# Why does time-domain convolution correspond to frequency-domain multiplication? (visual)

I seek a visual explanation of this. I've already seen the maths, and can derive the proofs - they amount to nill for an intuitive understanding. Any amount of math is welcome, as long as serving to ultimately explain it visually.

Examples of excellent explanations:

Continuous vs Discrete: both are welcome, but ultimately the discrete case must be explainable. I've long thought of the input-side algorithm myself for continuous convolution, but never completed the picture; real analysis is tricky.

Circular vs linear: a complete explanation ought to cover both, but I'm primarily interested in linear convolution (rather, how the circular of padded signals is equivalent to linear).

Duality: ideally, should be covered (conv in freq domain <=> mult in time domain).

Ideas:

1. Only right-padding works, and both inputs are padded; this looks like forcing inputs to correlate with lower and fractional frequencies (and more frequencies) relative to unpadded's frame.
2. Something about convolving with shifted deltas and observing modulations of complex sinusoids in other domain; awaiting clarification from @AndyWalls.
• I'm broke in terms of rep, but if there's a good enough answer, I'll be sure to drop a bounty sometime later. – OverLordGoldDragon Sep 14 at 8:53
• Your question titles and bodies are all false... Why convolution becomes a multiplication? is extremely easy to see using the simple math (given as an exercise) on any DSP / signal textbook. What you really want to ask is probably a Visual Demonstration of the convolution theorem... [I don't remember having seen it before] A visual demonstration (a visual aid) is extremely helpful in solidifying (and understanding) a lot of abstract concepts but will not be a substitude for a mathematical explanation. So please correct your question titles and bodies according to this. – Fat32 Sep 14 at 10:19
• This visualization is going to go something like: a) convolve with a shifted $\delta()$ in one domain and show that you get the other domain's function modulating a complex exponential in the other domain. b) pick another shifted $\delta()$ in the first domain and convolve and show that you get the other domain's function modulating a complex exponential of a different frequency c) add the results from a) & b) together d) repeat until things fill in. On another note: discrete convoltuion with DFTs is always circular. The flaw comes when using it incorrectly for linear convolution equivalence. – Andy Walls Sep 14 at 12:01
• @Fat32 If "the math" was sufficient to answer the "why", I wouldn't be asking, nor would the video or the chapter I linked in the question exist. And by "visual" I don't mean slap together some convenient approximations, I mean the real deal, but visually, just like my linked sources - so yes, they can very much substitute the walls of math that are useless for intuition. – OverLordGoldDragon Sep 14 at 15:51
• I retract statement on "convenient approx", as long as it's made clear it's inexact, and in what way; these can be powerful stepping stones to "the exact". – OverLordGoldDragon Sep 14 at 15:57