Chapter 6

# 6.1 First order systems

A first order system is described by the transfer function in Equation 6‑1:

 $G(s) = \frac{K}{s+a} = \frac{K_{dc}}{s\tau + 1}$ Equation 6‑1

$G(s)$ has only one pole, and no zeros. Its unit step response can be derived as shown in Equation 6‑2:

 $Y(s) = \frac{1}{s}G(s) = \frac{K}{s(s+a)} = \frac{K_{1}}{s} + \frac{K_{2}}{(s+a)}$ $K_1=\left .\begin{matrix} \frac{K}{s} \end{matrix}\right|_{s=0}=\frac{K}{a}$ $K_2=\left .\begin{matrix} \frac{K}{s+a} \end{matrix}\right|_{s=-a}=\frac{-K}{a}$ $Y(s) = \frac{K}{a}. \frac{1}{s} - \frac{K}{a}. \frac{1}{(s+a)}$ $y(t) = \frac{K}{a}. (1-e^{-at}. 1(t) = K_{dc}\left ( 1-e^{1\frac{t}{\tau}} \right ). 1(t)$ Equation 6‑2

Note that:

 $\frac{dy}{dt} = \frac{K_{dc}}{\tau}e^{-\frac{t}{\tau}}. 1(t)$ $\frac{dy}{dt} (0)= \frac{K_{dc}}{\tau} \neq 0$ Equation 6‑3

If the unit step input is used, the process DC gain and time constant can be evaluated directly from the graph, as shown in the following example.

### 6.1.1 Example

Consider a plot of the response of a certain unknown process, shown in Figure 6‑1. We would like to derive a model for this unknown system, i.e. a transfer function that would give a response closest to that of our system, let’s call it $G_{m}(s)$. The response looks like an exponential rise with a non-zero slope at t=0 and is therefore identified as the response of a first order process (system). As such, the response can be described by the following equation:

$y(t) = K_{dc}(1-e^{-\frac{t}{\tau}}). 1(t)$

## License

Introduction to Control Systems Copyright © by Malgorzata Zywno is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License, except where otherwise noted.