Color space conversion in video tonemapping (HDR to SDR) (bt2020 to bt709, YCbCr, YUV and RGB)

I press , Recently doing video TM Related work of , Specifically, given a HDR video （10bit Of YUV420 Format ）, To do it TM, Wrote an algorithm, but there is always a very outrageous color difference , Especially the red and blue channels . After careful examination, it was found that it was obtained HDR The color space of the video is BT2020 Of , But after the conversion SDR The video is BT709 Of , You need to downgrade the color space before processing . Here's a record of the process .

First post two addresses

[1] ​​​​​​RECOMMENDATION ITU-R BT.2087-0 - Colour conversion from Recommendation ITU-R BT.709 to Recommendation ITU-R BT.2020

[2] REPORT ITU-R BT.2407-0 - Colour gamut conversion from Recommendation ITU-R BT.2020 to Recommendation ITU-R BT.709

These are both ITU A standard programmatic document . The first document talks about how to BT709 go to 2020（ But only RGB The signal ）, The second one talks about how to BT2020 The switch to BT709（ Include YUV、RGB How to deal with it ）.

In the literature [2] in , A simple and crude diagram is given

E'RGB refer to “ Normalized nonlinearity RGB value ”, and E_RGB refer to “ Normalized linearity RGB value ”.

But in the algorithm , The data I got was 10bit Of YUV420 data , So we need to 10bitYUV420 Data conversion to E'RGB. How to turn ？ Not mentioned in the literature . But the literature [1] How will 709 The switch to 2020 when , Give a picture

In addition to here  E'RGB and E_RGB outside , We have the E'YCBCR and D'YCBCR. among E'YCBCR refer to “ The normalized YCBCR value ”, and D'YCBCR refer to “ Quantized YCBCR value ”

So I understand here , What I get at the input of the algorithm is called 10bitYUV420 data , It's actually this D'YCBCR value . So next , We combine two literatures , The corresponding process can be deduced .

Step 1. take D’YCbCr(2020) To E’YCbCr(2020)

In fact, it is the antiprocess of the following process

  Simple deduction , have to

        E_Y = ((D_Y / 4) - 16) / 219
        E_Cb = ((D_Cb / 4) - 128) / 224
        E_Cr = ((D_Cr / 4) - 128) / 224

Step 2. take E’YCbCr(2020) To E’RGB(2020)

The literature [1] The inverse process is given in

So here we just do a simple matrix inversion （ Here the inverse is given directly ）

It should be noted that , In the code given below E_YCbCr and E_RGB Such a variable , Their shapes are height x width x 3 Of , So here we do such a transformation of linear algebra . Use of is （AxB）T = BT x AT This is the nature of . This method is used in many places below , I won't explain it later .

        m_YCbCr2020_to_RGB2020 = np.array([[1.00000000e+00, -7.82308321e-18, 1.47460000e+00],
                                           [1.00000000e+00, -1.64553127e-01, -5.71353127e-01],
                                           [1.00000000e+00, 1.88140000e+00, 2.38961873e-17]])
        E_RGB = np.matmul(E_YCbCr, m_YCbCr2020_to_RGB2020.T)

Step 3.  take E’RGB(2020) To E RGB(2020)

        E_RGB = np.power(E_RGB, 2)
        #  Another version is 
        # E_RGB = np.power(E_RGB, 2.4)

Step 4. take E RGB(2020) To E RGB(709)

        m_RGB2020_to_RGB709 = np.array([[1.66051121, -0.58771059, -0.07280062],
                                        [-0.12456141,  1.13296051, -0.00839911],
                                        [-0.01816769, -0.1005606,  1.11872828]])
        E_RGB = np.matmul(E_RGB, m_RGB2020_to_RGB709.T)

Step 5. take E RGB(709) To E’RGB(709)

        E_RGB = np.power(E_RGB, 1/2)
        #  Another version is 
        # E_RGB = np.power(E_RGB, 1/2.4)

Step 6.  E’RGB(709) To D’RGB(709)

【 Here is to 10bit As an example of , If it is 8bit, Get rid of the back *4 that will do 】

D_RGB = ((E_RGB * 219 + 16) * 4).astype("uint16")

After this conversion , There is no color deviation problem .