I read the following in a tutorial:

The standard transfer function of a first order system is:

$$G(s) = \frac{k}{s + a}$$

Arranging this into unity constant form gives:

$$G(s) = \frac{\frac{k}{a}}{\frac{s}{a} + 1}$$

What is "unity constant form", and why do we turn it into this form?

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    $\begingroup$ Which tutorial? $\endgroup$ – David Z Mar 12 '16 at 11:04
  • $\begingroup$ I thought it would be a widely agreed term. It is about electronic first order system responses. $\endgroup$ – ergon Mar 12 '16 at 16:18
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    $\begingroup$ I thought it was pretty close to being a good question, actually. The only fault IMO, though it is a significant one, is that you didn't reference the tutorial - both for giving credit, and also for giving the context that people may need to answer the question more precisely. $\endgroup$ – David Z Mar 12 '16 at 20:11

I can't address this quesion specifically because you don't give enough context. However as a general rule the reason we often put things into the form $1+x$ is so we can approximate them using a binomial expansion.

In this case we can write your final equation as:

$$ G(s) = \frac{k}{a}\left(1 +\frac{s}{a}\right)^{-1} \tag{1} $$

and if $s \ll a$ the bracket expands to:

$$ \left(1 +\frac{s}{a}\right)^{-1} = 1 - \frac{s}{a} + \mathcal{O}\left(\frac{s}{a}\right)^2$$

and if $s/a$ is small then the squared terms are very small and to a good approximation we can ignore them. That means your equation (1) simplifies to:

$$ G(s) \approx \frac{k}{a}\left(1 -\frac{s}{a}\right) $$

Whether this is a useful approximation for your equation I can't say because I don't know the details of your system. However it is a massively useful approximation in many areas of physics and one I have used many times on this site.

  • $\begingroup$ No, this is seldom the case for a transfer function. $\endgroup$ – Massimo Ortolano Jul 23 '17 at 21:02

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