DVX100A's firewire output 4:1:1 or 5:1?

Crazy_Ideas

Crazy_Ideas

Active member
Hello everyone..

What is the DVX100A's firewire output compression-ratio if you were to use the firewire as "video-out" not capturing to tape, but using your PC's hard drive to capture video ... 4:1:1 or 5:1??

Thanks in advance,

Rob
 
Both. You get 5:1 & 4:1:1 out the firewire of all DV cameras (PAL = 5:1 & 4:2:0).

Both are types of compression.

4:1:1 is not "digital compression", per se, as referencing an algorithm for compressing/uncompressing, but it's a sort of "analog compression", where data is determined to be unnecessary and is just thrown away. 4:1:1 is straight chroma decimation, meaning that there is one red and one blue chroma sample for every 4 luminance samples.

5:1 refers to the compression ratio of the DV algorithm. Once the video has been reduced to 4:1:1 and quantized at 8 bits, then the DV algorithm compresses it to 1/5 its current size.

As expressed in terms of overall compression, as compared to a fully uncompressed 4:4:4 signal, all DV cameras actually operate at closer to 10:1 compression. A full uncompressed signal would take about 31 megabytes per second, DV puts it into about 3.1 megabytes per second.

None of this is specific to the DVX; all DV and DVCAM and DVCPRO25 cameras work this way.
 
Barry,

That was a great lesson! Thanks. I was under the impression that capturing via firewire to pc, that I would capture a less compressed signal.

Thanks again,

Rob
 
Barry, I am trying to understand where the 31MB/s comes from. Is this naive calc below for 4:1:1 anywhere near reality: (I assumed each data sample quantized at 8 bits)

For each frame:
Number of pixels = 720 x 480 = 345,600
Number of Luminance samples = 345,600
Number of Cr samples = 345,600 / 4 = 86,400
Number of Cb samples = 86,400
Total number of Y and C samples = 345,600 + 86,400 + 86,400 = 518400

Then, in one second (nominally 30 frames):
Number of samples: 518400 x 30 = 15,552,000 = 15.5 MBytes (if one byte for each sample)

This is off by a factor of two from the 31 MB figure you mention above. Where did I go wrong?
 
OK, thanks. When I do the math using 4:4:4 I come up with 31.1MB/s. I was not disputing Barry number, I was just trying to convince myself that I understand how it comes about. Is my method correct for coming up with the 31MB number, i.e., (#pixels) X 3 (one Y, two C samples) X (#frames/sec) = 31MB/sec
 
I was using 720 x 480 x 3 for frame size. 720x480 pixels, x 8 bit depth, = a representation of the image from one CCD. So you would have to do it three times for three CCDs. That would be for 4:4:4 color.

So 720 x 480 x 3 x 29.97 = 31,072,896 bytes per second, or about 31 megabytes, or, more properly expressed, 30.34 megabytes (dividing by 1024 instead of 1000).
 
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