Identifying First Order Transfer Function

In this section we shall determine the first order transer function model using the data obtained after performing step test experiment. Please note that this procedure is common for data obtained using both local and virtual experiments.

Identification of the transfer function of a system is important as it helps us to represent the physical system mathematically. Once the transfer function is obtained, one can acquire the response of the system for various inputs without actually applying them to the system. Consider the standard first order transfer function given below

$$G(s) = \frac{ C(s)}{ R(s)}$$ (3.1)

$$G(s) = \frac K{\tau s + 1}$$ (3.2)

Combining the previous two equations, we get

$$C(s) = K \left\{\frac {R(s)}{\tau s + 1}\right\}$$ (3.3)

Let us consider the case of giving a ramp input to this first order system. The Laplace transform of a ramp function with slope = $\upsilonup$ is $\frac \upsilonup {s^2}$. Substituting $R(s) = \frac \upsilonup {s^2}$ in equation 3.3, we obtain

$$C(s) = \frac K{\tau s + 1}\frac \upsilonup {s^2}$$ (3.4)

$$= \frac A{s} + \frac B{s^2} +\frac C{\tau s + 1}$$ (3.5)

Solving $C(s)$ using Heaviside expansion approach, we get

$$C(s) = K\upsilonup \left\{\frac1{s^2} - \frac \tau s + \frac {\tau^2}{\tau s + 1}\right\}$$ (3.6)

Taking the Inverse Laplace transform of the above equation, we get

$$c(t) = K\upsilonup \left\{t -\tau + \tau e^{\frac {-t}\tau }\right\}$$ (3.7)

The difference between the reference and output signal is the error signal $e(t)$. Therefore,

$$e(t) = r(t) - c(t)$$ (3.8)

$$e(t) = K\upsilonup t - K\upsilonup t + K\upsilonup \tau - K\upsilonup \tau e^\frac {-t}\tau$$ (3.9)

$$e(t) = K\upsilonup \tau (1 - e^{\frac {-t}\tau})$$ (3.10)

Normalizing equation 3.10 for $t>>\tau$, we get

$$e(t) = \tau$$ (3.11)

This means that the error in following the ramp input is equal to $\tau$ for large value of $t$ [3]. Hence, smaller the time constant $\tau$, smaller the steady state error.