Matlab and Aptina ApBaseCOM Integration

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Matías
Matías 2014-6-1
回答： Matías 2014-6-5
Hello,
I have a Micron Camera and I tried to take some images from it and processing them. I am using the ApBase interface from Aptina Software but, when I try to get the image, I received a raw matrix and rgb and I cannot draw the image again.
I based on the guide https://aptina.atlassian.net/wiki/display/DEVS/ApBase+COM+Development+Guide I executed:
apbase = actxserver('apbaseCom.ApBase');
camera = apbase.Create(0);
camera.LoadIniPreset('', '');
raw = camera.GrabFrame;
rgb = camera.ColorPipe(raw);
i = reshape(rgb,2592,1944);
I attached a capture about the workspace. I do not know how interpret the values of the matrix i and when I use imshow(i) on it I am not getting the image.
The guide said that the output is normally 32 bits per pixel with B – G – R – X byte order. What is mean the "X"?
Regards, Matías.

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采纳的回答

Image Analyst
Image Analyst 2014-6-2
It's a 32 bit (4 byte) image with the red, green, and blue values packed into the lower 3 bytes and the 4th byte unused and probably zero. You're going to have to mask and shift the value to extract the individual color channels. To mask, use the & operator to "AND" the array with 000000ff, 0000ff00, and 00ff0000. Then divide the upper byte images by 256 and 65536 to bring then down into the 0-255 range of uint8. But they may still be int32 or double at that point so you need to explicitly cast them to uint8 with the uint8() function.
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Image Analyst
Image Analyst 2014-6-2
Try this:
% Pack 200 in the lowest byte, 175 in the next lowest byte,
% 150 in byte 3, and the highest byte will be all zeros
int32Bit = uint32(150 * 2^16 + 175 * 2^8 + 200)
inBinary = dec2bin(int32Bit) % Just to show you the binary version
% Now recover the numbers from the bytes
redValue = bitand(int32Bit, 2^8-1)
greenValue = bitand(int32Bit, 2^16-1) / 2^8 - 1
blueValue = bitand(int32Bit, 2^32-1) / 2^16 - 1
Matías
Matías 2014-6-4
I do not understand you at all the part "int32Bit = uint32(150 * 2^16 + 175 * 2^8 + 200)" I mean where is my rgb matrix in the code?.
I found something else (because I don't want to bother you with a lot of questions) in the web and I executed:
apbase = actxserver('apbaseCom.ApBase');
camera = apbase.Create(0);
camera.LoadIniPreset('', '');
%camera.SetImageFormat(40,40,'');
raw = camera.GrabFrame;
rgb = camera.ColorPipe(raw);
i = reshape(rgb,2592,1944);
imgR = uint8((255/31)*bitshift(bitand(i,63488),-11));
imgG = uint8((255/63)*bitshift(bitand(i,2016),-5));
imgB = uint8((255/31)*bitand(i,31));
imgRGB = cat(3,imgR,imgG,imgB);
imshow(imgRGB)
And the output was the next image:
  1. Why I got those colors? I want to get the real image with real colors.
  2. Why used (255/31),(255/63),(255/31) on the uint8?
I don't know if I can share the link where I saw the example, so I didn't included it.

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更多回答（1 个）

Matías
Matías 2014-6-5
I could do it.
apbase = actxserver('apbaseCom.ApBase');
camera = apbase.Create(0);
camera.LoadIniPreset('', '');
%camera.SetImageFormat(40,40,'');
raw = camera.GrabFrame;
rgb = camera.ColorPipe(raw);
i = reshape(rgb,2592,1944);
imgB = uint8(bitshift(bitand(i,2^32-1),-16));
imgG = uint8(bitshift(bitand(i,65535),-8));
imgR = uint8(bitand(i,255));
imgRGB = cat(3,imgR,imgG,imgB);
imshow(imgRGB)
Here is the result:
I am really thankful.

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