reg. frame descriptor of USB 1.1 video class specification

Discussion in 'Windows Vista Drivers' started by muralitharan.perumal, Apr 5, 2006.

  1. Hi,

    YUV2 Packed data format

    It is mentioned that USB 1.1 video class specification supports YUV2
    (uncompressed) packed data format. What does that mean? what type of
    packing is supported i.e.,

    1. Y-Cb-Y-Cr or
    2. Y-Cr-Y-Cb or
    3. Cb-Y-Cr-Y or
    4. Cr-Y-Cb-Y

    Also, at the host side, how does it know which packing is supported as
    the Frame descriptor does not have any fields to mention about that?

    Frame Buffer Size

    One sample (pixel) is 16-bit size, suppose if the resolution is 1024 *
    768 then the video buffer size would be 1024 * 768 * 2 (1.6 MB). Is
    this correct?

    The buffer allocation at the host side would be

    Y - 1024 * 768 * 1 (8 -bit)
    U - 1024 * 384 * 1 (8-bit)
    V - 1024 * 384 * 1 (8-bit)

    In total it is 1.6 MB. Is this correct?

    Minimum and Maximum Bit rates

    The two fields dwMinBitRate and dwMaxBitRate are not clear to me. How
    will you set the value for it? For Uncompressed data, will both contain
    the same value?

    Suppose I have a frame size of 1024 * 768 * 2 (as above) and 15 frames
    per second then the bit rate would be 1024 * 768 * 2 * 15. is this

    Kindly provide me the answers asap.

    muralitharan.perumal, Apr 5, 2006
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  2. muralitharan.perumal

    Tim Roberts Guest

    The format is actually YUY2, not YUV2. This is a standard Microsoft
    It's #1. The bytes, in byte order, are Y0, U0, Y1, V0.
    The host knows because the format is in one of the class-specific
    No, not exactly. It's a packed format, so it's a single buffer of 1.6 MB,
    not separated like that.
    It depends on your frame rate. The frame size will be a constant; if you
    support a single frame rate, then the min and max bit rates will be the
    No, that's the BYTE rate. The bit rate is 1024 x 768 x 16 x 15.
    Tim Roberts, Apr 7, 2006
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