comparison of loop and vectorization in matlab

Question

let us consider following code for impulse function

function y=impulse_function(n);
y=0;
if n==0
    y=1;
end
end

this code

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Answer 1

In the first case you are comparing an array to the value 0. This will give the result [0 0 1 0 0], which is not a simple true or false. So the statement y = 0; will not get executed and f will be [0 0 0 0 0] as shown.

In the second you are iterating through the array value by value and passing it to the function. Since the array contains the value 0, then you will get 1 back from the function in the print out of f (or [0 0 1 0 0], which is an impulse).

You'll need to modify your function to take array inputs.

Perhaps this example will clarify the issue further:

cond = 0;
if cond == 0
    disp(cond) % This will print 0 since 0 == 0
end

cond = 1;
if cond == 0
    disp(cond) % This won't print since since 1 ~= 0 (not equal)
end

cond = [-2 -1 0 1 2];
if cond == 0
    disp(cond) % This won't print since since [-2 -1 0 1 2] ~= 0 (not equal)
end

Answer 2

Your function is not defined to handle vector input.

Modify your impluse function as follows:

function y=impulse_function(n)
    [a b]=size(n);
    y=zeros(a,b);
    y(n==0)=1;
end

In your definition of impulse_function, whole array is compared to zero and return value is only a single number instead of a vector.

Answer 3

You can write an even simpler function:

function y=impulse_function(n);
y = n==0;

Note that this will return y as a type logical array but that should not affect later numerical computations.

Answer 4

You could define your impulse function simply as this one -

impulse_function = @(n) (1:numel(n)).*n==0

Sample run -

>> n = -6:4
n =
    -6    -5    -4    -3    -2    -1     0     1     2     3     4
>> out = impulse_function(n)
out =
     0     0     0     0     0     0     1     0     0     0     0

Plot code -

plot(n,out,'o')    %// data points
hold on
line([0 0],[1 0])  %// impulse point

Plot result -