I have a simulink file, where I need to call a certain optimization function:
function u = MPC(r, x0, Thorizon, Q, W, cons)
R = [];
for s=1:Thorizon+1
R = vertcat(R, r);
end
ops = sdpsettings('solver','quadprog','verbose',0);
U = sdpvar(Thorizon+1,1);
cost = (R - (Q*U+W*x0))'*(R-(Q*U+W*x0))+U'*U;
optimize(cons,cost,ops);
u = value(U);
u = u(1);
end
However, I'm not sure how to implement this in Simulink.
I tried it with a matlab function block, but a couple of problems arise.
First of all, it returns some errors like
Function 'eval' is not supported for code generation. Consider adding
coder.extrinsic('eval') at the top of the function to bypass code
generation.
or like
Function 'sdpsettings.m' (#103.12741.12757), line 355, column 12:
"temporaryOptions" Launch diagnostic report.
It has (probably) to do with the fact that the optimizer (sdp) is an external function, one that is part of a seperately installed library.
The other possible issue, is one with the input arguments.
If I were to use a Matlab function block, there would be a problem with the cons argument, one that can take inequality forms, like
U<1
The function, does however, work in the Matlab environment itself.
But the output of this function, needs to be the input of a continuous non-linearm model present in simulink.
So my question is,
are there any suggestions, to either: Make such function in Simulink work, like via a Matlab function block, or use the function in the matlab runtime environment itself, and send the output value repeatedly to the simulink model?
Related
Using Matlab, I have the following code:
s = tf('s')
K=0.5;
H= 1/(s*(s+2));
Hcl=feedback(K*H,1)
ilaplace(Hcl)
rltool(H)
I want to get the inverse laplace transform of the unity closed-loop system.
rltool(H) generates automatically the unity closed-loop system. That's why I pass the open-loop system as an argument.
When I run this code, Matlab gives an error about ilaplace:
"Check for incorrect argument data type or missing argument in call to function 'ilaplace'."
Can someone help me how to use ilaplace and rltool concurrently
I have found a solution for this problem.
syms s;
K=1/2;
H= 1/(s*(s+2))
Hcl=simplify(K*H/(1+K*H))
P=poles(Hcl)
ilaplace(Hcl)
H = syms2tf(Hcl)
rltool(H)
ilaplace works with symbolic expressions and not with a transfer function. By converting the symbolic expression to a transfer function midway of the code, I can use both functions in the same code.
Notice I've added the function simplify and changed the beginning from s=tf('s') to syms s
The function syms2tf() can be found here
I was given the following protected Simulink model:
It works manually (I click on "Run" on the Simulink GUI and it gives me the three outputs).
I am writing a script using the Matlab SIMSET tool to define the necessary variables and to call this model in order to run it:
Constant=43;
Constant1=43;
Constant2=43;
Constant7=43;
Constant10=43;
Constant11=43;
t_stop = 100;
T_s = t_stop/1000;
options = simset('solver', 'ode5', 'fixedstep', T_s);
sim('test_lau.slx',t_stop,options)
I know that the interfaces "EC" and "UL" each take for input a vector, but even if I define two vectors in my script, how do I assign them to "EC" and "UL"?
Thank you in advance for your help!
I'm trying to use the MATLAB function fzero properly but my program keeps returning an error message. This is my code (made up of two m-files):
friction_zero.m
function fric_zero = friction_zero(reynolds)
fric_zero = 0.25*power(log10(5.74/(power(reynolds,0.9))),-2);
flow.m
function f = flow(fric)
f = 1/(sqrt(fric))-1.873*log10(reynolds*sqrt(fric))-233/((reynolds*sqrt(fric))^0.9)-0.2361;
f_initial = friction_zero(power(10,4));
z = fzero(#flow,f_initial)
The goal is to return z as the root for the equation specified by f when flow.m is run.
I believe I have the correct syntax as I have spent a couple of hours online looking at examples. What happens is that it returns the following error message:
"Undefined function or variable 'fric'."
(Of course it's undefined, it's the variable I'm trying to solve!)
Can someone point out to me what I've done wrong? Thanks
EDIT
Thanks to all who helped! You have assisted me to eventually figure out my problem.
I had to add another file. Here is a full summary of the completed code with output.
friction_zero.m
function fric_zero = friction_zero(re)
fric_zero = 0.25*power(log10(5.74/(power(re,0.9))),-2); %starting value for fric
flow.m
function z = flow(fric)
re = power(10,4);
z = 1/(sqrt(fric))-1.873*log10(re*sqrt(fric))-233/((re*sqrt(fric))^0.9)-0.2361;
flow2.m
f_initial = friction_zero(re); %arbitrary starting value (Reynolds)
x = #flow;
fric_root = fzero(x,f_initial)
This returns an output of:
fric_root = 0.0235
Which seems to be the correct answer (phew!)
I realised that (1) I didn't define reynolds (which is now just re) in the right place, and (2) I was trying to do too much and thus skipped out on the line x = #flow;, for some reason when I added the extra line in, MATLAB stopped complaining. Not sure why it wouldn't have just taken #flow straight into fzero().
Once again, thanks :)
You need to make sure that f is a function in your code. This is simply an expression with reynolds being a constant when it isn't defined. As such, wrap this as an anonymous function with fric as the input variable. Also, you need to make sure the output variable from your function is z, not f. Since you're solving for fric, you don't need to specify this as the input variable into flow. Also, you need to specify f as the input into fzero, not flow. flow is the name of your main function. In addition, reynolds in flow is not defined, so I'm going to assume that it's the same as what you specified to friction_zero. With these edits, try doing this:
function z = flow()
reynolds = power(10,4);
f = #(fric) 1/(sqrt(fric))-1.873*log10(reynolds*sqrt(fric))-233/((reynolds*sqrt(fric))^0.9)-0.2361;
f_initial = friction_zero(reynolds);
z = fzero(#f, f_initial); %// You're solving for `f`, not flow. flow is your function name
The reason that you have a problem is because flow is called without argument I think. You should read a little more about matlab functions. By the way, reynolds is not defined either.
I am afraid I cannot help you completely since I have not been doing fluid mechanics. However, I can tell you about functions.
A matlab function definition looks something like this:
function x0 = f(xGuess)
a = 2;
fcn =#(t) a*t.^3+t; % t must not be an input to f.
disp(fcn);
a = 3;
disp(fcn);
x0 = fsolve(fcn1,xGuess); % x0 is calculated here
The function can then ne called as myX0 = f(myGuess). When you define a matlab function with arguments and return values, you must tell matlab what to do with them. Matlab cannot guess that. In this function you tell matlab to use xGuess as an initial guess to fsolve, when solving the anonymous function fcn. Notice also that matlab does not assume that an undefined variable is an independent variable. You need to tell matlab that now I want to create an anonymous function fcn which have an independent variable t.
Observation 1: I use .^. This is since the function will take an argument an evaluate it and this argument can also be a vector. In this particulat case I want pointwise evaluation. This is not really necessary when using fsolve but it is good practice if f is not a matrix equation, since "vectorization" is often used in matlab.
Observation 2: notice that even if a changes its value the function does not change. This is since matlab passes the value of a variable when defining a function and not the variable itself. A c programmer would say that a variable is passed by its value and not by a pointer. This means that fcn is really defined as fcn = #(x) 2*t.^3+t;. Using the variable a is just a conveniance (constants can may also be complicated to find, but when found they are just a value).
Armed with this knowledge, you should be able to tackle the problem in front of you. Also, the recursive call to flow in your function will eventuallt cause a crash. When you write a function that calls itself like this you must have a stopping criterium, something to tell the program when to stop. As it is now, flow will call ifself in the last row, like z = fzero(#flow,f_initial) for 500 times and then crash. Alos it is possible as well to define functions with zero inputs:
function plancksConstant = h()
plancksConstant = 6.62606957e−34;
Where the call h or h() will return Plancks constant.
Good luck!
I'm trying to simulate a very simple model using an embedded matlab function that takes the input and add's 10 to the value using a constant block that inputs into the matlab function, which then outputs to a display block.
As soon as I press simulate I get an abundance of errors. First I get a huge paragraph in orange text stating a warning out the solver 'variableStepDiscrete' instead of solver 'ode45'
Here is the remaining lines that are echo'd from the command prompt:
Code Directory :
"/Users/dazgti/Documents/MATLAB/slprj/_sfprj/embeddedFunction/_self/sfun/src"
Machine (#32): "embeddedFunction" Target : "sfun"
Chart "MATLAB Function" (#49):
.
"c2_embeddedFunction.h"
"c2_embeddedFunction.c"
"embeddedFunction_sfun.h"
"embeddedFunction_sfun.c"
"embeddedFunction_sfun_debug_macros.h"
Interface and Support files:
"embeddedFunction_sfun_registry.c"
Code generation failed Attempt to execute SCRIPT union as a function:
/Users/dazgti/Documents/MATLAB/union.m
I have a script file within my matlab directory called union.m, but I have no idea why its mentioning it.
function y = fcn(u)
%#codegen
x = u + 10;
y = x;
MATLAB Function block works by generating "C" code for the MATLAB code you entered in the block. In the process of generating code there could have been a call to union function in MATLAB from MATLAB Function block infrastructure. Since you have overridden the union function instead of the built-in function MATLAB might have attempted to call your script which caused the error. It is better to avoid naming your functions same as MATLAB built-in functions.
I'm writing a program in MATLAB to solve integrals, and I have my function in a .M-file. Now I wonder how I can write a program in the .MAT-file that lets the user set a value that exists in the both files. The .M-file looks like this:
function fh = f(y)
fh = 62.5.*(b-y).*(40-20.*exp(-(0.01.*y).*(0.01.*y)));
and as you can see, the function depends on two variables, y and b. I want the user to set b. I tried putting b = input('Type in the value of b: ') in the .M-file but for some reason the user would then have to put in the same value four times.
Can I ask for the value of b in the .MAT-file?
Firstly, m-files store code (i.e. functions), while MAT-files store data (i.e. variables). You can save workspace variables to a MAT-file using the function SAVE and load them into a workspace from a file using the function LOAD. If you have a user choose a value for b, then save it to a MAT-file ('b_value.mat', for example), you can simply load the value from the MAT-file inside your m-file function like so:
function fh = f(y)
load('b_value.mat','b');
fh = 62.5.*(b-y).*(40-20.*exp(-(0.01.*y).*(0.01.*y)));
However, this is not a very good way to handle the larger problem I think you are having. It requires that you hardcode the name of the MAT-file in your function f, plus it will give you an error if the file doesn't exist or if b isn't present in the file.
Let's address what I think the larger underlying problem is, and how to better approach a solution...
You mention that you are solving integrals, and that probably means you are performing numerical integration using one or more of the various built-in integration functions, such as QUAD. As you've noticed, using these functions requires you to supply a function for the integrand which accepts a single vector argument and returns a single vector argument.
In your case, you have other additional parameters you want to pass to the function, which is complicated by the fact that the integration functions only accept integrand functions with a single input argument. There is actually a link in the documentation for QUAD (and the other integration functions) that shows you a couple of ways you can parameterize the integrand function without adding extra input arguments by using either nested functions or anonymous functions.
As an example, I'll show you how you can do this by writing f as an anonymous function instead of an m-file function. First you would have the user choose the parameter b, then you would construct your anonymous function as follows:
b = input('Type in the value of b: ');
f = #(y) 62.5.*(b-y).*(40-20.*exp(-(0.01.*y).^2));
Note that the value of b used by the anonymous function will be fixed at what it was at the time that the function was created. If b is later changed, you would need to re-make your anonymous function so that it uses the new value.
And here's an example of how to use f in a call to QUAD:
q = quad(f,lowerLimit,upperLimit);
In your m file declare b as a global
function fh = f(y)
global b
fh = 62.5.(b-y).(40-20.*exp(-(0.01.y).(0.01.*y)));
This allows the variable to be accessed from another file without having to create another function to set the value of b. You could also add b to the arguments of your fh function.