Beginner in matlab : syntax to recup value of a line matrice - matlab

i'm beginner in MATLAB. My problem is :
When i want to take the value of the line in my matrice L(1x2) to stock it in the a and b variables, I tried to use the following code
[a,b]=L;
It didn't work well. I don't understand why it's not possible like this. I used this syntax for the line ginput(n). I used another method to take the value but I want to understand my error in the code above.
My final code is this :
clf();
n=10;
axis([0 10 0 10]);
[px,py] = ginput(n);
Y = py';
X = ones(1,n);
X=[X ; px'];
L= Y*pinv(X);
a = L(1,1);
b = L(1,2);
x = 0:0.2:10;
plot(x,b*x+a, px,py,'r+');
grid;
Thanks a lot

You cannot assign in MATLAB using
[a,b]=L
Instead, you can do the following:
L=[1,2];
L=num2cell(L);
[a,b]=deal(L{:});
Then a=1 and b=2.

Related

MATLAB: Issue with Simulink "does not support code generation"

I have something similar to the following block diagram on Simulink, which looks rather messy especially with an increasing number of blocks.
I want to replace a 3-point summing block with a function block, while keeping the same output.
First I started by placing the code inside the function block:
function y = fcn(u)
sys1 = tf(0.5,[1 0 0 4]);
sys2 = tf([3 0.5],[1 0 15]);
sys3 = tf(1,[1 1]);
y = sys1 + sys2 + sys3;
However I was greeted with an error saying that Simulink does not support code generation.
"The 'tf' class does not support code generation."
I then came across a similar problem here: https://nl.mathworks.com/matlabcentral/answers/74770-is-there-any-way-to-disable-code-generation-in-simulink
I am trying to implement an extrinsic function or 'wrapper function' with some difficulty. I created a new script called myWrapper.m, containing the same code:
function y = myWrapper(u)
sys1 = tf(0.5,[1 0 0 0 4]);
sys2 = tf([3 5],[1 0 15]);
sys3 = tf(1,[1 1]);
y = sys1 + sys2 + sys3;
and the MATLAB Function edited to:
function y1 = fcn(u1)
y1 = myWrapper(u1);
The error persists.
I somehow want to access myWrapper.m file from the MATLAB Function block. Any pointers on how this should be done? Following the previous link given and the official docs I am ending up with something like this in my MATLAB Function block:
function y1 = fcn(u1)coder.extrinsic('myWrapper')
y1 = myWrapper(u1);
The last code above is syntactically incorrect and I am at a loss on how it should be done. MATLAB automaticaly corrects the above code to:
function y1 = fcn(u1,coder,extrinsic, myWrapper )
y1 = myWrapper(u1);
which is not what I want.
Any tips and/or suggestions on how this could be done would be appreciated.
A similar question was asked on the MathWorks forum here, two years ago, with no response.
I was going about tackling this problem completely wrong. Thanks to several helpful comments I realized that in order to replace the summing block, one must NOT remove the Transfer Function blocks which feed into the summing block.
A MATLAB Function does not support code generation (and rightly so) such that a transfer function may be implemented inside it. That is why the blocks simply feed into the MATLAB Function as follows.
The script would very simply be:
function y1 = fcn(u1, u2, u3)
x = (u1 + u2 +u3);
y1 = x;
end

interval spacing on matlab

Hi I'm new to matlab and programming in general and I was wondering if anyone could tell me whats wrong with my code.
These were my instructions and values
Complete the implementation of the dampedOsc function, which evaluates and plots a graph showing the function y=e-0.8xcos 3x at a
set of points spaced at intervals of 0.1 and ranging from 0 to 3π.
The curve should be displayed using a magenta dotted line with
'pentagram' markers.
and my code:
function y = dampedOsc(x)
fplot(#(x) (exp(-0.8*x)*cos(3*x)),[0 3*pi()],'mp:')
end
All tests passed except this one:
Test 1 (Test that function has been plotted):
Actual value does not have correct size:
expectedSize =
1 95
actualSize =
1 98
Test failed.
I honestly have no idea whats gone wrong, I thought it was because of the lack
of point spaced intervals but I have no idea how to input it.
This is an interesting question, so I'll give it a stab. I would probably do something like this.
function z = dampedOsc(x)
if isvector(x) == 0
return
else
z = zeros(1,length(x));
for n = 1 : length(x)
y = exp(-0.8 .* x(n)) .* cos(3 .* x(n));
z(1,n) = y;
end
end
z
end
Then, calling that dampedOsc function from another script, should give the figure.
plot([0 : 0.1 : (3*pi)],dampedOsc(x),'mp:')
I don't have MATLAB in front of me right now, but I think this should work. If it doesn't, I can fix it. Good luck with your project!

How to resolve MATLAB trapz function error?

I am working on an assignment that requires me to use the trapz function in MATLAB in order to evaluate an integral. I believe I have written the code correctly, but the program returns answers that are wildly incorrect. I am attempting to find the integral of e^(-x^2) from 0 to 1.
x = linspace(0,1,2000);
y = zeros(1,2000);
for iCnt = 1:2000
y(iCnt) = e.^(-(x(iCnt)^2));
end
a = trapz(y);
disp(a);
This code currently returns
1.4929e+03
What am I doing incorrectly?
You need to just specify also the x values:
x = linspace(0,1,2000);
y = exp(-x.^2);
a = trapz(x,y)
a =
0.7468
More details:
First of all, in MATLAB you can use vectors to avoid for-loops for performing operation on arrays (vectors). So the whole four lines of code
y = zeros(1,2000);
for iCnt = 1:2000
y(iCnt) = exp(-(x(iCnt)^2));
end
will be translated to one line:
y = exp(-x.^2)
You defined x = linspace(0,1,2000) it means that you need to calculate the integral of the given function in range [0 1]. So there is a mistake in the way you calculate y which returns it to be in range [1 2000] and that is why you got the big number as the result.
In addition, in MATLAB you should use exp there is not function as e in MATLAB.
Also, if you plot the function in the range, you will see that the result makes sense because the whole page has an area of 1x1.

How to write many functions depending on each other in matlab

I have some functions are depending on each ther , the functions are from this book page 136 http://www.cs.helsinki.fi/u/ahyvarin/papers/bookfinal_ICA.pdf .. I functions are presented below , How to write following functions in matlab ??
y(t) = W(t-1)*x(t)
h(t) = P(t-1)*y(t)
P(t)=(1/B)*Tri[P(t-1)-m(t)*h^T(t)]
m(t) = h(t)/(B+y^T(t))*h(t))
e(t) = x(t)-W^T(t-1)*y(t)
W(t) = W(t-1) + m(t)*e^T(t)
It is solving the weight matrix W(t) iteratively .. I tried to do like this in matlab but I did not work so may be you can advice to correct the code :
for i=1:10
e=randn(3,5000);
A=[1 0 0;-0.5 0.5 0;0.3 0.1 0.1];
x=A*e;
y(t) = W(t-1)*x(t)
h(t) = P(t-1)*y(t)
P(t)=(1/B)*Tri[P(t-1)-m(t)*h^T(t)]
m(t) = h(t)/(B+y^T(t))*h(t))
e(t) = x(t)-W^T(t-1)*y(t)
W(t) = W(t-1) + m(t)*e^T(t)
end
Thanks
Ok. I can't really understand what you want, but your code shows that you don't understand some moments. I will try to clarify some moments to you:
for i = 2:10
x = rand(3);
y = W(:,:,i-1)*x;
h = P(:,:,i-1)*y;
m=h/(1+y'*h);
P(:,:,i)=P(:,:,i-1)*m*h';
e=x-W(:,:,i-1)'*y;
W(:,:,i)=W(:,:,i-1)+m*e';
end
You must go something like this: 1. you calculate x and use it to calculate other functions.
2. all of them are matrices. So you need to define it first. For example y = ones(3) etc. 3.Thats not y^T or e^T. Its transposing. If you do not feel difference it's early for you to solve this task :)
And the last: Tri function will create a some kind of problems to you, but it's defined at 136 page.
P.S. i missed beta becouse of don't know what is it :)

troubles ploting double integral matlab

I want to plot the double integral
using MATLAB for s between 0 and 1. How can I do this?
I've tried the following code but since this is my first time using MATLAB it didn't worked after two hours:
syms r x s
fun=#(r,x)(r^8*sin(x)^8)/((r*sin(x)+1)^7)+....
f=int(fun,x,0,2*pi)
q=int(f,r,0,1)
Thanks for your help!
You can use function dblquad for each value of s on interval.For example(other parts add the feature yourself):
Int = zeros(1,101);
F = #(x,r) (r.^8.*(sin(x).^8)./(r.*sin(x)+1).^7);
for s = 1:101
Int(s) = dblquad(F,0,(s-1)*0.01,0,2*pi);
end
plot(0:0.01:1,Int)