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Homework Help: Defining a function in matlab

  1. Jan 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Ok so Im slowly learning the syntax and various functions of Matlab. I am triyng to design and plot an FIR bandpass filter. Below is a coefficent h(k). I need to define this h(k) in 3 parts from what I understand. for k=0:p-1, k=p, and k=p+1:m to satisfy.

    [itex]h(k)= \frac{sin(2 \pi(k-p)F_{1}T)-sin(2 \pi(k-p)F_{0}T)}{\pi(k-p)} [/itex]for o≤k≤m,k≠0
    [itex]h(k)=2(F_{1}-F_{0})T [/itex] for k=p
    [/itex] otherwise

    2. Relevant equations


    3. The attempt at a solution
    This is what I have

    This not fully correct. What I dont understand is how to put these ranges for the variable k within each string? Above I am defining it for the entire simulation. I dont even know how to properly search for this in google as I dont know what it would be called. I cant image it is too complicated but I just dont know the syntax. Ive tried a couple of different ways but nothing really worked correctly.

    Any help would be much appreciated!
  2. jcsd
  3. Jan 26, 2012 #2
    Now looking at what ive written maybe I wouldn't call it defining a function in matlab since this produces something different then I need.

    I just need to defin this h(k) at certain ranges. I would call h(k) a function this is why I used the word.

    Im still stuck on this. Im trying to google it but i dont know what to search for...
  4. Jan 26, 2012 #3

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    Hi Evo8! :smile:

    I do not know or have Matlab, but as a software engineer I can hazard a few educated guesses.

    Your attempt probably won't work, since your expression is not defined for k=p, which is included in the range 0:m.

    I'd try something like this:

    In my suggestion k is repeatedly used as a range variable and h is an array that is indexed using a range variable.
  5. Jan 26, 2012 #4


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    You can break the function up manually (like ILSe suggested), use if statements, or use conditionals like this:
    Code (Text):

    k = 0:m;
    h = ((0<k) && (k<=(p-1)) .* (sin(2*pi*(k-p).*f1)-sin(2*pi*(k-p).*f0.*T))./(pi*(k-p)) + ...
        (k==p) * 2 *(f1-f0).*T + 0;
    MATLAB will evaluate the conditions as 1 or 0, so if the conditional holds the term remains and if not it's multiplied by zero.
  6. Feb 1, 2012 #5

    Thanks for your response. I understand how you have everything written there and this may work. I will give something like this a try however i feel Matlab wont like it because the range of k is defined in two different areas. Maybe there is a hierarchy in matlab that i dont really understand. I will have to play with this. h(k) is also defined twice so it may not like this.

    jhae2.718, Thanks for your response as well. I am a little confused by some of the operators and syntax used in what youve posted below.

    First could you explain the use of? $$.*$$

    From looking this up I see that it is the element wise multiplication of arrays. I dont see why you have the first one in there?

    Also the
    These are like an and operator right? Why two of them?

    I also cant seem to find any information on the double equals sign.

    Last question is why the +0 at the end? Sorry for all of the questions but im trying to understand whats going on in the code.

    I appreciate the help!
  7. Feb 1, 2012 #6
    In addition Matlab doesnt seem to like the expression

    I get an error saying that the index must be a positive integer or logical. I see that this expression was supposed to satisfy my 0 otherwise statement.....
  8. Feb 1, 2012 #7

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    I expect that it will work, but please try it.

    k is a variable that can be assigned different values.
    If you assign it a different range (a range counts as a value), the value of k changes to this new range that has effect from then onward.

    h is an array that contains multiple values.
    You can access a single value in that array with for instance h(1).
    Or you can access a subset of values in the array with for instance h(1:3).
    In both cases the part of the array that is not accessed will remain unchanged.
  9. Feb 1, 2012 #8

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    Try h(0)=0 instead.
    You're assigning the variable "o" (small letter 'o'), but that variable is not defined.
    I intended the integer zero (0).
  10. Feb 1, 2012 #9
    opps! That was a typo here in the form but not in matlab. Sorry about that I did infact use h(0)=0 in my code. I just tried again and I get the same error. "Subscript indices must either be real positive integers or logicals."
  11. Feb 1, 2012 #10
    Should it be something like;

    This is an extra line of code but the same thing essentially.

    Edit: nevermind i just tried it and it does the same thing.

    I also found out what the == means. Its a logical if statement. Let me play with the code a little more and maybe i can figure it out. ill report back
    Last edited: Feb 1, 2012
  12. Feb 1, 2012 #11

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    Okay, the error means that the array h can not be indexed with position 0.
    Apparently matlab has arrays that start at the index 1.
    So just loose the h(0)=0 statement and you're done.
    Index position 0 is not relevant anyway.
  13. Feb 1, 2012 #12

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    "=" would mean assignment, where you assign an expression to a variable.
    "==" means a check on equality, where you compare 2 expressions and check whether they are equal or not.
  14. Feb 1, 2012 #13
    Ok this seems to work. I see what you mean with indexing at 0.

    I omitted this line and it seems to work. Now I just need to figure out why my plots looks weird. Forming this array was just part of the lab. A little more finagling and well see what I get.

    Thanks again for all of the help!
  15. Feb 1, 2012 #14


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    The dot does indeed indicate element wise operations. As for the first one, I have a habit of being overly explicit when doing element wise operations.

    In MATLAB, & and && are both symbols used for a logical AND. Using && results in MATLAB not bothering to check the second condition if the first is false.

    As it's been said, == is a comparison and = is for assignment.

    The + 0 isn't necessary, but just explicitly shows that the value is zero if the other conditions don't apply.
  16. Feb 5, 2012 #15
    Thank you for the clarification. I appreciate all of the help!
  17. Feb 5, 2012 #16

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    So do your plots look good now? :wink:
  18. Feb 5, 2012 #17
    They do indeed! Thanks again!
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