Harsh, "aliased" sound with digital TV converter box.

G-squared wrote:

You didn't look at PicScope - which is correct


Jim Mack wrote:

From the looks of this (lengthy) thread, I suspect there are a
few who might find this interesting. It's the beta version of a
"scope" pair that works on still images.

I actually do compute color difference signals (UV) to drive the
vectorscope, but I don't make them available at the line scope.
Still, the results can be instructive. A nice SMPTE color bar BMP
is included.

Take a look:

http://www.microdexterity.com/demos/PicScope.zip

--
Jim


Thanks Jim. How much does it cost?

GG


What was it the bartender said to the neutron?

We're treating this as an extended beta. Anyone can request a reg
code. We'd appreciate any feedback, and particularly any good sample
images that show the program's features to good effect.

Right now it's fully functional, with nags that the reg code removes.
When (if) it is available for sale, it'll be pretty cheap.

--
Jim

"Albert Manfredi" wrote in message
...

I keep trying to find the right combination of words to make this fact
click. If you can answer these questions specifically, and I mean
using simple algebra, you will notice without a shadow of a doubt that
all of the energy in an RGB signal is present also in the Y Pb Pr
signal, with no exception (unless one low-pass filters the difference
signals).

Of course that's true. So what? What does this have to do with the
discussion?

"jwvm" wrote in message
...

Color differences, however, are a function of both saturation
and luminance.

If they were, things wouldn't work the way they do.

On Mar 18, 12:14*pm, jwvm wrote:
On Mar 18, 3:10 pm, G-squared wrote:

snip

The difference signals ARE
strictly saturation / hue and the saturation CAN be high in low
luma
video. It's called illegal video.

You are absolutely correct that saturation can be high in dim areas
of
a scene. That is not what I was saying. Color differences, however,
are a function of both saturation and luminance. Differences will
be
higher in bright areas compared to dim areas even though identical
features in these areas have the same saturation. A strict
definition
of saturation such as for the HSI color model includes division by
the
intensity for normalization.

Changing saturation is very easy -
just change the amplitude of the differences.

I have no problem here because significant information about
saturation is present in the difference signals. Please remember
that
this thread got started when William stated that simply finding
color
differences reduced bandwidth and justified it partially on the
basis
that there is no luminance information with color differences. This
is
not true as noted in my preceding comment.

You were incorrect then and now. There is no luminance information in
the color difference signals. Look at Jim Mack's PicScope (way cool
BTW). He includes a line rate grey scale. You will see no vectors
meaning there is no color, just greys. Color differences are _not_ a
function of the luminance level. While it is true that typical images
have less saturation in lower luminance areas, that does _not_ mean
that they cannot or do not. Computer graphics/animation folks
frequently push the envelope on that issue. It's mainly a problem with
composite transmitted in analog and nearly a non-issue in digital
NOT because of digital but limits on amplitude of subcarrier for
composite. Subcarrier cannot exceed 110 IRE or -20 IRE but MAY exceed
100 IRE of chroma as long as the luma value does not make the
subcarrier exceed the upper / lower limits.

William is correct on his assertion that there is no luminance
information in the difference channels. His minor mistake was
including the bandpass filtering of the difference channels and
interpreting it as one operation when it is 2 distinct operations.

GG

On Mar 18, 8:58*pm, "William Sommerwerck"
wrote:
"Albert Manfredi" wrote in message

I keep trying to find the right combination of words to make this fact
click. If you can answer these questions specifically, and I mean
using simple algebra, you will notice without a shadow of a doubt that
all of the energy in an RGB signal is present also in the Y Pb Pr
signal, with no exception (unless one low-pass filters the difference
signals).

Of course that's true. So what? What does this have to do with the
discussion?

It has EVERYTHING to do with the discussion.

I wrote:

Yes, I think you don't appreciate that what information is taken away
from the difference signals is information that is added back in the
form of the new Y component. And yet, it's a simple concept.

And you replied, and this was recent was there's no excuse for
forgetting:

And it's wrong.

So, do the simple algebra and prove it to yourself. There is NOTHING
about Y Pb Pr that inherently saves bandwidth compared with RGB, as
you have repeatedly stated. Nothing at all. And the algebra will prove
it, but you have to do it or you'll never understand.

What happens with brilliant white? What are the Pr, Pb, and Y
components, and how do they differ from an RGB instance of brilliant
white?

What happens when R and B levels are zero? What does this say about
the bandwidth of the Pb and Pr components compared with the bandwidth
of Y?

If you answer these two questions, you will disprove everything you've
been so stubborn about.

Bert

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

William is correct on his assertion that there is no luminance
information in the difference channels. His minor mistake was
including the bandpass filtering of the difference channels and
interpreting it as one operation when it is 2 distinct operations.

It was rather central to the original point, early on in the thread,
that the reason for using YUV in NTSC was *not* for some inherent
bandwidth savings, but rather for (a) compatibility with black and
white, and (b) wasn't it fortunate that the necessary low-pass
filtering of the difference signals did not detract too much from the
final color image.

The supposed reasons for the "inherent" bandwidth savings when using
difference signals, which were not explanations at all it turns out,
created the distracting and irrelevant (to this point) side
discussions. IMO, of course.

Bert

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.

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

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

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.