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RC time constant

Discussion in 'Electronic Design' started by Mattias, Oct 28, 2003.

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  1. Mattias

    Mattias Guest

    Can someone please explain to us, why it is impossible to send a wave
    with frequency over 1/RC through a coax cable? This question has been
    driving us nuts. Explanation or links/references are very welcome!


  2. Because it is not impossible, its just that it isn't very efficient. The
    signal voltage at the T line end falls with frequency, but it dos not go
    to zero except at infinite frequency.

    Kevin Aylward
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
  3. Tim Shoppa

    Tim Shoppa Guest

    No wonder the question has been driving you nuts... it's a poorly-phrased
    question and implies something that simply isn't true.

    Interpreting R to be the impedance of the coax (again, I'm just trying
    to interpret a poorly-phrased question) and "C" to be the capacitance
    of the cable, you question implies that 100 feet of 50 ohm 30pF/ft coax
    being unable to carry frequencies over 10MHz. Nonsense.

    Coax is described by a number of parameters, including capacitance per
    unit length and inductance per unit length. But rarely are these relevant...
    almost always the impedance (independent of length) is what counts. That's
    not an accident... it's the entire purpose of a transmission line.

  4. Bill Sloman

    Bill Sloman Guest

    What is "R" and what is "C"?

    If the capacitance your are referring to is the capacitance of the
    cable (typically 100pF/metre) and the R is is characteristic impedance
    of the cable (50R for most applications, 75R for video cables, and 95R
    for some specialsied coaxial cables) you are probably trying to drive
    an unterminated cable from a source which is intended to drive a
    terminated cable.

    Try connecting a resistor equal to the characteristic impedance of
    your cable between the inner and outer of your cable at the far end -
    this will halve the signal level at all frequencies, but the high
    frequencies won't be attentuated by the cable capacitance.

    Many high frequency oscilloscopes have a built-in terminating resistor
    which you can switch in and out for doing this job.
  5. Mattias

    Mattias Guest

    Can someone please explain to us, why it is impossible to send a wave
    Thanks, Tim, for your answer. Yes, C is capacitance and R is
    resistance of the cable. We connect a function generator to the cable
    and watch the out voltage on an oscilloscope. This way, we can measure
    the RC time constant by watching how voltage drops. The question then
    is that for a certain (high) frequency, it is no longer possible to
    deduce the RC time constant. Supposedly, the answer is 1/RC and we
    can't figure out why!

  6. Fred Bloggs

    Fred Bloggs Guest

    You might be talking about a transmission line of characteristic
    impedance R terminated in load capacitor C. Then the reflection
    coefficient is ( 1/(jWC)-R)/(1/(jWC)+R)=EXP(-j2*ATN(WRC)). This is
    clearly a high pass characteristic with cut-in at W=1/RC, the reflection
    coefficient approaching -1 at higher frequency- thus no power transmission.
  7. We connect a function generator to the cable
    At frequencies that are "low", the cable can be treated as lumped
    parameters, being a capacitance primarily. Resistance is irrelevant
    if you are monitoring the same end of the cable that you drive, and
    possibly relevant if you monitor the other end.

    At frequencies that are "high", the cable is a transmission line
    and can NOT be treated as lumped parameter elements.

    "High" and "low" depend upon the relative length of the cable
    to the wavelength (not free space wavelength, but adjusted for
    transmission speed on the cable) of the frequency being used
    at the moment. If the physical length of the cable is much
    less than 1/4 wavelength, lumped parameter treatment is usually
    accurate enough. If the physical length is 1/4 or more wavelengths,
    then transmission line techniques must be used. THere is a
    gray area where the choice depends upon various secondary issues
    (what accuracy do you need, what are you looking at, &c).

    Modelling a cable as a RC circuit would be accurate ONLY at
    "low" frequencies.

    Since you do not mention cable length or frequency, we can only
  8. Don Pearce

    Don Pearce Guest

    At low frequencies the cable is "capacitive primarily" only if the
    load is greater than the cable impedance.

    If your cable terminates in an impedance lower than its characteristic
    impedance, then even at low frequencies the cable looks inductive.

    Far better to use the distributed model at all frequencies, so you
    don't need to worry whether you have it right or wrong - it is always


  9. Bill Sloman

    Bill Sloman Guest

    The resistance of a cable is not the same thing as the characteristic
    impedance of a coaxial cable or transmission line.

    A coaxial cable will have a resistance between each end of the outer
    (screen conductor) and a different resistance between the ends of the
    inner conductor.

    Both are usually less than the characteristic impedance, which happens
    to be the resistance that will critically damp the distributed
    inductance and capacitance of the cable, looked at as a transmission
  10. Far better to use the distributed model at all frequencies,

    Personally, I tend to agree; but the OP clearly had no
    idea of how to model distributed systems ...
  11. Mattias

    Mattias Guest

    We connect a function generator to the cable
    Thanks guys for taking time to help me out. The cable length is about
    3000 feet. Supposedly, the maximum frequency for this cable is 1/RC
    which equals about 115 kHz. Higher frequency will give no output on
    the oscilloscope. Resistance R is the sum of resistance of inner and
    outer cable. Capicitance C is standard capacitance for coax cable. Our
    teacher seemed to hint that it had something to do with the cable
    length and the time it takes for the signal to travel through the

  12. Joe Legris

    Joe Legris Guest

    Maybe your teacher's been using FTL coax.

    If this nonsense about 1/RC were true then how could cable TV be possible?

    And what is the definition of "no output on the oscilloscope". Even
    devices that were designed to completely block a signal (such as an open
    circuit) will still pass some of it - just crank up the gain on the 'scope.

    It sounds like you have been fed an oversimplification of a
    misunderstanding. Go back to your textbook for some facts.
  13. I read in that Mattias <>
    It will, if you actually do the experiment.
    I think you have been confused by poor explanations. A cable works only
    like a plain RC at low frequencies. The conductors have *inductance* as
    well as resistance, and at a sufficiently high frequency, this prevents
    the apparent RC low-pass filter actually existing.

    There isn't any 'standard capacitance for coaxial cable'. Cables of
    different dimensions and dielectrics have widely different capacitances
    per unit length.
    I think this is just more confusion. The cable is about 900 m long, and
    in free space a wavelength of 900 m belongs to a frequency of 333 kHz.
    But the wave travels more slowly in the cable, by a 'velocity factor'
    factor of around 1.6 normally, so the cable is a wavelength long at
    about 208 kHz. It looks to me as though your teacher has neglected the
    velocity factor (or has assumed a coaxial cable with vacuum dielectric!)
    and thinks that the cable is half a wavelength long at 115 kHz. Not that
    being half a wavelength long has any significant consequences if you are
    feeding the signal into the cable correctly.
  14. Fred

    Fred Guest

    If the coax is 1/2 wavelength, then the launch end will be seen as a short
    if the other end is o/c. Hence no power is launched into the cable.

    Is the end of the cable is terminated according to the impedance of the

    If not and the cable is now terminated is there any signal?
  15. Jim Meyer

    Jim Meyer Guest

    Ask your teacher if the inductance of the cable plays any part in
    how signals propogate through it.

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