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How can I obtain the time constant of the transfer function of a first order system, such as the example below?

$$ \frac{C(s)}{R(s)} = \frac{2}{s + 3}$$

Where $C(s)$ is the output of the system and $R(s)$ is the input of the system.

I'm not looking for an exact answer, I just would like to be pointed in the right direction as to how to solve for the time constant so that I can solve it myself.


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This doesn't seem to be a first order system, which has a t.f. of the form $K/(\tau s+1)$ and an exponential step response. This one's step response depends on the impulse function $\delta(t)$ –  Ganesh Oct 15 '12 at 19:15
@Ganesh: How is this NOT a 1st order system? I really do NOT understand your comment. –  Rod Carvalho Nov 14 '12 at 1:48
@Ganesh this is a first order system. If you want to get it in general form, you'll have to divide the top and bottom by 3, in which case you would find τ to be 1/3. –  chilemagic Oct 18 '13 at 3:00
Matt: Ah, see the revision history and look at the first revision. Then you'll understand. @RodCarvalho: Now that I see YOUR comment, I DO NOT understand your MANNERS. No need to FLAME. –  Ganesh Oct 18 '13 at 4:35

1 Answer 1

The LTI system whose transfer function is

$$H (s) = \frac{2}{s + 3}$$

has the impulse response

$$h (t) = 2 \, e^{-3 t} = 2 \, e^{- t / \tau}$$

where $t \geq 0$ and $\tau := \frac{1}{3}$ is the time constant. I am assuming that the system is causal, of course.

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