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For example, if your transfer function is defined as :

$$\frac{H(s)}{Q(s)} = \frac{1}{5s}$$

What would the gain and time constant be in this case (since it's usually in the form of $\dfrac{k}{Ts+1}$ where $k$ and $T$ are the time constant and gain, respectively)?

(This is the transfer function for a tank being filled, but the exit stream is a pump so it is constant. Where $H$ and $Q$ are deviation variables for output and input, respectively)

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$G(s) = \frac{1}{s}$ is called an integrator. It continuously integrates your signal and therefore it has no time constant, hence $\tau = 0$.

Your system $\frac{H(s)} {Q(s)} = \frac{1}{5s}$ can be written as an integrator multiplied with a gain of $\frac{1}{5}$.

Therefore the pole-zero gain is $k_{pz} = \frac{1}{5}$.

The static gain of your function $k_s = \lim_{s\to 0} \frac{1}{5s} = \infty$ or you might say there is no static gain.

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  • $\begingroup$ I talked to a prof at my school and he agreed that there is no time constant, but he said that the gain would actually be infinite since it has no value that it will eventually attain. $\endgroup$ – Curtis Oct 7 '16 at 2:24
  • $\begingroup$ @Curtis depends on your definition on gain. So I will update my post and add some more definitions off Gain. $\endgroup$ – useless-machine Oct 7 '16 at 9:49
  • $\begingroup$ @Curtis Now I included both the explanation of both gains. $\endgroup$ – useless-machine Oct 7 '16 at 11:44

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