Getting rid of the haze (technique)

[eNo]

I've posted an article on my blog on a technique I found useful in dealing with low contrast shots due to smoke-induced haze. I hope you find it useful: Getting rid of the haze

Note: pasting the article here wasn't practical due to the number of images that would need URL format conversion for a posting in this forum.

[Donald]

eNo

Good stuff. Thank you. Copied into the ever growing list of 'Important Notes'.

[rpcrowe]

Thank you for posting the link to this tutorial. I am traveling to China this April and with the generally terrible air quality present in many areas of that country, this technique might help save some shots for me!

[PopsPhotos]

Thank you.

Pops

[willgoss]

I used a similar Approach...except step one (curves) which I use (Un-Sharp Mask with High Radius (50-0.5-0 Gimp) and then a second Pass using High Pass Sharpen Set low (3)......now I duplicate and set Multiply Mode (Opacity to taste).......

This does work well on a hazy shot or on a flatly lit shot......I did use it on some China Pictures where it is very hazy......

[arith]

You learn something new everyday here. I've just tried it on an old image and it is by far a superior method of haze removal and the sharpening suggested by willgoss much better than ones I used previously. Why is the filter called 'high pass'? I read about it somewhere but can't remember a thing.

[rick55]

You learn something new everyday here. I've just tried it on an old image and it is by far a superior method of haze removal and the sharpening suggested by willgoss much better than ones I used previously. Why is the filter called 'high pass'? I read about it somewhere but can't remember a thing.

It refers to letting higher frequencies through in the frequency domain. If you transform an image to the frequency domain (fourier transform), the transitions will be higher frequencies. If you let those frequencies through (pass the high frequencies = high-pass), you're enhancing the edges, sharpening the image. Contrariwise, a low-pass filter smooths things by only letting the low frequencies through.

As far as I'm aware, the actual processing in real-world tools doesn't happen in the frequency domain, because the transforms to and from are expensive, but operations on a window of pixels can be shown mathematically to be equivalent.

Cheers,
Rick

[arith]

I was going to let this go sort of embarrased you know; don't want to look too stupid. But I'm not and I know I'm going to regret it. So firstly , what is frequency in an image?

What is a frequency domain? Fourier transform is that really important or just a function that will do. I mean Fourier transform is not any more something I'm familiar with. Actually I don't think it ever was but sort of crept in somewhere.

I am very grateful for your answer Rick, I can of course find the answers now just from what you have said. Thank you

[rick55]

I was going to let this go sort of embarrased you know; don't want to look too stupid. But I'm not and I know I'm going to regret it. So firstly , what is frequency in an image?

What is a frequency domain? Fourier transform is that really important or just a function that will do. I mean Fourier transform is not any more something I'm familiar with. Actually I don't think it ever was but sort of crept in somewhere.

I am very grateful for your answer Rick, I can of course find the answers now just from what you have said. Thank you

As I said, I don't think the real-world tools actually process things in the frequency domain. If you want to experiment with it, ImageJ is a good platform. I'm sure The GIMP provides it, as well, but it's been some years since I've used that.

Fourier transform, or specifically for our images, Digital Fourier Transform (DFT) is the mathematical technique that does the transform. The theory of image processing, especially for things like sharpening and blurring, is helped tremendously by the fact that an image, which is a spatial representation of intensity, can be mathematically transformed to a frequency representation. A frequency representation contains the rate of change of intensity over the distance (x,y pixel coordinates) of the image. So just as a "brighter" button can raise pixel intensities in the spatial domain, a "smooth" slider can conceptually decrease the rates of change in the frequency domain. Similarly, edge-detection is the process of leaving in only the greatest rate of change.

The best thing I could find on-line about image processing is this. Most references seem to assume that you already understand 1-D transforms (like for audio processing), and don't seem especially useful.

Cheers,
Rick

[arith]

Cheers for that; I'll have a look at it and the program looks useful. I will get some old books out and look up Fourier since I'm not familiar with this anymore. It will give me something to do.

[arith]

OK had a quick look and it is a bit like Laplace Transform but with an infinite series of periodic frequencies or harmonic analysis. It is 40 years almost since I did anything remotely similar. I don't know if I want to go down that road but just use the applications and get on with my photography.cheers

[rick55]

OK had a quick look and it is a bit like Laplace Transform but with an infinite series of periodic frequencies or harmonic analysis. It is 40 years almost since I did anything remotely similar. I don't know if I want to go down that road but just use the applications and get on with my photography.cheers

You're exactly right that the Laplace Transform and Fourier Transform are similar, used to work in the frequency domain. But you don't need to necessarily study the ins and outs to work on the processing: it was my fault for bring the terms in.

The original question was "high pass" and "low pass," and the point is that it's valid to consider an image as being made up of frequency components, where areas with intensity jumps contain high frequency, and areas of similar intensity are low frequency. If you only "pass" some of those frequencies through, you'll either sharpen or smooth the image.

Cheers,
Rick

[arith]

I can understand that one. cheers Rick