I was trying to capture the scanning process of a CRT TV screen in a photo. It gave me these dark, broad bands, shown below: CRT TV Screen Under Still-photo Fig. 1 a,b,c

I'm trying to understand the cause for this banding. However, I couldn't understand it clearly in very similar questions here, here, or here. All web sources I visited very briefly explained it as a difference between the scan-rate of the TV and the camera, and none provided a diagram.

I guess this may going on Fig. 2 a,b,c

I assume what is going on is in Fig. 2. The blue lines indicate the scan lines where cathode rays hit most recently (phosphorescence occurring), and black lines indicates where the cathode ray has not been in the image, so phosphorescence is all that is left and a dark band appears.

I assumed this, because out of 10 photographs I took, I found only the above 3 types of photos, and none of the type shown in Fig. 3 or Fig. 4:

Center Darkmany bands

Fig. 3. an Fig. 4.

I also assumed that the camera is much faster than the TV scan rate, so it could take picture in an instant.

My question is, Have I assumed correctly? or is something else going on?

Additional information:

TV:     Videocon 120 W colour CRT, Electricity power supply 50 Hz. 
Camera: Nikon Coolpix L24 digital camera.

(please also explain other possible situations, such as this one in Fig.5. I didn't experience it myself. It is similar to Fig. 4.

evidence for many-band

Fig 5. From this question

I'll be grateful if any explanations use diagrams.

Update: I've obtained several fig3 conditions, and though they are less frequent, they're not very rare.

fig3 condition

File properties show exposure time for this still-photo is 1/125 second.

I've also took some videos (in PAL and NTSC mode... both gave same result), file properties show frame rate is 30 frame/sec; and run them on slowest motion. I found similar, but much brighter and quite unclear bandings, and it seemed from slow-motion video that the alternating bright and dark band rising on the upward direction. From that videos I screenshoot some frames as-successive-as-possible. successive frames from slow-motion video. Red small bars added to indicate the margin of each banding

  • 1
    \$\begingroup\$ fig5 is unrelated. That's a TFT display, not a CRT. \$\endgroup\$ Aug 7, 2016 at 16:56
  • 1
    \$\begingroup\$ "I also assumed the camera is much faster than TV scan rate, so it could take picture at a moment." Whoops \$\endgroup\$ Aug 7, 2016 at 18:55
  • 1
    \$\begingroup\$ @LightnessRacesinOrbit Quotation followed by whoops. What does that mean? please write in realistic way. I accepted this-assumption because it led me to a solution .I did not claimed this-assumption is correct. However, if the shutter-speed would slower-than scanning-speed, I wouldn't get those intense dark bands. Isn't? \$\endgroup\$ Aug 7, 2016 at 19:02
  • \$\begingroup\$ Try taking pictures at higher and higher speeds. Note the speed with the illustration! \$\endgroup\$
    – JDługosz
    Aug 7, 2016 at 19:29
  • \$\begingroup\$ thanks for suggestion but my camera don't contain so-many speed-control features or maybe I don't know anysuch settings exist. \$\endgroup\$ Aug 7, 2016 at 20:01

4 Answers 4


Yes, you're basically on the right track. The bands appear because the camera shutter is not synchronized to the vertical scanning of the CRT.

A fast shutter speed will only capture part of a scan. A shutter speed exactly equal to the vertical scan period will capture a full scan, but there still might be a narrow band (either dark or light) somewhere in the image if there's any mismatch in the timing.

To minimize the banding, use a shutter speed that spans several vertical scans, but this only works if the image on the CRT is static.

  • 1
    \$\begingroup\$ But that's not OP's question, if I understand correctly – the problem is that he'd expect "fig. 3" to happen about as often as "fig 1b", if I understand correctly \$\endgroup\$ Aug 7, 2016 at 16:55
  • \$\begingroup\$ @MarcusMüller If the scheme I assumed is correct, then Fig.3 should not come. Carefully watch fig 1b and 2b. If fig 3 take place, then I'm wrong and there's going-on something more-complicated (including TV+Camera). \$\endgroup\$ Aug 7, 2016 at 17:04
  • 1
    \$\begingroup\$ ah, maybe I'm misunderstanding your overall question wrongly then, @AlwaysConfused: Does your assumption conflict with reality or not? \$\endgroup\$ Aug 7, 2016 at 17:05
  • \$\begingroup\$ In fig 2b still-photo, the oldest-portion (centre) lost phosphorescence, and new-portion before vertical-flyback (bottom) is glowing, and top-portion (slightly after vertical flyback event) is glowing. \$\endgroup\$ Aug 7, 2016 at 17:07
  • \$\begingroup\$ mmm. I wrote in the Question, I selected 10 photos. (which were not shaky). All-of them contained configuration similar to any 1 out of fig.1 a, b, c. and No fig 3 I found yet . so I guess my scheme wouldbe correct. However 10 is a litlle as sample number. \$\endgroup\$ Aug 7, 2016 at 17:10

The only reason you didn't see "Fig 3" (bright stripe in the middle) is that your shutter speed is too slow. You have a much higher chance of capturing a dark gap in the middle, because those pictures all include a vertical blanking interval (the time when the beam is offscreen and "moving" back to the top) as well as all of the vertical part of the picture hidden by overscan. You can see from Fig 1b that your shutter time captures all but 1/3rd of the visible part of one field. To get "Fig 3" would require pretty precise timing, and your bright stripe would be nearly the full height of the display.

  • \$\begingroup\$ hmm you mean if I could take the picture 1a in faster-speed (before the beam touch the bottom-of screen), I would get a fig3 . Is that? \$\endgroup\$ Aug 7, 2016 at 18:09
  • 1
    \$\begingroup\$ @AlwaysConfused Right. \$\endgroup\$ Aug 7, 2016 at 18:30

Your scanning beam description seems right on, to me. But my experience would be with a film camera, not digital. Your more complex images make me think of scan-on-scan of reading out the digital pixels vs cathode ray trace, presuming the readout speed might be similar to the scan rate on the CRT.

In film with a planar shutter, the choice of shutter speed could result in skew for the CRT image against the film. My first SLR had a left to right shutter. My better one did top to bottom. If you had an iris shutter, the results would not show such skewing because it opened from center out and back. The standing rule for taking a photo of a TV screen was to set the speed to 1/30 or lower to get the full interlaced image.


Not as a confirmed-answer, but hypothetical one,

the other-2 situations could be this:

2 . when Camera having speed synchronous with TV

Fig. 6 Out of phase , synchronous camera

Whether the camera (i)starts scanning with the TV, or (ii)start scanning out-of-phase (above picture); in both-cases we should get the accurate picture... neither dark band nor bright band.

3. when camera is slower than TV

Fig. 7 a, b, c, d, e from left to right. when camera slower than TV

conditions like fig 1 a, b or c , but with an extra shade of brightness could be obtained. So we could get a medium-bright band and a very-bright band (no dark-black band)

Fig7 a (leftmost) indicates ultimate photograph that should be obtained .

fig7 b -> c -> d-> e indicates step by step formation of 7a.

(gradients would take place if I consider the camera can detect the gradual loss of phosphorescence.)

(feedbacks welcome)

  • 1
    \$\begingroup\$ I think your illustrations are making it harder to understand. Draw the beam activity as a continuous timeline (maybe as several displays stacked on top of each other, with a gap for the vblank, and with a border to represent overscan), and then imagine your photo as a vertical slice of that timeline that's one shutter time long. \$\endgroup\$ Aug 8, 2016 at 20:38
  • \$\begingroup\$ @BenJackson I'll try. \$\endgroup\$ Aug 9, 2016 at 9:27
  • \$\begingroup\$ Thanks for informing me about role of vblank and overscan \$\endgroup\$ Aug 9, 2016 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.