mm-154
I need to use error function urgently in my program.
could you pleasegive me the links to websites explaining what
is an error functionwith worked out examples and code for
using it in a program.thank you very much!sunny
=You will
find code in Fortran for the error function at my web site
(seesignature below).--Alan
Millerhttp://users.bigpond.net.au/amillerRetired
Statistician>> i need to use error function urgently in my
program. could you please> give me the links to websites
explaining what is an error function> with worked out examples
and code for using it in a program.>> thank you very much!>
sunny
=I'm looking at the following problem:PA = LU (P a
permutation matrix, LU the usual factors)B = A + u e_i^T (
note this is only changing one column of A ) so thenP(A + u
e_i^T) = PB = L(U + w e_i^T) Lw = Puif we let V = U + w
e_i^Tand P1 V = L1 U1 then we can get the LU-factors of B ie.
P B = L2 U2as L2 = L P1^T L1 and U2 = U1.Obviously, the fact
that V is upper triangular except for the ithcolumn can be
exploited, ie. from the standpoint of applying
Gausstransforms, you will have to apply n-i transforms. So if
the ithcolumn is one of the 'rst columns, isn't
itthe case
that there isn't going to be much savings in work? It
looksto
me like it's going to be something like O(n(n-i)^2). Am I
missingsomething, or is this just the price of doing
business?Ryan
=An updated version of DMSUITE (A MATLAB
Differentiation MatrixSuite) by Satish Reddy and myself is now
available. Larry Shampineinformed us that several of the
functions failed to executeunder MATLAB Release 13, and these
corrections, along with a fewsmaller ones, have now been
made.The suite is freely available for download from the
MathWorks Web Site:please go to
http://www.mathworks.com/matlabcentral/'leexchange/and follow
the links MATHEMATICS -> DIFFERENTIAL EQUATIONS.If you 'nd
the
suite useful, please complete the review on that page.DMSUITE
also has a Web Site (and hopefully soon a guest book)
athttp://dip.sun.ac.za/~weideman/research/differ.htmlThis Web
Site replaces the old one at ucs.orst.edu, whichwill soon
cease to exist. Please update bookmarks accordingly.For those
readers unfamiliar with DMSUITE, here is an abstractfrom the
accompanying paper, which was published in ACM TOMS,Vol. 26,
pp. 465--519 (2000):Abstract: A software suite consisting of
seventeen MATLAB functionsfor solving differential equations
by the spectral collocation(a.k.a. pseudospectral) method is
presented. It includes functions forcomputing derivatives of
arbitrary order corresponding to Chebyshev,Hermite, Laguerre,
Fourier, and sinc interpolants. Auxiliary functionsare
included for incorporating boundary conditions,
performinginterpolation using barycentric formulas, and
computing roots oforthogonal polynomials. It is demonstrated
how to use the packagefor solving eigenvalue, boundary value,
and initial value problemsarising in the 'elds of special
functions, quantum mechanics, nonlinearwaves, and hydrodynamic
stability.JAC (Andre') WeidemanDepartment of Applied
MathematicsUniversity of StellenboschStellenbosch 7600South
Africa
=I have a positive semide'nite matrix A, which is
projected using an ortonormal basis Q build using Krylov
vectors [x | A^(-1)x | A^(-2)x | ...] . How to prove that Ritz
values after projection are concentrated around the smallest
eigenvaluea of A, i.e. that the norm (2) of Q^T * A * Q is
much smaller then norm (2) of A ?
= >I have a positive
semide'nite matrix A, which is projected using an >ortonormal
basis Q build using Krylov vectors [x | A^(-1)x | A^(-2)x |
>...] . How to prove that Ritz values after projection are
concentrated >around the smallest eigenvaluea of A, i.e. that
the norm (2) of Q^T * A >* Q is much smaller then norm (2) of
A ? > see B. Parlett: the symmetric eigenvalue problem
eigenvalue of Q^T(A^-1)Q = value of a Rayleighquotient of
Q^T(A^-1)Q = z^TQ^T(A^-1)Qz for some z with z^Tz=1 let y=Qz
then this is a Rayleighquotient y^TA^{-1}y approximating one
of the dominant eigenvalues of A^{-1} hence the reciprocal of
one of the small eigenvalues of A.hope this helpspeter
=ich
habe folgendes System:dx/dt=f(x,u)mitf(x,u)=u+k*sign(x)+r*xEs
ist nat.9frlich nicht-linear und deshalb versuche ich es in
matlab zusimulieren. Benutze das Runge-Kutta 4(5) mit step
size control. Dazuintegriere zun.8achst .9fber eine bestimmte
Zeit vorw.8arts, nehme mir denletzten wert und integriere dann
r.9fckw.8arts (dazu drehe ich einfach dasVorzeichen des step
size). Ich m.9fsste theoretisch wieder auf den
Endwertkommen.Das funktioniert soweit ganz gut, solange r und
k nicht gross genugsind. Sobald sie ungleich 0 sind, gibt es
schon Abweichungen aber zumBeispiel f.9fr ein r>0.5 bekomme
ich die Anfangsfunktion .9fberhaupt nichtmehr hin.Nun hab ich
gelesen, dass R.9fckw.8artsintegrationen nicht immer
eindeutigfunktionieren. Hat jemand eine Idee, wann es
funktionieren k.9annte undwann nicht?? Stichwort:
Controllability und Reachability, hat wer wasschriftliches,
wie man die reachability von einem nicht-linearen
Systemberechnen kann??Danke im voraus,omar
I'm working
on a project that requires sketch a surface knowing several>
(rather many but not enough) points on that surface. I need to
estimate> other points...> Some one told me that Fourier
analysis can handle this kind of problem.I don't think so.
What you want is called Oscattered data
interpolation';many
links are given in
http://www.mat.univie.ac.at/~neum/surface.html[~ is a
tilde]Arnold Neumaier
> I'm working on a project that
requires sketch a surface knowing several> (rather many but
not enough) points on that surface. I need to estimate>
other points...>> Some one told me that Fourier analysis
can handle this kind of problem.>> I don't think so.> What
you want is called Oscattered data
interpolation';> many links
are given in> http://www.mat.univie.ac.at/~neum/surface.html>
[~ is a tilde]>> Arnold NeumaierRadial basis functions (RBF)
spring to mind. Here one constructs the surfaceusing the
scattered shifts of some basis function. If you know the
surfaceat the points x_1,...,x_M then one could
use:sum_{j=1}^M a_j phi(x-x_j)as an approximation to the
surface given an appropriate basis function phi.The
coef'cients a_j being determined by the ensuing
interpolationequations:sum_{j=1}^M a_j phi(x_i-x_j) = f(x_i),
i=1,...,M.There are several popular choices for the basis
function. One could usephi(x)=exp(-|x|^2/2) for example in any
space dimension. In 2-dimensionsphi(x)=|x|^2ln|x| is popular
and the resulting interpolant is often calledthe thin plate
spline (TPS). Sometimes (for example with the TPS) apolynomial
term and additional constraints are added to the above
toguarantee the solvability of the system.RBF Interpolants are
a multivariate analogy of the well-loved naturalsplines from
1-d.R
=I am looking for an equation for a curve that is like
an ellipse butnot quite. It is not circle, either.I have a set
of points {(x0,y0), (x1,y1), (x2,y2), ..., (xN,yN)},where
(xN,yN) == (x0,y0), that look a lot of like an ellipse
whenplotted, and I would like to 't a closed curve on them.(I
have tried an ellipse but I want something better.)Any
ideas?
= >I am looking for an equation for a curve that is
like an ellipse but >not quite. It is not circle, either. >I
have a set of points {(x0,y0), (x1,y1), (x2,y2), ..., (xN,yN)},
>where (xN,yN) == (x0,y0), that look a lot of like an ellipse
when >plotted, and I would like to 't a closed curve on them.
>(I have tried an ellipse but I want something better.) >
>Any ideas?you can use a parametric periodic spline:invent an
arti'cial parameter (s) for the curve, for examplethe
arclength of the continuous piecewise linear interpolating
path.then interpolate the components x and y of your points
separately by anperiodic psline, using the 'rst data point
also as the last one. (that is you interpolate (s(i),x(i)) and
(s(i),y(i)) by two periodicsplines.) then your curve will be
closed C2 :C : (x(s),y(s)) : 0<= s <= smax hope this
helpspeter
I am looking for an equation for a curve
that is like an ellipse but>not quite. It is not circle,
either.>>I have a set of points {(x0,y0), (x1,y1),
(x2,y2), ..., (xN,yN)},>where (xN,yN) == (x0,y0), that look
a lot of like an ellipse when>plotted, and I would like to
't a closed curve on them.>>(I have tried an ellipse but I
want something better.)>>Any ideas?> you can use a
parametric periodic spline:> invent an arti'cial parameter
(s)
for the curve, for example> the arclength of the continuous
piecewise linear interpolating path.> then interpolate the
components x and y of your points separately by an> periodic
psline, using the 'rst data point also as the last one.>
(that
is you interpolate (s(i),x(i)) and (s(i),y(i)) by two periodic>
splines.) then your curve will be closed C2 :> C : (x(s),y(s))
: 0<= s <= smax> hope this helps> peterTry the family: (x/a)^n
+ (y/b)^n = 1with n > 2 . Of course, n=2 is an ellipse centered
at the origin.-- Julian V. NobleProfessor Emeritus of
^^^^^^^^^^^^^^^^^^http://galileo.phys.virginia.edu/~jvn/
Science knows only one commandment: contribute to science. --
Bertolt Brecht, Galileo.
=HeyAfter working on this problem
for a while now and with nosolution that works for all cases I
had to realize my math skillsare just not good enough. Maybe
someone here can shed some light onthis problem.Imagine a
right triangle divided into discreet elements, think texture
map. The triangle is n elements wide and m elements high.
Element (0,0) is in the lower left corner,(n-1,0) is lower
right and (0,m-1) upper left. The elements are numbered as
(0,0)=0 ,(1,0) = 1(0,1) = n. The complete number of elements
for the right triangle are n*m/2 + n/2 or m/2 depending on
which is largest,n or m.Every element that the hypotenuse
pierces belongs to the triangle. n and m are power of two.My
question is, how would one compute the height([0..m-1]) of an
element given its number, without using any iterative
method?Below is an example of an 2x4 triangle in ascii art, XX
is one element:XX XX XX XXXX XXElement number 0 has height 0.
Element number 4 has height 2.Element number 5 has height
3.The methods I have tried so far fails when width/heigh <
1/8. I'musing 32bit §oats so it isn't
really a precision
problem.Width andheight are 256 maximum.All help/comments
greatly appreciated!/Mathias
The methods I have tried so
far fails when width/heigh < 1/8.To be honest, when
width/height < 1 is where I have problems derivinga solution.
When width/height>=1 the height(y) can be written asy*width -
k(y * (y - 1))/2 = elementNbr ;where k=width/heightBut for k<1
I end up with step functions and who knows what. I must bedoing
something wrong./Mathias
= >kind of probabilistic
distribution (exponential, lognormal, or whatever) >that these
data points are exhibiting. If it 'ts one of the
>distributions, what are the parameters of the distribution?
Is there any >good software that allows me to perform this
task? I tried Matlab, and >- >Hai >in matlab (lsqnonlin you
must have the model already. try this :(from my
annotations)begin{citation} CurvFit (tm) ... v 3.01
announcement. -------------------------------------CurvFIT
(tm) is a curve'tting program; a Power, Exponential,and
Lorentzian series are available models to try to matchyour
data. Learn to Oread' eigenvalues of the
Jacobian andHessian
matrices. This is an example of Calculus levelprogramming ...
i.e. minutes to solve, days or years tounderstand solution and
what it implies (e.g. wrong model,sampling rate error, etc.)To
download this Shareware program (1.5 Mb 'le), point your
browser at the URL:
http://www.simtel.net/pub/pd/52762.shtml(zip version) or
http://www.Lanset.com/ecb/CurvFit.htm (ourweb page).Enjoy!Phil
Brubakerend{citation}hope it does what you want
peter
=However, Robert, I am not sure if all BA^T matrices
are symmetric andpostive de'nite.However, I have
'gured out
the solution.The least square solution requires that I solve
A^T A X = A^T BNow transposing the same equation and adding
it, we haveA^T A X + X A^T A = A^T B + B^T AThe above eqn is
Lyapunov equation, and the solution is guranteed if A^TA has
all eigen values <0 i.e. is stable and right hand side
issymmetric.Right hand side is symmetric. Now in my case, I
can also prove thatright hand side is postive de'nite and
left
hand matrix is stable. So Iwill always get the postive
de'nite
and symmetric solution of the aboveequation.Once the proof is
complete, the above equation can also be solved by'rst
replacing X by its SVD and then rearranging the terms.In this
case, we have someting like this:(A^TA d_1 - A^TB)r_i = 0,
where d_1 are the eigen values and r_i is therow vectors of R,
X = R^T D R. This is a pencil equation can be solvedusing QZ
decomposition.navneet>>I have to sovle AX= B for matrix X,>
>where A is nXd matrix, X = dXd matrix and B = nXd matrix and
n>>d>>The constraint is X should be symmetric and postive
de'nite.>> I think this may help:> Your equation implies>>
AXA^T = BA^T>> (^T = transpose)>> so C = BA^T must be
symmetric and positive semide'nite.>> Robert Israel
israel@math.ubc.ca> Department of Mathematics
http://www.math.ubc.ca/~israel> University of British
Columbia> Vancouver, BC, Canada V6T 1Z2
However, Robert,
I am not sure if all BA^T matrices are symmetric and>postive
de'nite.>However, I have 'gured out the
solution.>The least
square solution requires that I solve A^T A X = A^T B>>Now
transposing the same equation and adding it, we have>>A^T A X
+ X A^T A = A^T B + B^T A>>The above eqn is Lyapunov equation,
and the solution is guranteed if A^T>A has all eigen values <0
i.e. is stable and right hand side is>symmetric. if A^T A has
all eigen values < 0will be a bit of a problem as this is well
known to be positive de'nitewhen A is of full column rank.
Maybe you can 'nd spare minus signs tochange the signs of
both
sides of equation.>Right hand side is symmetric. Now in my
case, I can also prove that>right hand side is postive
de'nite
and left hand matrix is stable. So I>will always get the
postive de'nite and symmetric solution of the
above>equation.>>Once the proof is complete, the above
equation can also be solved by>'rst replacing X by its SVD
and
then rearranging the terms.>In this case, we have someting like
this:>(A^TA d_1 - A^TB)r_i = 0, where d_1 are the eigen values
and r_i is the>row vectors of R, X = R^T D R. This is a pencil
equation can be solved>using QZ decomposition.>>navneet>>
>I have to sovle AX= B for matrix X,>>where A is nXd matrix,
X = dXd matrix and B = nXd matrix and n>>d>>The constraint
is X should be symmetric and postive de'nite.>> I think this
may help:>> Your equation implies>> AXA^T = BA^T>> (^T =
transpose)>> so C = BA^T must be symmetric and positive
semide'nite.>> Robert Israel israel@math.ubc.ca>> Department
of Mathematics http://www.math.ubc.ca/~israel>> University of
British Columbia>> Vancouver, BC, Canada V6T 1Z2>
I
have to sovle AX= B for matrix X,> where A is nXd matrix, X =
dXd matrix and B = nXd matrix and n>>d> The constraint is X
should be symmetric and postive de'nite.> Can someone
Lagrange multipliers to adjoin the symmetric constraint:J =
(AX-B)'(AX-B)+ L(X-X'), and make J stationary
wrt
X,L.Adjoining the SPD constraint is much tougher since that
leadsto inequality constraints, and a NLP.
I'm not quite
sure what happens in general if A does not have full>column
rank. Things seem to be more complicated.Choosing the
appropriate orthogonal coordinate systems, we can write [ U 0
] [ B1 B2 ]A = [ 0 0 ], B = [ B3 B4 ]in block matrix form,
where the 'rst block of rows is for Ran(A^T),the second for
Ker(A), and the 'rst block of columns is for Ran(A),the
second
for Ker(A^T); U maps Ran(A^T) one-to-one onto Ran(A). [ X1 X2
]Then if X = [ X2^T X3 ], the equation AX = B becomes[ UX1 UX2
] [ B1 B2 ][ 0 0 ] = [ B3 B4 ],i.e. we need B3 = B4 = 0, X1 =
U^(-1) B1 and X2 = U^(-1) B2.In order for X1 to be symmetric,
U^(-1) B1 = B1^T (U^(-1))^Twhich is equivalent to B1 U^T = U
B1^T. For X1 to be positivede'nite, we need B1 U^T to be
positive de'nite. And then therewill be a condition on X4 to
make X positive de'nite, which I believeis that X4 - X2^T
X1^(-1) X2 must be positive de'nite.Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
> if A^T A has all eigen
values < 0>> will be a bit of a problem as this is well known
to be positive de'nite> when A is of full column rank. Maybe
you can 'nd spare minus signs to> change the signs of both
sides of equation.>This is exactly the case in my scenario. I
get a spare negtive sign from theright hand
>>> I have
to sovle AX= B for matrix X,>> where A is nXd matrix, X = dXd
matrix and B = nXd matrix and n>>d>>> The constraint is X
should be symmetric and postive
de'nite.>>Doesn't AX = B
already contrain X? I.e. constrain it to be A^{-1}*B.>>GC>
navneetI think what you say has some merit, although A is not
square. If Aand B have been speci'ed, then X is constrained.
How do we know thatthe two constraints do not con§ict? If you
read the posts that suggest taking a transpose and
doingsomething, I think this offers the most promise and
should be used tobring the rectangular matrices back to square
ones.Writing A~ for A transpose rather than A^t, and A'
instead
of A^(-1),(A~A) X = (A~B)Now we have 3 d-by-d matrices, and if
(A~A) has an inverse, what youare saying is true. If not,
write C = (A~A), D = (A~B) and try todiagonalize PCP' as
much
as possible by 'nding (invertible) matrix Pwhich will make
the
equationPCP' PXP' = PDP'This
should make 'nding a possible
solution for PXP' (in terms ofarbitrary contants) and hence
for X much more systematic. Thesepossible solutions must be
resubstituted in the original equation,since multiplying by A~
may have weakened the original constraint.I don't see any
guarantee that a symmetric solution for X can alwaysbe found,
though maybe someone can show me otherwise.
I have
to sovle AX= B for matrix X,> where A is nXd matrix, X = dXd
matrix and B = nXd matrix and n>>d> The constraint is X
should be symmetric and postive de'nite.> Can someone
Here is a LS + LM solution for a n=4, d=2 test example: A' =
[
9 3 7 6 ] B' = [ 6 4 1 5 ] [ 1 2 5 7 ] [ 1 8 6 7 ] X = [ X11
X12 ] [ X21 X22 ]Mathematica gives X11 = -810/1787 + X22 , X12
= X21 = 6010/5361 - X22;X is PD for X22 > 3612010/5141199 =
0.702562 ...The PD condition was checked graphically; for
bigger matrices a NLPapproach seems necessary.LS with no
multiplier followed by ad-hoc symmetrization (X+X')/2
givesX=
[ 3580/5361 - X12, (6010/5361 + X12 - X22)/2 ] [6010/5361 +
X12 - X22)/2 X22 ]which has 2 free parameters.
=1. i'm not a
mathematician, just a mechanical engineer.i have a
systemdx/dt=f(x,u)withf(x,u)=u+k*sign(x)+r*xit's nonlinear
and
i'm trying to simulate it in matlab. I'm
usingRunge-Kutta 4(5)
with step size control. I run the integration for aspeci'ed
time (let's say 20 sec) then take the resulting values (x)
andtry to backwards integrate over this function over the same
amount oftime. What i actually do is just turn the step size to
a negative one.The outcome should be the initial value.This
seems to work for small ks and rs (see equation). But if i try
toset the r and k to different values the two graphs aren't
equal at all.Now i read that backward integrated functions
aren't allways unique.Could someone help me out. Keywords
could be controllability andreachability, i'm still reading
into these but nothing usefull yet.thnx in advance.
= >1.
i'm not a mathematician, just a mechanical engineer. >i have
a system >dx/dt=f(x,u) >with >f(x,u)=u+k*sign(x)+r*x >it's
nonlinear and i'm trying to simulate it in matlab.
I'm using
>Runge-Kutta 4(5) with step size control. I run the
integration for a >speci'ed time (let's say 20
sec) then take
the resulting values (x) and >try to backwards integrate over
this function over the same amount of >time. What i actually
do is just turn the step size to a negative one. >The outcome
should be the initial value. >This seems to work for small
ks and rs (see equation). But if i try to >set the r and k to
different values the two graphs aren't equal at all. >Now
i read that backward integrated functions aren't allways
unique. >Could someone help me out. Keywords could be
controllability and >reachability, i'm still reading into
these but nothing usefull yet. >thnx in advance.the backward
per se is not the problem for an ordinary differential
equationsinitial value problem, but of course the stability of
the equation .your f is not lipschitz continuous in x, hence
does not satisfy the picard lindeloef unicity theorem, but
this is not the point here.say, if you integrate y'=lambda*y
with Re(lambda)<0 forward in time,then you have a stable
situation and if your stepsize sats'es the stabilityproperty
for ode4(5), (h*abs(lambda)<1 say, this is not at the
boundaryof the stability region, this latter depends on the
speci'c coeffs ofyour 4(5) formula, there are many) then
there
is a damping of old errorshence the global error is essentially
in the same order of magnitudeas the local error, hence small.
but if you reverse the direction, your solution grows
exponentially and of course the local errors madeat time t are
mani'ed at time t1<t by exp(-r*(t-t1), hence for large values
of r you obtainnaturally a large magni'cation of your small
local errors. but some other effect comes into mind looking at
your equation:in principle k enters as a constant external
force.hopefully you never experience a sign change in x. then
you will have a kink in x and the integrator will cause here a
rather large error(if not dying in the stepsize control).hope
this helpspeter
1. i'm not a mathematician, just a
mechanical engineer.>>i have a system>>dx/dt=f(x,u)>
>with>f(x,u)=u+k*sign(x)+r*x>>it's nonlinear and
i'm
trying to simulate it in matlab. I'm using>Runge-Kutta 4(5)
with step size control. I run the integration for a>speci'ed
time (let's say 20 sec) then take the resulting values (x)
and>
>try to backwards integrate over this function over the same
amount of>time. What i actually do is just turn the step
size to a negative one.>The outcome should be the initial
value.>>This seems to work for small ks and rs (see
equation). But if i try to>set the r and k to different
values the two graphs aren't equal at all.>>>Now i read
that backward integrated functions aren't allways unique.>
>Could someone help me out. Keywords could be controllability
and>reachability, i'm still reading into these but nothing
usefull yet.>>thnx in advance.> the backward per se is
not the problem for an ordinary differential equations>
initial value problem, but of course the stability of the
equation .> your f is not lipschitz continuous in x, hence
does not satisfy the > picard lindeloef unicity theorem, but
this is not the point here.> say, if you integrate >
y'=lambda*y with Re(lambda)<0 forward in time,> then you
have
a stable situation and if your stepsize sats'es the
stability>
property for ode4(5), (h*abs(lambda)<1 say, this is not at the
boundary> of the stability region, this latter depends on the
speci'c coeffs of> your 4(5) formula, there are many) then
there is a damping of old errors> hence the global error is
essentially in the same order of magnitude> as the local
error, hence small. but if you reverse the direction, your >
solution grows exponentially and of course the local errors
made> at time t are mani'ed at time t1<t by > exp(-r*(t-t1),
hence for large values of r you obtain> naturally a large
magni'cation of your small local errors. > but some other
effect comes into mind looking at your equation:> in principle
k enters as a constant external force.> hopefully you never
experience a sign change in x. then you will have > a kink in
x and the integrator will cause here a rather large error> (if
not dying in the stepsize control).> hope this helps>
peterAre you saying, i should be using a different runge-kutta
for backwardintergration or what would be the solution to this
problem??indeed, i have an error that looks like an e^(-x)
function. And you'reright about the k-factor, since at these
points (sign change in x) bothequations could diverge quite
alot.thnx
=I'm having a kind of a problem that I want to
share with you. Hopefully youcan helpme.In order to determine
the PID-settings of a process controller, I 'rst needto
determine the transfer function of the process I'm tuning.
The
transferfunction usually takes the (laplace-)form:Kp/(t*s+1) *
exp(-td*s)Where Kp is the gain of the process, t is the time
constant of the processand td is the lag-time. s is the
laplacian time.To determine Kp, t and td an experiment can be
done that is called astep-response. For example opening a
valve for a given amount and thenlooking how the process (§ow
through the pipe) responds to this change.When I do this I
plot the §ow as a funcion of time. As an example
here'ssome
typical data I collected earlier:t [s]| §ow [m3/s]1 | 1.02 |
1.03 | 1.04 | 1.15 | 1.36 | 1.67 | 2.58 | 3.99 | 5.410 | 7.011
| 8.412 | 9.313 | 9.614 | 9.815 | 9.916 | 10.017 | 10.018 |
10.019 | 10.020 | 10.0Next I determine the Kp, t and td
graphically as follows: I estimate thepoint of in§ection on
the graph and draw a tangent through it. The angle ofthe
tangent determines Kp: tan(alpha)=Kp. For the above example I
foundKp=1.58Then I determine td: Where the tangent crosses the
1m3/s line is thelag-time. td=6.1sFinally I can determine t:
Where the tangent crosses the asymptote (10m3/s)at 11.8s. So,
t= 11.8 - td =5.7sTherefore, the transfer function looks like
this:1.58/(5.7*s+1) * exp(-6.1*s)Who can help me?--
SomeOne
> I'm having a kind of a problem that I want to
share with you. Hopefullyyou> can help> me.>> In order to
determine the PID-settings of a process controller, I
'rstneed> to determine the transfer function of the process
I'm tuning. The transfer> function usually takes the
(laplace-)form:>> Kp/(t*s+1) * exp(-td*s)>> Where Kp is the
gain of the process, t is the time constant of the process>
and td is the lag-time. s is the laplacian time.>> To
determine Kp, t and td an experiment can be done that is
called a> step-response. For example opening a valve for a
given amount and then> looking how the process (§ow through
the pipe) responds to this change.>> When I do this I plot the
§ow as a funcion of time. As an example here's> some
typical
data I collected earlier:> t [s]| §ow [m3/s]> 1 | 1.0> 2 |
1.0> 3 | 1.0> 4 | 1.1> 5 | 1.3> 6 | 1.6> 7 | 2.5> 8 | 3.9> 9 |
5.4> 10 | 7.0> 11 | 8.4> 12 | 9.3> 13 | 9.6> 14 | 9.8> 15 |
9.9> 16 | 10.0> 17 | 10.0> 18 | 10.0> 19 | 10.0> 20 | 10.0>>
Next I determine the Kp, t and td graphically as follows: I
estimate the> point of in§ection on the graph and draw a
tangent through it. The angleof> the tangent determines Kp:
tan(alpha)=Kp. For the above example I found> Kp=1.58> Then I
determine td: Where the tangent crosses the 1m3/s line is the>
lag-time. td=6.1s> Finally I can determine t: Where the tangent
crosses the asymptote(10m3/s)> at 11.8s. So, t= 11.8 - td
=5.7s>> Therefore, the transfer function looks like this:>>
1.58/(5.7*s+1) * exp(-6.1*s)>> As you can imagine I 'nd all
this a bit tedious and I'd like to automate> this work. The
data is available electronically. I'm guessing that
itshould>
be possible to mathematically determine the tangent of the
curve in the> point of in§ection. If that would be possible I
could write a smallprogram> that can calculate the Kp, t and
td from the data gathered from> step-response experiments.>>
But, alas my knowledge of mathematics leaves me begging at
this point. I> have no idea how to 'nd the tangent in the
point of in§ection based onan> array of data.> Who can help
me?I wonder how you determined that Kp, the gradient at the
point of in§exion,was 1.58 .With your data the point of
in§exion will correspond to a maximum of thegradient. At
time=9 the §ow is 5.4 and at time=10 the §ow is 7.0 . 
Hencethe
average gradient of the curve during that second was 1.6 .
Hence it islikely that the gradient at the actual point of
in§exion was at least 1.6 .Anyway, you could just determine
the second with the maximum gradient andassume that the point
of in§exion occurs at the start of that second andthat the
slope at the point of in§exion is the slope during that
second.Adopting that approach with the above data gives:Point
of in§exion occurs at time=9Kp = slope = 7.0-5.4 = 1.6/(10-9)
= 1.6 [you had 1.58]So, the equation of the tangent is
y=1.6x+cBut the tangent is assumed to pass through the point
(9, 5.4)So, 5.4=1.6*9+cSo, c=-9So, the tangent is y=1.6x-9Or,
x=(y+9)/1.6When y=1, x = 10/1.6 = 6.25, so lagtime (td) = 6.25
[you had 6.1]When y=10, x= 19/1.6, so your t = 11.9 [you had
11.8]Given the likely errors in measuring the times and the
§ow rates, thismethod might be perfectly adequate.-- Clive
Toothhttp://www.clivetooth.dk
=The subjects: - The gentle
discerning approximation of line-like graphicpatterns. - The
uniform approximation of curves by segments with
non-sharp(fuzzy) ends. It is the base for most universal
tangent initialapproximation of arbitrary form curves. - The
unison of parametrical (pose) hypotheses as an indicator
oftemplate-matching.The
contents:http://shulga.tripod.com/geometry/E.htmThe procedural
collection created on this basis and the results of
itsexperimental probation is described. The investigation had
been doneon the contour preparations of actual half-tone
images of compositionsformed from arbitrary shape (piecewise
smooth) objects.
=We have a sequential application which
involves solving PDEs. We arenow looking at parallelizing the
code - starting with solving the PDEs(parallely) using Petsc.
Is there anyway I can integrate Petsc into myexisting
application - has anybody done this sort of thing? Iunderstand
it is possible to have Petsc interface to externalapplications,
can we also do it the other way around?Any light shed on the