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March 13th, 2005, 09:53 AM | #31 |
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Andre,
Sorry, I'm not following your explanation. Do you mean that some sharpener implementations have frequency responses proporional to higher powers of frequency? Sharpeners are usually simply differentiators (first order) with the impulse responses shaped (frequency response approximately proportional to frequency) for the magnitude of the enhancement we want to see and the pixel width over which we want to see them. I also don't understand the part about the shading. The transition from the doorway to the wall is of about the same magnitude as the transition from the wall to the lamp post. Shouldn't they be subject to the same amount of sharpening? The answer is that they should and in fact they are subject to the same amount though this is not easily perceived fromt he image itself. I've put a plot of a horizontal line of luminance (through the screwhole in the light swithchplate) at http://www.wetnewf.org/sharpening.jpg. This plot clearly shows the classic step response for differentiating sharpener: a droop before a transition from dark to light followed by an overshoot after the transition and converesely going from light to dark. I've marked some of these on the plot. Approximately equal transitions are approximately equally sharpened. Of course once you have looked at this plot and go back to the frame you'll see the sharpening in the frame as well. So at this point I think the "ghost" IS sharpening (in fact I'd say the luminance plot is a pretty strong proof). What I don't understand, however, is why changing the camera settings doesn't change the result. Now back for a momemnt to http://www.wetnewf.org/ghostshp.jpg (in which I took Frank's frame and sharpened it still further to the point where the over/undershoots were plainly visible). If you blow up that frame in the region to the right of the switch plate you will see a lovely illustration of what we are talking about with respect to the way the DCT works and the kind of artifacting it leads to. You can see the tiles I was talking about and you can see some of the basis functions that were illustrated in the article Lauri mentioned. This is for educational value only and is not to be construed as a statement on my part that I think the ghosting was from DCT's. I don't now and never did. |
March 13th, 2005, 11:30 AM | #32 |
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Unfortunately I cannot access yr pic. But on the other points:
1. Simple differentiation doesn't produce preshoots in the time domain, only overshoots( linear phase differentiator). The frequency response (magnitude) is indeed "proportional" to the frequency but this doesn't define the transient (time domain) properties as long as the phase characteristic of a filter isn't known. 2 On the "shade" issue: indeed the doorway/wall transient has (about) the same magnitude, but like I mentioned already it's the value of the second derivative that counts and this relates to magnitude AND steepness (focus). 3. DCT and other tranform based compressors introduce all kinds of artifacts, tiling is one of them if a rough quantization is used for the DCT components, but this doesn't relate to Frank's "ghost" pics. And b.t.w. DCT compression doesn't reduce bandwidth as such, it only skips low level DCT components (again depending on the Q table) and these are not always the higher freq components. |
March 13th, 2005, 02:03 PM | #33 |
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Sorry about the images. Try http://www.pbase.com/image/40759798 for the plot and http://www.pbase.com/agamid/image/40759802 if you want to look for the DCT tiles in my sharpened version of Frank's image. With repect to
1. Simple differentiators do produce some predroop but not in the way we'd like to see it. I think the best way to design a sharpener is to specify the desired step response (and I'm sure that's what the two controls on the typical USM implementation do) and then calculate the impulse response from that. Some tweaking may still need to be done to limit the size of the mask and keep the computation burden under control. A desireable response, like the one implemented in the XL2, in Photoshop etc has a frequency response that looks like a differentiator at lower frequencies i.e. there is a portion of the response that rolls up with omega but it looks like a differentiator followed by a low pass filter (and there is gain at dc). Given that the impulse repsonse is symmetric (which it would be if the step response is symmetric disregarding the dc component), phase is dead linear. 2. The second derivatives are the same: approximately 0. The transitions are from a low luma level to a high one in three points. The slopes are essentially constant. 3. I've said it before and I'll say it again. I don't believe the ghosting in Frank's frame has anything to do with tiling! The comment about tiling was for people who might be interested in what the DCT does. My doctored image shows it very well. As to whether DCT compression reduces bandwidth or not that would depend on your definition of bandwidth. It does not reduce bandwidth in the spatial domain as the resolution of the images is not effected though the amount of information conveyed about the high frequency part of the image is reduced. It does reduce the required capacity of the channel required to convey the image becuase fewer bits are required to describe each frame. When I bought this camera I never dreamed I'd be so far down into the bowels of it. Cheers, A.J. |
March 13th, 2005, 03:47 PM | #34 |
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Thx, I got the pics.
The luma drawing shows that both edges are enhanced. Because of the gamma correction applied before enhancement, white to black transients allways show somewhat more overshoot than the black to whit transients. That's why the pole has the black left border line. For the other points: 1. Just draw a squarewave signal and add some amount of its first derivative (differenciated squarewave) and try to find preshoot in the resulting signal... 3. The final amount of data conveyed about the high freq, part is, apart from the dynamic Q-table for the DV codec, only the result of the applied entropy (Huffman) coding. And further...I suspect that Frank's image got most of its artifacts (tiling and mosquito noise around the arrow) because of a strong second (re)compression. |
March 13th, 2005, 05:52 PM | #35 |
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Andre,
Can't do 1. because the first derivative of a square wave contains Dirac delta functions. But if I run a sampled square wave through a FIR differentiator (which is, of course, an approximation of a perfect differentiator) I do see some pre droop. Not much I grant you but it is there. The Huffman encoder deals with whatever bitstream it is handed. If the resultant rate is more than the channel can handle the process is throttled by tweaking the Q matrix. As its entries are powers of 2 each increment (e.g. from 1 to 2 or 2 to 4) throws away another bit.Not only does this present fewer bits to the encoder but as the number of bits per coeficient is reduced the probability of longer runs (and more efficiency for the encoder) is increased. Your point about the second compression is well taken. After all this analysis of Frank's frame I'll have to try some of my own! Cheers, A.J. |
March 14th, 2005, 04:30 AM | #36 |
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Just try a "real" sqaurewave in yr simulation (non zero rise time) in combination with a non zero time constant for the differentiator.
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March 14th, 2005, 10:04 AM | #37 |
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That's what I'm doing - designing differentiators with varying numbers of taps and transition bandwidths using the Parks-McLellan (Remez exchange) algorithm. The software computes the step response for each design. Each step response shows pre-ringing.
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March 14th, 2005, 10:32 AM | #38 |
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If you talk about "varying numbers of taps" you are no longer talking about a differentiators but recusive and/or non recursive filters. This has nothing to do with differentiation of the input function. And remenber first order differentiation is about determining dv/dt and this is only positive or negative for resp rising or falling edges, so no preshoot in a first order differentiator.
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March 14th, 2005, 11:07 AM | #39 |
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Messers Parks and McLellan think I am talking about differentiators. I suppose the question is how you would compute the first derivative of a stream of samples. The first difference? That is, in fact, a 2 tap (zero) FIR filter which only approximates a diffrentiator and not as well as a FIR filter with more taps.
Mathematically, a differentiator is a filter whose response is proportional to frequency and thus 0 at dc. A mathematical differentiator thus has infinite bandwidth and infinite gain (at infinite frequency). Such a differentiator has a step response with no predroop but is a mathematical abstraction (as is the step response - a pulse of 0 width, infinite height and unit area). In the DSP arena we can only approximate differentiators but, within the Nuyquist bandwidth do a damn good job if we use enough taps. I'm sure 99.99% of the readership has lost interest at this point and I'm sure that you can find someone on your side of the pond to discuss this with as opposed to turning this into an EE forum. I can recommend some DSP texts if you like. A.J. |
March 14th, 2005, 01:17 PM | #40 |
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Indeed nobody is maybe interested any further and certainly I am no longer interested in such discussions. After being involved for over 30 year in the design and implementation of real (not "mathematically") digital and analog signal processing that's sufficient for me. And no need for DSP literature. I wrote enough articles myself on the issue.
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March 15th, 2005, 12:39 AM | #41 |
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Thanks for all the info everyone.
En groetjes van een ex-antwerpenaar Andre ! :) Frank |
March 15th, 2005, 02:31 AM | #42 |
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I love it when you guys talk dirty :-)
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March 17th, 2005, 04:19 PM | #43 |
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Hallo Frank, Belgie zendt zijn zonen uit... Now back to the "rules".
I used to know the Oakland area pretty well and often visited Berkely university when I was for a post-doc (1974-75... long time ago) at the "nearby" Stanford University. I was involved in medical ultrasound imaging and processing at that time. Till 2003 I was with Barco(Belgium) as VP R&D. Now I am a "retired" engineer... |
March 18th, 2005, 12:35 AM | #44 |
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Heh ! I haven't been in the US that long. Only 8 years from january 1997.
I'm now working at Pixar. Came to the US to CG work. I have no plans at all to go back to Belgium. Too much fun here ! :-) Frank |
April 7th, 2005, 10:15 AM | #45 |
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It might be luma undershoot as described at this site:
http://www.dvcentral.org/DV-Beta.html I've noticed this artifact on some of my stuff. Mostly along vertical edges though. Its really a frustrating thing because you don't see it in the view-finder and you think you've got a pristine image, when you really don't. Every thing else is so great about this camera. I'ts really a shame Canon didn't do something about this problem. Michael Hamilton |
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