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Query regarding steady state error in a velocity control system

  1. Apr 26, 2008 #1
    Hi

    I was reading BC Kuo's Automatic Control Systems where I came across a solved problem (page 369 of 7th edition) regarding velocity control. I have a problem understanding how the steady state error has been computed. The original problem and its solution as given in the book are quoted below.

    Let the feed-forward transfer function be

    [tex]G(s) = \frac{1}{s^2(s+12)}[/tex]

    and the feedback transfer function be

    [tex]H(s) = K_{t}s[/tex]

    where [itex]K_{t}[/tex] is the tachometer constant.

    I am not clear about how the steady state error has been computed here. I understand that the dominating terms as [itex]t\rightarrow \infty[/itex] are the linear term and the constant term, but how does the limit of the (0.1t -y(t)) term represent steady state error? How is the reference signal equal to 0.1t?

    Thanks in advance.
     
  2. jcsd
  3. Apr 26, 2008 #2
  4. Apr 26, 2008 #3
    I haven't been able to figure this out yet, so I would be really grateful if someone could look into it for me and tell me what the mistake in my understanding/interpretation is?
     
  5. Apr 27, 2008 #4
    Got it...thanks to varunag :-)

    It should be 10V/rad/sec. The output is the angle measure.
     
  6. Apr 27, 2008 #5
    For the later occurrences, steady state means change of the function goes to zero when t goes to infinity right? That means if I take the derivative of the function it should go to zero since there is a steady state(existence must be checked of course!) then derivative means multiplying with s right?

    [tex]\lim_{t\to \infty} f(t)\star u(t) = \lim_{s\to 0}sf(s)u(s)\[/tex]

    If the limit exists it will give you the SS error.
     
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