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Open Loop Response

  • Thread starter Kayne
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  • #1
24
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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:

Answers and Replies

  • #2
Jmf
49
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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.
 
  • #3
24
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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
 

Attachments

Last edited:
  • #4
Jmf
49
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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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