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Homework Help: Graphs in matlab

  1. Nov 18, 2009 #1
    1. The problem statement, all variables and given/known data

    I have two functions in matlab that return y1 and y2 as the solutions and both functions depend on x.
    I plotted them on the same graph and there is a gap between the graphs as expected (i.e they are not on top of each other).
    Now, what i want to do is find the difference between these graphs.

    2. Relevant equations

    3. The attempt at a solution

    If the function for y1 is called f1 and that of y2 is called f2. i get an error if i do:

    so, how can i find the difference?

    thank you
  2. jcsd
  3. Nov 18, 2009 #2


    Staff: Mentor

    If a, f1, and f2, are all arrays of the same size, that should work. I would give a different name to a, though, something like diff.
  4. Nov 18, 2009 #3
    thats a good point and also my problem.
    I mean, if i have a function f1: y1=t^2+1, i can plot the graph like this:

    and f2 is an algorithm to approximately plot the function, as in it uses numerically methods. so, it depends on the mesh size, so if i pick an n, say n=50, i can plot the graph, and i can see that the graphs are similar.
    so the arrays of f1 and f2 wont necessarily be of the same size.
    yet i still need the difference.
    Is there a way to choose the t interval such that i can make them the same size?
  5. Nov 18, 2009 #4


    User Avatar
    Gold Member

    one way is to go back after you've made f2 and redefine f1 by redefining t:

    t= 0:1/length(f2):length(f2)

    or you may have to do:
    t = 1/length(f2):1/length(f2):length(f2)


    f1 = (some function of t)
  6. Nov 18, 2009 #5
    I have defined the function f1 as:

    which works. then i plot f2(10) where the 10 is chosen for the mesh size, which also works. then i do:
    now im getting the error:

    ??? Attempt to execute SCRIPT f1 as a function:
    Last edited: Nov 18, 2009
  7. Nov 18, 2009 #6


    Staff: Mentor

    f2(10) is just one element of your array, whereas f2 represents the entire array. Also notice that your assignment statement for t is different from what Pythagorean had. In one of his versions it was
    Code (Text):
    t= 0:1/length(f2):length(f2)
    In yours it is
    Code (Text):
  8. Nov 18, 2009 #7
    I corrected it,
    thanks for pointing it out.
  9. Nov 19, 2009 #8


    Staff: Mentor

    Is everything copacetic now?
  10. Nov 19, 2009 #9
    no, because the arrays are still different sizes.
  11. Nov 19, 2009 #10


    Staff: Mentor

    OK, here's a thought. Suppose your y1 array has 100 elements and your y2 array has 400 elements. You can create a new array for some of the y2 data, but with only 100 elements.
    Code (Text):

    for i=4:4:400
      y2_new(i/4) = y1(i)
    I think this will work. The idea is that y1(4) --> y2_new(1), y1(8) --> y2_new(2), ..., y1(400) --> y2_new(100). After that you can plot the values of y1 and y2_new, and both arrays are the same size.
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