[VFKD] Wird bei Rückwegstörungen überhaupt etwas unternommen?

[Flole]

Entkopplung wirkt aber mit und ohne Abschlusswiderstand genau gleich. Der ändert daran genau gar nichts, sie wird dadurch weder stärker noch schwächer.

[denunge]

How is that supposed to work? As I said, the signal from modem A has to go back out through the tap INPUT. I don't see how you could prevent that from being propagated to all tap ports, as it is on a frequency that HAS to be propagated from the tap input to all tap ports, for the downstream signal.

The tap has isolation of many dBs between tap port and out.

Going away from the node: A small amount of upstream signal will leak out through the out port.

Going towards the node: A small amount of signal will leak into the tap port from the out port.

How much of the signal will leak? It depends on the tap and its tap value. Higher value taps have more isolation than lower value taps. The coax itself will also add to the attenuation.

By using a sounding scheme (one modem is talking while the rest see if they can hear it), the system will know who interferes with who.

[Fabian]

Entkopplung wirkt aber mit und ohne Abschlusswiderstand genau gleich. Der ändert daran genau gar nichts, sie wird dadurch weder stärker noch schwächer.

Die Entkopplung des Verteilers zwischen den Ausgangsports gilt für korrekt abgeschlossene Ports.
Bei fehlenden Abschlusswiderständen in Baumverteilungen wird Leistung an der Enddose reflektiert. die auf den entsprechenden Ports gebracht wird, sodaß dann mehr Leistung auf den Nachbarport gekoppelt wird.

[Flole]

Die Entkopplung wird in dB angegeben, du redest jetzt von Leistung, die nicht in dB angegeben werden kann. Die Entkopplung wirkt immer gleich, egal ob mit oder ohne Abschlusswiderstand, es ist aber ein Unterschied ob diese Entkopplung auf 10dBmV oder 100dBmV wirkt.

[Wechsler]

Mir ist nicht klar, wie das funktionieren sollte. Irgendwo muss ja ein Verteiler sitzen, an dessen einem Ausgang ein Kabelmodem auf einer Frequenz sendet, die er sowohl zum Eingang als auch zu allen anderen Ausgängen durchlassen muss. Wie soll es da möglich sein, das Signal von dem Kabelmodem zu isolieren?

Das ganze wird für Amerika mit seinen suburbanen oberirdischen Kabel-TV-Verteilungen entwickelt, nicht für deutsche MFH mit seinen Baum-Netzen. Dort kann man auch jeden potenziellen Störer von außen anfahren und direkt auf der Straße abklemmen.

Das ist genau das Problem, warum eine importierte Technologie wie DOCSIS in Deutschland grundsätzlich nur schlecht funktioniert.

[Fabian]

The tap has isolation of many dBs between tap port and out.

Going away from the node: A small amount of upstream signal will leak out through the out port.

Going towards the node: A small amount of signal will leak into the tap port from the out port.

How much of the signal will leak? It depends on the tap and its tap value. Higher value taps have more isolation than lower value taps. The coax itself will also add to the attenuation.

By using a sounding scheme (one modem is talking while the rest see if they can hear it), the system will know who interferes with who.

How about splitters ?
They have isolation below 30 dB, taps up to 36 dB.

[denunge]

Isolation is one factor. Another factor is power levels: How strong is the downstream feeding the splitter vs how strong is the leaked upstream. Insufficient isolation will be detected by the sounding process (grouping of modems), which will prevent the scheduler from telling these modems to transmit on top of each other's download.

Another concept is "soft FDX". Here, the spectrum never overlaps, but you get the advantage of the dynamic spectrum allocation: When there is high demand for upstream, it will take some spectrum away from downstream - and give it back again, when there is no longer demand for it.

If we look at Comcast: They use this dynamic feature to offer much higher upload tiers, without upgrading the taps. There is an issue with insufficent isolation in current low value taps, so when Comcast begins to deploy FDX amplifiers, each node leg will be "soft FDX" - a single interference/transmission group.

[robert_s]

The tap has isolation of many dBs between tap port and out.

Going away from the node: A small amount of upstream signal will leak out through the out port.

Going towards the node: A small amount of signal will leak into the tap port from the out port.

Are you sure that works for FDX? For frequency division, I can see how that works: Lowpass and highpass filters split the signal, and then you make sure each signal path only goes one-way.

But when it comes to FDX, there is no direction to a frequency, everything is bidirectional. How would the tap achieve separating the signals? It has to "mirror" the mixed signals from its outputs to the input, without attenuation, as otherwise the FDX upstream would not work. The downstream-only signal is not available, so how would the tap output only that on the other ports?

[denunge]

I'm sure. That is how FDX is designed to work.

FDX doesn't change the nature of RF and how it propagates through the coax and the taps (in either direction). The node and the amplifiers have traditionally used diplex filters to separate upstream and downstream. That must change. Echo cancellation is also invoved in this.

When the modem transmits something to the node, it goes from the tap port to the in port. When it reaches the next tap, it goes from the out port to the in port (insertion loss). You don't get hit by the isolation between out and tap port. FDX doesn't change this.