MATLAB Answers

0

Convolution of absolute magnitude =/= absolute magnitude of convolution?

Asked by Nathan Jessurun on 4 Jun 2019
Latest activity Answered by Nathan Jessurun on 14 Jun 2019
I'd like to figure out what I'm doing wrong to cause this issue.
Consider the code below:
A = [0.008912 0.0008934 -0.002584 0.003039 -0.002209 0.0007542 0.0005372 -0.0003424 -0.009147 0.03952 -0.05169 -0.07655 0.3637 -0.5018 0.1533 0.4625 -0.7134 0.3894 0.06538 -0.2249 0.1327 -0.02629 -0.006015 0.003311 0.000262 -0.0007913 0.0005139 0 -0.0004398 0.0005598 -0.0003478];
B = [-0.0017 0.002354 -0.001407 -0.0003384 0.001932 -0.002653 0.002321 -0.0002952 -0.003325 -0.01342 0.09965 -0.219 0.149 0.2581 -0.682 0.597 -0.02179 -0.4432 0.42 -0.1383 -0.04043 0.05345 -0.01858 0.002102 0 0.0003302 0 0 0 0 0];
H = [0.1986 0.4934 0.6589 0.4934 0.1986];
C = A + B*1j;
plot(abs(conv(C,H,'same')));
hold on;
plot(conv(abs(C),H,'same'));
Why are the two plots different? I thought
Am I missing something simple? Thanks for the help!

  0 Comments

Sign in to comment.

3 Answers

Answer by Rik
on 12 Jun 2019
 Accepted Answer

Let's consider a minimal example:
data=[-3 0 3];
kernel=[1 1 1]/3;
Now it is easy to see that convolving first and then taking the absolute value will do something different than doing them in reversed order.
data=[-3 0 3];
kernel=[1 1 1]/3;
conv_first=conv(data,kernel,'same');%result: [-1 0 1]
conv_first=abs(conv_first)
%conv_first=[1 0 1]
data=abs(data);
abs_first=conv(data,kernel,'same')
%abs_first=[1 2 1]
I can't pinpoint where your math is wrong, but it must be wrong somewhere. Probably the second step, as you're doing multiple operations there at once, not all of which I understand.

  0 Comments

Sign in to comment.


Answer by Shivam Sardana on 12 Jun 2019

Please look at More About section of conv function.

  0 Comments

Sign in to comment.


Answer by Nathan Jessurun on 14 Jun 2019

That makes sense, thanks. It looks like I messed up when I put the magnitude operator in the summation.

  0 Comments

Sign in to comment.