A cable fault location estimation

Discussion in 'General Electronics Chat' started by sajidkaleem, Jun 20, 2012.

  1. sajidkaleem

    Thread Starter New Member

    Jun 20, 2012
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    My final project is as mentioned above in my Title – but not in time domain measurement but by amplitude measurement (positive half sine wave).
    I would like to know what actual relationship between bandwidth and frequency in measurement ? How does the frequency differ in increase in length of coaxial 50 ohms cable ?
    I am using VCO, precision rectifier (op-amp) which operates above 1 Mhz (please guide me this circuit of precision rectifier) and microcontroller (Arduino).
    Kindly explain me in details and also show me its relationship in equation like – (beta2 x length) – (beta1 x length) = pie
    ….
    …. in terms of frequency
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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  3. Sparky49

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    Jul 16, 2011
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  4. chuckey

    Well-Known Member

    Jun 4, 2007
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    If you get a piece of cable that is open circuit and apply RF to it at a frequency where the wave length is 1/4 the cable reflects a short circuit. So if you measure the voltage at this point you will find that the voltage dips to nothing. if you up the frequency by three times another short (voltage dip) this is at 3/4 of the wavelength. In fact the short occurs at every ODD multiple of a 1/4 wave length.
    Wavelength (metres in free space) = 300/F in megahertz. i.e. 150 MHZ = 2m therefore 1/4 wavelength = .5m.
    One complicating factor is that RF travels slower in a cable, this is called the "velocity factor" and is .6 or so, this makes the physical wavelengths as measured on a cable increase by 1/.6 = 1.6 or so.
    So put a voltmeter and the generator into a Tpiece, connecting the cable to the open end, log the frequencies at which the voltage dips. If you have a sample of the cable that is being tested, measure its length and put it in series with the cable you are testing, re-run the tests. You should get another set of frequencies. By doing the maths you should be able to calculate how long the broken cable is.
    If the broken cable has a short at its far end, the cable relects a short circuit at multiples of 1/2 wavelengths.
    Frank
     
    Last edited: Jun 20, 2012
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  5. sajidkaleem

    Thread Starter New Member

    Jun 20, 2012
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    can you please guide me what circuit design shud i use in precision rectifier op-amp based operating above 1 MHz that converts ADC and trnasmit to microcontroller (Arduino) which is interphased with LCD.

    Can you guide me how can i decide the frequency from the below equations:

    β x l = π .....(1) l = length of cable(minimum)
    (β2 x l) - (β1 x l) = π β = 2π/λ
    ((2π/λ2) x l) - ((2π/λ1) x l) =π (v = fλ)
    solving
    l (f2 - f1) = 10^8

    further how can i find frequency for my VCO and circuit design for precision rectifier.

    Please help me
     
  6. sajidkaleem

    Thread Starter New Member

    Jun 20, 2012
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    can you explain how frequency differs with increase in length of cable (50 ohms coaxial ) (note:experiment is not on time domain system instead it is carried on by scalar measurement of amplitude measurement : +ve half sine wave) what is the relationship of bandwidth and frequency
     
  7. chuckey

    Well-Known Member

    Jun 4, 2007
    75
    10
    If the piece of cable is .8m long for dip in voltage, then multiply by the velocity factor, .8 X .6 = .48 ~.5m which is 1/4 wave length at 150MHZ. if your piece of cable is 1.2 m long the dip frequency is .8/1.2 X 150 = 100MHZ. Conversely if the dip frequency is 40 MHZ, then the cable length is ~ 100/40 X 1.2 = 3m.
    Frank
     
  8. sajidkaleem

    Thread Starter New Member

    Jun 20, 2012
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    Hello

    Could you please help me in building up a precision circuit op-amp based circuit diagram for my cable fault location estimator project (not TDR method) i am measuring full wave rectifier amplitude
     
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