If I were to connect computers together using coaxial cable and BNC connectors, I would have to ensure a terminator is fixed at either end to prevent reflections up and down the line. Why are the not needed using twisted pair into RJ45?
There's no difference in the principle. The NIC at each end provides the proper impedance termination to match the characteristic impedance of the CAT3/5/5e/whatever cable, which is what the terminator did for coax.
Thanks for the reply, if I unplug my RJ45 from the Telecoms outlet socket on the wall, I am therefore removing the NIC from the network and therefore the terminating effect, why dont I have to plug a terminator in its place?
In theory, you should. And one into the NIC. And you would have to if there were other stations on the segment. But each link is isolated buy the hub/switch, so it doesn't much matter if the drivers see opens, or drive voltage to Vcc. No-one is listening. I don't even think this worsens EMI, because there's less current.
If I have a video distribution system (call it a hub) going out to several coax cables, I have to terminate the unused cables with a 75 ohm termination, otherwise I'll get ghosting in the monitors that are connected to the "hub". The reason is because all the video feeds are using analog technology which does isolate them from each other, whereas an ethernet hub using digital switching chips has isolation.
Anyway, the coaxial Ethernet is a bus where all nodes share the one line whereas twisted-pair is hub-and-spoke where each node connects to a hub on a dedicated pair. If you unplug one node from a hub-and-spoke the other nodes are still electrically matched with the hub.
The data on a bus (coaxial Ethernet) is messed up by the signal reflections of an unterminated node. An unterminated node will act as a mirror and energy reaching that open connector will reflect back into the coax at pretty much full strength. The original signal will electrically add to the reflected signal, which in many instances will reduce, cancel, or increase the original signal. Not good. The proper resistance terminator absorbs the energy, making the cable carry only the signal, not signal+reflection.
In twisted pair ethernet system there are separate wire pairs used for transmitting and receving (applies 10Base-T and 100base-TX). If there is no NIC or termination on the other end of cable, signal gets there, gets reflected back etc... the signa waveform on the cable itself is useless, but no problem nobody is listening. Sending data to pairs that do not have NIC connected does not cause network problems, the signal sent there does not get anywhere. The signal coming back from NIC to hub/switch is transported on the different wire pair...
The EMI issue is real. For this reason Ethernet hub/switch hardware is designed in siwch way that it sends data to only those ports that have an active NIC connected to them. In 10Base-T network, the hubswitch knows that NIC is connected to it when it receives either link pulses or real data from the NIC. In 10Base-TX system there is first link nagotiation (sending link pulses every now and them) and then when link is established a constant flow of bits (keeps the link up and clocks syncronized).
A coaxial system is unbalanced to begin with. What do you mean by "an imbalance on one leg?" Do you mean an impedance mismatch?
An unterminated leg causes reflections, which results in standing wave. In the digital world, said standing wave can cause packet loss. If the standing wave is severe enough, service can be disrupted completely. In the analog world, said reflections can cause multiple images. If severe enough, those same reflections can render a picture unwatchable.
Please elaborate on the statement "A true distribution amp has one active amp per channel and won't suffer this fate."
A passive splitter is bidirectional; while the intended effect is the input signal appears at the output port; the converse is also true. Not correctly terminating the output affects the input to some extent.
An active amplifier has FAR greater isolation in the reverse direction. You can hold a rain dance on the output & the input will never wake up. If your distribution amp is multi-port, and has a separate amp for each channel, then you'll have excellent isolation between outputs as well.
But most of them are ONE amp, followed by a passive splitter. That will offer little more output-to-output isolation than a fully passive system.
 Subject to the attenuation ratio, and other factors, of course.
With the return path being utilized for DCTs, cable modems, and other advanced services, most distribution amplifiers have a return path, either passive or active.
A splitter on the output of a distribution amplifier? Yes. Splitters have 25-40dB port-to-port isolation(depending on frequency and manufacturer. We're also talking real devices, not the ones you pick up at Wally World or Home Depot.). How else are you going to get the signal to several outlets? Manufacturers don't know the design of your system. Perhaps you need a DC-16 to feed the 3 outlets that are close to the amp, but the majority of the outlets are some distance away. Perhaps there are 23 outlets that are homerun to the closet. Perhaps it is a hotel where the system consists of a DC in the closet on each floor, with RG 6 and taps running down the hallways.
Perhaps, perhaps, perhaps...
In order for there to be an amplifier for each individual leg, there would need to be a splitter prior to the amps, thus increasing the input signal requirements to the amplifier, which would be different for an amp with 1 output than an amp with 2 outputs. A 4-output amp would be different still, as would an 8-output. Not to mention the cost of an amp with 4 amplifiers would be ridiculous.
Splitters have the same specs, whether inserted at the output or input of an amplifier, which means that a 2-way splitter inserted at the input of 2 amplifiers that have passive return paths will have the same
25-40dB port-to-port isolation that it would inserted at the output of an amp. The distribution legs would see a slight increase in isolation (~1-2dB) because of the insertion loss of the passive return paths. If
1 or both of the amps have active return, the isolation goes right down the toilet.
The best solution all around is to make the amplifier and let the end user worry about the distribution.
Coaxial ethernet is much more sensitive to the voltage on the line than UTP ethernet. UTP is terminated at 100 ohms, but for 10baseT that is probably good for up to 150 ohms.
When I was first doing ethernet wiring with thick (10base5) cable we had some two port and four port transceivers. It seems that the design of the two ports wasn't quite right. If you put an AUI cable on one port, but didn't connect it to anything, the other port would not work. After complaints to the company they sent AUI terminators to connect to the unused cable end.
10baseT is designed with unterminated cables in mind, and so must ignore any reflections from unterminated cables.
Also, 10Base-2 coax (like CCTV) was designed with "daisy-chaining" via tee's in mind. In that case you don't want to "terminate" the signal except at the two ends of the daisy-chain. 10Base-T does not allow for daisy-chaining, and the terminating is done inside the device's port.
A) The terminators go at the ends. With UTP, there's no daisy-chaining so the NIC's are the endpoints and have the termination within...
B) ThinNet used a trick for collision detection, as I recall. The active talker sourced/sank a given level, through a resister. If TWO ports with talking, the voltage on the coax would be half if 2 were active.
But unconnected cables aren't terminated. If the signal was sent out such cables, and without careful design of the transmitters, it would be possible for the reflection to cause problems with other ports.
I would expect that part of the design was that it had to work with unterminated cables.
But a port transmits on one pair, and receives on another.. The 'sending' electronics will block anything coming back. Any sort of discontinuity (short, open, mangled cable) will induce a reflection, but as the hub/switch/port is 'blowing' down the cable, anything coming back is ignored. Co-ax was/is a bi-directional system.
Yep, otherwise every time a port was patched to an 'open' jack, the network would crash! ;-)
Well, anything coming back goes to the output driver of the transmitter. If it isn't isolated enough, it can get back to the power supply.
As I wrote previously, I had this exact problem with two port 10base5 transceivers. It might be that they drove both outputs with the same driver, though probably with resistors in between. Enough came back to interfere with the other port if an unterminated AUI cable was connected. Four port transceivers did not have this problem.