function r = fun(x, params, type)
h = #(x) x(1)-x(8);
x= sym('x',[1,8]);
gamma = 16;
if type == 'linear'
r = gamma*h(x);
elseif type == 'sin'
r = gamma*sin(h(x));
end
end
When I run this function for type 'sin' I always get this error
Arrays have incompatible sizes for this operation.
Error in fun (line 7)
if type == 'linear'
How to fix this?
I just one to pass type and depending on that create my ouput, I though string could be okay, but it doesn't work.
I want to multiply my function handle with gamma in case of linear and multiply and take sin in case of sine.
You may mean
if strcmp(type,'linear')
type is a char array, and matlab does not support that type of direct array to array comparison for text. It tries to compare each letter with each other, but if length(type) is not the same as length('linear') it errors.
Related
I have written a function called "tension.m" in which I have used if else condition as shown below.
function [T,T_earlyvalues,T_latervalues] = tension(u,sigma,G,N,K)
%the values of sigma,G,N,K can be taken arbitrary.
sigma=2; G=3;N=8;K=1; v=1;
w=2.2;
if u<w
T =v*sqrt(sigma+G^2/(N-K));
T_earlyvalues=T;
else
T=(2*v)*sqrt(sigma+G^2/(N+K));
T_latervalues=T;
end
Now in another script "myvalues.m" I need to call T_earlyvalues and T_latervalues separately.
%have some coding before this part
sigma0=2400; lambda=1.3; v=2; sigma=2; G=3;N=8;K=1;
u=0:0.01:5;
T=tension(u,sigma,G,N,K);
T_earlyvalues=tension(u,sigma,G,N,K);
T_latervalues=tension(u,sigma,G,N,K);
deltaA=T_earlyvalues*sigma0*pi;
deltaB=T_latervalue*lambda*pi/2;
%have some coding after this part
How could I call the said values which are under if-else statement from tension.m function to myvalues.m script?
You have defined the tension function such that it returns three outputs.
If you call that function by requiring only one output, the function returns the first value, in your case, T
This implies that
T=tension(u,sigma,G,N,K);
shoud work since T is the first output parameter
T_earlyvalues=tension(u,sigma,G,N,K);
T_latervalues=tension(u,sigma,G,N,K);
are not working, since, actually tension returns the first value (T, whjikle you are expecting the second and the third respectively.)
You can cahnge the two above calls this way:
[~,T_earlyvalues,~]=tension(u,sigma,G,N,K);
[~,~,T_latervalues]=tension(u,sigma,G,N,K);
The ~ allows to avoid the function return the output paraemter.
You can find additional information here
Notice that in your function T_earlyvalue is not set in the else block, same for T_latervalue in the if block.
This will generate an error such as
Output argument T_earlyvalue (and maybe others) not assigned during call to tension
or
Output argument T_latervalues (and maybe others) not assigned during call to tension
You can initialize the output values to default values, at the beginning of the function, for example:
T=NaN
T_earlyvalue=NaN
T_latervalues=NaN
You can then use these special values (or any other you want to use) to trace, for example, if the if block has been executed or the else.
There seem to be a number of issues here, not the least of which is some confusion about how output argument lists work when defining or calling functions. I suggest starting with this documentation to better understand how to create and call functions. However, this issue is somewhat moot because the larger problem is how you are using your conditional statement...
You are trying to pass a vector u to your function tension, and from what I can tell you want to return a vector T, where the values of T for u < w are computed with a different formula than the values of T for u >= w. Your conditional statement will not accomplish this for you. Instead, you will want to use logical indexing to write your function like so:
function [T, index] = tension(u, sigma, G, N, K)
T = zeros(size(u)); % Initialize T to a vector of zeroes
w = 2.2;
index = (u < w); % A logical vector, with true where u < w, false where u >= w
T(index) = u(index)*v*sqrt(sigma+G^2/(N-K)); % Formula for u < w
T(~index) = 2*(u(~index)-v)*sqrt(sigma+G^2/(N+K)); % Formula for u >= w
end
Now you can call this function, capturing the second output argument to use for identifying "early" versus "late" values:
sigma0 = 2400; lambda = 1.3; v = 2; sigma = 2; G = 3; N = 8; K = 1;
u = 0:0.01:5;
[T, earlyIndex] = tension(u, sigma, G, N, K); % Call function
T_earlyvalues = T(earlyIndex); % Use logical index to get early T values
T_latervalues = T(~earlyIndex); % Use negated logical index to get later T values
And you can then use the subvectors T_earlyvalues and T_latervalues however you like.
I want to evaluate the simple example of integral
a = max(solve(x^3 - 2*x^2 + x ==0 , x));
fun = #(x) exp(-x.^2).*log(x).^2;
q = integral(fun,0,a)
and the error is
Error using integral (line 85)
A and B must be floating-point scalars.
Any tips? The lower limit of my integral must be a function, not a number.
The Matlab command solve returns symbolic result. integral accepts only numeric input. Use double to convert symbolic to numeric. As your code is written now, already max should throw an error due to symbolic input. The following works.
syms x;
a = max(double(solve(x^3 - 2*x^2 + x)));
fun = #(x) exp(-x.^2).*log(x).^2;
q = integral(fun,0,a)
Output: 1.9331.
the lower limit of my integral must be a function, not a number
integral is a numeric integration routine; the limits of integration must be numeric.
Check values of a by mouse over in breakpoint or removing the ; from the end of the line so it prints a. Based on the error, a is not a scalar float. You might need another max() or double() statement to transform the vector to a single value.
Solve Help : http://www.mathworks.com/help/symbolic/solve.html
Integral Help : http://www.mathworks.com/help/ref/integral.html
Im new to MATLAB. Want to use integral2 as follows
function num = numer(x)
fun=#(p,w) prod((p+1-p).*(1-w).*exp(w.*x.*x/2))
num= integral2(fun ,0,1,0,1)
end
I get several errors starting with
Error using .*
Matrix dimensions must agree.
Error in numer>#(p,w)prod(p+(1-w).*exp(w.*x.*x/2)) (line 5)
fun=#(p,w) prod(p+(1-w).*exp(w.*x.*x/2))
Can you please tell me what I do wrong.
Thanks
From the help for integral2:
All input functions must accept arrays as input and operate
elementwise. The function Z = FUN(X,Y) must accept arrays X and Y of
the same size and return an array of corresponding values.
When x was non-scalar, your function fun did not do this. By wrapping everything in prod, the function always returned a scalar. Assuming that your prod is in the right place to begin with and taking advantage of the properties of the exponential, I believe this version will do what you need for vector x:
x = [0 1];
lx = length(x);
fun = #(p,w)(p+1-p).^lx.*(1-w).^lx.*exp(w).^sum(x.*x/2);
num = integral2(fun,0,1,0,1)
Alternatively, fun = #(p,w)(p+1-p).^lx.*(1-w).^lx.*exp(sum(x.*x/2)).^w; could be used.
I have the following matlab code
v is an array of one dimention.
v = getV(x,y,z);
if isempty(v)
fail_code = 1;
elseif max(v) <= vmax % <============== error is here
fail_code = 0;
vplus = max(v);
else
vplus = vmax;
end
this working fine, however when I try to convert it into a c code in matlab coder I get the following error:
Expected a scalar. Non-scalars are not supported in IF or WHILE statements, or with logical operators. Instead, use ALL to convert matrix logicals to their scalar equivalents.
I am not fully familiar with matlab data types, maybe why I am missing something.
I would put:
elseif all(max(v) <= vmax)
or
elseif max(v(:)) <= vmax
In MATLAB, if test can pass even if test is not a scalar. If test is an array of logicals, it will pass if all the elements are non-zero.
However, that is not supported by MATLAB Coder when converting to C. So, you would need to explicitly ensure that you get a scalar, either by inserting an all, or comparing v to its maximum as a vector.
Why not follow the advice within the error-message and try:
elseif all(max(v) <= vmax)
In your special case the all() might be superfluous, but I assume that the coder tries to respect the possibility that the comparison could in principle result in an array.
You probably want to test for v being a vector with isvector.
if isvector(v)
%true case
vplus = max(v); % returns a scalar
vplus(vplus>= vmax) = vmax;
else
%false case (matrix)
error('something wrong - v dimension');
end
the dimension get mixed up, the compiler telling me to use all function, however
I did the following, and it was the fix
v = v(:);
before passing it to max, and all solved
I got this problem since I have a variable with flexible size as
if a==b
c = 1;
else
c = [1,1]
end
To solve this problem, I have to define c in advance as [0,0].
It seems that in Matlab Coder the flexible size is not supported.
i'm stuck with this error:
In an assignment A(I) = B, the number of elements in B and I must be the same.
yres(1)=((u - uc).^2) + ((y - yc).^2) -(d.^2);
i don't understand, why this won't get a skalar?since the elements are all scalar. what should be changed to get a scalar?
best regards
edit: thanks sloede, all inputs are scalar, but i still get this error
In an assignment A(I) = B, the number of elements in B and I must be the
same.
Error in myfun (line 7)
yres(1)=sqrt(((u - uc).^2) + ((y - yc).^2) ) -d;
Error in fsolve (line 241)
fuser = feval(funfcn{3},x,varargin{:});
Error in modfsolve (line 26)
x= fsolve(#myfun,x0,options,uc,d,spacing_amplitude,spacing_width);
Caused by:
Failure in initial user-supplied objective function evaluation. FSOLVE
cannot continue.*
The "." before an operator means that the following operation should be applied element-wise and not on the vector as a whole. Thus
a = b.^2
will give you as a result all elements of b squared and saved back to a. Therefore, in your code statement above, if any of u, uc, y, yc, d are not scalar but a vector, your result will be a vector as well.
Otherwise there seems to be nothing wrong with your code.
read the documentation of fsolve: http://www.mathworks.nl/help/toolbox/optim/ug/fsolve.html
it states:
fun
The nonlinear system of equations to solve. fun is a function that accepts a vector x and returns a vector F, the nonlinear equations evaluated at x.
Obviously your function myfun doesn't handle vector input.
You can solve this by adding the following construction inside your function (and of course change it to your needs/your parameters):
function out = myfun(in)
if ~isscalar(in)
% assuming it's a matrix or vector
out = reshape(arrayfun(#myfun,in(:)),size(in));
else
% your actual function execution statements
out = dostuffon(in);
end
end
or properly vectorize your function (if that's possible)