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Thread: Help with final and initial value

  1. #1
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    Help with final and initial value

    The function is (s-10)/((s^2 +20s)). I am trying to understand how to compute the final and initial value of the function with a unit step impulse. If I take the lim @ inf, using l'hopital's rule I get 0 this should be the final value. However, if I take the lim @ 0 I also get 0 and this should be my initial value. So I am confused. Does l'hopital's rule not apply to this function? I do get an indeterminate form of inf/inf.

    If I put the function into wolfram I get an output plot that looks like what I would expect from 1/(s+a) if I were to check the limits at inf and 0 with a unit step input. However, if I create the transfer function in matlab and apply the unit step I get an output that looks like 1/(s+a) had I applied a ramp function to it.

    Can someone tell me why taking the limit of this function at inf leads to 0 when it should lead to -inf according to matlab with a unit step input but according to wolfram it leads to 0. Is the output that matlab gave me just the time domain output that wolfram gave me but multiplied by 1/s and why doesn't take the lim @ 0 give me the value .2 which is the amplitude of the impulse displayed on the wolfram output. For instance, lim @ 0 of 1/(s+a) = 1/a which would be the final value but also the amplitude of the initial impulse given that my input is a step.

    Please help. I am obviously confused.
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  2. #2
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    Re: Help with final and initial value

    you need to find

    $\displaystyle{\lim_{t \to \infty}~f(t) = \lim_{s\to 0}}~ s F(s)$

    not

    $\displaystyle{\lim_{t \to \infty}~f(t) = \lim_{s\to 0}}~ F(s)$
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  3. #3
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    Re: Help with final and initial value

    OH I see I must have missed that part. So is the same for lim @ 0? Do I multiply by s for that as well?
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  4. #4
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    Re: Help with final and initial value

    Quote Originally Posted by vysero View Post
    OH I see I must have missed that part. So is the same for lim @ 0? Do I multiply by s for that as well?
    yep
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  5. #5
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    Re: Help with final and initial value

    Okay still a bit confused. If I multiply the function by s so that I have (s+10)/(s+20) and, using l'hopital's rule, I take the limit @ inf then I get 1 and the limit at 0 I get -.5. However, if I create the function in Matlab and run a step on it I get a ramp output that starts at zero and has a slope of -1 but goes off to -inf.

    So here is what I have determined. If I have a function say 2/(s+3) and I do not multiply it by s but instead just evaluate the limit @ inf and 0 then I can determine the initial and final value of that function given a unit step input. However, given a unit step input, I cannot determine the final value of the same function if I take the limit at 0 after multiplying by s. Also, given the original function: (s-10)/((s^2 +20s)) I cannot determine the initial value or final value regardless of if I multiply by s or not, given a unit step input or (according to matlab) a simple time domain sketch of the function.
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  6. #6
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    Re: Help with final and initial value

    I suspect you're are neglecting to include the $\dfrac 1 s$ factor due to exciting your system with a unit step rather than a unit impulse.

    Peruse this sheet.

    Help with final and initial value-clipboard01.jpg
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  7. #7
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    Re: Help with final and initial value

    Here is what I put in matlab:

    Code:
    >> a = tf([1 -10 ],[1 20 0])
    
    a =
     
        s - 10
      ----------
      s^2 + 20 s
     
    Continuous-time transfer function.
    
    >> step(a)
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