Problems with Plotting Matlab Function

Question

I am Beginner in Matlab, i would like to plot system concentration vs time plot at a certain time interval following is the code that i have written

%Input function of 9 samples with activity and time calibrated with Well %counter value approx : 1.856 from all 9 input values of 3 patients

function c_o = Sample_function(td,t_max,A,B)

   t   =(0 : 100  :5000); % time of the sample post injection in mins
   c   =(0 : 2275.3 :113765);

   A_max= max(c);   %Max value of Concentration (Peak of the curve)

   if (t >=0 && t <= td)
      c_o(t)=0;
   else if(td <=t && t<=t_max)
      c_o(t)= A_max*(t-td);
   else if(t >= t_max)
      c_o(t)=(A(1)*exp(-B(1)*(t-t_max)))+(A(2)*exp(-B(2)*(t- t_max)))+...
             (A(3)*exp(-B(3)*(t-t_max)));
   end

   fprintf('plotting Data ...\n');
   hold on;
   figure;
   plot(c_o);
   xlabel('Activity of the sample Ba/ml ');
   ylabel('time of the sample in minutes');
   title (' Input function: Activity sample VS time ');
   pause;
   end

I am getting following error

Operands to the || and && operators must be convertible to logical scalar values.

Error in Sample_function (line 18)
if (t >=0 && t <= td)

Kindly .Let me know if my logic is incorrect

See what I meant in my comment to your previous question? In this case, other people have taken the time to edit your question and format/indent your code properly, so that everyone can read it. For subsequent questions, please consider doing the formatting yourself to save other people's time. Oh and thanks for caring about my students :-) — Luis Mendo
– Luis Mendo, Commented Feb 16, 2014 at 3:21

CodeMonkey · Accepted Answer · 2014-02-14 17:48:50Z

2

Your t is not a single value to compare with 0 so it cannot evaluate to true or false.

answered Feb 14, 2014 at 17:48

CodeMonkey

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Comments

craigim · Accepted Answer · 2014-02-14 19:08:38Z

1

You want to do this with logical indexing

c_o = zeros(size(t));

c_o(t>=0 & t<=td) = 0; % this line is actually redundant and unnecessary since we initialized the vector to zeros
c_o(t>td & t<=t_max) = A_max*(t(t>td & t<=t_max)-td);
c_o(t>t_max) = (A(1)*exp(-B(1)*(t(t>t_max)-t_max)))+(A(2)*exp(-B(2)*(t(t>t_max)- t_max)))...
      +  (A(3)*exp(-B(3)*(t(t>t_max)-t_max)));

You could also make this a little prettier (and easier to read) by assigning the logical indexes to variables:

reg1 = (t>=0 & t<=td);
reg2 = (t>td & t<=t_max);
reg3 = (t>t_max);

Then, for instance, the second assignment becomes the much more readable:

c_o(reg2) = A_max*(t(reg2)-td);

answered Feb 14, 2014 at 19:08

craigim

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Comments

Ankush · Accepted Answer · 2014-02-14 17:53:52Z

1

t is written as a array of numbers. So, it can't be compared with a scalar value ex. 0. Try it in a for loop

for i=1:length(t)
   if (t(i) >=0 && t(i) <= td)
      c_o(t(i))=0;
   else if(td <=t(i) && t(i)<=t_max)
      c_o(t(i)))= A_max*(t(i)-td);
   else if(t(i) >= t_max)
      c_o(t)=(A(1)*exp(-B(1)*(t(i)-t_max)))+(A(2)*exp(-B(2)*(t(i)- t_max)))...
      +  (A(3)*exp(-B(3)*(t(i)-t_max)));

   end
end

answered Feb 14, 2014 at 17:53

Ankush

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