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Physical explanation of bandwidth (analog)

  1. Aug 7, 2011 #1
    Hello friends,

    I am finding it a bit difficult to comprehend what the author of a famous text book has written about TV signal band width. Here is the excerpt from his book. Could any of you please explain what exactly he means? I don’t get why the required bandwidth is around 5.5 MHz.



    "Another example of a nonperiodic composite signal is
    the signal received by an old-fashioned analog blackand-
    white TV. A TV screen is made up of pixels. If we
    assume a resolution of 525 × 700, we have 367,500
    pixels per screen. If we scan the screen 30 times per
    second, this is 367,500 × 30 = 11,025,000 pixels per
    second. The worst-case scenario is alternating black and
    white pixels. We can send 2 pixels per cycle. Therefore,
    we need 11,025,000 / 2 = 5,512,500 cycles per second, or
    Hz. The bandwidth needed is 5.5125 MHz."
     
  2. jcsd
  3. Aug 7, 2011 #2
    I think it's easier if you think of it in terms of cycles (single waveform).

    This means you should have a 11,025,000 cycles per second, that is 11.025 MHz.

    If we send within a single waveform two signals (2 pixels per cycle), we have a 2 sign per cycle: this halves the number of required cycles to 5,512,500 per second, hence we have 5.5125 MHz.

    I'm quite confident about this explanation, despite having studied only digital transmission. The principle here is quite similar. If you are interested in seeing how a single waveform can send two or more "bits" (here, pixels), you can read something about PAM modulation http://en.wikipedia.org/wiki/Pulse-amplitude_modulation" [Broken]
     
    Last edited by a moderator: May 5, 2017
  4. Aug 7, 2011 #3
    I think it takes two pixels to make one cycle because the worst case is one high and one low and this is consider one cycle. so for 11.0125 mega pixels, the 5.512MHz should be the correct one.
     
  5. Aug 7, 2011 #4
    Suppose you want to represent an analog waveform with N 'values' per second. In general, this sequence has to be sampled N times per second, so the sampling frequency is [itex]f_{s} = 1/N \; \mathrm{Hz}[/itex]. But, according to the sampling theorem, the sampling frequency has to be at least twice the highest frequency component (bandwidth). Therefore:
    [tex]
    f_{s} = 2 f_{b} \Rightarrow f_{b} = \frac{1}{2 \, N} \, \mathrm{Hz}
    [/tex]
     
  6. Aug 7, 2011 #5
    I don't think this is sampling. For digital recording, you need twice the pixel speed to sample all the information. For example if you want to digitize a 1MHz sine wave, you need a sample rate of 2MHz to capture all the information. I think this is call Nyquest or something!!! It been 30 years and I forgot most of it already.


    but when you display, the fastest speed is only from alternate 1 and 0 which is half the bit rate. They are not related.

    I worked 3 years for LeCroy that produce Digital scope designing transient recorders in the early 80s, I know.
     
    Last edited: Aug 7, 2011
  7. Aug 7, 2011 #6
    yungman is right, it does not involve sampling, as it is an analog transmission. The trick that halves the required bandwidth is carrying two pixels in each waveform.

    In digitalization, since sampling in time means obtaining a periodic waveform in frequencies, to avoid aliasing you need to sample with a frequency that it is at least double than the highest frequency of the signal. This is the Nyquist-Shannon sampling theorem http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem" [Broken].

    However, this is not the case, since manulal said that the signal is analog.
     
    Last edited by a moderator: May 5, 2017
  8. Aug 7, 2011 #7
    It doesn't matter if the signal is analog or not.
     
    Last edited by a moderator: May 5, 2017
  9. Aug 7, 2011 #8
    Thank you very much to all for your replies.

    I think I have made a mistake of not phrasing my question correctly and not stating what exactly the question is for which I seek the answer.

    I am a mechanical engineer and I am trying to grasp the basics of electrical engineering on my own at my own pace.

    When I reached AC circuits and its applications, I stumbled upon the above quoted paragraph and this doubt started bothering me since then.

    I clearly understood the logic behind 11,025,000 pixels per second and since a cycle can have two pixels 5,512,500 cycles per second or 5.5125 MHz.

    So, if I have understood correctly any signal with a frequency exceeding 5.5125 or let us say 6 MHz can theoretically contain all the information required for the TV transmission in the worst case scenario.

    Now, let me mention an example.

    North American channel 2 occupies the spectrum from 54 MHz to 60 MHz. A bandwidth of 6 MHz.

    My question is, provided that any frequency more than 6 MHz can contain the all information in the TV transmission, why can’t Channel 2 use a particular frequency above 6MHz, say 57 MHz, instead of using the spectrum from 54 to 60 MHz?

    I am sure that the reason for my doubt is my ignorance about electronic communication fundamentals. But this doubt is pestering me and I would be grateful to any one kind enough to clear it.

    Could you please also suggest some good books on this subject which covers the subject from basics?
     
  10. Aug 8, 2011 #9
    You should re-phrase "the BW is more than 6MHz".

    I don't understand your second question:

     
  11. Aug 8, 2011 #10
     
  12. Aug 8, 2011 #11
    @Dickfore: Shannon-Nyquist theorem applies to digital signals, or better to system of transmission that implies the digitalization of the signal, since the sampling frequency required to digitalize the signal is responsible for the possible aliasing.
    If you are dealing only with analog circuits that do not sample anything, you can't have aliasing, so the shannon-nyquist theorem does not apply.


    I'm not sure if the two questions -the one related to the TV signal and the one related to channel 2- are related.
    However, channel 2 can't use only 57 MHz because the signal usually is composed of several frequencies, so when you say that its bandwidth is 6MHz you are saying that the distance (in term of frequency) between the lowest and the highest frequency is 6MHz.
    Even if the signal has one frequency, the interference and the disturb along the path from source to receiver broadens the band of the signal transmitted.

    Thus, channel 2 must use 6MHz to transmit its signal, but it has no limitation about the central frequency: it could be 20-26MHz as well as 30-36MHz.

    However, I am sure there are other reason for the band being around 57MHz, but either I didn't study them, or I can't remember :D
     
  13. Aug 8, 2011 #12
    And what kind of signal is the one representing [itex]N[/itex] 'black' or 'white' pixels per second?
     
  14. Aug 8, 2011 #13
     
  15. Aug 8, 2011 #14
    Well, I'm not a technician, I think an oversimplified example could be
    [itex]-|\sin(t/2)|[/itex] if the pixel is white
    [itex]|\sin(t/2)|[/itex] if the pixel is black

    or something like this.
    This doesn't need any sampling of the signal. The waveform source generate an analog sine and then you just make it positive or negative (i'm not EE, so i don't know actually how).

    It's just that Shannon theorem applies to digital signals, so if you are dealing with analog signals it means nothing applying it.
     
  16. Aug 8, 2011 #15
    And what is the value of [itex]t[/itex]?
     
  17. Aug 8, 2011 #16
    t is a continuous variable
     
  18. Aug 8, 2011 #17
    So, how does one distunguish between, let's say, the 1st and the 3rd white pixel?
     
  19. Aug 8, 2011 #18

    AlephZero

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    Science Advisor
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    In an analog TV system, there aren't any real "pixels". The electron beam just scans across the screen and its brightness varies as it goes along, following the continuous analog signal.

    (To keep things simple, let's ignore practical details like how the TV set "knows" where each line of the picture starts, etc.)

    However, since there are a fixed number of vertical lines (525 for the US system) and the proportions of the screen are 4 wide to 3 high, it wouldn't make much sense for one horizontal line to display more than about 525 x 4/3 = 700 separate "pixels", because there is no sense having a higher resolution of the picture horizontally than vertically. That's where the idea of the "525 x 700 = 367500 pixels in the screen image" comes from.
     
  20. Aug 8, 2011 #19
    This does not answer my question. What you are calculating has been done in the first post.
     
  21. Aug 8, 2011 #20
    For pure analog tv, there is really no pixels to talk about, it is the frequency components we are talking about. You don't distinguish between pixels. You really care about the transient response of the tv reciever circuits being able to distinguish the transient ( as small edge detail ) on the screen.

    It is not the pixels that matter, it's the ability of the tv to resolve the highest frequency component of the signal that is important. That is where the bandwidth come it. From the original example, the BW is 6MHz, so the tv circuit has to have the BW of 6MHz to resolve the most detail signal of the channel.
     
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