User defined function to output matrix

I am in need of some help with a user defined function. I have it so it will output a column array of zeros however I need to function to fill the array dependant on a number of conditions and I have no idea how to get from where I am to where I need to be.

function [Q] = O_R_B(H)
%Operating rule B for varying dam heights
%   Determining the usage rates for the dam for different dam heights
H_max = 30.9; %dam height in m (scaled 1:300)
Q_out = 138; %normal output rate
Q = zeros(length(H),1);
for k = 1:length(H)
if H >= 1*H_max;
    Q = string('issue warning');
elseif H >= 0.9*H_max;
    Q(k) = 4 * Q_out;
elseif H < 0.9*H_max & H>= 0.85*H_max;
    Q(k) = 2* Q_out;
elseif H < 0.85*H_max & H>= 0.4*H_max;
    Q(k) = Q_out;
elseif H < 0.4*H_max & H>= 0.2*H_max;
    Q(k) = 0.5*Q_out;
else H< 0.2*H_max;
    Q(k) = 0;
end    
end
end

Answer 1

[James Tursa]

You need to index H also in your code. E.g.,

if H(k) >= 1*H_max;
    error('H too large');
elseif H(k) >= 0.9*H_max;
    Q(k) = 4 * Q_out;
        :
       etc

[Rhys Webb]

thank you this has gotten me closer to the answer however it is only outputting values in the first and last location (for testing purposes the range I am using is 10:10:30) and the centre value is being produced as a 0. Thank you for your assistance with this.

ans =

    69
     0
   552

[James Tursa]

Please post your current code. Did you change both of the H's into H(k)'s in the middle tests?

Also, since these are scalar tests you are doing, it would be more appropriate to use the && operator instead of the & operator.

[Rhys Webb]

sorry should have made sure to post my code in my previous comment.

function [Q] = O_R_B(H)
%Operating rule B for varying dam heights
%   Determining the usage rates for the dam for different dam heights
H_max = 30.9; %dam height in m (scaled 1:300)
Q_out = 138; %normal output rate
Q = zeros(length(H),1);
for k = 1:length(H)
if H(k) >= 1*H_max;
      Q = string('issue warning');
elseif H(k) >= 0.9*H_max;
    Q(k) = 4 * Q_out;
elseif H(k) < 0.9*H_max & H>= 0.85*H_max;
    Q(k) = 2* Q_out;
elseif H(k) < 0.85*H_max & H>= 0.4*H_max;
    Q(k) = Q_out;
elseif H(k) < 0.4*H_max & H>= 0.2*H_max;
    Q(k) = 0.5*Q_out;
else H(k) < 0.2*H_max;
    Q(k) = 0;
end    
end
end

This is where I am at currently. The other thing I also need to achieve is that if a value greater than H_max occurs i need an output of issue warning in the matrix as well and I have no clue on how to achieve this.

[James Tursa]

You didn't change both of the H's to H(k)'s. E.g., this line

elseif H(k) < 0.9*H_max & H>= 0.85*H_max;

should be this instead:

elseif H(k) < 0.9*H_max & H(k) >= 0.85*H_max

And, to simplify things, since your tests cover the entire range of 0 to H_max, you don't even need the < part of the tests. You know you will pass that part of the test guaranteed because you just failed the previous test. So you only need the >= part of the test. E.g.,

if H(k) >= 1*H_max;
    Q = string('issue warning');  % <-- YOU NEED TO CHANGE THIS!
elseif H(k) >= 0.9*H_max;
    Q(k) = 4 * Q_out;
elseif H(k) >= 0.85*H_max;
    Q(k) = 2* Q_out;
elseif H(k) >= 0.4*H_max;
    Q(k) = Q_out;
elseif H(k) >= 0.2*H_max;
    Q(k) = 0.5*Q_out;
else 
    Q(k) = 0;
end

You need to change your warning line. You can't mix a string in with your double data. I would either put a NaN there or maybe just error() the function at that point.

[Rhys Webb]

You sir are a life saver I have been looking at this for 2 days trying to figure it out.

[James Tursa]

You could issue a warning and fill the element with NaN. E.g.,

if H(k) >= 1*H_max;
    Q(k) = NaN;
    warning(sprintf('H(%d) element too large',k));