I read many resources on the internet talking about NTSC, PAL, timing and voltage levels. But I have specific practical question, and I can not find definite answer to it. There're very helpful posts on Ataris, but they also more theoretical than practical.

The situation: the video display processor (VDP) outputs 1368 color clocks per line in NTSC mode, with 1024 in the visible "active screen" area. It has dot clock of 3.579545*6=21.477270 MHz, and thus display of whole line takes 21477270/1368 ~ 63.669 us (15700 Hz).

VDP has graphics modes of 256 pixels per line, and another of 512 pixels per line - displayed within 1024 clock dot periods. I assume that in former mode each pixel is clocked 4 times, and in latter 2 times.

In RGB mode receiver is free to choose its dot clock, performing oversampling or detection of the underlying a dot clock, or just use clock in accordance to its screen resolution.

But for me it is not clear what happens in NTSC mode - in both composite and S-video (with luma and chroma separated).

As I understand, NTSC's "transmission speed" is 3.579545 MHz, thus it will be performing undersampling of the color (and luma) data generated by the VDP.

Simple calculation gives the following:

  • Dot clock = 3579545 Hz
  • Line display time = 63695.24618 us
  • Dots displayed on the line = 228.

Using the similar calculations for 1024 displayable area I get 170.6666667 dots.

Even for 256 pixels per line, with 3.579545 MHz clock, the media will be able to sample first dot of first group of 4, then third dot of the second group, and will miss third group completely. And so on, with each third dot missed completely.

From this I conclude that composite can not achieve 256 dots/pixels, and image will be displayed with color data loss?

Then how old good CRT monitors were able to display very decent 256 horizontal dot picture over composite?

  • \$\begingroup\$ "very decent", well... :) \$\endgroup\$
    – pipe
    Jun 26, 2019 at 19:09
  • \$\begingroup\$ Actually I don't understand your question now after reading it again. You're talking about digital sampling at specific frequencies throughout the text, but a video signal is analog, as is the monitor. What is the sampling device here? Is your question how a modern digital video decoder works? \$\endgroup\$
    – pipe
    Jun 26, 2019 at 19:12
  • \$\begingroup\$ CRT’s hurled the dot “.” Symbol , at the speed of light into the phosphor and turned it into a splat “*” and barely did 80 column fonts with a VGA 25.175 MHz pixel clock on an NTSC TV. Higher end monitors like Iliad(sp?), Hitachi, Sony supported up to 250MHz of video BW for higher resolution. NTSC easily supports 4Mbps NRZ as well in old TV tuners. \$\endgroup\$ Jun 26, 2019 at 20:06

2 Answers 2


From this I conclude that composite can not achieve 256 dots/pixels, and image will be displayed with color data loss?

Correct. Chroma bandwidth is typically only 600kHz, much less than the luma bandwidth. The color signal is phase modulated so a digital decoder may have to sample it at a frequency several times higher than 3.58Mhz, but the phase cannot change that fast due to the low bandwidth. This is not generally a problem for TV programs because the human eye's color resolution is also low, and strong colors are rare in Nature.

enter image description here

For computers it's a problem because fully saturated colors are often used (particularly in early designs which had a limited color palette). Certain color combinations such as green and magenta or red and cyan are not only blurry, but also hard to view because the eye has difficulty resolving them.

This vectorscope display shows the phases for all combinations of fully saturated red, green, and blue.

enter image description here

To switch from red to cyan, green to magenta, or yellow to blue the phase has to change by 180°, which takes a while. Where the phase change is less the transition can be quicker. In the composite color bar display below going from green to magenta causes severe fringing, but magenta to red is quite sharp. From cyan to green is also quite sharp, but doesn't look it due to the low contrast.

enter image description here

The luma signal has a bandwidth of around 4MHz, so it should be able to fully resolve alternating white and black dots with a pixel clock of 7.16MHz. However the chroma signal is also transmitted in the upper part of the luma band, and separating them out causes some loss of luma at the higher frequencies.

Artists working on graphics for composite displays learn which color combinations are good and which to avoid. Using muted colors and/or different brightnesses helps. Small details lose color, but the contrast makes up for it. The fringing on certain color changes can also be used to create shadowing effects.

The bandwidth limitations in NTSC are inherent and cannot be eliminated. However modern TVs often apply techniques such as edge enhancement and text recognition to try to make the image look better. This can significantly improve the apparent sharpness, but may fail on certain images (I had to turn text mode off on my TV because it was making 8's look like B's, not good when you are working with Hex numbers!).


NTSC's "transmission speed" is 3.579545 MHz

Not at all. That just happens to be the color burst frequency. Color information is carried in sidebands both above and below this carrier frequency.

The actual sample rate typically used in modern receivers for both NTSC and PAL is either 13.5 or 27 MHz.

For NTSC, this corresponds to 858 total pixels per line: 30 × 525 × 858 / 1.001 = 13.5 MHz

And for PAL, this corresponds to 864 total pixels per line: 25 × 625 × 864 = 13.5 MHz

Either one allows 512 displayed pixels in the line. In fact, we normally think of NTSC as being capable of 640 × 480 pixels (interlaced), and this was the basis for the original VGA standard (progressive).

  • \$\begingroup\$ With both cases (PAL and NTSC), in digital domain there typically will be 720 active digital samples per line. Sounds enough for sampling a signal that was generated with 512 active samples. \$\endgroup\$
    – Justme
    Jun 27, 2019 at 0:23

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