What happens when we deconvolve an audio signal before convolving it?Started by 5 years ago●3 replies●latest reply 5 years ago●744 views
This may seem like an odd question, but what happens if we deconvolve an audio signal before convolving it?
Normally, we deconvolve an audio signal that has already been convolved by some IR (in order to recover the original unconvolved signal).
In audio mixing / post-production, this can be used to remove reverb from an audio recording.
However, what happens if we were to reverse the process?
For example, let's say we took an unconvolved audio signal N, and the performed the deconvolution of N using some IR (impulse response) A.
If we were to then convolve the resultant audio signal from our deconvolution of N by A, would we arrive at N?
In other words, are convolution and deconvolution mathematical inverse functions (like division is the mathematical inverse of multiplication), such that the following two scenarios hold true?
Convolve(N) by IR(A) = A(N)
Deconvolve A(N) by IR (A) = N
Deconvolve (N) by IR(A) = -A(N)
Convolve -A(N) by IR(A) = N
Note: For those who want a bit more information, please see my earlier post here:
It seems like you’ve got it right, although your terminology is a bit strange. As you know, if you convolve a signal and then deconvolve it you get the original signal. The reverse is also true, because deconvolution is the same as convolution and the order of operations can be changed without changing the result.
With regards to “what happens”, keep in mind that again, convolution and deconvolution are the same. For any impulse response, h, there is an impulse response g, such that h convolved with g yields a unit pulse. The caviat being that if h has a zero in its Fourier spectrum, that g may not be realizable.
I can't thank you enough for your help.
I had just two more questions I was hoping you could answer.
Also, please excuse any incorrect terminology (as I don't come from a DSP background).
Question 1: Assuming we have an audio signal N and two IR's A and B.
If we represent the convolution of N with IR A as A(N), and the deconvolution of N by A as -A(N). Would the following also hold told true (I suspect it would).
1. Let's say we first deconvolve N by IR A, resulting in -A(N).
2. Let's say we then convolve -A(N) by IR B, resulting in B(-A(N)).
3. If we then convolve B(-A(N)) with IR A, will the result by B(N)?
I suspect the answer is yes, but I just wanted to be certain.
Question 2: In regards to your following quote:
"...The caveat being that if h has a zero in its Fourier spectrum, that g may not be realizable."
Can I assume that as long h has a non-zero in its Fourier spectrum, that g will be realizable?
In other words, as long as h has some data in its Fourier spectrum (in places where g also has data in its Fourier spectrum), then there will be no loss of frequency data when h convolves or deconvolves g?
Question 3: If it's the case that the answer to question 2 is "yes / true", can we simply add data (via EQ - aka Equalization) where no such data exists in h but does exist in g (in order to fill in the gaps in h's Fourier spectrum)?
If so, a big thanks for all your help!
3: not exactly. Eq is a filter, so it is a scalar in the Fourier domain. A zero multiplied by any number is also zero. You would need to add data to h using some other method