SAW gradients at various Gammas

[Bill Ravens]

as I understand what you're asking...
the gradient you see as the SAW image, really is a true geometric(as opposed to linear) gradient...no flat spots. Take an frame grab of the SAW image into Photoshop and look at it with the eyedropper. I think you'll see a continuous gradient, no "flat spots" no constant values as you move across the gradient. It's a true gradient, non-linear, to match the gamma of the sensor. In fact, I think it is the true inverse of the way the sensor sees. Add the gradient of the SAW to the gradient of the sensor, and you may well get a linear line, something like what you drew, with a perfect slope of 45 degrees, or 1:1.

[Randy Strome]

I am sure I will feel silly for this, but this is my confusion:

In Photoshop, the related areas indicated in the two pics below are at a static 207,207,207 in photoshop. My question, was this area on the "untouched" test pattern originally gradient that has been mapped to flat?

Thanks for your good humor.

http://www.dvinfo.net/gallery/showim...3&userid=50923
http://www.dvinfo.net/gallery/showim...3&userid=50923

[Bill Ravens]

hmmm...how did you get the image into photoshop?
IOW, what PP setting were you using? If you used settings that limited super white to a value less than 100 IRE, your image capture would reflect that max value. Since it measures 207, then 207/255= ~80%. Using a gain setting of -3dB will limit your super white like this.

[Randy Strome]

Originally Posted by Bill Ravens

hmmm...how did you get the image into photoshop?
IOW, what PP setting were you using? If you used settings that limited super white to a value less than 100 IRE, your image capture would reflect that max value. Since it measures 207, then 207/255= ~80%. Using a gain setting of -3dB will limit your super white like this.

Hi Bill,

That image and the resulting waveform was pirated from Prunes post here which was done at +3 db:

http://www.lecentre.net/blog/pmw-ex1/gamma

I chose just one (Cine 2) for my question, but each of the gradients has a long solid value at its end, and each of these values is confirmed by the corresponding flat spots on the waveform curve.

I simply do not know if this is a feature of SAW (I am brand new to the term and any of its properties) that would for instance apply a long solid white tone at input to give a more sizable referance for how input white would be handled or if this was initially gradient that has been processed to white.

Apologies to belabor this.

[Bill Ravens]

Randy...

Cine2 gamma forces whites to be legal. If one assumes that the EX1 puts superwhite at 108%(as I believe it does), then the Cine2 gamma will force superwhite to be 100%. Superwhite is RGB255, white is 235. If white is remapped 8% lower by Cine2, that would be RGB216. Now, if Prune did a screen capture or frame grab of this image of the SAW to bring it into his software wfm, 207 seems close. I'm only guessing, here. I should probably try to duplicate what he did on my cam to answer the question.

I think I understand what you're getting at. Are you suggesting the flat spot on the SAW curve is the result of a flat spot on the calibration gradient and not really due to clipping?

[Greg Boston]

Originally Posted by Bill Ravens

gamma exists because the human eye doesn't see linearly. a gamma curve also represents the way a sensor(ccd or cmos) "sees" light. typically, a sensor sees the inverse of the way an eye sees. when the image is presented on a monitor, a "reverse" gamma is applied to make the way a sensor sees consistent with the way a human eye sees. there is no "linear" involved and really doesn't represent anything.

Bill, I respectfully disagree to some extent.

In the early days of television, it was determined that the response curve of a CRT was not linear. Given the ratio of cameras to TV sets, the engineers decided to cancel the gamma curve of the CRT by applying an equal and opposite curve to the camera's processing. They figured it would be easier to adjust a few cameras than the many television sets. The resultant algebraic curve would then be a straight line slope. It was more about getting the picture on your tv set to be represented as the camera saw it.

As we wean ourselves away from CRT technology, this necessity for gamma curve compensation may become obsolete, except for the express purpose of emulating film stocks.

regards,

-gb-

[Bill Ravens]

Greg..
you are right. ;o)
I was close, tho'.

[Greg Boston]

Originally Posted by Randy Strome

Thanks to all of you for your thought and effort on this. Just to clarify, do you envision the unaltered input "curve" of the SAW gradient to be what I have represented by the escalating straight line in the link below?

http://www.dvinfo.net/gallery/showimage.php?i=861&c=3

Yes Randy, that straight line slope in a repeating fashion is what's known as a 'sawtooth' waveform. Other common waveforms are sine, square, and triangle. These waveforms in the range of audible frequencies are the basic building blocks for sound synthesis. If you were to look at the control panel of a vintage analog synthesizer, you would be presented with oscillators that give you a choice of these waveform types. Actually, they are still used on modern day synths for the LFO (low frequency oscillator). See the screen grab from my voice editor.

-gb-

[Randy Strome]

Originally Posted by Bill Ravens

Randy...

Cine2 gamma forces whites to be legal. If one assumes that the EX1 puts superwhite at 108%(as I believe it does), then the Cine2 gamma will force superwhite to be 100%. Superwhite is RGB255, white is 235. If white is remapped 8% lower by Cine2, that would be RGB216. Now, if Prune did a screen capture or frame grab of this image of the SAW to bring it into his software wfm, 207 seems close. I'm only guessing, here. I should probably try to duplicate what he did on my cam to answer the question.

Again, thanks for lending your knowledge to this.

Bill, my mistake, I should have written Cine1, as that is the example I used, but actually, all of the Gammas including the STD's have the same flat spot. Additionally, it shows on the waveform through HDSDI as well, so no real need to bring it to photoshop at all, except to confirm.

I think I understand what you're getting at. Are you suggesting the flat spot on the SAW curve is the result of a flat spot on the calibration gradient and not really due to clipping?

I am wondering that.

As I see it, the flat spots can only be one of two things, a flat spot on the calibration gradient similar to the flat blacks that surround the gradient *or* gradient that has been converted to flat.

[Randy Strome]

I think it is possible that the applied processing, in all instances, is mapping at least all of the incoming superwhites to a flat value.

[Bill Ravens]

If that were true, Sony is a lot more ignorant than I think they are. Why use a SAW curve that doesn't extend from superblack to superwhite? That would defeat the purpose of having a calibration gradient built in to the camera. Like making a colorbar pattern that is wrong.

[Randy Strome]

Originally Posted by Greg Boston

Yes Randy, that straight line slope in a repeating fashion is what's known as a 'sawtooth' waveform.
-gb-

Thanks for the added info Greg,

My question was actually intended to be less about the form of the unaltered gradient's curve (or straight line as it may be), but of its end point positions. Which is correct?

http://www.dvinfo.net/gallery/showimage.php?i=864&c=3

[Randy Strome]

Originally Posted by Bill Ravens

If that were true, Sony is a lot more ignorant than I think they are. Why use a SAW curve that doesn't extend from superblack to superwhite? That would defeat the purpose of having a calibration gradient built in to the camera. Like making a colorbar pattern that is wrong.

In either case it would extend from superblack to superwhite (0-255). The top value would still be 255, the question would be did they include an addition straight white reference tone to the right of the gradient?

Maybe this will help with my question

http://www.dvinfo.net/gallery/showimage.php?i=864&c=3

[Greg Boston]

Originally Posted by Randy Strome

Thanks for the added info Greg,

My question was actually intended to be less about the form of the unaltered gradient's curve (or straight line as it may be), but of its end point positions. Which is correct?

http://www.dvinfo.net/gallery/showimage.php?i=864&c=3

Number 1 is correct as it lines up over the base at the bottom. All scopes will show this type of behavior. A fast transition from white to black (or vise versa) is faster than the response time of the scope and will show a blank between the high and low, you have to fill in the blank with a straight line. Same thing happens with square waves, they look like a set of alternating high and low straight lines on a scope because it can't register the fast vertical transition of the waveform.

-gb-

[Randy Strome]

Originally Posted by Greg Boston

Number 1 is correct as it lines up over the base at the bottom. All scopes will show this type of behavior. A fast transition from white to black (or vise versa) is faster than the response time of the scope and will show a blank between the high and low, you have to fill in the blank with a straight line. Same thing happens with square waves, they look like a set of alternating high and low straight lines on a scope because it can't register the fast vertical transition of the waveform.

-gb-

Thanks Greg, but you have lost me :)

Are you saying that #1 does not indicate that all values above (in this case 207) are being mapped to a flat value?