1
$\begingroup$

The deblurring problem can be modelled as follows

$$ f = \phi u + \epsilon, \; \epsilon \sim N(0, \sigma) $$

where $\phi$ is a filter (e.g. a low-pass filter) and $\epsilon$ is a Gaussian noise.

In computer vision, what is the relation between deblurring and deconvolution?

|improve this question
$\endgroup$
  • 2
    $\begingroup$ Possible duplicate of Why can convolution only be applied to compute the output of a linear filter? $\endgroup$ – Marcus Müller May 19 at 16:50
  • $\begingroup$ @MarcusMüller Why would that be a duplicate of this question? $\endgroup$ – nbro May 19 at 16:51
  • $\begingroup$ As you've asked (and got an explanation in the above question), filtering is convolution. So, if you want to invert a blur, and you define blur to be a filter, then the question "what is the relationship of deblurring and deconvolution" becomes "what is the relationship between reversing a convolution and deconvolution" and the answer to that is: it's the same. It's literally the definition of deconvolution. $\endgroup$ – Marcus Müller May 19 at 16:52
  • $\begingroup$ @MarcusMüller This is a completely different question. It's not a duplicate. My other question was: "why can convolution only be applied to compute the output of a linear filter?". I don't care if the answer is the same. The question is different. You should flag a question as duplicate if the question has a duplicate (not the answer). $\endgroup$ – nbro May 19 at 16:54
  • $\begingroup$ Marcus, in general deblurring does not only concern lti filters. It is only the most primitive approaches which do. $\endgroup$ – mathreadler May 19 at 19:03
2
$\begingroup$

Let me present the following Diagram:

enter image description here

So, both Deblurring and Deconvolution are operations within the family of Image Restoration (Which is a subset of Inverse Problem set).

Basically we build the Image Restoration set by different Degradation Models.
The one related to the question are:

  1. Linear Degradation Model
    Namely, the degradation is made by a Linear Operator.
  2. Spatially Invariant Model
    A Model where the degradation is the same for any place in the image.
    We created it as a subset of Linear Model though it doesn't have to be. But for clarity.
    Any operator which is both Linear and Spatially Invariant can be defined by a Convolution Operation. Hence it can be reversed by Deconvolution.
  3. Low Pass Operator
    A set of images degraded by a Low Pass Filter. Namely by a convolution with a Low Pass Filter.

Now, there is a set of degradation which basically create a blurry image.
Reversing this operation is called Deblurring.
In case the blurring is made by a Low Pass Filter applied by Convolution the Deconvolution of this operation is also a Deblurring process.

|improve this answer
$\endgroup$
2
$\begingroup$

In the context of image processing (and machine vision as well), blurring is an operation that reduces the sharpness of an image by some lowpass filtering applied on it.

There are different causes of blurring such as lens blur, motion blur, or just LSI (linear shift invariant) lowpass filtering.

Deblurring refers to any restoration performed on the image that try to remove the effect of a previous blurring, by outputting a sharper (similar to original) version of the image.

When blurring can be mathematically defined as an LSI convolution operation (aka LSI filtering), then the operation of deblurring can be defined as a deconvolution (i.e; inverse of convolution), and that's the sole relation between deblurring and deconvolution.

|improve this answer
$\endgroup$
  • $\begingroup$ In image processing, LTI becomes LSI - Linear Spatially Invariant. $\endgroup$ – Royi May 20 at 4:27
  • $\begingroup$ Ah thank you! so easy to ignore. Although LSI stands for Linear Shift Invariant to my knowledge..? But spatial invariance is also quite fitting... ;-) $\endgroup$ – Fat32 May 20 at 18:20
  • $\begingroup$ I think shift invariant would be generalization of both :-). Just like we assume in 1D the coordinates are time and we call it Time Invariant, I think in 2D the assumption is the coordinates are spatial -> Spatial Invariant. $\endgroup$ – Royi May 21 at 2:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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