`sol = pdepe(m,#ParticleDiffusionpde,#ParticleDiffusionic,#ParticleDiffusionbc,x,t);
% Extract the first solution component as u.
u = sol(:,:,:);
function [c,f,s] = ParticleDiffusionpde(x,t,u,DuDx)
global Ds
c = 1/Ds;
f = DuDx;
s = 0;
function u0 = ParticleDiffusionic(x)
global qo
u0 = qo;
function [pl,ql,pr,qr] = ParticleDiffusionbc(xl,ul,xr,ur,t,x)
global Ds K n
global Amo Gc kf rhop
global uavg
global dr R nr
sum = 0;
for i = 1:1:nr-1
r1 = (i-1)*dr; % radius at i
r2 = i * dr; % radius at i+1
r1 = double(r1); % convert to double precision
r2 = double(r2);
sum = sum + (dr / 2 * (r1*ul+ r2*ur));
end;
uavg = 3/R^3 * sum;
ql = 1;
pl = 0;
qr = 1;
pr = -((kf/(Ds.*rhop)).*(Amo - Gc.*uavg - ((double(ur/K)).^2).^(n/2) ));`
dq(r,t)/dt = Ds( d2q(r,t)/dr2 + (2/r)*dq(r,t)/dr )
q(r, t=0) = 0
dq(r=0, t)/dr = 0
dq(r=dp/2, t)/dr = (kf/Ds*rhop) [C(t) - Cp(at r = dp/2)]
q = solid phase concentration of trace compound in a particle with radius dp/2
C = bulk liquid concentration of trace compound
Cp = trace compound concentration at particle surface
I want to solve the above pde with initial and boundary conditions given. Tried Matlab's pdepe, but does not work satisfactorily. Maybe the boundary conditions is creating problem for me. I also used this isotherm equation for equilibrium: q = K*Cp^(1/n). This is convection-diffusion equation but i could not find any write ups that addresses solving this type of equation properly.
There are two problems with the current implementation.
Incorrect Source Term
The PDE you are attempting to solve has the form
which has the equivalent form
where the last term arises due to the factor of 2 in the original PDE.
The last term needs to be incorporated into pdepe via a source term.
Calculation of q average
The current implementation attempts to calculate the average value of q using the left and right values of q passed to the boundary condition function.
This is incorrect.
The average value of q needs to be calculated from a vector of up-to-date values of the quantity.
However, we have the complication that the only function to receive all mesh values is ParticleDiffusionpde; however, the mesh values passed to that function are not guaranteed to be from the mesh we provided.
Solution: use events (as described in the pdepe documentation).
This is a hack since the event function is meant to detect zero-crossings, but it has the advantage that the function is given all values of q on the mesh we provide.
So, the working example below (you'll notice I set all of the parameters to 1 since I didn't know better) uses the events function to update a variable qStore that can be accessed by the boundary condition function (see here for an explanation), and the boundary condition function performs a vectorized trapezoidal integration for the average calculation.
Working Example
function [] = ParticleDiffusion()
% Parameters
Ds = 1;
q0 = 0;
K = 1;
n = 1;
Amo = 1;
Gc = 1;
kf = 1;
rhop = 1;
% Space
rMesh = linspace(0,1,10);
rMesh = rMesh(:) ;
dr = rMesh(2) - rMesh(1) ;
% Time
tSpan = linspace(0,1,10);
% Vector to store current u-value
qStore = zeros(size(rMesh));
options.Events = #(m,t,x,y) events(m,t,x,y);
% Solve
[sol,~,~,~,~] = pdepe(1,#ParticleDiffusionpde,#ParticleDiffusionic,#ParticleDiffusionbc,rMesh,tSpan,options);
% Use the events function to update qStore
function [value,isterminal,direction] = events(m,~,~,y)
qStore = y; % Value of q on rMesh
value = m; % Since m is constant, it will never be zero (no event detection)
isterminal = 0; % Continue integration
direction = 0; % Detect all zero crossings (not important)
end
function [c,f,s] = ParticleDiffusionpde(r,~,~,DqDr)
% Define the capacity, flux, and source
c = 1/Ds;
f = DqDr;
s = DqDr./r;
end
function u0 = ParticleDiffusionic(~)
u0 = q0;
end
function [pl,ql,pr,qr] = ParticleDiffusionbc(~,~,R,ur,~)
% Calculate average value of current solution
qL = qStore(1:end-1);
qR = qStore(2: end );
total = sum((qL.*rMesh(1:end-1) + qR.*rMesh(2:end))) * dr/2;
qavg = 3/R^3 * total;
% Left boundary
pl = 0;
ql = 1;
% Right boundary
qr = 1;
pr = -(kf/(Ds.*rhop)).*(Amo - Gc.*qavg - (ur/K).^n);
end
end
Related
I am very new to Scilab, but so far have not been able to find an answer (either here or via google) to my question. I'm sure it's a simple solution, but I'm at a loss. I have a lot of MATLAB scripts I wrote in grad school, but now that I'm out of school, I no longer have access to MATLAB (and can't justify the cost). Scilab looked like the best open alternative. I'm trying to convert my .m files to Scilab compatible versions using mfile2sci, but when running the mfile2sci GUI, I get the error/message shown below. Attached is the original code from the M-file, in case it's relevant.
I Searched Stack Overflow and companion sites, Google, Scilab documentation.
The M-file code follows (it's a super basic MATLAB script as part of an old homework question -- I chose it as it's the shortest, most straightforward M-file I had):
Mmax = 15;
N = 20;
T = 2000;
%define upper limit for sparsity of signal
smax = 15;
mNE = zeros(smax,Mmax);
mESR= zeros(smax,Mmax);
for M = 1:Mmax
aNormErr = zeros(smax,1);
aSz = zeros(smax,1);
ESR = zeros(smax,1);
for s=1:smax % for-loop to loop script smax times
normErr = zeros(1,T);
vESR = zeros(1,T);
sz = zeros(1,T);
for t=1:T %for-loop to carry out 2000 trials per s-value
esr = 0;
A = randn(M,N); % generate random MxN matrix
[M,N] = size(A);
An = zeros(M,N); % initialize normalized matrix
for h = 1:size(A,2) % normalize columns of matrix A
V = A(:,h)/norm(A(:,h));
An(:,h) = V;
end
A = An; % replace A with its column-normalized counterpart
c = randperm(N,s); % create random support vector with s entries
x = zeros(N,1); % initialize vector x
for i = 1:size(c,2)
val = (10-1)*rand + 1;% generate interval [1,10]
neg = mod(randi(10),2); % include [-10,-1]
if neg~=0
val = -1*val;
end
x(c(i)) = val; %replace c(i)th value of x with the nonzero value
end
y = A*x; % generate measurement vector (y)
R = y;
S = []; % initialize array to store selected columns of A
indx = []; % vector to store indices of selected columns
coeff = zeros(1,s); % vector to store coefficients of approx.
stop = 10; % init. stop condition
in = 0; % index variable
esr = 0;
xhat = zeros(N,1); % intialize estimated x signal
while (stop>0.5 && size(S,2)<smax)
%MAX = abs(A(:,1)'*R);
maxV = zeros(1,N);
for i = 1:size(A,2)
maxV(i) = abs(A(:,i)'*R);
end
in = find(maxV == max(maxV));
indx = [indx in];
S = [S A(:,in)];
coeff = [coeff R'*S(:,size(S,2))]; % update coefficient vector
for w=1:size(S,2)
r = y - ((R'*S(:,w))*S(:,w)); % update residuals
if norm(r)<norm(R)
index = w;
end
R = r;
stop = norm(R); % update stop condition
end
for j=1:size(S,2) % place coefficients into xhat at correct indices
xhat(indx(j))=coeff(j);
end
nE = norm(x-xhat)/norm(x); % calculate normalized error for this estimate
%esr = 0;
indx = sort(indx);
c = sort(c);
if isequal(indx,c)
esr = esr+1;
end
end
vESR(t) = esr;
sz(t) = size(S,2);
normErr(t) = nE;
end
%avsz = sum(sz)/T;
aSz(s) = sum(sz)/T;
%aESR = sum(vESR)/T;
ESR(s) = sum(vESR)/T;
%avnormErr = sum(normErr)/T; % produce average normalized error for these run
aNormErr(s) = sum(normErr)/T; % add new avnormErr to vector of all av norm errors
end
% just put this here to view the vector
mNE(:,M) = aNormErr;
mESR(:,M) = ESR;
% had an 'end' placed here, might've been unmatched
mNE%reshape(mNE,[],Mmax)
mESR%reshape(mESR,[],Mmax)]
figure
dimx = [1 Mmax];
dimy = [1 smax];
imagesc(dimx,dimy,mESR)
colormap gray
strESR = sprintf('Average ESR, N=%d',N);
title(strESR);
xlabel('M');
ylabel('s');
strNE = sprintf('Average Normed Error, N=%d',N);
figure
imagesc(dimx,dimy,mNE)
colormap gray
title(strNE)
xlabel('M');
ylabel('s');
The command used (and results) follow:
--> mfile2sci
ans =
[]
****** Beginning of mfile2sci() session ******
File to convert: C:/Users/User/Downloads/WTF_new.m
Result file path: C:/Users/User/DOWNLO~1/
Recursive mode: OFF
Only double values used in M-file: NO
Verbose mode: 3
Generate formatted code: NO
M-file reading...
M-file reading: Done
Syntax modification...
Syntax modification: Done
File contains no instruction, no translation made...
****** End of mfile2sci() session ******
To convert the foo.m file one has to enter
mfile2sci <path>/foo.m
where stands for the path of the directoty where foo.m is. The result is written in /foo.sci
Remove the ```` at the begining of each line, the conversion will proceed normally ?. However, don't expect to obtain a working .sci file as the m2sci converter is (to me) still an experimental tool !
I have model with "matlab function block" in which I have recursive least square method. Recursive algorithm needs to know length of incoming signal in order to work correctly. But when I use command N=length(y) it returns me length N= 1. But I think it should give me higher length.
Simulink model
Matlab function block code "rls_iden6"
function [P,N] = fcn(u,y)
%%
N = length(y);
sigma=1;
C = sigma*eye(2); %p
P = ones(2,1);
z= [y; u];
lamda=1;
for n=1:N
sample_out = y(n);
C = (C - ( (C*z*z'*C)/( lamda+(z'*C*z) ) ))/lamda;
P = P + (C*z* (sample_out - (z'*P)));
end
My final code should look like it's shown below, because it works in matlab workspace. Simulink should give me 5 parameters instead of just 2.
load data_cela.mat
u=U; %input
y=Y; %output
%%
input = 3;
output = 2;
system = input + output;
N = length(y);
%initial conditions
sigma = 1;
C = sigma*eye(system);
P = ones(system,1);
lamda = 1; %forgetting factor
for n=3:N
for i=1:2
W(i) = y(n-i); %output
end
for i=1:3
V(i) = u(n-i+1); %input
end
z = [V';W'];
sample_out = y(n);
pom(n)= z' * P;
error(n) = y(n) - pom(n);
C = (C - ( (C*z*z'*C)/( lamda+(z'*C*z) ) ))/lamda;
P = P + (C*z* (sample_out - (z'*P) ) );
change(1:system,n) = P;
end
f_param = [P(1:3);-P(4:5)];
num = [P(1:3,1)];
den = [1;-P(4:5,1)];
num1 = num(3,1);
trasferfunction = tf(num1,den',1)
Result:
0.002879
----------------------
z^2 - 1.883 z + 0.8873
You will need to add a buffer before signal to convert the scalar to matrix. Then after the buffer has been added set the buffer size to the amount of data you want, i.e. by setting it to 2 will make 2 rows and 1 column. This will help you to get the data however, for setting delay properly you will require to set buffer overlap to 1.
Hope this helps.
I am using MATLAB to calculate the numerical integral of a complex function including natural exponent.
I get a warning:
Infinite or Not-a-Number value encountered
if I use the function integral, while another error is thrown:
Output of the function must be the same size as the input
if I use the function quadgk.
I think the reason could be that the integrand is infinite when the variable ep is near zero.
Code shown below. Hope you guys can help me figure it out.
close all
clear
clc
%%
N = 10^5;
edot = 10^8;
yita = N/edot;
kB = 8.6173324*10^(-5);
T = 300;
gamainf = 0.115;
dTol = 3;
K0 = 180;
K = K0/160.21766208;
nu = 3*10^12;
i = 1;
data = [];
%% lambda = ec/ef < 1
for ef = 0.01:0.01:0.1
for lambda = 0.01:0.01:0.08
ec = lambda*ef;
f = #(ep) exp(-((32/3)*pi*gamainf^3*(0.5+0.5*sqrt(1+2*dTol*K*(ep-ec)/gamainf)-dTol*K*(ep-ec)/gamainf).^3/(K*(ep-ec)).^2-16*pi*gamainf^3*(0.5+0.5*sqrt(1+2*dTol*K*(ep-ec)/gamainf)-dTol*K*(ep-ec)/gamainf).^2/((1+dTol*K*(ep-ec)/(gamainf*(0.5+0.5*sqrt(1+2*dTol*K*(ep-ec)/gamainf)-dTol*K*(ep-ec)/gamainf)))*(K*(ep-ec)).^2))/(kB*T));
q = integral(f,0,ef,'ArrayValued',true);
% q = quadgk(f,0,ef);
prob = 1-exp(-yita*nu*q);
data(i,1) = ef;
data(i,2) = lambda;
data(i,3) = q;
i = i+1;
end
end
I've rewritten your equations so that a human can actually understand it:
function integration
N = 1e5;
edot = 1e8;
yita = N/edot;
kB = 8.6173324e-5;
T = 300;
gamainf = 0.115;
dTol = 3;
K0 = 180;
K = K0/160.21766208;
nu = 3e12;
i = 1;
data = [];
%% lambda = ec/ef < 1
for ef = 0.01:0.01:0.1
for lambda = 0.01:0.01:0.08
ec = lambda*ef;
q = integral(#f,0,ef,'ArrayValued',true);
% q = quadgk(f,0,ef);
prob = 1 - exp(-yita*nu*q);
data(i,:) = [ef lambda q];
i = i+1;
end
end
function y = f(ep)
G = K*(ep - ec);
r = dTol*G/gamainf;
S = sqrt(1 + 2*r);
x = (1 + S)/2 - r;
Z = 16*pi*gamainf^3;
y = exp( -Z*x.^2.*( 2*x/(3*G.^2) - 1/(G.^2*(1 + r/x))) ) /...
(kB*T));
end
end
Now, for the first iteration, ep = 0.01, the value of the argument of the exp() function inside f is huge. In fact, if I rework the function to return the argument to the exponent (not the value):
function y = f(ep)
% ... all of the above
% NOTE: removed the exp() to return the argument
y = -Z*x.^2.*( 2*x/(3*G.^2) - 1/(G.^2*(1 + r/x))) ) /...
(kB*T);
end
and print its value at some example nodes like so:
for ef = 0.01 : 0.01 : 0.1
for lambda = 0.01 : 0.01 : 0.08
ec = lambda*ef;
zzz(i,:) = [f(0) f(ef/4) f(ef)];
i = i+1;
end
end
zzz
I get this:
% f(0) f(ef/4) f(ef)
zzz =
7.878426438111721e+07 1.093627454284284e+05 3.091140080273912e+03
1.986962280947140e+07 1.201698288371587e+05 3.187767404903769e+03
8.908646053687230e+06 1.325435523124976e+05 3.288027743119838e+03
5.055141696747510e+06 1.467952125661714e+05 3.392088351112798e+03
...
3.601790797707676e+04 2.897200140791236e+02 2.577170427480841e+01
2.869829209254144e+04 3.673888685004256e+02 2.404148067956737e+01
2.381082059148755e+04 4.671147785149462e+02 2.238181495716831e+01
So, integral() has to deal with things like exp(10^7). This may not be a problem per se if the argument would fall off quickly enough, but as shown above, it doesn't.
So basically you're asking for the integral of a function that ranges in value between exp(~10^7) and exp(~10^3). Needless to say, The d(ef) in the order of 0.01 isn't going to compensate for that, and it'll be non-finite in floating point arithmetic.
I suspect you have a scaling problem. Judging from the names of your variables as well as the equations, I would think that this has something to do with thermodynamics; a reworked form of Planck's law? In that case, I'd check if you're working in nanometers; a few factors of 10^(-9) will creep in, rescaling your integrand to the compfortably computable range.
In any case, it'll be wise to check all your units, because it's something like that that's messing up the numbers.
NB: the maximum exp() you can compute is around exp(709.7827128933840)
I have originally written the following Matlab code to find intersection between a set of Axes Aligned Bounding Boxes (AABB) and space partitions (here 8 partitions). I believe it is readable by itself, moreover, I have added some comments for even more clarity.
function [A,B] = AABBPart(bbx,it) % bbx: aabb, it: iteration
global F
IT = it+1;
n = size(bbx,1);
F = cell(n,it);
A = Part([min(bbx(:,1:3)),max(bbx(:,4:6))],it,0); % recursive partitioning
B = F; % matlab does not allow
function s = Part(bx,it,J) % output to be global
s = {};
if it < 1; return; end
s = cell(8,1);
p = bx(1:3);
q = bx(4:6);
h = 0.5*(p+q);
prt = [p,h;... % 8 sub-parts (octa)
h(1),p(2:3),q(1),h(2:3);...
p(1),h(2),p(3),h(1),q(2),h(3);...
h(1:2),p(3),q(1:2),h(3);...
p(1:2),h(1),h(1:2),q(3);...
h(1),p(2),h(3),q(1),h(2),q(3);...
p(1),h(2:3),h(1),q(2:3);...
h,q];
for j=1:8 % check for each sub-part
k = 0;
t = zeros(0,1);
for i=1:n
if all(bbx(i,1:3) <= prt(j,4:6)) && ... % interscetion test for
all(prt(j,1:3) <= bbx(i,4:6)) % every aabb and sub-parts
k = k+1;
t(k) = i;
end
end
if ~isempty(t)
s{j,1} = [t; Part(prt(j,:),it-1,j)]; % recursive call
for i=1:numel(t) % collecting the results
if isempty(F{t(i),IT-it})
F{t(i),IT-it} = [-J,j];
else
F{t(i),IT-it} = [F{t(i),IT-it}; [-J,j]];
end
end
end
end
end
end
Concerns:
In my tests, it seems that probably few intersections are missing, say, 10 or so for 1000 or more setup. So I would be glad if you could help to find out any problematic parts in the code.
I am also concerned about using global F. I prefer to get rid of it.
Any other better solution in terms of speed, will be loved.
Note that the code is complete. And you can easily try it by some following setup.
n = 10000; % in the original application, n would be millions
bbx = rand(n,6);
it = 3;
[A,B] = AABBPart(bbx,it);
I've got at State System, with "forced" inputs at bounds. My SS equation is: zp = A*z * B. (A is a square matrix, and B colunm)
If B is a step (along the time of experience), there is no problem, because I can use
tevent = 2;
tmax= 5*tevent;
n =100;
dT = n/tmax;
t = linspace(0,tmax,n);
u0 = 1 * ones(size(z'));
B = zeros(nz,n);
B(1,1)= utop(1)';
A = eye(nz,nz);
[tt,u]=ode23('SS',t,u0);
and SS is:
function zp = SS(t,z)
global A B
zp = A*z + B;
end
My problem is when I applied a slop, So B will be time dependent.
utop_init= 20;
utop_final = 50;
utop(1)=utop_init;
utop(tevent * dT)=utop_final;
for k = 2: tevent*dT -1
utop(k) = utop(k-1) +(( utop(tevent * dT) - utop(1))/(tevent * dT));
end
for k = (tevent * dT) +1 :(tmax*dT)
utop(k) = utop(k-1);
end
global A B
B = zeros(nz,1);
B(1,1:n) = utop(:)';
A = eye(nz,nz);
I wrote a new equation (to trying to solve), the problem, but I can't adjust "time step", and I don't get a u with 22x100 (which is the objective).
for k = 2 : n
u=solveSS(t,k,u0);
end
SolveSS has the code:
function [ u ] = solveSS( t,k,u0)
tspan = [t(k-1) t(k)];
[t,u] = ode15s(#SS,tspan,u0);
function zp = SS(t,z)
global A B
zp = A*z + B(:,k-1);
end
end
I hope that you can help!
You should define a function B that is continuously varying with t and pass it as a function handle. This way you will allow the ODE solver to adjust time steps efficiently (your use of ode15s, a stiff ODE solver, suggests that variable time stepping is even more crucial)
The form of your code will be something like this:
function [ u ] = solveSS( t,k,u0)
tspan = [t(k-1) t(k)];
[t,u] = ode15s(#SS,tspan,u0,#B);
function y = B(x)
%% insert B calculation
end
function zp = SS(t,z,B)
global A
zp = A*z + B(t);
end
end