I can't seem to find the error in my program.

My program looks like this, its supposed to take the input signal and output signal after convolution and graph them versus a variable.

function graphconv(~)  
L=200;
K=50;
n=0:L-1;
z = double(rem(n,K) < K/2);
h=@(n) 1/15 .* n>=0&&n<=14;
y = conv(h,z);
plot(n,z,'r',n,y,'b')

but i keep getting these errors and i can't seem to understand what they mean. I already tried googling it but to no avail.

ERROR CODES:

>> graphconv

Input arguments to function include colon operator. To input the colon character,
use ':' instead.

Error in conv (line 43)
c = conv2(a(:),b(:),shape);

Error in graphconv (line 11)
y = conv(h,z);

Answer 1

[Star Strider]

You are passing a function handle as an argument to conv. That is throwing the error.

Pass the function with an argument (so it returns a value) instead:

h=@(n) 1/15 .* ((n>=0) & (n<=14));
y = conv(h(n),z, 'same');

Use only one ‘&’ in ‘h’ to test vectors. (Using ‘&&’ requires logical arguments, and you have a vector of numeric arguments.) I added the 'same' argument because that then works in your plot call without problems. If you want a different plot, you will have to call it with either one argument (probably ‘z’), or two equal-length arguments.

[Santos Tapia]

This response was really close to what I was looking for but it was cut short. So perhaps im missing something in my range of n.

[Star Strider]

If you want the x-axis to go to 200, ‘L’ must be 200.

This slight revision should do what you want:

L=400;
K=50;
n=0:L-1;
z = double(rem(n,K) < K/2);

h=@(n) 1/15 .* ((n>=0) & (n<=14));
y = conv(h(n),z, 'same');

h=@(n) 1/15 .* ((n>=0) & (n<=14));
y = conv(h(n),z, 'same');

plot(n,z,'--r', n,y,'-b')
axis([0  L/2    -0.5  +1.5])
legend('Input','Output')

All I added here was to increase ‘L’, add an axis call after the plot call, and the legend call. The code I posted earlier is unchanged.

[Santos Tapia]

This one! Thank you so much, I have some questions if you don't mind answering them.

Why did you post the function h and convolution function twice does it change something or was it a typo?

What is a function handle and why did it mess up the program is there a more intuitive way at looking at it?

If I wanted to use a variable in the function say like "10^n" is it as simple and replacing the "1/15" with it?

[Star Strider]

As always, my pleasure.

It was a typo.

A function handle is the way MATLAB refers to functions in some contexts. You can call a function by calling it with an argument, for example, sin(x), and here h(n). If you want to pass a function reference as an argument to a function, it is necessary to pass it as a function handle, with the function name preceded by an ‘@’ sign. Here, ‘h’ is a function handle by definition. The difference between your original code and my revision is that you passed the function handle ‘h’ to conv, and I passed the function call h(n) to conv.

The documentation explains it much better than I can, so I refer you to Function Basics (link) and Anonymous Functions (link) for an extended discussion.

I apologise for the delay in responding.

[Santos Tapia]

No need to apologize, thank you very much you've been a great deal of help and have helped alleviate any confusion I previously had!

[Star Strider]

As always, my pleasure!

Answer 2

[Walter Roberson]

What it is really getting at is that you cannot conv() a function handle on a matrix. The firs two arguments to conv must be numeric vectors or arrays.

It appears to me that you probably want

conv(z, ones(1,15)/15)

and that possibly you might want to add the 'same' or 'valid' option in the third parameter.

[Santos Tapia]

Do i eliminate the lines "h=...." and "y=conv(.." and replace it with that?

[Walter Roberson]

Yes,

y = conv(z, ones(1,15)/15);

or more likely,

 y = conv(z, ones(1,15)/15, 'same');

However you should also be looking at filter() instead, as filter() gives more control over boundary conditions.

[Santos Tapia]

I tried the suggested code but the output isn't quite correct perhaps it might be some of the parameters I set but i'm not sure

[Walter Roberson]

   y = conv(z, [ones(1,15)/15, zeros(1, L-15)], 'same');

Answer 3

[Image Analyst]

h needs to be an actual numerical array, not a function defnintion. Try

h = 1/15 .* n>=0&&n<=14;

instead.

[Santos Tapia]

I end up getting an error like this:

>> graphconv

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

Error in graphconv (line 9) h=1/15 .* n>=0&&n<=14;

[Santos Tapia]

I'm attempting to recreate this function on matlab but I can't find out how to do it properly.

[Walter Roberson]

h = @(n) 1/15 .* (n>=0 & n<=14);

However, you still cannot convolve with that.