1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You're welcome.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: System Modelling/Dynamics question (time settling)
  1. Dynamical Systems (Replies: 0)

Loading...