Having downloaded a copy of the raw data (hlogj3w9k.txt) my first action was to re-plot the Hlog graph so that there was more detail on the scale of the X-axis. (A copy of the graph is attached, below.)
Taking note of the approximate location of each minimum (from the graph), I then turned to the tabular data and attempted to specify each minimum by sub-carrier (tone). It was quite clear for the first, not quite so obvious for the second & third (neither were required for the calculations) and somewhat difficult for the fourth. For the latter minimum, I eventually decided to take note of the two extreme sub-carriers and perform the calculations twice.
Here is the table of the data used in the calculations --
53 -25.5000
54 -25.6250 Minimum 1
55 -25.5625
56 -25.5000
176 -34.6875
177 -34.7500
178 -34.8125
179 -34.8125 Minimum 2 (approximate)
180 -34.8125
181 -34.8125
182 -34.6875
298 -42.5625
299 -42.6875
300 -42.6250
301 -42.6250 Minimum 3 (approximate)
302 -42.6875
303 -42.6875
304 -42.5625
409 -50.2500
410 -50.0000
411 -50.3750 Minimum 4 ?
412 -50.2500
413 -50.3750
414 -50.4375
415 -50.4375 Minimum 4 ?
416 -50.3750
How can we be sure that we are looking at the effect of a bridging tap on that circuit's transfer function?
(A) Look at the shape of the lowest frequency minimum. Notice how it is similar to an exponential growth (or decay) curve that has been rotated in the plane and an image of itself reflected along the vertical axis. In this particular case it is not too clear but in other examples it is a distinct diagnostic "give away".
(B) Consider the delta between each successive minimum. Is it consistent?
The delta between minimum 1 & minimum 2 is (approximately) 125.
The delta between minimum 2 & minimum 3 is (approximately) 122.
The delta between minimum 3 & minimum 4 is (approximately) either 110 or 114.
The three deltas are of similar order of magnitude hence, coupled with the shape of minimum 1, we can conclude that we are looking the effect of a bridging tap.
The calculations will be performed using minima 1 and 4, noting that there are two intervening minima.
Now working with the relevant data from Table 2 of the JDSU Application Note ("Detecting Bridged Tap and Noise Interference in VDSL2 Access Networks using the JDSU SmartClassTM TPS"):
Using the minima at 54 & 411. Delta is 357.
According to Table 2, for two sub-minima . . .
350 198.8
400 173.9
Interpolation for 357 gives either . . .
350 / 357 = N / 198.8
So N = 350 x 198.8 / 357 = 194.90
Or . . .
357 / 400 = 173.9 / N
So N = 173.9 x 400 / 357 = 194.85
Using the minima at 54 & 415. Delta is 361.
According to Table 2, once again, for two sub-minima . . .
350 198.8
400 173.9
Interpolation for 361 gives either . . .
350 / 361 = N / 198.8
So N = 350 x 198.8 / 361 = 192.74
Or . . .
361 / 400 = 173.9 / N
So N = 173.9 x 400 / 361 = 192.69
The arithmetic mean of the four above values is --
(194.90 + 194.85 + 192.74 + 192.69) / 4 = 193.80
The delta of 194.90 & 193.80 is 1.10
The delta of 193.80 & 192.69 is 1.11
Accepting the larger of the two deltas, 1.1, (to one decimal place) as the error, I propose that the bridging tap responsible for the effect shown in the Hlog plot has a length of 193.8 +/- 1.1 metres.
We cannot deduce the location of the start of the bridging tap (relative to a known location in the circuit) by considering the Hlog plot. Other techniques (Time Domain Reflectometry, for example) would have to be deployed to obtain the required datum point.
