Help with the output of my function

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Hi, I have written the following function which works fine when I pass it a value such as 2. But when I say something like:
>> i = -2.5:0.1:2.5;
>> p = myfun1(i);
I get there error:
Error in myfun1 (line 3)
if abs(t) <= 1
Output argument "result" (and maybe others) not assigned during call to "/Users/mbxs3/Desktop/VCU/VCU Spring 2014/EGRE 335 - Signals and Systems/MATLAB/myfun1.m>myfun1".
Here is my function code:
function result = myfun1(t)
if abs(t) <= 1
t1 = (2-abs(2*t));
result = t1;
elseif t < -1
t2 = 0;
result = t2;
elseif t > 1
t3 = 1;
result = t3;
end

采纳的回答

Walter Roberson
Walter Roberson 2014-2-1
You should learn how to use logical indexing.
function result = myfun1(t)
result = zeros(size(t));
idx = abs(t) <= 1;
result(idx) = (2-abs(2*t(idx)));
idx = t < -1;
result(idx) = 0;
idx = t > 1
result(idx) = 1;

更多回答(3 个)

Amit
Amit 2014-2-1
编辑:Amit 2014-2-1
This is because when you input t as a vector, unless all of the values in vector satisfy the condition, it will not go through any of the if ... end statements.
When this happens, the value for result does not get declared and thats why you'll get error.
If you tell us, what you're trying to do in this code, maybe people can help you in fixing the issue

Wayne King
Wayne King 2014-2-1
编辑:Wayne King 2014-2-1
First don't use i, i the unit imaginary. That can often cause problems.
Next, you are giving it a vector input. What do you actually want to know when you write
abs(t) <=1
Do you want to know if there are ANY elements in the input that are less than 1, or whether ALL elements are?
If so use, any() and all()
if all(abs(t)<=1)
result = (2-abs(2*t));
end
or
if any(abs(t)<=1);
result = (2-abs(2*t));
end

Stephen
Stephen 2014-2-1
I want to plot the function with respect to i, from i =-2.5 to i = 2.5. The output of my function should be 0 for all values of i < -1. The output should be (2-abs(2*i)) for values of -1 < i <= 1. The output should be 1 for all values of i > than 1.
On a side note, does MATLAB recognize i and an imaginary #? Because I am used to using the variable i for iterations and what not. I use j to represent imaginary values.
  2 个评论
Amit
Amit 2014-2-1
You should try Walter's method. That will give you the result.
And using i and j are choices by the user And Wayne's comment was just a reminder to you just in case you dont know.
Walter Roberson
Walter Roberson 2014-2-1
The code I gave in my answer will output the values you would like.
By default, i and j are both initialized to sqrt(-1), as variables. You can change either or both. However, people tend to confuse "i" used as a loop variable and "i" used as sqrt(-1), so it is usually clearer to avoid using "i" as a variable.

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