3
\$\begingroup\$

How many transistors/logic gates are used in the signal path between a TV studio and the restitution of the image on my HD-TV ?

You see what I mean ? I need a rough estimation...:-)

I especially want to estimate the number of purely digital-purpose transistors (logic gates) : that means how many SoCs (system-on-chip) are needed in the capture->mpeg encoding->transmission(source,channel coding)->etc..->mpeg decoding, and how many gates are in these SoCs.

Maybe some of you know partially (on one of those chips) how many gates are involved.

Together, maybe we can try to count them ?

\$\endgroup\$
8
  • \$\begingroup\$ I'm gonna guess a few thermionic valves got involved too, especially in the video transmitting parts. \$\endgroup\$
    – Thomas O
    Jan 22, 2011 at 18:16
  • \$\begingroup\$ Are you including things that don't directly manipulate the video, for example, the power station's electronics for the TV station? \$\endgroup\$
    – Thomas O
    Jan 22, 2011 at 18:18
  • \$\begingroup\$ Please add much more information about the intended system. Basically, everything you've got. I suspect an XY problem... \$\endgroup\$
    – tyblu
    Jan 22, 2011 at 18:18
  • 5
    \$\begingroup\$ Without a lot more details the only answer is "Billions of transistors" \$\endgroup\$
    – Mark
    Jan 22, 2011 at 18:52
  • \$\begingroup\$ Why? - - - - - - \$\endgroup\$
    – starblue
    Jan 23, 2011 at 9:26

2 Answers 2

8
\$\begingroup\$

Here are the functional blocks:

  1. Video camera

    • 1920x1080 pixels, CCD, modeled as DRAM for simplicity, giving 1920x1080=2073600T
    • columns dump to analog multiplexors, which I'll guess to have 1080x10=10800T
    • 1080/2^3=135 ADC blocks -- will make them 24-bit delta-sigma ADCs, giving it around 135x(24x2+50)=13230T
    • memory buffer should hold at least 2 frames, giving 24x1920x1080=49766400T
    • high speed bus driver will add another 10^3T or so
  2. Encoding DSP

    • x264 codec requires 32-bit DSP processor with ??? T
    • should buffer (SRAM) at least 10 frames for codec compression, giving 10*9*49766400=4478976000T
  3. Transmission

    • Going with 2 Windows PCs to prep the stream and send it to a server, then 5 LAMP servers to get it to the end user, giving 7x2000000000=14000000000T
    • modems and switches ignored
  4. Decoding DSP: make equal to encoding

  5. Display: make equal to capture

TOTAL: 23061682060 transistors
ERROR: ±2628%, possible error in error calculation

\$\endgroup\$
8
  • 2
    \$\begingroup\$ How can you have -2628% error? That would be −13,976,786,195,591,899,169,808 transistors... \$\endgroup\$
    – Thomas O
    Jan 22, 2011 at 22:29
  • 1
    \$\begingroup\$ @ThomasO, Yep, which goes along well with my theory of negative resistance resistors. \$\endgroup\$
    – tyblu
    Jan 22, 2011 at 22:47
  • \$\begingroup\$ @tyblu They exist! It's just a conspiracy by the NWO, and the CIA, and big oil!!! \$\endgroup\$
    – Thomas O
    Jan 22, 2011 at 22:59
  • \$\begingroup\$ @ThomasO if you made vacuum tubes negative numbers... Personally I like the idea of big oil keeping the practical implementation of negative resistors from coming to a product. \$\endgroup\$
    – W5VO
    Jan 23, 2011 at 5:32
  • 1
    \$\begingroup\$ If you connect two negative resistors in parallel, at 88 mph, don't you rip a hole in universe, and go back in time? \$\endgroup\$ May 2, 2011 at 7:48
2
\$\begingroup\$

In the days of "classical" analog video, one could meaningfully count the number of transistors (or vacuum tubes) between the video camera tube and the video display tube, since there would generally be a single path for the signal to take. Even if a video was captured by a CCD or had to undergo something like analog or digital NTSC/PAL conversion, one could still produce a meaningful figure since every portion of the video image would gone through an identifiable path (even though different portions of the video image would go through different paths). A video signal might get converted to video, but all the digital stages it went through were purely combinatorial (sequential circuits would be used to keep track of horizontal and vertical scanning, but the picture data would be handled by combinatorial circuits) and, at any given time, one could point to a circuit and identify that it was handling some particular portion of the original picture.

With compressed digital video, it's impossible to meaningfully count how many transistors the signal goes through, since it goes through a many sequential circuits which munge and reconstitute it. Often times some of these circuits will be running in parallel. If the data from 64 pixels are munged together so as to yield 64 numbers, and then those numbers are processed in parallel, and the results are then reconstituted so as to yield 64 pixels, how many transistors did the pixel data go through? I don't think any particular answer is really "right".

\$\endgroup\$
1
  • \$\begingroup\$ We are inviting experts for specialized questions like this. Now there is a new proposal at Areas 51: Broadcast and Media Technologies which focuses on Television and Broadcast in depth. Please join and invite others. \$\endgroup\$ Mar 1, 2012 at 14:06

Your Answer

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

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