System Modelling/Dynamics question (time settling)

[KingDaniel]

Homework Statement

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:

t_0.1% = t_1% + 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
t_s = -Tln(0.001)
t_s = Tln(1000)
t_s = 6.9T
t_0.1% = 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.

Homework Equations

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

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:
t_0.1% = t_1% - ln(0.991)T
t_0.1% = t_1% + ln(1.01)T

But I guess I was wrong because clearly ln(10)T is not equal to ln(1.01)T.

 

[BvU]

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.

 

[KingDaniel]

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 realized that ln(1000) = ln(100) + ln(10), ie:

t_0.1% = -Tln(0.001)
t_0.1% = Tln(1000) = T[ln(100) + ln(10)]
BUT, we already know that ln(100) = t_1%
so, t_0.1% = t_1% + ln(10)

If you have another explanation, please share it so I can get this off my chest.

 

[BvU]

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

 

[KingDaniel]

Hi Dan,

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 !

@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^-t_0.1%/T) - H(1 - e^-t_1%/T)
0.009 = 1 - e^-t_0.1%/T -1 + e^-t_1%/T
0.009 = - e^-t_0.1%/T + e^-t_1%/T
ln(0.009) = - ln(e^-t_0.1%/T) + ln(e^-t_1%/T)
ln(0.009) = t_0.1%/T - t_1%/T
Tln(0.009) = t_0.1% - t_1%
t_0.1% = t_1% + Tln(0.009)

Working seems to make sense, but still completely wrong answer...unless I can't take logs like that.

 

[BvU]

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 )

 

[KingDaniel]

@BvU , Alright. Cheers mate

 

[BvU]

You're welcome.