I was wondering if anyone had any tips or tricks in terms of dealing with jumps/discontinuities in Modelica. I'm using OpenModelica, and a simplified example code for my problem is shown below.
model PowerGenerator
Modelica.SIunits.Power P(start=0);
output Modelica.SIunits.Energy E;
equation
if (5 < time) and (time < 15) then P = 3;
else P = 0;
end if;
der(E) = P;
end PowerGenerator;
How can I make the jumps at 5 and 15 sec into continuous transitions, where the derivative of the slope is finite? I've tried the noEvent and smooth functions, but I haven't been able to make them do what I need.
Edit:
The issue in my main model is that these events induce chattering, and so I also need it to work in real-time. Also in my full model, the events are state events, and so the time is not known.
Perhaps a behavior like the code below is what you were looking?
model PowerGenerator
Modelica.SIunits.Power Plow;
Modelica.SIunits.Power Phigh;
Modelica.SIunits.Power P(start=0);
output Modelica.SIunits.Energy E;
Modelica.SIunits.Power Pold(start=0);
output Modelica.SIunits.Energy Eold;
equation
if (5 < time) and (time < 15) then
Pold = 3;
else
Pold = 0;
end if;
der(Eold) = Pold;
Plow = Modelica.Media.Air.MoistAir.Utilities.spliceFunction(3,0,time-5,1);
Phigh = Modelica.Media.Air.MoistAir.Utilities.spliceFunction(0,3,time-15,1);
P = Modelica.Media.Air.MoistAir.Utilities.spliceFunction(Phigh,Plow,time-10,1);
der(E) = P;
end Power Generator;
Below is the plot of the outputs:
The answer by Scott G is good, but it only works well if the change just depends on time. A more general solution is to use a low-pass filter:
model PowerGenerator
Modelica.SIunits.Power P(start=0);
output Modelica.SIunits.Energy E;
Modelica.Blocks.Continuous.LowpassButterworth lowpassButterworth(f=1)
annotation (Placement(transformation(extent={{-40,20},{-20,40}})));
equation
if (5 < time) and (time < 15) then
P = 3;
else
P = 0;
end if;
lowpassButterworth.u=P;
der(E) = lowpassButterworth.y;
end PowerGenerator;
(I would recommend using connect-statements instead - but the above should also be legal Modelica.)
Related
How can I simulate this question using MATLAB?
Out of 100 apples, 10 are rotten. We randomly choose 5 apples without
replacement. What is the probability that there is at least one
rotten?
The Expected Answer
0.4162476
My Attempt:
r=0
for i=1:10000
for c=1:5
a = randi(1,100);
if a < 11
r=r+1;
end
end
end
r/10000
but it didn't work, so what would be a better way of doing it?
Use randperm to choose randomly without replacement:
A = false(1, 100);
A(1:10) = true;
r = 0;
for k = 1:10000
a = randperm(100, 5);
r = r + any(A(a));
end
result = r/10000;
Short answer:
Your problem follow an hypergeometric distribution (similar to a binomial distribution but without replacement), if you have the necessary toolbox you can simply use the probability density function of the hypergeometric distribution:
r = 1-hygepdf(0,100,10,5) % r = 0.4162
Since P(x>=1) = P(x=1) + P(x=2) + P(x=3) + P(x=4) + P(x=5) = 1-P(x=0)
Of course, here I calculate the exact probability, this is not an experimental result.
To get further:
Noticed that if you do not have access to hygepdf, you can easily write the function yourself by using binomial coefficient:
N = 100; K = 10;
n = 5; k = 0;
r = 1-(nchoosek(K,k)*nchoosek(N-K,n-k))/nchoosek(N,n) % r = 0.4162
You can also use the binomial probability density function, it is a bit more tricky (but also more intuitive):
r = 1-prod(binopdf(0,1,10./(100-[0:4])))
Here we compute the probability to obtain 0 rotten apple five time in a row, the probabily increase at every step since we remove 1 good juicy apple each time. And then, according to the above explaination, we take 1-P(x=0).
There are a couple of issues with your code. First of all, implicitly in what you wrote, you replace the apple after you look at it. When you generate the random number, you need to eliminate the possibility of choosing that number again.
I've rewritten your code to include better practices:
clear
n_runs = 1000;
success = zeros(n_runs, 1);
failure = zeros(n_runs, 1);
approach = zeros(n_runs, 1);
for ii = 1:n_runs
apples = 1:100;
a = randperm(100, 5);
if any(a < 11)
success(ii) = 1;
elseif a >= 11
failure(ii) = 1;
end
approach(ii) = sum(success)/(sum(success)+sum(failure));
end
figure; hold on
plot(approach)
title("r = "+ approach(end))
hold off
The results are stored in an array (called approach), rather than a single number being updated every time, which means you can see how quickly you approach the end value of r.
Another good habit is including clear at the beginning of any script, which reduces the possibility of an error occurring due to variables stored in the workspace.
I'm trying to create a model where one Modelica variable is a triangular wave of another variable. First I tried the floor() function as below:
model test1
final constant Real pi=2*Modelica.Math.asin(1.0);
parameter Real b = 1;
parameter Real a = 1;
Real x,p,u;
equation
if sign(sin(x*pi/b))>=0 then
p=a*(x-b*floor(x/b));
else
p=a*(b-(x-b*floor(x/b)));
end if;
x=time;
u = floor(x/b);
end test1
(x=time; is arbitrary so the model compiles)
but the result is weird, as you can see below
zoom in:
somehow 0.005 seconds before the next step floor function behaves unexpectedly and becomes a linear function ending by the next value.
then I tried the ceil() function. everything seemed right till I realised the same problem happens with ceil() function at other values (e.g. x=13)
I would appreciate if you could:
help me understand why this "glitch" happens and if it is intentional by design or a bug?
how I can fix this?
are there any alternatives to create a triangular wave function?
P.S. I am using this "wave function" to model the interaction between two jagged bodies"
If you are allowed to utilize the Modelica Standard Library, you can build up a parametrized, time-based zigzag signal using the CombiTimeTable block with linear interpolation and periodic extrapolation. For example,
model Test4
parameter Real a=2 "Amplitude";
parameter Real b=3 "Period";
Real y=zigzag.y[1] "Zigzag";
Modelica.Blocks.Sources.CombiTimeTable zigzag(
table=[0,0;b/4,a;b/4,a;b/2,0;b/2,0;3*b/4,-a;3*b/4,-a;b,0],
extrapolation=Modelica.Blocks.Types.Extrapolation.Periodic)
annotation(Placement(transformation(extent={{-80,60},{-60,80}})));
Modelica.Blocks.Sources.Trapezoid trapezoid(
amplitude=2*a,
rising=b/2,
width=0,
falling=b/2,
period=b,
offset=-a)
annotation(Placement(transformation(extent={{-80,25},{-60,45}})));
annotation(uses(Modelica(version="3.2.2")));
end Test4;
I don't have an explanation for the glitches in your simulation.
However, I would take another approach to the sawtooth function: I see it as an integrator integrating +1 and -1 upwards and downwards. The integration time determines the amplitude and period of the sawtooth function.
The pictures below show an implementation using MSL blocks and one using code. The simulation results below are the same for both implementations.
Best regards,
Rene Just Nielsen
Block diagram:
Code:
model test3
parameter Real a=2 "amplitude";
parameter Real b=3 "period";
Real u, y;
initial equation
u = 1;
y = 0;
equation
4*a/b*u = der(y);
when y > a then
u = -1;
elsewhen y < -a then
u = 1;
end when;
end test3;
Simulation result:
The problem I guess is due to floating point representation and events not occurring at exact times.
Consider x-floor(x) and 1-(x-floor(x)) at time=0.99, they are 0.99 and 0.01; at time=1.00 they are 0.0 and 1.0, which causes your problems.
For a=b=1, you can use the following equation for p:
p=min(mod(x,2),2-mod(x,2));. You can even add noEvent to it, and you can consider the signal continuous (but not differentiable).
model test
parameter Real b = 1;
parameter Real a = 3;
Real x, p;
equation
p = 2*a*min(1 / b * mod(x, b ),1 - 1/b * mod(x, b));
x = time;
end test;
My first advise would be to remove the sign-function, since there is no benefit of doing sign(foo)>=0 compared to foo>=0.
Interesting enough that seems to fix the problem in Dymola - and I assume also in OpenModelica:
model test1 "almost original"
final constant Real pi=2*Modelica.Math.asin(1.0);
parameter Real b = 1;
parameter Real a = 1;
Real x,p,u;
equation
if sin(x*pi/b)>=0 then
p=a*(x-b*floor(x/b));
else
p=a*(b-(x-b*floor(x/b)));
end if;
x=time;
u = floor(x/b);
end test1;
Now I only have to explain that - and the reason is that sin(x*pi/b) is slightly out of sync with the floor-function, but if you use sin(x*pi/b)>=0 that is within the root-finding epsilon and nothing strange happen.
When you use sign(sin(x*pi/b))>=0 that is no longer possible, instead of having sin(x*pi/b) an epsilon below zero it is now -1, and instead of epsilon above zero it is 1.
The real solution is thus slightly more complicated:
model test2 "working"
parameter Real b = 1;
parameter Real a = 1;
Real x,p,u;
Real phase=mod(x,b*2);
equation
if phase<b then
p=a/b*phase;
else
p=a-a/b*(phase-b);
end if;
x=time;
u = floor(x/b);
end test2;
which was improved based on a suggested solution:
model test3 "almost working"
parameter Real b = 1;
parameter Real a = 1;
Real x,p,u;
equation
if mod(x,2*b)<b then
p=a/b*mod(x,b);
else
p=a-a/b*mod(x,b);
end if;
x=time;
u = floor(x/b);
end test3;
The key point in this solution, test2, is that there is only one problematic event generating expression mod(x,2*b) - and the < will not get out of sync with this.
In practice test3 will almost certainly also work, but in unlikely cases the event generation might get out of sync between mod(x,2*b) and mod(x,b); with unknown consequences.
Note that all three examples are now modified to generate output that looks similar.
I implemented the following user defined function in MATLAB:
function Q = Calc_Q(Head, freq)
b6 = [3.7572E-07 -1.5707E-05 6.0490E-03 5.0018E-02 2.1180E-01];
b5 = [-9.0927E-06 8.9033E-04 -3.2415E-02 5.4525E-01 -8.1649E+00] / 10e2;
b4 = [7.5172E-06 -5.6565E-04 1.0024E-02 3.5888E-01 3.8894E-02] / 10e5;
b3 = [-4.8767E-06 4.8787E-04 -1.3311E-02 -1.2189E-01 -5.3522E+00] / 10e8;
b2 = [5.9227E-06 -8.1716E-04 3.5392E-02 -4.5413E-01 1.9547E+00] / 10e11;
b1 = [-2.0004E-06 2.9027E-04 -1.3754E-02 2.3490E-01 -1.2363E+00] / 10e14;
a = [polyval(b1,abs(freq)), polyval(b2, abs(freq)), polyval(b3, abs(freq)), polyval(b4, abs(freq)), polyval(b5, abs(freq)), polyval(b6, abs(freq)) - Head];
Q_roots = roots(a);
%Delete roots with imaginary part
i = 1;
while i <= length(Q_roots)
if(imag(Q_roots(i)) ~= 0)
Q_roots(i) = [];
i = i - 1;
end
i = i + 1;
end
%Delete roots with real part greater then 3100
i = 1;
while i <= length(Q_roots)
if(Q_roots(i) >= 3100 || Q_roots(i) < 0)
Q_roots(i) = [];
i = i - 1;
end
i = i +1;
end
if freq < 0
Q = real(Q_roots(1)) * -1;
else
Q = real(Q_roots(1));
end
end
When I Call this function in Matlab it works fine. However if I use this exact code as a MATLAB function in simulink it stop's working. (actually it works, but the ouput is always zero.)
I do have a suspicion of what the problem might be. When running the script in de-bugging mode, I cannot view a result for Q_roots (It just doesn't display anything).
Q_roots = roots(a);
Any ideas ?
The problem is most likely due to your logic that eliminates any roots that do not have exactly zero in the imaginary part. This is a mathematical way of thinking that does not really work well numerically, at least not in general. All the roots are probably being found in both cases (there is no limitation that implies otherwise), but in Simulink and in code generation the problem is treated as a complex one, and some roots might be coming back with tiny imaginary parts. Instead of deleting roots if their imaginary parts are not exactly zero, eliminate the roots with imaginary parts that are numerically insignificant, either very small relative to the real part or very small altogether. Something like
tol = 10*eps(class(Q_roots));
keepers = abs(imag(Q_roots)) < tol*max(abs(real(Q_roots)),1) & ...
real(Q_roots) >= 0 & real(Q_roots) <= 3100;
Q_roots = Q_roots(keepers);
would take care of all the deletions in one fell swoop. I used 10*eps as a tolerance here.
But if you only need the first qualifying root, then you could just do this:
Q = nan('like',a);
tol = 10*eps(class(a));
for k = 1:numel(Q_roots)
r = real(Q_roots(k));
if abs(imag(Q_roots(k))) < tol*max(abs(r),1) && r >= 0 && r <= 3100;
Q = r;
break
end
end
if freq < 0
Q = -Q;
end
Oke I found the problem.
From a different forum:
Hi Cosmin,
I took a look at the implementation of roots for the Embedded MATLAB
Function block (\toolbox\eml\lib\matlab\polyfun\roots.m).
It's stated there:
% Limitations: % Output is always variable size. % Output is
always complex. % Roots may not be in the same order as MATLAB. %
Roots of poorly conditioned polynomials may not match MATLAB. The last
sentence is what makes you the headache (and yes, your polynomial is
badly conditioned!). If you take a look at the plot you will see, that
the curve hardly touches the x-axis.
I have a suggestion though: the value -z/b is a (very) good
approximation of the root you are looking for ...?
Titus
http://www.mathworks.com/matlabcentral/answers/25624-roots-in-simulink
Apparently the root function in simulink does not always found all the roots of a given polynominal.
This is unfortunate and not easily solvable. I did however found a solution.
For all the different polynomials I have to solve, I know the interval of the root I am interested in ( [-3000, 3000]).
I just basically takes steps of 50 from -3000 to 3000, until the function drops below 0. I then know the approximate solution of the root. I use this approximation as seed for the Newton-Raphson method.
Straight implementation of the Newton raphson method with a given seed for all the polynomials I have to solve did not work because sometimes it iterated to a different root (one which i was not interested in.)
Here's the code:
function Q = Calc_Q(Head, freq)
b6 = [3.7572E-07 -1.5707E-05 6.0490E-03 5.0018E-02 2.1180E-01];
b5 = [-9.0927E-06 8.9033E-04 -3.2415E-02 5.4525E-01 -8.1649E+00] / 10e2;
b4 = [7.5172E-06 -5.6565E-04 1.0024E-02 3.5888E-01 3.8894E-02] / 10e5;
b3 = [-4.8767E-06 4.8787E-04 -1.3311E-02 -1.2189E-01 -5.3522E+00] / 10e8;
b2 = [5.9227E-06 -8.1716E-04 3.5392E-02 -4.5413E-01 1.9547E+00] / 10e11;
b1 = [-2.0004E-06 2.9027E-04 -1.3754E-02 2.3490E-01 -1.2363E+00] / 10e14;
%coeff for the polynominal
a = [polyval(b1,abs(freq)), polyval(b2, abs(freq)), polyval(b3, abs(freq)), polyval(b4, abs(freq)), polyval(b5, abs(freq)), polyval(b6, abs(freq)) - Head];
%coeff for the derrivative of polynominal
da = [5*a(1) 4*a(2) 3*a(3) 2*a(4) a(5)];
Q = -3000;
%Search for point where function goes below 0
while (polyval(a, Q) > 0)
Q = Q + 25;
end
error_max = 0.01
iter_counter = 1;
while abs(polyval(a,Q)) >= error_max && iter_counter <= 1000
Q = Q - polyval(a, Q)/polyval(da, Q);
iter_counter = iter_counter + 1;
error = abs(polyval(a,Q));
end
if(freq < 0)
Q = Q * - 1;
end
end
I'm trying to run this loop as a parfor-loop:
correlations = zeros(1,N);
parfor i = 1:(size(timestamps,1)-1)
j = i+1;
dts = timestamps(j) - timestamps(i);
while (dts < T) && (j <= size(timestamps,1))
if dts == 0 && detectors(i) ~= detectors(j)
correlations(1) = correlations(1) + 2;
elseif detectors(i) ~= detectors(j)
dts = floor(dts/binning)+1;
correlations(dts) = correlations(dts) + 1;
end
j = j + 1;
if j <= size(timestamps,1)
dts = timestamps(j) - timestamps(i);
end
end
end
Matlab gives me the following error:
Error: File: correlate_xcorr.m Line: 18 Column: 17
The variable correlations in a parfor cannot be classified.
See Parallel for Loops in MATLAB, "Overview".
Line 18 is the following:
correlations(1) = correlations(1) + 2;
I can not understand why this shouldn'n be possible. The final value of correlations doesn't depend on the order in which the loop is executed, but only dts and detectors. I found similar examples in the documentation which work fine.
Why can't Matlab execute this code and how can I fix it?
I found the following solution and it seems to work. The program looks a litte different, but it has the same shape. This way Matlab is forced to think x/correlations is a reduction variable.
X = zeros(1,5);
parfor i= 1:1000
a = zeros(1,5);
dts = randi(10)-1;
if dts == 0
a(1) = (a(1) + 2);
elseif dts <= 5
a(dts) = a(dts) +1;
end
X = X + a;
end
MATLAB cannot determine that your loop is order independent because of the way you're accessing correlations(1) from multiple iterations of the PARFOR loop. It looks like this value is in some way 'special', it should work to make a 'reduction' variable, i.e. replace correlations(1) with correlations_1 or something.
The next problem you'll encounter is that you are not correctly 'slicing' the remainder of correlations. For MATLAB to analyse a PARFOR loop, it needs to be able to tell that each loop iteration is writing only to its 'slice' of the output variables. In practice, this means that you must index the outputs using literally the loop index.
More about PARFOR variable classification here: http://www.mathworks.com/help/distcomp/advanced-topics.html#bq_of7_-1
EDIT:
If you want correlations to behave strictly as a reduction variable as I believe you imply in your comments, you need to give PARFOR a clue that's what you mean: in particular, you need to add to the whole variable each time you need to. In other words, something more like:
parfor ...
dummyVec = zeros(1, N);
dummyVec(elementToIncrement) = 1;
correlations = correlations + dummyVec;
end
I agree this is non-obvious. See http://blogs.mathworks.com/cleve/2012/11/26/magic-squares-meet-supercomputing/ for more.
function [y]=AmericanPutClassic (St,t)
% S0 = 100;
K = 100;
r = 0.05;
sigma = 0.3;
T = 2;
nsteps = 5;
% St
dt = T / nsteps;
u=exp(sigma*sqrt(dt));
d=1/u;
Pu=(exp(r*dt)-d)/(u-d);
Pd=1-Pu;
if t==T
y=max(K-St,0);
return
elseif t<T
upPrice=AmericanPutClassic(St*u,t+dt);
downPrice=AmericanPutClassic(St*d,t+dt);
PrevPrice=(Pu*upPrice+Pd*downPrice)*exp(-r*dt);
if max(K-St,0) > PrevPrice
y=max(K-St,0);
else
y=PrevPrice;
end
return
end
end
I think my code does the job, but when I make 'nsteps' higher than 5, it crushes... I keep getting different errors... now it just says there is problem with my code when its higher than 5... before it would say:
"??? Maximum recursion limit of 500 reached. Use set(0,'RecursionLimit',N)
to change the limit. Be aware that exceeding your available stack space can
crash MATLAB and/or your computer."
Can anybody spot the problem?
I start by calling AmericanPutClassic(100,0)...
I don't know what you're trying to do but this problem has been described so many time on SO it's not funny anymore.
Let's paint a picture:
you start off with
dt = T / nsteps;
which is my first red flag.
then you do:
if t==T
where t = t + dt
Why is that wrong? Because of this wonderful thing called quantization. In other words there will be a time when because of super small micro difference, the result will not be correct. The bigger nsteps is, the worse this will be.
And then to add insult to injury, you put this line in
elseif t<T ... end
which means your code will skip everything, and you'll return nothing, causing your code to crash.
How does one solve this? If you can move integers around instead of floating values, you'll be in a better position. So instead of dt, you can have currentStep, such that:
function [y]=AmericanPutClassic (St, currentStep)
if nargin < 2
currentStep = 0;
end
...
if currentStep>=nsteps % if t==T
...
else
...
upPrice=AmericanPutClassic(St*u,currentStep+1);
...
end