqueness of the state-space realizations from the transformdomain model arises because the transform-domain representation inherently contains Jess information. The information about the state variables, missing from the transform-domain form but required in the state-space form, can be incorporated in an arbitrary number of equally acceptable ways as shown in Eq. (4.63).
- In all realizations, the relationship between the input u(t) and output y(t) is unique even if the definition of the state variables is not.

Regarding the actual “mechanics” of the procedure we note:

- Even though we utilized the inverse of the process of Laplace transformation in arriving at Eq. (4.62) (and Eq. (4.63)) from Eq. (4.60), the utility was only partial. A complete Laplace inversion would have provided the solution to the differential equation, not merely recover the differential equation in its unsolved form.
- We may therefore consider the process of realization as a “reverse” Laplace transform, which produces differential equations from transform-domain transfer functions, not to be confused with the process of Laplace inversion (or inverse Laplace transformation) which provides the solution to the differential equation.
- In realization, the desired end result is the (unsolved) differential equation itself; in Laplace inversion, it is the solution of the differential equation that is desired.