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The key ”trick” of the operation seems to be this revelation: > In other words, performing the convolution of two signals in time domain is equivalent to multi
by eranation 2y ago
The key ”trick” of the operation seems to be this revelation:
> In other words, performing the convolution of two signals in time domain is equivalent to multiplying them in the frequency domain.
Great article, as it breaks down a complex idea into much smaller steps that even my math challenged mind can somehow grasp, but did I miss a step? Or is it left as an exercise to the reader to look up? I was already stretching my math ability to that point, but it felt a little bit like - “and then, draw the rest of the F-ing owl” to me. Is it just me?
Great writing otherwise.
- clearprop 2y agoThanks! In case it helps: * Multiplying two polynomials as taught in school is in reality a convolution. * "performing the convolution of two signals in the time domain is equivalent to multiplying them in the frequency domain" * FFT allows us to convert from time domain to frequency domain * We use FFT to convert our polynomial to frequency domain. * If we are now in the frequency domain, we just need to multiply. Faster than a convolution. Does it clarify the missing step? Happy to update the post with what's missing.
- eranation 2y agoYep, it definitely helps! What I'm struggling to understand is why "performing the convolution of two signals in the time domain is equivalent to multiplying them in the frequency domain" (I'm going to google/gpt it, this is more of an exercise left for me to dive into, your post is perfect as it is)