Electro-Optical Transfer Function (EOTF)
An EOTF maps an encoded picture signal to the light a display should produce.
What light should a screen produce for a signal value of 0.5? An electro-optical transfer function, or EOTF, supplies the relationship needed to answer. The signal is electrical or numerical; the result is light.
For an ideal SDR display with zero black, 100-nit white, and a 2.4 power response, use L = 100 × V^2.4. V is the normalized signal and L is neutral-gray luminance:
| Signal V | Calculated light |
|---|---|
| 0.25 | 3.6 nits |
| 0.50 | 18.9 nits |
| 0.75 | 50.1 nits |
| 1.00 | 100 nits |
Halfway through the signal range therefore produces about 19% of white's light in this example. Equal signal steps do not create equal light steps. These are the zero-black conditions of BT.1886; another EOTF gives another relationship.
Apply the function to the right input
The stored video code first needs the correct numerical-range interpretation. Nominal black and white may occupy only part of the available integer range. Interpreting those endpoints produces a normalized signal; the EOTF then supplies its light meaning.
The order is significant. If nominal black is mistakenly normalized to a positive signal, the EOTF produces light for a sample that was meant to be black. If a nonlinear sample is treated as already linear, later arithmetic also begins from the wrong light value.
A color-processing path applies the relevant component relationships and transfer interpretation to obtain a suitable working representation. A neutral-gray calculation keeps the example above simple; colored pictures additionally need their primaries and component matrix interpreted correctly.
Reconstruct light, then choose what the output can show
PQ gives HDR samples an absolute reference-light interpretation. Recovering that interpretation can reveal a 1,000-nit highlight, but a destination with a smaller usable range still needs a mapping decision. The EOTF identifies the requested light. Tone mapping adapts it.
After a light-domain operation, the output needs an encoding appropriate to its own display relationship. In the ideal SDR example, a desired 50-nit gray requires V = (50/100)^(1/2.4), approximately 0.749. Writing 0.5 instead would request 18.9 nits. That inverse operation explains why light values and final video signal values cannot be interchanged.
An Emby Direct Play session identifies how the server delivers the file. The app and display still render the samples using their output path. Comparing displayed brightness therefore requires the source interpretation and the destination response alongside the delivery mode.