Chapter 6

6.1 First order systems

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

[latex]G(s) = \frac{K}{s+a} = \frac{K_{dc}}{s\tau + 1}[/latex]

Equation 6‑1

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

[latex]Y(s) = \frac{1}{s}G(s) = \frac{K}{s(s+a)} = \frac{K_{1}}{s} + \frac{K_{2}}{(s+a)}[/latex]

[latex]K_1=\left .\begin{matrix} \frac{K}{s} \end{matrix}\right|_{s=0}=\frac{K}{a}[/latex]

[latex]K_2=\left .\begin{matrix} \frac{K}{s+a} \end{matrix}\right|_{s=-a}=\frac{-K}{a}[/latex]

[latex]Y(s) = \frac{K}{a}. \frac{1}{s} - \frac{K}{a}. \frac{1}{(s+a)}[/latex]

[latex]y(t) = \frac{K}{a}. (1-e^{-at}. 1(t) = K_{dc}\left ( 1-e^{1\frac{t}{\tau}} \right ). 1(t)[/latex]

Equation 6‑2

Note that:

[latex]\frac{dy}{dt} = \frac{K_{dc}}{\tau}e^{-\frac{t}{\tau}}. 1(t)[/latex]

[latex]\frac{dy}{dt} (0)= \frac{K_{dc}}{\tau} \neq 0[/latex]

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 [latex]G_{m}(s)[/latex]. 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:

[latex]y(t) = K_{dc}(1-e^{-\frac{t}{\tau}}). 1(t)[/latex]

Figure 6-1: First Order System response
Figure 6-1: First Order System response

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