Curve problem
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From: dirt on 3 May 2010 03:00 "dirt " <posto86(a)hotmail.com> wrote in message <hrlg98$aje$1(a)fred.mathworks.com>... > Walter Roberson <roberson(a)hushmail.com> wrote in message <tSqDn.118852$kj3.75722(a)newsfe08.iad>... > > dirt wrote: > > > > > I have generated a curve that is consisting of transient response which > > > is followed by steady state in a form of data points. Is there a way to > > > find out whether the curve has reached steady state or eitherwise... Is > > > it possible to do a loop from the last element of the data? > > > > I'm not sure if this is what you are after, but: > > > > for K = length(V):-1:1 > > .... V(K) > > end > > > > would loop backwards on V. > > > > > > > > I can think of more than one variety of "steady state". In some of them, > > you can check find the end of the unsteadiness via a test similar to > > > > find( abs(diff(V)) > NoiseLevel, 1, 'last' ) > > > > In other varieties of "steady state", > > > > find( abs(diff(log(V))) > NoiseLevel, 1, 'last' ) > > > > would be appropriate > > May i ask what is 'NoiseLevel',1,'last' refers to? > > regards > dirt Oh my question is that is it possible to do a for loop from the data points with respect to the crest and through values of the curve to determine steady state?
From: TideMan on 3 May 2010 03:46 On May 3, 7:00 pm, "dirt " <post...(a)hotmail.com> wrote: > "dirt " <post...(a)hotmail.com> wrote in message <hrlg98$aj...(a)fred.mathworks.com>... > > Walter Roberson <rober...(a)hushmail.com> wrote in message <tSqDn.118852$kj3.75...(a)newsfe08.iad>... > > > dirt wrote: > > > > > I have generated a curve that is consisting of transient response which > > > > is followed by steady state in a form of data points. Is there a way to > > > > find out whether the curve has reached steady state or eitherwise.... Is > > > > it possible to do a loop from the last element of the data? > > > > I'm not sure if this is what you are after, but: > > > > for K = length(V):-1:1 > > > .... V(K) > > > end > > > > would loop backwards on V. > > > > I can think of more than one variety of "steady state". In some of them, > > > you can check find the end of the unsteadiness via a test similar to > > > > find( abs(diff(V)) > NoiseLevel, 1, 'last' ) > > > > In other varieties of "steady state", > > > > find( abs(diff(log(V))) > NoiseLevel, 1, 'last' ) > > > > would be appropriate > > > May i ask what is 'NoiseLevel',1,'last' refers to? > > > regards > > dirt > > Oh my question is that is it possible to do a for loop from the data points with respect to the crest and through values of the curve to determine steady state? Questions, questions, always questions................ But no indication that you have done anything whatsoever yourself.
From: John D'Errico on 3 May 2010 06:52
"dirt " <posto86(a)hotmail.com> wrote in message <hrln8h$9mo$1(a)fred.mathworks.com>... > "John D'Errico" <woodchips(a)rochester.rr.com> wrote in message <hrlgq3$dls$1(a)fred.mathworks.com>... > > "dirt " <posto86(a)hotmail.com> wrote in message <hrld7o$qdu$1(a)fred.mathworks.com>... > > > Hello, > > > > > > I have generated a curve that is consisting of transient response which is followed by steady state in a form of data points. Is there a way to find out whether the curve has reached steady state or eitherwise... Is it possible to do a loop from the last element of the data? > > > > > > > No. You can never KNOW what will happen, simply from > > a sampling of values. > > > > John > Why is that so? Can u give a brief explaination? A simple counterexample is all that is necessary. Consider the function f(t), described by this simple test... if t < 1e6, f_t = 0; else f_t = inf; end f_t is zero as long as t is less than 1e6. After 1e6, it is inf. Now, suppose you sample the function many times. In fact, you can take nearly a million samples. Every one of them will be identically the same. All zero. Is this function at "steady-state"? Surely it must be, since we have 999999 samples, all of which are identically zero. Yeah, right. Case closed. John |