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November 10th, 2003, 01:18 PM  #1 
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More on 1CCD vs. 3CCD (technical)
In order to further illustrate the disadvantages of 1CCD systems versus 3CCD systems, I figured I would post exactly how 1CCD systems work, and explain how the approximation is done. (This in particular is exactly how Sony 1CCD systems work, other systems might be slightly different.)
First of all, some symbols. As you might know, colors can be represented in terms of Red, Green and Blue components(RGB), or they can be represented in terms of Luminance(Y) and Chromiance(C) So: Cy=Cyan=G+B Ye=Yellow=R+G Mg=Magenta=R+B R=Red=R G=Green=G B=Blue=B Y=Luminance=R+G+B C=Chroma difference=RY Now, Sony 1CCD systems have a onecolor filter in front of each CCD sensor or pixel. The filter is either Cy, Ye, Mg or G. Thus, a Cy(Cyan) filter, would let the Green and Blue components through, while blocking off the Red component of the light. Here is the the arangement of the filters, along a trivial 4pixel by 4pixel area of the CCD: Line1= Cy Ye Cy Ye Line2= G Mg G Mg Line3= Cy Ye Cy Ye Line4= Mg G Mg G You get the picture. So, how do we turn this mosaic of filtered pixels into relevant color? First of all, to approximate the actual color of one pixel, we need 4 adjacent pixels. The CCD itself, doesn't output the value of each pixel, but actually outputs the SUM of the value of TWO vertically adjacent pixels. For example, (Line 1 + Line 2) and (Line 3+Line4) would be output for the EVEN fields of the interlaced image, and (Line2+Line3) would be a line in the ODD field. The read out from the CCD for the first line(part of the EVEN FIELD) would be: (Cy+G), (Ye+Mg), (Cy+G), (Ye+Mg) Now, if we have these values, how do we get color from them? The truth is, we can't get perfect R,G,B color but we can approximate it. Like I said, we need values from 4 sensors so the first thing we do is add the value we get for the pixel we are interested in, with the next value for the sum of the two adjacent pixels. Suppose we are interested in the very first pixel on the EVEN line above, and we want to approximate the luminance(R+G+B) of the pixel: (Cy+G)+(Ye+Mg) = (2G+B)+(2R+B+G) = 2R+3G+2B if we now divide this by 2 we get: R+1.5G+B This is the best we can do, which is an approximation for the actual luminance(R+G+B). So we see, that one aspect of the approximation is the fact that the colors are not taken from the actual pixel we are interested in, but rather surrounding pixels. Yet, the other aspect of the apprixmation is the mathematical one above, in which we cannot obtain an exact measure of Y. To fully characterize the color we not only need Y but also C(chroma difference) which is defined as RY. However, it is clear that since Y in itself is an approximation and not accurate, RY will also be an approximation since it uses our value for Y calculated above. In a 3CCD system, we have one CCD for each color R,G,B. Thus, colors are accurately represented and do not have to be approximated. Manufacturers usually increase the pixel count of 1CCD systems in order to obtain a better approximation. Hope this helps, Juan 
November 14th, 2003, 11:18 AM  #2 
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Panasonic PVDV852 1CCD
On my Panasonic PVDV852 1CCD, there are 4 CCD photosites for every pixel stored on the MiniDV tape. I wonder if they even do the typical Bayer interpolation in this case. I'm suspecting that they just combine the values from the 4 photosites and make one true color. The colors are claimed to be quite close to a 3CCD camcorder.
True each colored photosite covers only part of the area of the final pixel. But that's much better than guessing (interpolating) the true colors when only one photosite is used for each final pixel. But most consumer 1CCD camcorders now have at least as many pixels as the Panasonic PVDV852 but many have poor colors. Also do you think that manufacturers can actually line up the photosites on the 3CCD systems so final pixels on each of the CCDs are exactly on top of each other. Maybe the photosites are off by 1/2 or more of a photosite. In this case the 4 photosites combined into one final pixel 1CCD system might even be better than 3CCD systems. I'm wondering if the major reason why 3CCD camcorders have better colors than the Megapixel 1CCD camcorders is because 3CCD camcorders cost more and so they spend the money to make the colors right. Most 3CCD cameras also have much larger photosites. Even if a 3CCD camcorder has a 1/6" CCD, there are three of these. My Nikon 5400 digital still camera with 1CCD Bayer sensor has incredibly much better picture quality than any video or still picture from my Panasonic PVDV852 1CCD at identical resolution. In this case I suspect all the quality difference is due to different JPG compression used, not the CCD or lense. The Nikon set on FINE has hardly any noticeable compression. I even copy magazine articles, hardly any fuzzy edges of the black on white text. The Panasonic seems to use either heavier compression or poorer JPG techniques which results in a huge loss of picture quality compared to the Nikon. So who cares about the incremental quality increases of the CCD when most of it is often destroyed in the processing. So I have to wonder if most of the quality issues are not 1CCD Bayer vs 3CCD, but rather in the processing software used? Or am I missing something? I'm listening. 
November 14th, 2003, 01:35 PM  #3 
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Juan, your conclusion w.r.t. 1 vs 3CCD is a bit "easy". Processing is indeed inportant for the 4 and also for the 3 color filter structures on single CCD's as well. Four color filter layers have some advantages w.r.t. horizontal luma resolution as compaired to 3 filter architectures, but there is much more in the game...

November 14th, 2003, 02:39 PM  #4 
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My conclusion was with respect to the Sony arrangement. Having one color filter per CCD element is going to yield reduced resolution and the color is indeed only an approximation. This is mathematical.
Right now I am doing an experiment piping out the raw A/D output from a TRV19 into my computer. I am decoding the images with software I wrote, so I am aware of the math involved. A 3CCD setup is superior to this in every way. Juan 