Fault Finding with TDR

Discussion in 'General Electronics Chat' started by Biggsy100, Apr 7, 2014.

Apr 7, 2014
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Measurements performed using a TDR tester show that the time delay resulting from a power-line fault is 72.5 mS. The line-under- test has a velocity-factor of 0.67 and the echo pulse is non-inverted. Determine the distance to the power-line fault and discuss the significance of the measurement results.

Have I been 'sold a pup' with this question? I have no distance to measure so how can I get a location. As I understand it PL Fault/Velocity factor = X Which would then either be the distance or the power? 2. w2aew Active Member

Jan 3, 2012
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Sure, you *have* been given a distance. You know the speed of the TDR signal (0.67 * speed of light), and the round-trip time (72.5ms). And, since the echo is inverted, this tells you the nature of the fault.

Of course, we're not going to do your homework here, but just point you in the right direction.

For reference, you might want to view these videos on TDR. These videos show testing of coaxial cable, but the same principles apply in your case:

3. crutschow Expert

Mar 14, 2008
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The OP stated the echo was non-inverted, which still tells the nature of the fault, of course.

Apr 7, 2014
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After looking into this a little further today I now understand it as “Speed X Time = Distance” with one change.
Since the subject is a TDR the 72.5mSec delay is most likely the round trip time for the pulse leave the TDR, reflect on the fault or end of cable, and return to the TDR’s detector. So divide the time by two. Next to obtain the distance in meters rather than a fraction of kilometers multiply the speed of light by 1000. To get the most accurate results look up the true speed of light in km/seconds on Wikipedia or some other reference. The common quoted speed is generally a rounded figure. Now multiply the speed of light by the 67% figure in the question to obtain the cable’s signal velocity. Take those two results and apply the equation “Speed X Time = Distance” to get the answer.

Apr 7, 2014
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So the mats does this?

Time = 72.5/2 = 36.25
Speed = 300,000 x 1000 = 3000000

Distance = 3000000 x 36.25 = 108750000

6. RichardO Late Member

May 4, 2013
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Since WBahn is not here, I will ask for him...

What are your units? Does the result make sense?

Apr 7, 2014
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2 errors noticed first of all I should put 036.25 as they are m seconds?

And I have rounded up the figure and of course left a series of 0's out of what should be three million. By looking at several references on line speed of light is 299,792,458, 000 ?

Units are metres

8. RichardO Late Member

May 4, 2013
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Yes, the delay ws given in milliseconds.

300 x10^6 m/s is good enough for me. No TDR made is going to measure to 9 digits of precision! Your other numbers are only given to 2 and 3 digits of precision. Your result can not be any better than that.

You need to track your units through all of your calculations. Also you need to do a reality check when you are done.

You also forgot the include the propagation velocity in your calculation.

Apr 7, 2014
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So as I understand, this should be written as the following?

0.03225 msecond X (300 x10^6 meter’s/second X 0.67 Velocity) = Distance?

With regards to propagation, I always understood this to be associated with coaxials? This should be measured agaianst Speed of Light or in Parallel?

10. RichardO Late Member

May 4, 2013
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Yes, much better. There is still one oops. It appears that you are converting to milliseconds twice. 72.5 milliseconds should be 0.0725 seconds.

With this correction, does your result seem reasonable?

(Don't forget the second part of the question. Crutschow has given a good hint).

Not just coax cables. Any time a signal travels down a wire, it is slowed down. A wire can be viewed as a large number of little inductors and capacitors. These L's and C's create a delay. The inductance is caused by the magnetic field of the wire. The capacitance is formed by the wire coupling to the return path of the signal. In a coax cable, this return path is the shield. In the case of the power line, the other wire on the power pole or earth ground is the return path.

You have have used the velocity-factor given in the question correctly. It is the multiplier to get the effective fraction of the speed of light that the signal travels down the wire.

Apr 7, 2014
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Hmmm, ok a little confused now. I thought at the beginning I was to told to divide by two?

12. alfacliff Well-Known Member

Dec 13, 2013
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the devide by two is to let you know that its a two way path. out, reflect, and back. and a simple check, the speed of light is the maximum velocity, propagation DELAY will never be negative.

Apr 7, 2014
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So either way is ok? I'll think I will demonstrate both ways on the paper or should I not bother?

By doing sum 0.0725 x (300 x 10^6* x 0.67) = 1.45725 x 10^5**

* I am assuming 10 to minus 6 is what you ment ?
** 10 to minus 5 ?

With regards to the distance, looking at it I should take the first 4 digits as other wise this is one seriously long cable?