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Old 10th Nov 2017, 12:22 am   #15
julie_m
Dekatron
 
Join Date: May 2008
Location: Derby, UK.
Posts: 7,735
Default Re: Question about scanning and "sync pulses"

The best description I ever heard of the TV transmission process was that it's as though you take a white knitted jumper; screen-print a design right onto the knitting with deep-pigment inks that will soak right into the wool, colouring it all the way through; and then undo the stitches. Now you have a single, continuous length of wool that changes colour all along its length according to the printed pattern. This is your video signal. You can then pass the end of this wool through a keyhole; and another person, on the other side of the locked door, picks it up and begins knitting with it.

As long as your friend casts on the correct number of stitches in the first row and uses the same size needles so each stitch will use the same amount of wool as before, then the end result will be identical to the original, printed sweater.

Now, you already agreed in advance on the number of stitches per row, the diameter of the knitting needles and the amount of tension in the wool, which between them determine the amount of wool that should be used in each stitch. Synchronising pulses in a video signal are like the slight kinks in the wool where the original stitches used to have been, that the person behind the door can feel and keep fine-tuning the tension as they go, thereby ensuring the re-knitted sweater matches the original perfectly. Even if you cannot feel each individual stitch, as long as your assistant uses the right number of centimetres of wool in each row and finds some way of dealing with extra or missing stitches, you will get a reasonable enough approximation to the original picture. And in real life, the error is only goimg to be visible for 1/50 or 1/60 of a second anyway .....

The slight downward slope of the scanning lines is self-compensating if the camera is an analogue device, since the lines it sees will have the same slope as the lines drawn on the screen. You shoild be able to show this with just GCSE-level maths. But even if the source is digital and there is effectively no slope to the source lines, the difference is so tiny in practice (just 1 part in 287.5, in 625-line countries with 575 visible lines in the picture -- it takes all of 25 lines' worth of time to scan the beam back up again, at the bottom of each field -- or 1 part in 241.5 in 525-line countries with 483 visible lines) that it will never be noticeable in practice.
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