Coursework : Root Locus

  • Thread starter Altairs
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  • #26
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I get following plot for Derivative only controller. Is it correct ? If yes then there isn't going to be any rise time but there should be some settling time ? And if the plot is correct can you please explain the characteristics of the graph ?

If incorrect, following is the code I am using. Where am I making mistake ?

Code:
Kp=0;
Ki=0;
Kd=1000;
m=1000;
b=50;
u=10;
num=[1];
den=[m b];
cruise=tf(num,den);
contr=tf([Kd Kp Ki],[1 0]);	
sys_cl=feedback(contr*cruise,1);
t=0:0.1:20;
step(u*sys_cl,t)
axis([0 20 0 10])
http://www.freeimagehosting.net/uploads/b0495eb59b.jpg
 
  • #27
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And how do I find rise time etc for a TF of the form [tex]\frac{K_iS}{(1000+K_i)s + 50}[/tex] ? Even if I divide it I still cant get to solve it to get the risetime etc using the formulae for 1st order systems because a constant term comes..what should I do with it ?
 
Last edited:
  • #28
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And how do I find rise time etc for a TF of the form [tex]\frac{K_iS}{(1000+K_i)s + 50}[/tex] ? Even if I divide it I still cant get to solve it to get the risetime etc using the formulae for 1st order systems because a constant term comes..what should I do with it ?
As you can see by the plot, the rise time is zero. The response goes instantaneously from 0 to 5, beacause of the derivative term [tex]K_is[/tex] and then decays exponentially because of the pole at -50/(1000+Ki).
 
  • #29
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As you can see by the plot, the rise time is zero. The response goes instantaneously from 0 to 5, because of the derivative term [tex]K_is[/tex] and then decays exponentially because of the pole at -50/(1000+Ki).
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 ?
 
  • #30
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And what about PID controller how do I find rise time etc for a PID controller...It has to has some...
 
  • #31
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656
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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?
 
  • #32
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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.
 
  • #33
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In a negative exponent exponential, what is the time when this happens?
Indefinite ?
 
  • #34
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Indefinite ?
When is[tex]Ke^{-\alpha t}[/tex] equal to 0.01K or 0.02K?
 
  • #35
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When is[tex]Ke^{-\alpha t}[/tex] equal to 0.01K or 0.02K?
[tex]\frac{-ln(0.01)}{\alpha}[/tex]
 
  • #36
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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 doesnt work..Any approximation ? I dont want to go for any tedious method...
 
  • #37
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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 doesnt work..Any approximation ? I dont 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.
 

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