3
$\begingroup$

I've noticed that the gamma value used in level adjustment of Photoshop and Canvas is 'opposite' of gamma values defined in Wikipedia or MATLAB imadjust documentation.

In Wikipedia or MATLAB, high gamma (>1) will make an image darker, whereas in Photoshop high gamma (>1) will make an image brighter.

I've also noticed that, even among image processing software, Zeiss' Zen software uses high gamma (>1) to make an image darker.

There seem to be two definitions of gamma value. This is just confusing... I assume that this is related to 'gamma encoding' and 'gamma correction'. But can anyone explain the rationale of these two defintions and when to use which one?

Vout = Vin ^gamma % Wikipedia or MATLAB

Vout = Vin ^1/gamma % Photoshop or Canvas
Improve this question
$\endgroup$

2 Answers 2

Reset to default
4
$\begingroup$

The concept of image gamma (and its correction) has emerged as a means of controlling and correcting the nonlinear voltage-brightness characteristics of the legacy input cameras and legacy display CRTs which are both analog devices.

Both of them had nonlinear relationships between the optical brightness levels being captured (or displayed) and the corresponding voltage levels that result of (or driving) it.

In order to reproduce the image brightness levels (and RGB channel intensity levels as well) as faithfull as possible to the original content, the exponential distortions introduced by the CRT displays are pre-corrected by the studio processing equipment before the content is being transmitted for broadcasting. The correction being applied at the studio is the inverse of the distortion being created at the CRT display, hence they will cancel each other, yielding a faithful, brightness neutralized, image at the CRT output.

Furthermore, for adequately displaying video content recorded with different standards and/or captured with different gamma correction parameters, the TV sets and monitors are available with gamma selection options as well.

Hence you can apply a gamma correction either at the camera input stage before it's transmitted or gamma adjustment at the display (CRT) stage during it's played back. So the difference you percieve possibly refers to the stage at which it's meant to be applied. So as you have stated, the same gamma parameter value will create two opposing results on the same image according to which stage it was applied to. Or stated in other words, a given value $\gamma$ will create the same effect with $1/ \gamma$ when applied at opposite sides.

Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ Great answer Fat32. Just to add to this, the gamma correction is also meant to compensate for the differences between physiological perception of humans to visible light and the ability of optoelectrical components to faithfully reproduce it. The sensitivity of human eyes is actually very good at low light intensities and gets less good at distinguishing subtle differences in brightness as overall scene brightness increases. This is the exact opposite for imaging sensors and displays which is why the corrections are non-linear. combining inverse at sensor & crt results in linear overall output $\endgroup$
    – Inflexionist
    Dec 14, 2017 at 6:32
  • $\begingroup$ Thanks! In the image titled "Gamma versus Gamma Correction", gamma = 2 results in darker output, whereas gamma correction = 2 results in the brighter output. Perhaps, Photoshop and Canvas are meant to be used in post processing of images, so they adopted "gamma correction" implicitly. But the distinction about the stage, earlier or later, is still a bit ambiguous... $\endgroup$
    – Kouichi C. Nakamura
    Dec 14, 2017 at 9:01
1
$\begingroup$

In the days of analog CRTs and TVs, normally one would adjust the gamma of the input signal one way, and the physical characteristics of the CRT would do the inverse in converting voltage to luminance.

Improve this answer
$\endgroup$

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.