The purpose of this model is to provide the indexed elements of the "vector" array the values of the array "tester", which are increasing numbers from 1 to 12.
This shall happen every time the BooleanPulse is true. The arrays are indexed by the Integer counter, which increases by 1 every boolean pulse.
The model simulates but gives every "vector" element the value 0. Also the warning message is "Failed to solve linear system of equations (no. 21) at time 0.000000. Residual norm is 0.5."
model Array_Test_simple
Real vector[12];
Integer counter;
Real tester[12]=1:1:12;
Modelica.Blocks.Sources.BooleanPulse booleanPulse(period = 0.1);
initial algorithm
counter:=1;
algorithm
when booleanPulse.y then
vector[counter]:=tester[counter];
counter:=pre(counter)+1;
end when;
equation
annotation(
uses(Modelica(version = "4.0.0")));
end Array_Test_simple;
This is a known bug in OpenModelica and was recently fixed in PR #8541.
Try out the latest nightly-build of OpenModelica and check if that solves your issue.
But four your example it should also work to use equations and not algorithm equations. In general they are supported much better and the compiler can try to optimize them. Algorithms stay basically untouched.
Related
Using unspecified array dimensions (:) is an essential feature to design flexible components for reuse. I am well aware that the actual dimension has to be fixed when the model is compiled. To my knowledge binding a variable with unspecified array dimensions to one that has clearly defined dimensions should suffice.
So I am a bit confused why the following model Test will not validate in either OpenModelica or the Wolfram System Modeler:
package VectorFunctions
model Test
VectorSum converter "Component taking the sum of a vector input";
InformationSource source "Vector input";
equation
connect( source.y, converter.u );
end Test;
block VectorSum "Take the sum of an input with unspecified dimension"
Modelica.Blocks.Interfaces.RealInput u[:];
Modelica.Blocks.Interfaces.RealOutput y;
equation
y = sum(u);
end VectorSum;
block InformationSource "Provide some vector output"
Modelica.Blocks.Interfaces.RealOutput y[3];
equation
y = ones( 3 );
end InformationSource;
end VectorFunctions;
How can something like this be done then?
My guess would be that the Modelica Spec does not specify, that vector sizes can be automatically detected from connections, so the tools don't support that.
I think you have to set the vector size somehow by yourself, e.g. with a parameter which is set in your Test model as follows:
model Test
VectorSum converter(nu=size(source.y, 1)) "Pass in the vector size";
InformationSource source "Vector input";
equation
connect(source.y, converter.u);
end Test;
block VectorSum "Take the sum of an input with unspecified dimension"
Modelica.Blocks.Interfaces.RealInput u[nu];
parameter Integer nu(min=0)=0;
output Real y;
equation
y = sum(u);
end VectorSum;
Note that Dymola complains in your example code that connect statements can only be applied to connectors. Therefore I changed input Real to Modelica.Blocks.Interfaces.RealInput(and similar in InformationSource)
I have been given (inofficial) feedback on Wolfram Community by someone from Wolfram MathCore (e.g. the developers of the System Modeler):
Hi, I agree with your interpretation, I think we should support it. I have filed a bug to keep track of this issue internally, unfortunately I do not see any work around. We will come back to you when we have fixed this problem.
So, hopefully flexbile array sizes will be supported for blocks as they are for functions.
Currently, I'm working on a school project involving buck converters. As current increase through an inductor, its inductance decrease (most likely). Each phase is adding an inductor. By adding an inductor, I divide the current by each added inductor. The current is ramped from 0 to 500.
My issue with the following code is that it does not use each array value of i_L(i,j) correctly. I receive some negative values, which is absolutely wrong.
In example...
At 500 Amps with 10 phases, each inductor uses 50 amps. Now L will be designed after
i_L(i,j)=current(j)./phases(i)= 500/10=50amps
L(i,j)= (-9.22297516731983*10^(-16).*(50^(4)))+(9.96260934359008*10^(-14).*(50^(3)))-(3.6355216850551*10^(-12).*(50^(2)))+(9.0205832462444*10^(-12).*(50^(1)))+1.06054781561763E-07 = 1.04106*10^(-7)
and so on
creating 10x10 = 100 cells
clc; clear all;
phases=linspace(1,10,10);
current=linspace(0,500,10);
for j = 1:10
for i=1:10
i_L(i,j)=current(j)./phases(i);
L(i,j)=(-0.000000000000000922297516731983*(i_L(i,j).^(4)))+(0.000000000000099626093435900800*(i_L(i,j).^(3)))-(0.000000000003635521685055100000*(i_L(i,j).^(2)))+(0.000000000009020583246244400000*(i_L(i,j).^(1)))+0.000000106054781561763000000000;
end
end
Thank you!
Your matrix i_L(i,j) got values up to 500=500(current)/1(amp).
The polynomial you're using is generating negative solutions for values greater than 130.
So the operation is using each array value correct.
Maybe you should reevaluate the polynomial, if you're dissatisfied with the solution.
Try:
x=[0:1:500];
y=(-9.22297516731983*10^(-16).*(x.^(4)))+(9.96260934359008*10^(-14).*(x.^(3)))-(3.6355216850551*10^(-12).*(x.^(2)))+(9.0205832462444*10^(-12).*(x.^(1)))+1.06054781561763E-07;
plot(x,y)
You will see the polynomial will diverge against negative infinite for positive values.
I want to know if a model can be inversed in modelica. (here inverse means: if in causal statement y= x +a; x and a are input and y is output; but if I want to find 'x' as output and 'y' and 'a' as input, the model is called reversed/inversed model) For example, if I have compressor with input air port and output air port, and port has variables associated with it are pressure(P), temperature(T) and mass flow rate(mdot). I have simple steady state model containing three equations as follow:
OutPort.mdot = InPort.mdot
OutPort.P = rc * InPort.P
OutPort.T = InPort.T * (1 + rc[ (gamma-1)/gamma) - 1][/sup] / eta);
Here, rc, gamma and eta are compression ratio, ratio of specific heat capacitites and efficiency of compressor respectively.
I want to know, if I know values of : gamma, eta, OutPort.mdot, OutPort.P and OutPort.T and InPort.P and InPort.T, can I find the value of rc.
Can I find values of rc and how should be the model of compressor with above equation in Modelica. As far as I know, there are some variables designated as parameters which can not be changed during simulation. How the modelica model should be with above equations
Thanks
Yes, this should not be a problem as long as you make sure that rc is not a parameter, but a normal variable, and you supply the appropriate number of known quantities to achieve a balanced system (roughly, number of unknowns matches number of equations).
E.g. in your case if you know/supply OutPort.P and InPort.P, rc is already determined from eq 2. Then, in the third equation, there are no unknowns left, so either the temperature values are consistent with the equation or you (preferably) leave one temperature value undetermined.
In addition if you only want to compute the parameter rc during steady-state initialization i.e. that nothing changes with time that is also possible:
...
parameter Real rc(fixed=false);
initial equation
Inport.mdot=12; // Or something else indirectly determining rc.
The fixed=false means that rc is indirectly determined from the initialization. However, if the model is not completely stationary it will only find the correct rc during the initialization and then use that afterwards.
I am trying to do a Monte Carlo simulation of a local volatility model, i.e.
dSt = sigma(St,t) * St dWt .
Unfortunately the Matlab package class sde can not be applied, as the function is rather complex.
For this reason I am simulating this SDE manually with the Euler-Mayurama method. More specifically I used Ito's formula to get an SDE for the log-process Xt=log(St)
dXt = -1/2 sigma^2(exp(Xt),t) dt + sigma(exp(Xt),t) dWt
The code for this is the following:
function [S]=geom_bb(sigma,T,N,m)
% T.. Time horizon, sigma.. standard deviation, N.. timesteps, m.. dimensions
X=zeros(N+1,m);
dt=T/N;
t=(0:N)'*dt;
dW=randn(N,m);
for j=1:N
X(j+1,:)=X(j,:) - 1/2* sigma(exp(X(j,:)),t(j))^2 * sqrt(dt) + sigma(exp(X(j,:)),t(j))*dW(j,:);
end
S=exp(X*sqrt(dt));
end
This code works rather good for small sigma, however for sigma around 10 the process S always tends to zero. This should not happen as S is a martingale, and therefore has expectation =1 (at least for constant sigma).
However X should be simulated correctly, as the mean is exact.
Can anyone help me with this issue? Is this only due to numerical rounding errors? Is there another simulation method that should be preferred to solve this problem?
First are you sure S=exp(X*sqrt(dt)) outside the loop is doing what you want ? Why not have it inside the loop to start with ? You're using the exp(X) for sigma() inside the loop in any case, which is now missing the sqrt(dt).
Beyond that, suggested ways to improve behavior: use the Milstein scheme instead, increase the number of timesteps, make sure your sigma() value is commensurate with your timestep. Sigma of 10 means 1000% volatility, i.e. moves of 60% per day. Assuming dt is more than a few minutes, this simply can't be good.
This question is somewhat related to a previous question of mine, where I didn't quite get the right solution. Link: Earlier SO-thread
I am solving PDEs which are time variant with one spatial dimension (e.g. the heat equation - see link below). I'm using the numerical method of lines, i.e. discretizing the spatial derivatives yielding a system of ODEs which are readily solved in Modelica (using the Dymola tool). My problems arise when I simulate the system, or when I plot the results, to be precise. The equations themselves appear to be solved correctly, but I want to express the spatial changes in all the discretized state variables at specific points in time rather than the individual time-varying behavior of each discrete state.
The strategy leading up to my problems is illustrated in this Youtube tutorial, which by the way is not made by me. As you can see at the very end of the tutorial, the time-varying behavior of the temperature is plotted for all the discrete points in the rod, individually. What I would like is a plot showing the temperature through the rod at a specific time, that is the temperature as a function of the spatial coordinate. My strategy to achieve this, which I'm struggling with, is: Given a state vector of N entries:
Real[N] T "Temperature";
..I would use the plotArray Dymola function as shown below.
plotArray( {i for i in 1:N}, {T[i] for i in 1:N} )
Intuitively, this would yield a plot showing the temperature as a function of the spatial coordiate, or the number in the line of discrete units, to be precise. Although this command yields a result, all T-values appear to be 0 in the plot, which is definitely not the case. My question is: How can I successfully obtain and plot the temperatures at all the discrete points at a given time? Thanks in advance for your help.
The code for the problem is as indicated below.
model conduction
parameter Real rho = 1;
parameter Real Cp = 1;
parameter Real L = 1;
parameter Real k = 1;
parameter Real Tlo = 0;
parameter Real Thi = 100;
parameter Real Tinit = 30;
parameter Integer N = 10 "Number of discrete segments";
Real T[N-1] "Temperatures";
Real deltaX = L/N;
initial equation
for i in 1:N-1 loop
T[i] = Tinit;
end for;
equation
rho*Cp*der(T[1]) = k*( T[2] - 2*T[1] + Thi) /deltaX^2;
rho*Cp*der(T[N-1]) = k*( Tlo - 2*T[N-1] + T[N-2]) /deltaX^2;
for i in 2:N-2 loop
rho*Cp*der(T[i]) = k*( T[i+1] - 2*T[i] + T[i-1]) /deltaX^2;
end for
annotation (uses(Modelica(version="3.2")));
end conduction;
Additional edit: The simulations show clearly that for example T[3], that is the temperature of discrete segment no. 3, starts out from 30 and ends up at 70 degrees. When I write T[3] in my command window, however, I get T3 = 0.0 in return. Why is that? This is at the heart of the problem, because the plotArray function would be working if I managed to extract the actual variable values at specific times and not just 0.0.
Suggested solution: This is a rather tedious solution to achieve what I want, and I hope someone knows a better solution. When I run the simulation in Dymola, the software generates a .mat-file containing the values of the variables throughout the time of the simulation. I am able to load this file into MATLAB and manually extract the variables of my choice for plotting. For the problem above, I wrote the following command:
plot( [1:9]' , data_2(2:2:18 , 10)' )
This command will plot the temperatures (as the temperatures are stored together with their derivates in the data_2 array in the .mat-file) against the respetive number of the discrete segment/element. I was really hoping to do this inside Dymola, that is avoid using MATLAB for this. For this specific problem, the amount of variables was low on account of the simplicity of this problem, but I can easily image a .mat-file which is signifanctly harder to navigate through manually like I just did.
Although you do not mention it explicitly I assume that you enter your plotArray command in Dymola's command window. That won't work directly, since the variables you see there do not include your simulation results: If I simulate your model, and then enter T[:] in Dymola's command window, then the printed result is
T[:]
= {0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0}
I'm not a Dymola expert, and the only solution I've found (to actively store and load the desired simulation results) is quite cumbersome:
simulateModel("conduction", resultFile="conduction.mat")
n = readTrajectorySize("conduction.mat")
X = readTrajectory("conduction.mat", {"Time"}, n)
Y = readTrajectory("conduction.mat", {"T[1]", "T[2]", "T[3]"}, n)
plotArrays(X[1, :], transpose(Y))