First, just consider just the classic telephony circuit.There is a pair which leaves the exchange building and, as part of a cable bundle, proceeds to the PCP. That is the E-side for your circuit. Let us call the physical length of your E-side
LE.
In the PCP, the E-side is connected to the D-side of your circuit. Let us call the physical length of your D-side
LD.
The total circuit length, from telephony equipment within the exchange building to the NTE5/A in your home, will therefore be the sum of the E-side length plus the sum of the D-side length plus the sum of the final drop (or service feed) into your home plus any internal wiring (on the
Openreach side of the NTE5/A). Thus --
L
Total = L
E + L
D + L
Otherwhere
LOther is the sum of the final drop (or service feed) plus any internal wiring to the NTE5/A.
A Broadband Internet (VDSL2) service is now added to the above.Within your home, let us assume that a centralised filter is fitted to the NTE5/A. The low frequencies (300 - 3400 Hz) now only reach the telephony socket and the higher frequencies (~26 kHz and upwards) only reach the modem (or modem/router). (See
xDSL Frequency Plan for Annex A.)
In the PCP, at the junction of the D- & E-side cables, the direct connection of those two parts of the circuit are severed. They are each connected to a tie cable pair to link the "fibre cabinet's" DSLAM into the circuit. Within the "fibre cabinet", those two pairs are connected to the same point -- the line card port of the DSLAM. In the case of the D-side pair, they are connected directly. In the case of the E-side pair, they are connected via a low pass filter.
The length of the metallic pathway for the Broadband Internet service is --
L
Other + L
D + L
D-tie (1)
where
LD-tie is the length of the D-side tie cable.
The length of the metallic pathway for the telephony service is thus lengthened to --
L
Other + L
D + L
D-tie + L
E-tie + L
E (2)
where
LE-tie is the length of the E-side tie cable.
In terms of the location, the low pass filter should "appear" between L
D-tie + L
E-tie, thus --
L
Other + L
D + L
D-tie +
Low Pass Filter + L
E-tie + L
E (3)
where I now make use of the formula that defines the telephony total circuit length as indicators of the wires of the circuit pair.
Consider the telephony circuit defined in equation (2); all is good the telephone operates satisfactorily.
Consider the telephony circuit defined in equation (3); all is good the telephone operates satisfactorily.
Equation (3) is the normal situation when a VDSL2 service is provisioned to share the pair (D-side + Other) to your home.
If the situation depicted by equation (2) becomes true (the low pass filter becomes faulty, falls out or has never been correctly inserted) then there is no effect on the telephony service.
However, for the VDSL2 service, which is normally depicted by equation (1) the situation becomes sub-optimal (or even disastrous). The VDSL2 service now "sees" the situation as depicted by equation (2). The E-side of the pair, which just brings the telephony service to the cabinet, has become a bridging tap of length
LE-tie +
LE.
It is my current speculation that
LE-tie +
LE is of the order of approximately 744 metres. The only cause for this defect would be a problem with the low pass filter within the "fibre cabinet".
[Edited to correct the approximate length, as a result of reworking my earlier calculations to include the last dip (the twelfth) that had been missed from the original calculations.]