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Homework Help: System Modelling/Dynamics question (time settling)

  1. Aug 3, 2015 #1
    1. The problem statement, all variables and given/known data
    This is for a first order step function response, where T is the time constant and x(t) is H...(input).
    For settling time, the percentage most often taken is 1% within the final value, which gives the settling time, ts, from:

    0.99H = H(1 - e-ts/T)

    We get, ts = 4.6T.

    Now, in some applications it may be appropriate to take another value of output. For example: time to get within 0.1% of final value.
    I saw this example somewhere but I don't understand their working method which was:

    t0.1% = t1% + ln(10)T

    I understand that the time to get within 0.1% of the final value will be "the time to get within 1% of the final value (H)" + something.
    But I don't understand why it's plus ln(10)T. It's been bothering me since last night and I need to know why they did that and still got the right answer of 6.9T.

    When you do it the normal way, you get 6.9T too. ie:

    0.999H = H(1 - e-ts/T)
    e-ts/T = 0.001
    ts = -Tln(0.001)
    ts = Tln(1000)
    ts = 6.9T
    t0.1% = 6.9T

    My question is, why did they add "ln(10)T" to "the time it takes to get within 1% of the final value"?
    I have a feeling I might just be overlooking something small.

    2. Relevant equations

    y(t) = H(1 - e-ts/T)

    3. The attempt at a solution

    At first I thought that the ln(10)T was the time it takes to cover 0.009H, because:

    0.999H = 0.99H + 0.009H, which is:
    t0.1% = t1% - ln(0.991)T
    t0.1% = t1% + ln(1.01)T

    But I guess I was wrong because clearly ln(10)T is not equal to ln(1.01)T.
    Last edited: Aug 3, 2015
  2. jcsd
  3. Aug 3, 2015 #2


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    Hi Dan,

    You correctly guess that jumping from 0.999H = 0.99H + 0.009H to t 0.1% = t 1% - ln(0.991)T isn't right. So my question is: where did you get the idea that it is ?
    If I write out the difference, I get:$$
    y(t_{0.1\%}) - y(t_{1\%}) = H(1-e^{ \; t_{0.1\%}/T } ) - H(1-e^{ \; t_{1\%}/T } )
    $$with, on the left 0.009 H allright, but on the right I am stuck with a difference, so pretty awkward to solve !

    The advice is: you want to end up sith something that looks like $$ t_{0.1\%} - t_{1\%} = \ln(10)T\ , \rm {or} \ \ t_{0.1\%}/T - t_{1\%}/T = \ln(10)\ ,$$ so exponentiate left and right and see where that gets you.
  4. Aug 3, 2015 #3
    Hi @BvU , I thought it was correct because 0.99 + 0.009 = 0.999.

    I think the author of the text skipped a few steps. I just realised that ln(1000) = ln(100) + ln(10), ie:

    t0.1% = -Tln(0.001)
    t0.1% = Tln(1000) = T[ln(100) + ln(10)]
    BUT, we already know that ln(100) = t1%
    so, t0.1% = t1% + ln(10)

    If you have another explanation, please share it so I can get this off my chest.
  5. Aug 3, 2015 #4


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    Yes, that's all there is to it: instead of working out the difference ##y(t_{0.1\%}) - y(t_{1\%})## , focus on the ratio of (1-the difference):$$
    {1 - y(t_{0.1\%}) \over 1- y(t_{1\%}) } = {1\over 10} $$and take logarithms
  6. Aug 3, 2015 #5
    @BvU , I solved this by taking ln of both sides but I got ln(0.009) instead of ln(10) as the other term:

    0.009H = H(1 - e-t0.1%/T) - H(1 - e-t1%/T)
    0.009 = 1 - e-t0.1%/T -1 + e-t1%/T
    0.009 = - e-t0.1%/T + e-t1%/T
    ln(0.009) = - ln(e-t0.1%/T) + ln(e-t1%/T)
    ln(0.009) = t0.1%/T - t1%/T
    Tln(0.009) = t0.1% - t1%
    t0.1% = t1% + Tln(0.009)

    Working seems to make sense, but still completely wrong answer...unless I can't take logs like that.
  7. Aug 3, 2015 #6


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    Oops ! Taking the logarithm of a sum does NOT give the sum of the logarithms !!!!

    (instead, taking the logarithm of a product does give the sum of the logarithms..., and for a fraction you use ln(1/x) = -ln(x), so you get a difference )
  8. Aug 3, 2015 #7
    @BvU , Alright. Cheers mate
  9. Aug 3, 2015 #8


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    You're welcome.
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