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Conversion from HSI to RGB

Given , and , we can first find , , and on the RGB triangle, and the find , , and . Consider three possible cases in terms of the hue angle :

( and , i.e., inside )
From , solving for we get

$\begin{displaymath}b=\frac{1-S}{3} \end{displaymath}$

To find , consider the triangle :

Here vector is the projection of vector onto the line , and all points , , and are on the RGB triangle. Now consider the following steps:
- On the triangle , we have
  
  $\begin{displaymath} \frac{\vert P_rQ_r\vert}{\vert P_rO\vert}=\frac{\vert P_rQ_... ....}\;\;\;\;\;\; d=\frac{\vert P_rP''\vert}{\vert P_rQ_r\vert} \end{displaymath}$
  
  and
  
  $\begin{displaymath} r=\vert P_rO\vert-d=1-d =1-\frac{\vert P_rP''\vert}{\vert P... ...l{*}{=}\frac{1}{3}+\frac{\vert WP''\vert}{\vert P_rQ_r\vert} \end{displaymath}$
  
  (* $\vert WQ_r\vert=1/\sqrt{6}$ and $\vert P_rQ_r\vert=\sqrt{6}/2$ )
- The length of is $\vert WP''\vert=\vert WP\vert\cos\angle H$ , and the saturation is $S=\vert PW\vert/\vert WP'\vert$ , i.e., $\vert PW\vert=S\vert WP'\vert$ , we have:
  
  $\begin{displaymath}\vert WP''\vert=\vert WP\vert\cos H=\vert WP'\vert S\cos H \end{displaymath}$
  
  and we get
  
  $\begin{displaymath}r=\frac{1}{3}+\frac{\vert WP''\vert}{\vert P_rQ_r\vert} =\frac{1}{3}+S\cos H \frac{\vert WP'\vert}{\vert P_rQ_r\vert} \end{displaymath}$
- As $\vert P_rQ_r\vert/\vert WQ_b\vert=(\sqrt{6}/2)/(1/\sqrt{6})=3$ , i.e., $\vert P_rQ_r\vert/3=\vert WQ_b\vert$ , and also $\cos(60-H)=\vert WQ_b\vert/\vert WP'\vert$ , we have
  
  $\begin{displaymath} \frac{\vert WP'\vert}{\vert P_rQ_r\vert}=\frac{1}{3}\frac{\... ...rac{\vert WP'\vert}{\vert WQ_b\vert} =\frac{1}{3\cos(60- H)} \end{displaymath}$
  
  Substituting this into the expression of above, we get:
  
  $\begin{displaymath}r =\frac{1}{3}\left[1+\frac{S\cos H}{\cos(60- H)}\right] \end{displaymath}$
- Finally we can get
  
  $\begin{displaymath}g=1-b-r \end{displaymath}$
  
  Given and , we can get :
  
  $\begin{displaymath}R=3Ir,\;\;\;G=3Ig,\;\;\;B=3Ib \end{displaymath}$
(p inside $\triangle P_gwP_b$ )

$\begin{displaymath}H=H-120 \end{displaymath}$

$\begin{displaymath}r=\frac{1-S}{3},\;\;\;\;\;\; g=\frac{1}{3}\left[1+\frac{S \cos H}{\cos(60-H)}\right], \;\;\;\;\;\;\;\;\;b=1-g-r \end{displaymath}$
(p inside $\triangle P_bwP_r$ )

$\begin{displaymath}H=H-240 \end{displaymath}$

$\begin{displaymath}g=\frac{1-S}{3},\;\;\;\;\;\;\;\; b=\frac{1}{3}\left[1+\frac{S \cos H}{\cos(60-H)}\right], \;\;\;\;\;\;\;\;\; r=1-g-b \end{displaymath}$

In summary, given the H, S, and I values, we can obtain the R, G, and B components by

$\begin{displaymath}b=(1-S)/3 \end{displaymath}$

$\begin{displaymath}r =\frac{1}{3}\left[1+\frac{S\cos H}{\cos(60- H)}\right] \end{displaymath}$

$\begin{displaymath}g=1-b-r \end{displaymath}$

and

$\begin{displaymath}R=3Ir,\;\;\;\;\;G=3Ig,\;\;\;\;\;B=3Ib \end{displaymath}$

Note that after the HSI to RGB conversion, we need to convert the R, G, and B values from the range of $(0,\;1)$ back to the three components for a pixel in a color image in the range of $(0,\;255)$ .

Next: Color processing examples Up: ColorProcessing Previous: Conversion from RGB to

Ruye Wang 2014-10-10