## Chapter 4

## Q. 4.4

**OBTAINING A STATE-SPACE MODEL FROM AN IMPULSE-RESPONSE MODEL.**

Find an equivalent state-space representation of the following impulse-response model:

y(t) = \frac{K}{\tau} \int_{0}^{t}{e^{- (t – \sigma)/ \tau} u(\sigma) d \sigma} (4.75)

## Step-by-Step

## Verified Solution

Clearly, in this case, the impulse-response function g(t) is given by:

g(t) = \frac{K}{\tau} e^{-t/ \tau} (4.76)

which is differentiable. If we now differentiate Eq. (4.75) with respect to t (with the aid of Eq. (4.74)), we obtain:

\frac{d}{dx} \left\lgroup\int_{A(x)}^{B(x)}{f(x, r)dr} \right\rgroup = \int_{A}^{B}{\frac{\partial f(x, r)}{\partial r} dr} + f(x, B)\frac{dB}{dx} – f(x, A)\frac{dA}{dx} (4.74)

\frac{dy}{dt} = \frac{K}{\tau}\int_{0}^{t}{- \frac{1}{\tau} e^{- (t – \sigma) / \tau} u(\sigma) d \sigma + \frac{K}{\tau} u(t) – 0}

which simplifies further (using Eq. (4.75) to eliminate the integral) to give:

\frac{dy}{dt} = – \frac{1}{\tau} y(t) + \frac{K}{\tau} u(t)

or

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

Again, we may now specify y = x, to obtain:

\tau \frac{dx}{dt} + x(t) = K u(t) (4.62a)

y(t) = x(t) (4.62b)

as required state-space representation.

It is important to recognize that the impulse-response model is a time-domain input/ output model form that also does not explicitly contain information about the state variable x(t); therefore as with the transform-domain model, conversions to the state space will not be unique.