Process Dynamic and Control (Dale, Thomas and Duncan) - Graphical fitting of 1st order model using step test

If I can estimate K and ? from an actual step test data for an assumed first order system, how can I fit the test data using equation below;
y(t) = KM(1 - e^(t/?))
I mean, how can I use the equation to predict output response using the equ ation above?
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On Fri, 07 Dec 2012 23:22:40 -0800, The Romanov wrote:

Your question makes no sense. If you can estimate K and tau, you _already have_ fit your test data to the above equation.

If a linear time invariant system has the step response that you give above, then what must the transfer function of that system be?
(Hint: take the Laplace transform of y(t), and see if you can divide out the Laplace transform of the step.)
--
Tim Wescott
Control system and signal processing consulting
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Tim.
I think I may have worded the question wrongly. Allow me to rephrase it as;
Can I use the equation to determine the output response for each time inter val by using back the step input having estimated K and ?? I intend to calculate square error between the predicted and actual step response valu e.
The transfer function for the step response would be in the form of K/( s + 1) right?
On Saturday, December 8, 2012 10:29:18 AM UTC+3, Tim Wescott wrote:

umed

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On Fri, 21 Dec 2012 09:07:30 -0800, The Romanov wrote: (top posting fixed, for my sanity's sake)

Maybe. You can certainly use the equation you give to predict what the response would be if your chosen values of tau and K were correct.
If you wanted to say something like "I'm going to pretend that my system model is correct, I have noisy measurements of y(t), and an exact knowledge of the input, and I want to make my best estimate of the _real_ y(t)", then you can do that, too, only with more difficulty.

As a point of terminology a signal has a Laplace _transform_, while systems have a _transfer function_.
The transfer function of a system with that given step response would be K*M / (s + 1/tau) (unless your step has amplitude M, in which case the transfer function would be K / (s + 1/tau).
The laplace transform (I don't know why I'm doing your work for you tonight, by the way) of y(t) is
Y(s) = K * M / (s * (s + 1/tau))
--
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My conservative friends think I'm a liberal kook.
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On Sunday, December 23, 2012 8:06:22 AM UTC+3, Tim Wescott wrote:

assumed

;
intend

e
I may have jumped too far ahead in the book trying to understand the mathem atics of curve fitting that leads to determination of correct transfer func tion. So I decided to move back several chapters and now stuck at the 'part ial fraction expansion' on page 52. There is an example of solving the part ial fraction as follows;
Y(s) = s + 5/{(s + 1)(s + 4)} which is expanded to;
s + 5/{(s + 1)(s + 4)} = [?1/s + 1] + [?2/s + 4]
One of the method (method 2 actually) to solve ?1 and ?2 is speci fying two values of 's' since the above equation hold for all values of 's' . The example specified s = -5 and s = -3. I don't seems to understand this because when I specified other values of s, I got different result fro m the book.
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On Sat, 12 Jan 2013 08:47:00 -0800, The Romanov wrote:

Something is wrong there. If you start with s + 5/((s + 1)(s + 4)) and lump everything together, then the numerator must be third degree, and that's not a proper transfer function (nekkid derivatives just don't exist in nature). But if you start with the right side of that equation, then the numerator is second degree. That's just not going to work.
s + 5 / ((s + 1)(s + 4)) = a_0 s + a_1/(s + 1) + a_2/(s + 4)
would work.
Are you sure about the leading 's'? Could you mean
Y(s) = (s + 5) / ((s + 1)(s + 4)) ?
I get consistent results with my version of Y(s).
--
Tim Wescott
Control system and signal processing consulting
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On Sunday, January 13, 2013 1:27:01 AM UTC+3, Tim Wescott wrote:

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Sorry Tim. Mistake on my part due to typo error. The correct Y(s) is the on e that you have presented above with s + 5 in bracket. So how does the exam ple simply chose s = -5 and s = -3 solving the ?1 = 4/3 and 2 = -1/3?
On personal note, as I mentioned that I'm trying to understand the mathemat ics behind curve or step response fitting deriving to transfer function, do you think that the "Process Dynamics and Control" book as a good start?
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On Sun, 13 Jan 2013 10:10:13 -0800, The Romanov wrote:

I think the idea is that you can choose lots of different things and still get an answer.
At s = -5, Y(s) = 0, a_1/(s+1) = -a_1/4, and a_2/(s+4) = -a_2
At s = -3, Y(s) = -1, a_1/(s+1) = -a_1/2, and a_2/(s+4) = a_2
Now you have two equations in two unknowns which you then solve for a_1 and a_2.

I have no clue: I don't have that book, and worse, I don't have any recommendations. Possibly the best thing to do is to start a new topic asking what the best book to read is.
--
Tim Wescott
Control system and signal processing consulting
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On Monday, January 14, 2013 12:09:15 AM UTC+3, Tim Wescott wrote:

w
s
tand

o > >> work.

e
e
and ?2 = -1/3.

Thanks for the explanation above. However, I still don't understand how or why the 's' was chosen to be -5 and -3. If I chose 's' to be say 1 and 7, I will get different answer from the example question.
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On Mon, 14 Jan 2013 09:43:25 -0800, The Romanov wrote:

s = 7: Y(s) = 3/22, a_1/(s+1) = a_1/8, a_2/(s+4) = a_2/11; s = 1: Y(s) = 3/5, a_1/(s+1) = a_1/2, a_2/(s+4) = a_2/5;
I get the same answer each time, no matter what I use for s. This makes sense, because you can show algebraically that the basic equation you're solving must hold.
(You can, in fact, simply add up the a_1/(s+1) + a_2/(s+4) and find that you have the correct denominator, and a 1st-order polynomial in the numerator whose coefficients are linearly independent expressions in a_1 and a_2, meaning that you can match _any_ arbitrary numerator in a polynomial with those roots).
You've got to be doing something wrong. Instead of repeating that you get different answers, why not show what you _are_ doing, and maybe we can make some progress.
--
Tim Wescott
Control system and signal processing consulting
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On Tuesday, January 15, 2013 7:48:22 AM UTC+3, Tim Wescott wrote:

e
2 is

l
1 = 4/3

1
I substituted s with any two chosen numbers but my mistake again because I claimed I will get different answer simply because I ended with two equatio ns that is not similar to equations below;
?1 + ?2 = 1 4?1 + ?2 = 5
and I stop short of trying to complete solving ?1 and ?2 for any selection of s.
By the way, those two equations above are derived from solving ?1 and ?2 by 1st method presented in the book.
This frigid winter temperature froze by brain I guess....
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On 18.01.2013 15:18, The Romanov wrote:

claimed I will get different answer simply because I ended with two equations that is not similar to equations below;

I'd like to see the final anwer to
'Plant Transfer Function on Page 122, Figure 7.8'
Thanks JCH
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On 14.01.2013 18:43, The Romanov wrote:

> >> work.

the 's' was chosen to be -5 and -3. If I chose 's' to be say 1 and 7, I will get different answer from the example question.

I think that is the solution:
http://home.arcor.de/janch/20130116-control/default.html
JCH
http://home.arcor.de/janch/20130116-control/default.html
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"The Romanov" schrieb im Newsbeitrag
On Sunday, January 13, 2013 1:27:01 AM UTC+3, Tim Wescott wrote:

wrote:

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Sorry Tim. Mistake on my part due to typo error. The correct Y(s) is the one that you have presented above with s + 5 in bracket. So how does the example simply chose s = -5 and s = -3 solving the ?1 = 4/3 and ?2 = -1/3?
On personal note, as I mentioned that I'm trying to understand the mathematics behind curve or step response fitting deriving to transfer function, do you think that the "Process Dynamics and Control" book as a good start?
____________________________________________
I had a look into that book
See
http://home.arcor.de/janch/20130114-control/default.html
JCH
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On 08.12.2012 08:22, The Romanov wrote:

Algebraic Solution:
Solving 3 nonlinear equations:
Example Data
t_1    =    0,000000000000000 t_2    =    0,500000000000000 t_3    =    1,000000000000000 y_1    =    1,000000000000000 y_2    =    0,800000000000000 y_3    =    0,700000000000000
y_1 - a_1 + a_2*EXP(1 - t_1/a_3) = 0 y_2 - a_1 + a_2*EXP(1 - t_2/a_3) = 0 y_3 - a_1 + a_2*EXP(1 - t_3/a_3) = 0
a_1    =     0,599999999999997 a_2    =    -0,147151776468579 a_3    =     0,721347520444526
ODE Solution: See
http://home.arcor.de/janch/20130119-control/default.html
JCH
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On Friday, December 7, 2012 11:22:40 PM UTC-8, The Romanov wrote:

ed first order system, how can I fit the test data using equation below;

quation above?
Obviously you must find the value of KM, I am assuming KM is one constant, and tau. You do this by using a least squares fit routine like Levenberg M arquardt to find the value of KM and tau that minimizes the least squared e rror. Then one uses the value of KM and Tau to build an observer so then you can estimate values that are smooth relative to the noisy measured feed back.
http://www.designnews.com/document.asp?doc_id "9767&dfpPParams=ind_182 ,aid_229767&dfpLayout=article
If you build a non-linear model like I did one can model non-linear and sys tems with dead time too. The data in the article is from a real valve. Yo u can see the estimated velocity is very smooth compared to the computed ve locity from measured and truncated data that was then digitally differentia ted.
I don't visit here much anymore.
Peter Nachtwey
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