# RC time constant

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

1. ### MattiasGuest

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!

Thanks,

Mattias

2. ### Kevin AylwardGuest

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

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

3. ### Tim ShoppaGuest

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.

Tim.

4. ### Bill SlomanGuest

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. ### MattiasGuest

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!

Mattias

6. ### Fred BloggsGuest

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.

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
speculate.....

8. ### Don PearceGuest

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
right.

d

_____________________________

http://www.pearce.uk.com

9. ### Bill SlomanGuest

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
line.

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. ### MattiasGuest

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
cable?

Mattias

12. ### Joe LegrisGuest

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. ### John WoodgateGuest

I read in sci.electronics.design 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. ### FredGuest

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
coax?

If not and the cable is now terminated is there any signal?

15. ### Jim MeyerGuest

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

Jim