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I have a TDR with a sampling frequency of 200MHz that equals 1 sample very 5 ns. When I use a really long square wave the response will look like the left plot. Where delta t is the time of propagation through the cable * 2. When I use a short needle pulse the response will look like the right plot. These plots are ideal plots without considering attenuation of the cable.

But lets say I have a little damage in the middle of the cable. The damage changes the line impedance by 10%. The damage has a length of 30cm. With a NVP of 2/3 that means the signal travels 30cm in 1.5ns.

enter image description here

When using a long square wave the width of the reflection will be 3ns?

Using a needle pulse or square wave won't show me the damage in the cable or am I mistaken?

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  • \$\begingroup\$ What specifically do you mean by sampling rate? \$\endgroup\$
    – Andy aka
    Dec 18 '20 at 17:57
  • \$\begingroup\$ @Andyaka the sampling rate can be compared to the effective resolution of the device. For example an digital oscilloscope or an ADC with a sampling rate of 1GHz is able to sample 1000 000 000 values a second from a signal. \$\endgroup\$
    – Yoomo
    Dec 18 '20 at 18:01
  • \$\begingroup\$ A needle pulse also requires high time resolution to start the pulse and end the pulse so it doesn't make a lot of sense that something capable of "needle-shaped" resolution is then going to squander all that good work on the generation side by using a sh1tty resolution. \$\endgroup\$
    – Andy aka
    Dec 18 '20 at 18:06
  • \$\begingroup\$ @Andyaka lets say the width of the needle pulse is 10ns. But the important thing here is the time resolution of 5ns. I will be able to see the ingoing pulse but I think the small damage can't be seen \$\endgroup\$
    – Yoomo
    Dec 18 '20 at 18:09
  • \$\begingroup\$ @Andyaka I think either the time resolution has to be bigger or the damage has to span a wider length to be able to see it. \$\endgroup\$
    – Yoomo
    Dec 18 '20 at 18:12
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@Andyaka lets say the width of the needle pulse is 10ns. But the important thing here is the time resolution of 5ns.

And,

@Andyaka I think either the time resolution has to be bigger or the damage has to span a wider length to be able to see it.

See additions further below for a different anomalous scenario

If the pulse originally emitted is wider than the time-resolution of the measurement circuit, then you will see a reflection from a single point cable anomaly (such as a bridge of some conducting medium). Consider a 10 ns pulse (V1) being generated at the left-end of the simulated cable: -

enter image description here

At the midpoint area, R1 is disconnected (not representing a cable "anomaly") and, we get no reflection: -

enter image description here

I've terminated the far right end in 50 Ω so as not to confuse things with its reflection. Now, if I connect R1 (500 Ω) we see this: -

enter image description here

Now there is a pulse (also 10 ns) seen as a reflection from R1 and, this tells us that a single-physical-point deformity in the cable (represented by a slight increase of the mid-point conductance) produces a reflection that has the same length as the original transmitted pulse.

If there was an "extra length" to the received pulse, then that is due to the cable conductance deformity being physically longer than just a single-point.


However, if you were to insert a short middle section of cable with a different (lower) characteristic impedance, then the reflections will have pulse widths corresponding with the short length of the inserted cable. For instance, here's the model: -

enter image description here

Notice in the above model that the middle section is 0.25 metres long and has a significantly higher capacitance per metre than the two 5 metre sections on either side. The characteristic impedance of this added short middle section is: -

$$\sqrt{\dfrac{250\text{ nH}}{150\text{ pF}}} = 40.8\text{ Ω}$$

This is the reflection: -

enter image description here

If the inserted cable was halved in length to 0.125 metres we would get thinner reflections: -

enter image description here

If the inserted middle section were removed and replaced by a rogue 1 pF capacitor we would see this: -

enter image description here

So, take your pick - is the cable fault some rogue conductance bridging the forward and return wires or, is the cable fault because someone has inserted a short length of "the wrong cable"? Maybe it's just been squeezed locally at one point and the capacitance has increased a tad?

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  • \$\begingroup\$ I think the question is whether the spacing (the time difference) can be determined if it is below the scope resolution. \$\endgroup\$
    – P2000
    Dec 18 '20 at 19:15
  • \$\begingroup\$ @P2000 read the comments under the question - my discussion established that the transmit pulse was wider than the physical measurement resolution. \$\endgroup\$
    – Andy aka
    Dec 18 '20 at 19:16
  • \$\begingroup\$ Hmm indeed, but this confuses me more: if the "needle" pulse is already 10ns wide, i.e. above the reflection delay, then his response is his left diagram not right and he has a "really long square wave". \$\endgroup\$
    – P2000
    Dec 18 '20 at 19:21
  • \$\begingroup\$ @Andyaka thanks for your post. This was really helpful. Can I ask you which program you used for that type of simulation? \$\endgroup\$
    – Yoomo
    Dec 20 '20 at 7:41
  • \$\begingroup\$ @Yoomo I used microcap V12.2.0.4 - it's now free here - download the executable on the top line of the download table. \$\endgroup\$
    – Andy aka
    Dec 20 '20 at 10:20
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I think your math is correct.

I also agree with the comments by @Math

The first reflection occurs at the first change in impedance (Z0 -> Zdamage), and the second when at the end of the damage (Zdamage -> Z0). I suppose you know how to calculate Gamma and that you know you are dealing with very small reflections that might be difficult to "pin"-point after attenuation and dispersion, but that is a practical matter and it isn't your question.

The reflection from the second impedance change is 60cm behind the first, or 2x 1.5ns = 3ns. Also note there will be more reflections than just those too. In fact, if you swept with a sinusoid you might see a peak in reflected amplitude (or an extreme max/min in impedance) proportionate to the length of the damage, attributed to a standing wave (albeit with poor quality) between the two impedance changes.

To obtain such fine time resolution with sub-sampled (fs=5ns) needle-pulses is of course not theoretically correct. However, your 200MHz scope which perhaps samples at 1Gsps, combined with dispersion applied to the pulses, might still provide you with reflections that can be compared in time.

If you have the possibility to apply a train of pulses, you can try using PN (pseudo-noise) sequences, even short sequences or sinusoids, and cross correlate to find the mutual delay. Or you could model the delay and perform a maximum-likelihood comparison with the shape of the received pulse.

The dispersion will in fact provide you with slopes, and sloped signals can have their delay estimated even if sub-sampled. If you have the equipment, you could take short captures and process them off-line.

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  • \$\begingroup\$ "The reflection from the second impedance change is 60cm behind the first, or 2x 1.5ns = 3ns." Ah, yes, I forgot to count the time twice. \$\endgroup\$ Dec 18 '20 at 19:41

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