# The small red peak in the CIE standard observer

The human eye has 3 types of color cones: red, green and blue. Their spectral sensitivity peaks in short (S, 420–440 nm), middle (M, 530–540 nm), and long (L, 560–580 nm) wavelengths. See diagram below.

Then we have the CIE standard observer. Wikipedia says it can be thought as the spectral sensitivity curves of three linear light detectors yielding the CIE tristimulus values X, Y and Z.

I don't get why the red line in the CIE standard observer color matching function also peaks at 440 nm. To be more specific: What causes the x-line in a CIE standard observer color matching function to peak at around 440nm?

Can someone explain me that?

Probably you've noticed that primarities are $\mathbf{X}$, $\mathbf{Y}$, $\mathbf{Z}$, not $\mathbf{R}$, $\mathbf{G}$, $\mathbf{B}$ (which are corresponding to the color values $R$,$G$,$B$). This is the aftermath of original work conducted by Wright and Guild yielded the Color Matching functions: $r(\lambda)$, $g(\lambda)$, and $b(\lambda)$ having a negative value of the red primarity around $522 \;\texttt{nm}$ (figure below). Everyone will agree that it is non-intuitive and somewhat confusing.

Why values of $r(\lambda)$ are negative? Well after tristimulus color matching experiments, it turned out that not every color can be created with the preset primarities ($700\; \mathtt{nm}$, $564.1\; \mathtt{nm}$, and $435.8\; \mathtt{nm}$). So for example, if you wanted to create the blue-green color of wavelength $\lambda = 490 \; \mathtt{nm}$ it cannot be done only by adding blue and green primarity alone. You also need to subtract the red one. Mathematically speaking, the color equation:

$$\mathbf{C}=r\mathbf{R} + g\mathbf{G} + b\mathbf{B}$$

Must by modified by adding the outer mixture $-r_{out}\mathbf{R}$ to both side of the above equation:

$$\mathbf{C} + r_{out}\mathbf{R} = g\mathbf{G} + b\mathbf{B}$$

That is why CIE introduced new set of so-called virtual primarities: $\mathbf{X}$, $\mathbf{Y}$, $\mathbf{Z}$. In order to obtain only positive or null values of the Color Matching Functions. That's how $2^{\circ}$ observer Standard Color Matching Functions (also called CIE 1931) were created. By doing so they ensured that (amongst other assumptions):

• New SCMF's $x(\lambda)$, $y(\lambda)$, $z(\lambda)$ have values $\ge 0$ (not like $r(\lambda)$).
• Are related to previous CMF's ($x$,$y$,$z$) by a appropriate transform.
• $\mathbf{X}$, $\mathbf{Y}$, $\mathbf{Z}$ must be independent (kind of orthogonal).
• Only color value $Y$ is proportional the the lightness of a colour.

Newly obtained color quantities $X$,$Y$,$Z$ are called Standard Color Values. The relation between CMF's and SCMF's is given by a following linear transform: $$\left( \begin{array}{c} x(\lambda) \\ y(\lambda) \\ z(\lambda) \\ \end{array} \right) = \left( \begin{array}{ccc} 2.7689 & 1.7517 & 1.1302 \\ 1 & 4.5907 & 0.0601 \\ 0 & 0.0565 & 5.5943 \end{array} \right) \cdot \left( \begin{array}{c} r(\lambda) \\ g(\lambda) \\ b(\lambda) \\ \end{array} \right)$$

After over a three decades people from CIE had realised that $2^{\circ}$ observation angle might not be enough for adequate color assessment. That's why in 1946 they introduced the $10^{\circ}$ observer, also known as CIE 1964 observer. Latter is recommended for visual angles above $4^{\circ}$, whereas CIE 1931 should be used for any angle below that. Of course there is no sudden change in perception of colour at $4^{\circ}$, this number is just arbitrary. Comparison of these two standards is depicted of a following plot, on which you can see the 'bump' mentioned by you for the $x(\lambda)$ standard color value:

• 10° observer is from 1964, not 1946. – Ruslan Sep 29 '18 at 12:27
• Thanks for that, Czech mistake. Luckily CIE standard has a correct date, later in that sentence. – jojek Sep 29 '18 at 12:50