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

On Mar 8, 6:51 am, "William Sommerwerck"
wrote:
"Smarty" wrote in messagenews:[email protected]
Imagine a blue screen with a gradient or checkerboard of blue values
only, as an example. Alternating squares of light and dark blue, or a
smooth transition from light to dark, or any conceivable variation of
blues
including steps, ramps, wedges, bars, etc.
This blue signal B will contain, before conversion into NTSC, lots of
high frequency content, if the transitions are abrupt / small, and little
or no high frequency content if it the blue field is solid or smoothly
transitioning.
Now imagine what the luminance component Y of this signal would look
like, namely, an envelope exactly the same in amplitude (since red and
green are zero) and exactly the same in spectral / frequency components
as the blue signal.
NOW........subtract the two (as in B-Y).........
Note that the difference would be precisely, exactly zero at all times.
Their instantaneous values, being precisely the same, cancel each
other entirely.

Your mental experiment is wrong. As the color-difference signal represents
saturation, and a blue field has at least _some_ saturation, the resulting
signal _can't_ be zero.

Here's the error in your reasoning... You're ignoring the way the Y signal
is derived.

Ignoring absolute luminance levels, the situation you describe is B = 1, R =
G = 0. The Y signal isn't 1, it's

0.30 R + 0.59 G + 0.11 B

Therefore, the B-Y signal must have an amplitude of 0.89, not zero.

Color difference signals are not the same as saturation! Please
provide a citation that states otherwise.

On Mar 8, 10:36*am, jwvm wrote:
On Mar 8, 6:51 am, "William Sommerwerck"
wrote:





"Smarty" wrote in messagenews:[email protected]
Imagine a blue screen with a gradient or checkerboard of blue values
only, as an example. Alternating squares of light and dark blue, or a
smooth transition from light to dark, or any conceivable variation of
blues
including steps, ramps, wedges, bars, etc.
This blue signal B will contain, before conversion into NTSC, lots of
high frequency content, if the transitions are abrupt / small, and little
or no high frequency content if it the blue field is solid or smoothly
transitioning.
Now imagine what the luminance component Y of this signal would look
like, namely, an envelope exactly the same in amplitude (since red and
green are zero) and exactly the same in spectral / frequency components
as the blue signal.
NOW........subtract the two (as in B-Y).........
Note that the difference would be precisely, exactly zero at all times..
Their instantaneous values, being precisely the same, cancel each
other entirely.

Your mental experiment is wrong. As the color-difference signal represents
saturation, and a blue field has at least _some_ saturation, the resulting
signal _can't_ be zero.

Here's the error in your reasoning... You're ignoring the way the Y signal
is derived.

Ignoring absolute luminance levels, the situation you describe is B = 1, R =
G = 0. The Y signal isn't 1, it's

* * *0.30 R + 0.59 G + 0.11 B

Therefore, the B-Y signal must have an amplitude of 0.89, not zero.

Color difference signals are not the same as saturation! Please
provide a citation that states otherwise.- Hide quoted text -

- Show quoted text -

the amplitude of the color subcarrier signals represent the color
saturation.

the phase of the color subcarrier signals (relative to the burst)
represents the color hue.

for an image with no color, the subcarrier signals are zero and the
individual RGB signals are in a fixed PROPORTION (fixed difference)
to produce white or grey or black depending upon the Y (luma)
amplitude.

My citation is Video Demystifed as posted earlier.

Mark

On Mar 8, 2:43 pm, Mark wrote:
On Mar 8, 10:36 am, jwvm wrote:



On Mar 8, 6:51 am, "William Sommerwerck"
wrote:

"Smarty" wrote in messagenews:[email protected]
Imagine a blue screen with a gradient or checkerboard of blue values
only, as an example. Alternating squares of light and dark blue, or a
smooth transition from light to dark, or any conceivable variation of
blues
including steps, ramps, wedges, bars, etc.
This blue signal B will contain, before conversion into NTSC, lots of
high frequency content, if the transitions are abrupt / small, and little
or no high frequency content if it the blue field is solid or smoothly
transitioning.
Now imagine what the luminance component Y of this signal would look
like, namely, an envelope exactly the same in amplitude (since red and
green are zero) and exactly the same in spectral / frequency components
as the blue signal.
NOW........subtract the two (as in B-Y).........
Note that the difference would be precisely, exactly zero at all times.
Their instantaneous values, being precisely the same, cancel each
other entirely.

Your mental experiment is wrong. As the color-difference signal represents
saturation, and a blue field has at least _some_ saturation, the resulting
signal _can't_ be zero.

Here's the error in your reasoning... You're ignoring the way the Y signal
is derived.

Ignoring absolute luminance levels, the situation you describe is B = 1, R =
G = 0. The Y signal isn't 1, it's

0.30 R + 0.59 G + 0.11 B

Therefore, the B-Y signal must have an amplitude of 0.89, not zero.

Color difference signals are not the same as saturation! Please
provide a citation that states otherwise.- Hide quoted text -

- Show quoted text -

the amplitude of the color subcarrier signals represent the color
saturation.

the phase of the color subcarrier signals (relative to the burst)
represents the color hue.

for an image with no color, the subcarrier signals are zero and the
individual RGB signals are in a fixed PROPORTION (fixed difference)
to produce white or grey or black depending upon the Y (luma)
amplitude.

My citation is Video Demystifed as posted earlier.

Mark

I was asking William to cite a reference. I have no problem at all
with your posts. :-)

On Mar 8, 5:11*am, "Mike" wrote:

One also has to take into consideration the amount of kit that has been
produced with SCART conectivity since its introduction, maybe hundreds of
millions of devices, therefore in all honesty I don't see it disappearing
for a very very long time to come.

The only possible cloud on that horizon is the Hollywood studios'
paranoia about providing HD content unprotected.

Realistically, leaving aside the nonsensical marketing blather,
component or RGB analog video is every bit as good as HDMI or DVI for
the purpose of providing HD content to a display. Matter of fact, I
have read many reviews that preferred the analog to the digital
interface, for image quality. There was only one reason to introduce
the digital interfaces (DVI and HDMI): copy protection.

So it is possible that the Hollywood types will eventually make it so
their HD signals won't be preserved as HD on any analog baseband
interface. That's when HDMI or DVI with HDCP will become de rigeur.

Bert

On Mar 8, 6:44*am, "William Sommerwerck"
wrote:
"Albert Manfredi" wrote in message

Even assuming your assumption to be correct, all of the black and
white luminance information NOT sent with the B-Y and R-Y must still
be transferred. Moving it over to the Y does not mean you have reduced
any bandwidth requirements. I fail to understand why you keep
confusing yourself with tangential discussions.

Whether _my_ reasoning is correct has nothing to do with your reasoning. The
issue is not the bandwidth of the Y signal, but the bandwidth of the color
signals -- and whether one can choose color signals that make better use of
the available bandwidth.

Sorry, that doesn't make any sense.

If you want to "inherently" reduce the amount of bandwidth required to
transfer the signal, ignoring the fact that you have created a Y
component is sort of like trying to ignore the elephant in the room.

Again, if you DO NOT low-pass filter the color difference signals,
then transferring RGB or transferring Y plus the two difference
signals would take up the same amount of bandwidth.

QED.

Bert

"Don Pearce" wrote in message
...
snip

I haven't been following this thread, but has anybody yet mentioned
the ratio of rods to cones in the eye?

snip

Save yourself, Don. Run away now!

I said almost the same thing a week ago and had it contradicted. I then
ducked out the door, so to speak, and have been peeking in through a window.
This fracas has taken on a life of its own.

"Sal"

"Don Pearce" wrote ...
I haven't been following this thread,...

And you haven't missed anything.
The only thing most of us have learned is how little
some people know about video.

On Sat, 8 Mar 2008 21:51:01 -0800, "Sal M. Onella"
wrote:


"Don Pearce" wrote in message
...
snip

I haven't been following this thread, but has anybody yet mentioned
the ratio of rods to cones in the eye?

snip

Save yourself, Don. Run away now!

I said almost the same thing a week ago and had it contradicted. I then
ducked out the door, so to speak, and have been peeking in through a window.
This fracas has taken on a life of its own.

"Sal"



Thanks Sal. Don't worry, I'm already gone.

d

--
Pearce Consulting
http://www.pearce.uk.com

In article "William Sommerwerck" writes:

And as I've repeatedly pointed out, the color signals could have been
primaries, rather than color-difference signals, and still fit within the
required bandwidth.

And how were you going to get 12.5 MHz of bandwidth in 4.2 MHz of spectrum?

Alan

"Alan" wrote
in message ...
In article "William Sommerwerck" writes:

And as I've repeatedly pointed out, the color signals could have
been primaries, rather than color-difference signals, and still fit
within the required bandwidth.

And how were you going to get 12.5 MHz of bandwidth in 4.2 MHz of
spectrum?

What kind of a stupid question is that? Shall I ask how you'd get the 12.5
MHz luminance signal derived from 12.5 MHz color primaries in 4.2 MHz of
sepctrum?

We're talking about analog NTSC/PAL, okay?