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Homework Help: Open Loop Response

  1. Mar 15, 2010 #1
    Hi,


    I have the equation G(s)=C(s)/M(s)=10e^(ts)/s+3 and I have to find the open loop response c(t) with a unit step which is 1/s



    The explanation in the Text book is quite confusing, does anyone know if there are easier explanations like worked exampled on the internet so I can work though then solve this equation




    Thanks for your help
     
    Last edited: Mar 15, 2010
  2. jcsd
  3. Mar 15, 2010 #2

    Jmf

    User Avatar

    Have you studied Laplace Transforms?

    if so, then you can get the Laplace transform of the response by multiplying the transfer function of the system - which you have - with the LT of your input - a unit step, which has transform 1/s.

    Then you need to take the inverse transform of this, which will yield the response (with respect to time). To do this you'll probably need to make use of a partial fraction expansion.

    I don't know any good websites with examples of this though, sorry.
     
  4. Mar 19, 2010 #3
    Thanks for your help I have since worked though and found an answer, whether its correct or not that is another story. If anyone understands Laplace transforms my answers are attached, if you can have a look and see if i am wrong or right,

    The question I have done is a little different to the above G(s)=C(s)/M(s)=5e^-(ts)/s+5 and I have to find the open loop response c(t) with a unit step which is 1/s.

    My answers are attached

    Thanks
     

    Attached Files:

    Last edited: Mar 19, 2010
  5. Mar 19, 2010 #4

    Jmf

    User Avatar

    I got:

    [tex]c(t) = u(t-T) - u(t-T)e^{-5(t-T)}[/tex]

    where u(t) is the Heaviside step function, and the T is the constant 't' in your transfer function (I changed it to a big T since it's a bit dodgy to use the same symbol for two different things)

    i.e. I took:

    [tex]G(s) = e^{-Ts}\frac{5}{s+5}[/tex]

    Though I didn't take much care over it, so I'm not sure I'm correct. The Heaviside function will almost certainly be in your answer though, since there's a time delay in the transfer function.

    I'd recommend going and looking up the t-shifting theorem for Laplace Transforms.
     
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