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

[The Romanov]

If I can estimate K and τ 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/τ))

I mean, how can I use the equation to predict output response using the equation above?

[Tim Wescott]

On Fri, 07 Dec 2012 23:22:40 -0800, The Romanov wrote:

> If I can estimate K and τ 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/τ))

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

> I mean, how can I use the equation to predict output response using the
> equation above?

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
www.wescottdesign.com

[The Romanov]

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 interval by using back the step input having estimated K and τ? I intend to calculate square error between the predicted and actual step response value.

The transfer function for the step response would be in the form of K/(τs + 1) right?

[Tim Wescott]

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

> On Saturday, December 8, 2012 10:29:18 AM UTC+3, Tim Wescott wrote:
>> On Fri, 07 Dec 2012 23:22:40 -0800, The Romanov wrote:
>>
>> > If I can estimate K and τ 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/τ))
>>
>> Your question makes no sense. If you can estimate K and tau, you
>>
>> _already have_ fit your test data to the above equation.
>
>> > I mean, how can I use the equation to predict output response using
>> > the
>>
>> > equation above?
>>
>>
>>
>> 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.
>
> 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
> interval by using back the step input having estimated K and τ? I intend
> to calculate square error between the predicted and actual step response
> value.

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.

> The transfer function for the step response would be in the form of
> K/(τs + 1) right?

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))

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

[The Romanov]

I may have jumped too far ahead in the book trying to understand the mathematics of curve fitting that leads to determination of correct transfer function. So I decided to move back several chapters and now stuck at the 'partial fraction expansion' on page 52. There is an example of solving the partial fraction as follows;

Y(s) = s + 5/{(s + 1)(s + 4)} which is expanded to;

s + 5/{(s + 1)(s + 4)} = [α1/s + 1] + [α2/s + 4]

One of the method (method 2 actually) to solve α1 and α2 is specifying 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 from the book.

[Tim Wescott]

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).

[The Romanov]

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 α1 = 4/3 and α2 = -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?

[Tim Wescott]

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.

> 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 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.

[JCH]

"The Romanov" schrieb im Newsbeitrag
news:daaccc28-d0ad-4885...@googlegroups.com...

____________________________________________


I had a look into that book

See

http://home.arcor.de/janch/20130114-control/default.html

JCH

[The Romanov]

> > example simply chose s = -5 and s = -3 solving the α1 = 4/3 and α2 = -1/3.

>
> 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.

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.

[Tim Wescott]

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.

[JCH]

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

[The Romanov]

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 equations that is not similar to equations below;

α1 + α2 = 1
4α1 + α2 = 5

and I stop short of trying to complete solving α1 and α2 for any selection of s.

By the way, those two equations above are derived from solving α1 and α2 by 1st method presented in the book.

This frigid winter temperature froze by brain I guess....

[JCH]

I'd like to see the final anwer to

'Plant Transfer Function on Page 122, Figure 7.8'

Thanks JCH

[JCH]

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

[pnac...@gmail.com]

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 Marquardt to find the value of KM and tau that minimizes the least squared error. 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 feedback.

http://www.designnews.com/document.asp?doc_id=229767&dfpPParams=ind_182,aid_229767&dfpLayout=article

If you build a non-linear model like I did one can model non-linear and systems with dead time too. The data in the article is from a real valve. You can see the estimated velocity is very smooth compared to the computed velocity from measured and truncated data that was then digitally differentiated.

I don't visit here much anymore.

Peter Nachtwey