View Full Version : SAW gradients at various Gammas


Randy Strome
March 18th, 2008, 09:43 PM
I started a new post here so as not to further hijack anyone elses post with matters that many may wish to avoid. For those who have interest, here are some photographs taken of a the waveform produced by the EX1 SAW Gradient as viewed on a TV logic monitor connected via HDSDI.

**apologies in advance for the perspective issues - these were hand shot**

The color temp was set to 3200 (color temp does effect the curve)

Gain at 0 on row 1 and 2.

Row 1, left to right: STD1,2,3,4

Row 2, left to right: Cine1,2,3,4

Row 3 left to right: STD3 Slope at +99, STD3 Slope at -99, STD3 Gain at -3, STD 3 Gain at +3

I hope this is helpful, and I would love to hear some comments on the issue of the mapping of such a wide ranche of brights to a flat value. This baffles me. Possibly the unadjusted SAW pattern includes an area of straight white just to the right of the gradient???

http://www.dvinfo.net/gallery/showimage.php?i=860&c=3

Leonard Levy
March 19th, 2008, 12:01 AM
OK so what is a saw gradient anyway?

Craig Seeman
March 19th, 2008, 12:32 AM
I would love to have a saw gradient as test chart to point camera at and tweak to see how various settings impact the gamma curve, etc.

Can you actually adjust PP settings and see how the curve changes?

You don't get to see how this is handled through the optical system though.

Sebastien Thomas
March 19th, 2008, 02:56 AM
Just hope this will help you to understand :

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

Be carfull as pictures are with a +3 Gain. I hope to put picture at 0 gain tonight.

Also check the thread on hightlight clipping and picture profile receipe. They are full of information about that and how gamma curves change the picture look.

Randy Strome
March 19th, 2008, 08:50 AM
I would love to have a saw gradient as test chart to point camera at and tweak to see how various settings impact the gamma curve, etc.

Can you actually adjust PP settings and see how the curve changes?


Yes. Every change made in PP settings live updates the curve on the monitor, just as if you were pulling a curve in an editing program.

There are some very interesting thing to be learned here. For instance, black is not at zero on the waveform until PP "black" is set to -4. Also, it is not hard to "kink" the curves with combinations of settings, so operating without a visual reference might be ill advised.

Randy Strome
March 19th, 2008, 08:59 AM
OK so what is a saw gradient anyway?

This is brand new to me. I learned of its presence from Sebastien's post.

What makes sense to me (and what it looks like), is that the camera generates a constant black to white gradient (ie 0-255) that would mimic the sensor having recieved those values prior to processing. The camera settings are applied to this gradient before it shows on the LCD, HDSDI monitor, or SXS file. So, what you get when you view it on a wavefrom monitor is the actual curve that the camera is applying.

In short, these are your processing curves.

That is my understanding. Slam me if I am wrong. I want to know.

Bill Ravens
March 19th, 2008, 11:51 AM
Great work, Randy!!!!
I'm off to look at SAW.

Michael H. Stevens
March 19th, 2008, 12:04 PM
OK so what is a saw gradient anyway?

Please respond someone? What is SAW?

Greg Boston
March 19th, 2008, 12:23 PM
Please respond someone? What is SAW?

Well it depends. If you are in the military, it stands for Squad Automatic Weapon.

If you were to view the SAW pattern in video on an oscilloscope, the black to white gradient would show as a SAWTOOTH type waveform.

-gb-

Bill Ravens
March 19th, 2008, 12:46 PM
Well, after some initial runs thru HDRack, a few things are apparent to me.
1-I'm running BLACK way too low. This is confirmed by the noise problems I'm seeing in my shadows. The WFM shows BLACK should be at -3 to -4 for max dynamic range and no more.

2-Running at a gain of -3dB limits the max IRE to no greater than 90%. Running at a gain of 0dB allows max IRE to go up to 100% in the Std mode. Cine modes, of course, are restricted to max IRE of ~80%. So, if one is wanting max dynamic range, don't run at -3dB.

3-As someone said earlier, there is a definite hard knee at the top of the gamma curve. The knee can be tailored to smooth the hard corner transition.When operating with this hard knee, watch out for clipping, it can be disastrous. Tailoring the knee for a smooth transition to hard clip can have undesireable effects on the knee itself.

This is a way cool discovery. Mucho thanx to "prune" at lecentre and to Randy for posting.

Sebastien Thomas
March 19th, 2008, 12:50 PM
This is brand new to me. I learned of its presence from Sebastien's post.

What makes sense to me (and what it looks like), is that the camera generates a constant black to white gradient (ie 0-255) that would mimic the sensor having recieved those values prior to processing. The camera settings are applied to this gradient before it shows on the LCD, HDSDI monitor, or SXS file. So, what you get when you view it on a wavefrom monitor is the actual curve that the camera is applying.

In short, these are your processing curves.

That is my understanding. Slam me if I am wrong. I want to know.

The SAW should work the same way the color bars do. It should effectively simulate what the sensor view.
I don't know for the EX1, but I worked a lot with Sony F900, and if you see the internal architecture diagram (it may exist for EX1 also ?) you will see that :
1) gamma curve is the first thing applied, with knee function
2) then detail, white..
2) matrix is the last, with color correction, as long as I remeber

Of course, before the gamma, you have the shading options, which compensate the levels directly out from the sensor.

What you learn here is that matrix don't change the way the camcorder get the image, only how it save it. this means it is equal as doing it in post. So try to get balanced colors while shooting, then grade in post.

you see, SAW can teach a lot :)

and yes, if you look at the parade on my screengrab, you'll see the 3 toothsaw, where the name comes from.

Greg Boston
March 19th, 2008, 01:12 PM
matrix don't change the way the camcorder get the image, only how it save it.

The custom matrix adjustments can be thought of as an artist's pallete with a RED, GREEN, BLUE choice. Matrix is the equivalent of the artist dipping their brush in one well, and mixing it into the other to alter the color output from that well.

Both analog and digital SAW waveforms are on the F350, but they are in the service menu. The XDCAM SD cameras offer the SAW pattern in the user menus.

-gb-

Randy Strome
March 19th, 2008, 06:50 PM
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? I have set it over the output curve for Cine2 just for reference.

Thanks again,
Randy

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

Bill Ravens
March 19th, 2008, 07:23 PM
no.
a linear line isn't consistent with why the gamma exists. unlerss you plot on a log scale, which you're not.
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.

in mathematical terms, gamma is really an exponent. it represents a geometric value rather than a linear value. In very fundamental terms, there's 3 qualities a gamma curve has.
1-the clip at the foot
2-the clip at the head
3-the camber of the knee.(which can be related to the slope at the shadow, mid, and hi-lights), which is the geometric progression.

So, a sensor sees in the way you could envision a concave gamma curve would look. The eye sees in the opposite way, as a convex gamma curve. The "gamma" curve you apply is the correction to the sensor characteristics to match the human eye characteristic. Of course, playing one against the other allows certain "corrections" to the displayed image that enhances the image but doesn't represent the true inverse of what the camera sensor captured. In practice, a camera sensor is much more sensitive to changes in highlights than changes in shadows. Hence, more data is captured in the highlights than the shadows. Black stretch is an attempt to correct this non-linearity in the way a sensor sees light.

I suppose if you added one to the other, a true inverse added to the original gamma, would produce a linear line, but, it is rather meaningless except to denote a true inverse, or how much the vector sum varies from a true inverse vector sum.

Randy Strome
March 19th, 2008, 07:51 PM
Thanks Bill for the thorough answer. It will take me some time to digest.

Maybe I can simplify my question to get at the root of why I am asking. Do you envision the digitally produced test image (prior to any curves, etc being applied) to have an prolonged area of straight white just to the right of the gradient itself, or is the flat spot on each of the curves actually representing a portion of the initial gradient itself now mapped to a single value?

Thanks as always

Bill Ravens
March 19th, 2008, 08:11 PM
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
March 19th, 2008, 09:52 PM
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/showimage.php?i=862&c=3&userid=50923
http://www.dvinfo.net/gallery/showimage.php?i=863&c=3&userid=50923

Bill Ravens
March 20th, 2008, 07:59 AM
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
March 20th, 2008, 08:21 AM
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
March 20th, 2008, 08:29 AM
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
March 20th, 2008, 08:33 AM
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
March 20th, 2008, 08:42 AM
Greg..
you are right. ;o)
I was close, tho'.

Greg Boston
March 20th, 2008, 08:48 AM
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
March 20th, 2008, 08:49 AM
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
March 20th, 2008, 09:06 AM
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
March 20th, 2008, 09:06 AM
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
March 20th, 2008, 09:16 AM
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
March 20th, 2008, 09:19 AM
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
March 20th, 2008, 09:21 AM
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
March 20th, 2008, 09:32 AM
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?

Bill Ravens
March 20th, 2008, 09:36 AM
thanx, Greg.
wow, did I miss your question somewhere, Randy?
I just plugged my EX1 into HDRACK and brought up the SAW gradient. I measure a full range in the gradient, from RGB0 to RGB255, across the gradient. There are "flat spots" of super black and super white at either extreme of the gradient.
Hope this answers your q.

Greg Boston
March 20th, 2008, 09:55 AM
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?

I am saying that I don't know where slope #2 came from, but #1 is a 'true' sawtooth wave because of its endpoint being directly above the base.

I originally entered this discussion to explain what 'SAW' meant as it applies to that test gradient, in that it will appear on a scope as a sawtooth wave.

-gb-

Randy Strome
March 20th, 2008, 10:19 AM
thanx, Greg.
wow, did I miss your question somewhere, Randy?
I just plugged my EX1 into HDRACK and brought up the SAW gradient. I measure a full range in the gradient, from RGB0 to RGB255, across the gradient. There are "flat spots" of super black and super white at either extreme of the gradient.
Hope this answers your q.

Thanks again guys. I genuinely appreciate your input, although, if I am understanding Greg correctly, it has resulted in two contradictory opinions.

Edit: Or possibly Greg is saying that this sawtooth wave result is based on an input that included an area of gradient followed by an area of straight white-in which case, no disagreement at all.