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[burakkucat]

Over the years, our friend konrado5 has posted many links to various documents that he has come across in his own research. I suspect that I am not the only one to always make a point of reading the documents so linked. At times, my reading list becomes rather long and requires a concerted effort to reduce its length.

One recent document was pages from the Analogue Applications Journal. This was an article titled "Evaluation criteria for ADSL analog front end" by John Z. Wu and C.R. Teeple. At the time of its publication, the authors were then employees of Texas Instruments, Incorporated.

Two sections of that document caught my eye. The first was at the last bullet-point on page 17 --

• The telephone loop noise is assumed to be 31.62 nV/Hz^1/2, or -140 dBm/Hz. It is specified as the quiet line noise PSD N(f) for a particular subcarrier in G.992.3-ADSL2; and it is the rms level of the noise present on the telephone line when no ADSL signals are present on the line.

It was the presence of the value -140 dBm/Hz . . . a value I routinely consider (and quote) to be the average noise floor of an xDSL circuit. Its stated equivalence with 31.62 nV/Hz^1/2 caused me to have an "ah" moment. With a little bit of manipulation, those two values will allow the resistance (I should really say, more correctly, impedance) of the standard xDSL circuit to be evaluated. Having read much about the alignment of audio circuits in the era of the GPO, and then Post Office, telephone service of the U.K. the value of 600 Ohms is a regular occurrence. Does that value also hold for higher frequency xDSL circuits?

Starting from first principles . . .

The power, in Watts, of a circuit --

P = V x I

Ohm's law --

V = I x R

By combining the above two equations --

P = V² / R

Let M be the power, in milli-Watts, of a circuit. Thus --

M = V² x 10³ / R          (eqn 1)

Let D be the power, in deci-Bels relative to one milli-Watt, of a circuit. Thus --

D = 10 x log₁₀ M

Rearranging so that M is to the left of the equality symbol --

M = 10^(D/10)               (eqn 2)

Now substituting equation (2) into equation (1) --

10^(D/10) = V² x 10³ / R

Rearranging so that R is to the left of the equality symbol --

R = V² x 10³ / 10^(D/10)    (eqn 3)

Substituting the known values into equation (3) --

R = (31.62 x 10^-9)² x 10³ / 10^(-140/10)

  = (31.62)² x 10^-18 x 10³ / 10^-14

  = (31.62)² x 10^-1

  = 99.98 Ohms

Hence 100 Ohms for an xDSL circuit and not 600 Ohms, as used for a telephony audio circuit.

I continued to read the document and saw that the first complete paragraph on page 19 stated --

The actual twisted-pair loop impedance Z(f) is the frequency function of the telephone line. The magnitude of the twisted-pair impedance is about 600 Ohms for the POTS bandwidth; but for the ADSL signal bandwidth, it is about 100 Ohms without bridged taps and varies from 60 to 100 Ohms with bridged taps.

I feel that my earlier calculation is, therefore, correct.

[boost]

I struggle to count the change in my pocket

What does all this ultimately mean?

[burakkucat]

What does all this ultimately mean?

In as few words as possible . . . "We should remember to regard the metallic pathway, used by an xDSL circuit, as a radio frequency transmission line and not as a telephony audio circuit, from which it is derived."

[Weaver]

Good advice!