Designing a Cruise Control System Controller with Root Locus Method | Coursework

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Altairs said:
Does that mean that the terms risetime, settling time and overshoot etc have no meaning at all for a pure derivative controller ? But when I think again I get confused that there should be some settling time formula or for overshoot etc. If yes then what is it ?

So what is going to be steady state value in case of pure derivative controller ? Zero, always ?
The settling time is zero.
The ss value can be obtained from the final value theorem. The settling time is the time when response attains 1% or 2% of its final value. In a negative exponent exponential, what is the time when this happens?
 
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Altairs said:
And what about PID controller how do I find rise time etc for a PID controller...It has to has some...

Obtain the time response and verify how long does it take from 10% to 90% of ss.
 
CEL said:
In a negative exponent exponential, what is the time when this happens?

Indefinite ?
 
Altairs said:
Indefinite ?

When is[tex]Ke^{-\alpha t}[/tex] equal to 0.01K or 0.02K?
 
CEL said:
When is[tex]Ke^{-\alpha t}[/tex] equal to 0.01K or 0.02K?

[tex]\frac{-ln(0.01)}{\alpha}[/tex]
 
CEL said:
Obtain the time response and verify how long does it take from 10% to 90% of ss.

I have got the steady state value but how do I get the time response ? Laplace Transform ? Isn't there some shorter method ? This is because I have got a zero as well in the PI controller so the general formulae doesn't work..Any approximation ? I don't want to go for any tedious method...
 
Altairs said:
I have got the steady state value but how do I get the time response ? Laplace Transform ? Isn't there some shorter method ? This is because I have got a zero as well in the PI controller so the general formulae doesn't work..Any approximation ? I don't want to go for any tedious method...

You must take the inverse tarnsform of the response Y(s). Don't forget that the transform of the step is 1/s, so the s in the denominator cancels the s in the numerator.