1
$\begingroup$

Problem/Question

After taking the convolution of two discrete sine waves of the same frequency and length the result appears to be a sinc function but I was expecting to get a sine wave since this should be the same as multiplying two impulses in the frequency domain right? There is a similar question posted here -> What is convolution of two sine waves (tones)? but the answers there suggest that the result should have been a sine wave since both have the same frequency and length. But that doesn't seem to be the case when I actually perform the operation. Can anyone help me understand why this isn't acting as a simple low pass filter allowing the 2nd sine function through unchanged?

MATLAB steps

t = 0:20
x1 = sin((2 * pi * t) / 21)
x2 = 100*sin((2 * pi * t) / 21)
y = conv(x1,x2)
plot(y)

enter image description here

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Hint: re-run your code with t=0:200. What do you see? $\endgroup$
    – jojeck ♦
    Commented Sep 1, 2023 at 19:56
  • 1
    $\begingroup$ @jojeck Thanks. I see sine wave with some tremelo. Are you saying what I'm seeing is not really a sinc but a piece of a shifted version of a sine wave? $\endgroup$
    – Usman Mutawakil
    Commented Sep 1, 2023 at 20:02
  • $\begingroup$ While we know that the premise to this question is false, an interesting question is, to me, why the convolution of two identical sinc functions is another identical sinc function. It's easy to see in the frequency domain. $\endgroup$
    – robert bristow-johnson
    Commented Sep 2, 2023 at 17:17

1 Answer 1

Reset to default
17
$\begingroup$

You are not convolving two sine waves, you are convolving two short snippets of sine waves.

A snippet of a sine wave is the same as a real (infinite) sine wave multiplied with a rectangular window. The convolution of two rectangular windows is a triangle. That's why you end up with a sine wave with a triangular envelope.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.