# Zo of a transmission line

Discussion in 'Electronic Design' started by mook johnson, Jul 6, 2004.

1. ### mook johnsonGuest

If I were to measure the Zo of a transmission line (long sucker at 5 miles
was a scope and function generator (50 ohm output) and few various passive
components (resistors, pots, etc.) and transformers. What method would be
best to use to measure such a beast?

take a guess a the range based in the construction of the cable (~120
ohmsfor twisted pair, 50-75 ohms for coax) then adjust the terminating
resistance until the square wave input in one end looks good on the other?

Or do I use input voltage and current (phase and amplitude) to calculate the
Zo for several frequencies across this range. It should flatten out as the
line becomes a transmission line.

Any other methods of corrections to the above methods?

The reason for the sparse equipment list would be that this will be done in
a remote location.

2. ### Jim BackusGuest

Because the line is 5 miles long, the minimum transmission time will
be > 25 microseconds. So for more than 50 microseconds after a signal
is applied the line will appear to be an infinite length. If you apply
a squre wave with 50 ohm source impedance and observe the voltage on
the line at the start of the square wave, Zo can be readily
determined.

E.G.

If Vo from the generator = 10 volts, Zo of generator = 50 ohms and
voltage on line, VL = 7.06 volts then Zo (line) = VL x Ro / (V0-VL) =
2.94 x 50 / 7.06 = 120 ohms.

3. ### Ken SmithGuest

The cable loss increases with increasing frequency. Any squarewave you
put in will look rounded when it comes out.

If both ends are handy to the operator, you can just load the far end with
a near match and use Ohms law at the near end. Once you have a first
guess, you can trim the loading pot to that value and measure again. It
should converge quickly because the cable loss will be quite high.

4. ### mikeGuest

I'd use a pencil.
If it's twisted pair, write down 120 ohms.
For a 5 mile cable, I'd guess that the error in this "measurement"
is well swamped by a bunch of other issues...losses, radiation, crosstalk...

So, satisfy my curiosity...why do you need to know the exact value?
mike

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5. ### John WoodgateGuest

The formula for Zo is Zo = sqrt{(R + jwL)/(G + jwC)}, where all the
quantities are per unit length.

For almost any practicable cable, at such low frequencies as 10 to 50
kHz, L and G are negligible. (G might not be entirely negligible at 50
kHz if the insulation is PVC.) The result is that Zo is not resistive
but can be represented by a capacitor (perhaps lossy).

If you have a 10 m or so length of the cable, you can measure R, L and
C, and maybe G if its not too low. If it's too low to measure, it's
negligible. Ideally, you would use a good RLC bridge at 50 kHz, but
bridges seem to be regarded as stone-age technology now.

6. ### K WilliamsGuest

If your transmission line were infinitely long, you could measure Z0
with a multi-meter. Scale your multi-meter timing for the round-trip
length of the transmission line. i.e. apply your voltage and measure
the current within the time it takes to transverse the cable from
That works. ...or use a series resistor for the same goal.
Too complicated. If your signals transverse end-to-end in the time of
your measurement you'll get standing waves that make measurement more
difficult (not impossible, just more difficult).
As has been noted, launch a low-duty-cycle (less than one pulse per
several line-lengths) square wave at one end and measure the
voltage/current of the driver. Or similarly, drive the line with a
source of a known voltage and impedance, then measure the voltage
response at the source. This is the essence of a
Time-Domain-Reflectometer.

7. ### Roy McCammonGuest

You measure current, voltage and phase at one end with the
other end shorted. Call that Zsh. Do it again with the
other end open. Call that Zop.

Then Zo = sqrt (Zsh * Zop)

where all those quanties are complex.

8. ### mook johnsonGuest

The Zo is being calculated to determine the necessity of using a matching
transformer when using this customer supplied line (could be any kind of
line matched only in voltage and current capability) to extend our known
instrumentation cable which measures 110 ohms Zo through that range.

These cables will be used for 2-way telemetry that will occupy the 10Khz to
50Khz frequency range and we want to reduce the reflections by terminating
the cable on both sides and driving the cable with source impedances that
are matched to the Zo of the cable. I have never tested what the tolerance
of mismatch can be before our telemetry is scrambled by standing waves
(bi-phase Manchester) so, if the effort is reasonable, I prefer to match it
as close as possible.

thanks for the replies.

9. ### John WoodgateGuest

(in <HWQGc.7\$>) about 'Zo of a transmission line',
I'll bet you 0.01 Turkish Lira that it isn't 110 ohms resistive in the
frequency range 10 kHz to 50 kHz. It would have to be a very strange
cable for that to be true.

10. ### John LarkinGuest

Such a 5-mile cable is almost surely dominated by resistive losses. 5
miles of #20 twisted pair will have about 500 ohm resistance, so much
loss there will be no significant reflections or standing waves. The
phone companies have traditionally considered their lines to be 600
ohms, even though the actual impedance of a short length of a
telephone pair is closer to 100 ohms. Don't worry about reflections,
with frequency.

John

11. ### Tam/WB2TTGuest

John,
We assumed unloaded twisted pair was 100 Ohms at 1 MHz, 135 Ohms around 50
KHz, and 600 - 900 Ohms at voice band. The thing I worked on required a
bandwidth of 100 Hz to 50 KHz; we terminated it in 135 Ohms.

Tam

12. ### mook johnsonGuest

You are correct sir. It has a negative slope and the 110 ohm number is the
50K reading where it is leveling out.

I'll had you that .01 Turkish Lira when I figure out how to slice up
pennies.

thanks for the input.

13. ### Jim BackusGuest

Which is why telephone lines (used to have) have loading coils at
regular intervals. It flattens the frequency response by making the
line look more like a true transmission line with the correct value of
L for the required impedance. The downside is that the line behaves as
a low pass filter. It's too long ago for me to remember and too late
coils and their value determines the cut-off frequency of the line.
IIRC the normal value of loading coils was 88 mH which is why they
used to be so popular in amateur teleprinter (teletype) decoders.

14. ### John LarkinGuest

That's right for an ideal transmission line, but if the line is
seriously resistive, the apparent impedance will ramp up from Zo as
time goes on.

John

15. ### BFoelschGuest

Well, let's see. If you have a charged, uniform transmission line, open at
both ends, and you suddenly connect one end to a resistance equal to the
characteristic impedance of the line, you will see (measured across the
resistance) a constant amplitude pulse of one-half the charge voltage and of
duration equal to twice the electrical length of the line.

Open the line at the distant end. Arrange a power supply to charge the line
through a resistor that is high compared to the expected impedance, say 5000
ohms. Use a relay to connect the line to a pot of 1000 ohms. Put the scope
across the pot and adjust the pot until you see a nice flat pulse of about
one-half the supply voltage each time the relay closes. Measure the
resistance of the pot. That will be the characteristic impedance of the
line.

This method only requires access to one end of the line, as ling as the
other end is known to be open.

16. ### KeithGuest

Fair enough, but most transmission lines are pretty nearly perfect, at
least until you get into the *miles* long stuff. ;-)

17. ### J M NoedingGuest

I've shown some measurements for one specific subscriber cable on
http://www.noding.com/la8ak/12345/n92.htm for 200-6000Hz and above
this it approaxes 120 ohm around 150kHz

Jan-Martin