[jwvm]

On Mar 18, 9:12 pm, G-squared wrote:

snip

You were incorrect then and now. There is no luminance information in
the color difference signals.

All right, then please explain the following paradox. Suppose that
there are two identical objects in a scene with varying illumination.
In a bright part of the scene, the object has a red channel value of
100 and the other two channels are (artificially) zero. This won't
happen in the real world but the principle remains the same. Using the
LUV coordinate system, luminance is defined as

Y = 0.299R + 0.587G + 0.114B

Y = .299(100) = 29.9

Cr = R-Y = 100-29.9 = 70.1

Now suppose that in a dimmer area we have R=50 and the other channels
are still zero.

Y = .299(50) or about 15 rounded

Cr = R-Y = 35

The color difference has to be less in dimmer areas of an image. Why
else would saturation be normalized in the HSI coordinate by the
luminance?

As a test, I created two all-red bmp files. In one case, the red value
was 127 and in the other case it was 255. Clearly the saturation is
100% in both cases. Using Jim Mack's vectorscope program, the dimmer
image vector was about 2/3 of the way to the circle while the bright
image vector actually went a bit outside of the circle. Again, this
indicates that color differences are clearly related to saturation but
there is still luminance information there.

For another test, I took the color bars image and reduced the
brightness while maintaining the same proportions of red, green and
blue. The vectors were always much shorter in the dim image even
though the saturation was essentially unchanged.

[G-squared]

On Mar 18, 6:27*pm, jwvm wrote:
On Mar 18, 9:12 pm, G-squared wrote:

snip

You were incorrect then and now. There is no luminance information in
the color difference signals.

All right, then please explain the following paradox. Suppose that
there are two identical objects in a scene with varying illumination.
In a bright part of the scene, the object has a red channel value of
100 and the other two channels are (artificially) zero. This won't
happen in the real world but the principle remains the same. Using the
LUV coordinate system, luminance is defined as

Y = 0.299R + 0.587G + 0.114B

Y = .299(100) = 29.9

Cr = R-Y = 100-29.9 = 70.1

Now suppose that in a dimmer area we have R=50 and the other channels
are still zero.

Y = .299(50) or about 15 rounded

Cr = R-Y = 35

The color difference has to be less in dimmer areas of an image. Why
else would saturation be normalized in the HSI coordinate by the
luminance?

As a test, I created two all-red bmp files. In one case, the red value
was 127 and in the other case it was 255. Clearly the saturation is
100% in both cases. Using Jim Mack's vectorscope program, the dimmer
image vector was about 2/3 of the way to the circle while the bright
image vector actually went a bit outside of the circle. Again, this
indicates that color differences are clearly related to saturation but
there is still luminance information there.

For another test, I took the color bars image and reduced the
brightness while maintaining the same proportions of red, green and
blue. The vectors were always much shorter in the dim image even
though the saturation was essentially unchanged.

I finally fiigured out what our problem is. You're using the wrong
tems to express yourself. You claim that the color difference signal
changes as you change the value of the Red while Blue and Green stay
at 0. Of course it works that way but you're calling it luminance.
Luminance to a TV engineer is the 'Y' or black and white portion of
the signal. while you change the Red, it is the only portion of the
luma so of course all 3 components track together even though you're
only changing the one color. For a given luma value when all 3 colors
are contributing, changing the 2 difference signals up/down to
increase/decrease saturation will change the _relative_ values of R/G/
B such that the value of luma which is as you stated, Y = 0.299R +
0.587G + 0.114B

This is only valid if all RGB values remain positive (no negative
light - yet) AND less than max value. As soon as 1 of the colors
reaches min or max, the difference signals will start to scale.

GG

[Alan[_2_]]

In article G-squared writes:

I finally fiigured out what our problem is. You're using the wrong
tems to express yourself. You claim that the color difference signal
changes as you change the value of the Red while Blue and Green stay
at 0. Of course it works that way but you're calling it luminance.

I don't think he is. He is simply claiming that for this image,
the R-Y signal will vary at the same time as the R does. Since the
luminance (Y) is composed of 0.299 R, the luminance is varying.

Luminance to a TV engineer is the 'Y' or black and white portion of
the signal.

yes, and as both of you noted, it is a function of the R G and B
inputs.

He is showing that for some signals (with fixed saturation) the
R-Y signal will vary when the luminance varys. The example he used
was constructed with input only in the red channel.

It doesn't seem like he is mis-using the term luminance.

You could test this by disconnecting the green and blue inputs
to an encoder, and looking at the R-Y component generated. From
his example, it will vary.

Since Y = 0.299 R + 0.587 G + 0.114 B - and G and B are zero,
Y = 0.299 R
or
R = Y / 0.299
finding the R - Y difference signal
R - Y = R - 0.299 R
= 0.701 R
substituting for R
R - Y = (0.701 / 0.299) Y

Thus, it appears that R - Y varies linearly with Y for signals with
100 percent red saturation.


Alan

[jwvm]

On Mar 19, 3:31 am, (Alan) wrote:
In article G-squared writes:

I finally fiigured out what our problem is. You're using the wrong
tems to express yourself. You claim that the color difference signal
changes as you change the value of the Red while Blue and Green stay
at 0. Of course it works that way but you're calling it luminance.

I don't think he is. He is simply claiming that for this image,
the R-Y signal will vary at the same time as the R does. Since the
luminance (Y) is composed of 0.299 R, the luminance is varying.

Luminance to a TV engineer is the 'Y' or black and white portion of
the signal.


It is also a common concept in image analysis.

yes, and as both of you noted, it is a function of the R G and B
inputs.

He is showing that for some signals (with fixed saturation) the
R-Y signal will vary when the luminance varys. The example he used
was constructed with input only in the red channel.

The assumption here was that full spectrum illumination was used but
the object only reflected red light. No such object exists of course
but I used it to illustrate the concept.

snip

Thus, it appears that R - Y varies linearly with Y for signals with
100 percent red saturation.

Alan

As I pointed out earlier, scaling the color bar image also showed that
vectorscope vectors are strongly affected by illumination. Whenever
at least one color was low, the vector differed markedly in length
between the bright and dim images. In both cases the saturation was
unchanged but the luminance changed. When all colors had the same
amplitude, the vector had zero length as expected because saturation
was zero.

[Albert Manfredi[_2_]]

On Mar 19, 3:31*am, (Alan) wrote:

* You could test this by disconnecting the green and blue inputs
to an encoder, and looking at the R-Y component generated. *From
his example, it will vary.

* Since Y = 0.299 R + 0.587 G + 0.114 B * - and G and B are zero,
* * *Y = 0.299 R
or
* * *R = Y / 0.299
finding the R - Y difference signal
* * *R - Y = R - 0.299 R
* * * * * *= 0.701 R
substituting for R
* * *R - Y = (0.701 / 0.299) Y

* Thus, it appears that R - Y varies linearly with Y for signals with
100 percent red saturation.

You can show that the difference signals vary as luminance varies in
other ways too, which is why I found this subthread to be a bit of a
distraction to the original point.

Start with bright white and then gradually remove all the red and all
the blue, while retaining a full strength green.

In RGB, you start with full strength R, G, and B levels, and notice
gradually decreasing R and B levels. The G remains constant at full
strength.

To reconstruct RGB from Y Pb Pr,

R = Y + Pr = Y + (R-Y)
B = Y + Pb = Y + (B-Y)
G = (Y - 0.3R - 0.11B) / 0.59

So what is the effect of starting with full white, then gradually
reducing the R and B levels, while reatining full green, on the Y Pb
Pr interface?

Since Y = 0.3R + 0.59G + 0.11B, The Y component will gradually reduce
from 1 to 0.59.

Since Pr = R-Y and Pb = B-Y, they will both start at 0 for the white
(because R = B = Y = 1 for max white), and they will both end up at -
Y, or -0.59, when R and B are at a zero level.

So, does this mean that "the difference signals carry luminance
information?" I don't know the semantics here, but they certainly DO
vary as luminance varies. There may be some issues with the jargon
that I'm not aware of.

Bert

[G-squared]

On Mar 19, 8:47*am, jwvm wrote:
On Mar 19, 3:31 am, (Alan) wrote:

In article
G-
squared writes:

I finally fiigured out what our problem is. You're using the
wrong
tems to express yourself. You claim that the color difference
signal
changes as you change the value of the Red while Blue and Green
stay
at 0. Of course it works that way but you're calling it
luminance.

* I don't think he is. *He is simply claiming that for this
image,
the R-Y signal will vary at the same time as the R does. *Since
the
luminance (Y) is composed of 0.299 R, the luminance is varying.

Luminance to a TV engineer is the 'Y' or black and white portion
of
the signal.

It is also a common concept in image analysis.

* yes, and as both of you noted, it is a function of the R G and
B
inputs.

* He is showing that for some signals (with fixed saturation) the
R-Y signal will vary when the luminance varys. *The example he
used
was constructed with input only in the red channel.

The assumption here was that full spectrum illumination was used
but
the object only reflected red light. No such object exists of
course
but I used it to illustrate the concept.

snip

* Thus, it appears that R - Y varies linearly with Y for signals
with
100 percent red saturation.

* * * * Alan

As I pointed out earlier, scaling the color bar image also showed
that
vectorscope vectors *are strongly affected by illumination.
Whenever
at least one color was low, the vector differed markedly in length
between the bright and dim images. In both cases the saturation was
unchanged but the luminance changed. When all colors had the same
amplitude, the vector had zero length as expected because
saturation
was zero.

I stand corrected. Since 'Y' is part of all 3 equations, changing 1
color only by of course will change all 3 signals. I was looking at
the same data froma different standpoint and drew the wrong
conclusion. Thanks.

I was looking at it from the standpoint of varying hue/sat by
manipulating the color difference signals without changing the luma
value. It's also easy to vary the luma of individual colors working
with delta R-Y, delta B-Y and delta Y where delta is modulated decoded
RGBYCM.

GG

GG

[Alan[_2_]]

In article Albert Manfredi writes:

So what is the effect of starting with full white, then gradually
reducing the R and B levels, while reatining full green, on the Y Pb
Pr interface?

Since Y =3D 0.3R + 0.59G + 0.11B, The Y component will gradually reduce
from 1 to 0.59.

Since Pr =3D R-Y and Pb =3D B-Y, they will both start at 0 for the white
(because R =3D B =3D Y =3D 1 for max white), and they will both end up at -
Y, or -0.59, when R and B are at a zero level.

So, does this mean that "the difference signals carry luminance
information?" I don't know the semantics here, but they certainly DO
vary as luminance varies. There may be some issues with the jargon
that I'm not aware of.

This time the chrominance varied, as the signal went from zero saturation
to fully saturated green. You expect the difference signals to change.

That differs from previous examples where the hue and saturation remained
constant, but the luminance varied.

Alan

[Albert Manfredi[_2_]]

On Mar 19, 11:54*pm, (Alan) wrote:
In article Albert Manfredi

[When going from white to saturated green]

Since Y = 0.3R + 0.59G + 0.11B, The Y component will gradually reduce
from 1 to 0.59.

Since Pr = R-Y and Pb = B-Y, they will both start at 0 for the white
(because R = B = Y = 1 for max white), and they will both end up at -
Y, or -0.59, when R and B are at a zero level.

* This time the chrominance varied, as the signal went from zero saturation
to fully saturated green. *You expect the difference signals to change.

* That differs from previous examples where the hue and saturation remained
constant, but the luminance varied.

My intent when asking the two questions (show how bright white is
transmitted in Y Pb Pr, and show how saturated green is transmitted in
Y Pb Pr) was to cover everything that had been argued through the
thread in one fell swoop.

Sure enough, a white signal creates zero bandwidth in the two color
difference signals Pb and Pr.

But in the case of only saturated green, which is an extreme case, the
general answer I showed above was that:

Pb equals -Y
Pr equals -Y

So this answers several points in the long thread.

1. The color difference signals can vary directly as Y varies (180
degrees out of phase).

2. The bandwidth of Pb and Pr may be identical to the bandwidth of Y.

3. The energy of the RGB signal gets redistributed when using Y Pb Pr.
To wit, pure green only exercises the G line of RGB, but creates
signals in all three lines of Y Pb Pr. Conversely, pure white does the
opposite.

4. It's really hard to correctly understand the underlying principles
if one goes merely by the seat of the pants. You really have to do the
math to see what's going on.

Bert

[Sal M. Onella]

"Albert Manfredi" wrote in message
...

snip

It's really hard to correctly understand the underlying principles
if one goes merely by the seat of the pants. You really have to do the
math to see what's going on.


... which is why we benefited from being dragged kicking and screaming
through all that calculus.

For example, as a tech, I had learned the circuit functionality of
wave-shaping circuits like differentiators and integrators. I saw their
behavior in the sync-separator circuits in television sweep circuits, for
example and understood.

However, it wasn't until decades later, when I went to college, that I
saw the pure math of differentials and integrals. It all made such perfect
sense. I could kick myself for waiting so long.