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I thought the Ethernet is logically a one-line communication bus (for argument's sake, I am excluding hubs). All machines attached on the bus hears the same signals and the machines themselves try to avoid collisions by randomly backing off.

http://computer.howstuffworks.com/ethernet6.htm

If so, why would splitting one Ethernet line from my home router into two and connecting two computers not work? Why do I have to add a switch to it?

*What the Internet said would not work.

[4 port home router] ------[one Ethernet cable]-----[simple splitter]======[two computers]

*What the Internet said I should do

[4 port home router] ------[one Ethernet cable]-----[switch]======[two computers]

Is this because of the signal degradation (reduced electric current)?


Thank you for all the answers! The reason why I did not just use the two ports of my home router is...

The 4-port gigabit router is in my room, and I had put a computer in another room (also my room, though). Since a wired network is far more reliable and secure, I had bought a long Ethernet cable and and connected the computer to the router. Now I was thinking about adding another computer to that room. I could buy another long Ethernet cable, but then there will be two cables between the rooms. The one line already is a minor annoyance, so I thought if I could share the one line between the two computers in that room. A switch would work, but it requires power and is a little bit pricey. That is why I wondered why it would not work to simply split the physical Ethernet cable.

Apparently I do not completely understand how Ethernet and a switch work. I just have some bit of knowledge I heard in my college class.

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If you understand that ethernet is one-line communication, and you understand how a Switch works, I don't understand your question. If you split an Ethernet cable it no longer would result in one-line communicaton because two devices would be on the exact same line. You understand how a switch right? –  Ramhound Jun 6 at 17:49
    
You can do with your 4 ports router and two cables. You don't need a switch, because what you call router is actually a router+4ports_switch+(in some cases)modem. To put it differently, if your router has more than one exit port, it also has some switch capabilities. –  Bruno9779 Jun 6 at 17:54
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Actually, it does work, sorta, sometimes. Only not very reliably and the transmitters are operated outside of their parameters. Plus you can't communicate with your "twin", only the other end, because the transmitted data will not be reflected back up your (shared) cable. –  Daniel R Hicks Jun 6 at 18:59
    
I don't see why you just don't lay down two cables if you don't want another hub: [4 port home router] =========================== [two Ethernet cable]======[two computers] –  RoboKaren Jun 7 at 5:55
    
Ethernet has been reinvented a few times. Only the first version was based on a single bus that all computers are connected to. –  user20574 Jun 7 at 8:09

6 Answers 6

up vote 44 down vote accepted

In 10BASE-T and 100BASE-TX, one pair of wires is used for transmitting, and one for receiving. That is, one pair is the pair the Ethernet host transmits on, and the hub or switch receives on, and the other pair is the pair that the the hub/switch transmits on, and the Ethernet host receives on.

If you split the cable with a simple passive splitter, you're hooking up those two Ethernet hosts transmitter-to-transmitter and receiver-to-receiver. That's like holding the phone handset upside down and trying to speak into the speaker and listen to the microphone—it just doesn't work. So even if both were in half-duplex mode (like they were hooked to a hub, not a switch), neither of the Ethernet hosts would be able to sense when the other was transmitting, because neither one's receiver was hooked up to the other one's transmitter. So they would have undetectable collisions. Not to mention that they'd both be connected to the same port of the hub, probably confusing the hub's autonegotiation ability, because hubs don't expect to autonegotiate with two separate hosts on the same port.

In many ways, things are even worse in your case of hooking them both up to a switch, because they could both end up thinking they can do full-duplex, which means even more undetectable collisions, on what's supposed to be a collision-free link (properly-wired full duplex links can't possibly have collisions).

With 1000BASE-T (Gigabit Ethernet over Cat5 or better UTP copper cabling), the situation is even worse, because all 4 pairs of wires are used for both transmit and receive (simultaneous, full-duplex), and the transceivers are sophisticated enough to enable that. But if you suddenly have a third party on the line transmitting and receiving all at the same time, it completely blows away the way the simultaneous bidirectional signaling scheme works. With three devices all transmitting at the same time, even when you subtract out your own transmission, you can't differentiate the other two devices' transmissions in the signal you're receiving.

Some early flavors of Ethernet, such as 10BASE-2 a.k.a. "thinnet" a.k.a "cheapernet", featured a bus topology where all the hosts on the LAN literally shared the same wire (the same coaxial cable). Because the same wire was used for both Tx and Rx and there could be any number of hosts on the bus, it had to be half-duplex. But a 10BASE-2 transceiver was expecting it to be that way. And since all the transmitters and receivers were hooked up to the same wire, everyone could hear each other (unlike your split 10/100/1000BASE-T example).

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Would anyone downvoting this care to explain why? –  Spiff Jun 6 at 21:28
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+1 from me, excellent explanation –  Chris McKeown Jun 6 at 22:12
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Thank you for your comprehensive explanation. It seems the the concept of Ethernet I had learned in my college was the "cheapernet" you mentioned. –  Sin Jeong-hun Jun 7 at 1:28
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Might be worth mentioning too that while not really Ethernet, 802.11 must also contend with collisions as well as an arbitrary number of hosts on the "bus" (radio channel) -- even more so than 10base-2 and friends, because there might be several separate networks even using the same channel all within range of each other. –  Michael Kjörling Jun 7 at 19:23

The original Ethernet spec called for coaxial cables that were tapped (split) to each workstation (hence the "ether" in ethernet). But we're talking ancient history here. Technically it's still possible with RJ-45 cables since the ethernet protocol still supports the collision detection mechanisms, but why in God's name would you want to set it up that way? Especially since your router has 4 ports to work with in the first place.

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I don't think this really answers his question. –  Spiff Jun 6 at 18:02
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@Spiff, as I read it, this is exactly the thing the op seems to be missing (eg that you can split coax cause its a single wire analog cable, but that the same is not true of cat-X cabling). –  Frank Thomas Jun 6 at 18:38
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I think this is better that an actual answer. If the OP only has 4 hosts why wouldn't he just use one port for every host? If the two in the example were only hypothetical and they were more, then the collision rate must be a PITA. It just wouldn't be efficient, maybe the best thing would be to buy a switch. –  arielnmz Jun 6 at 20:55
    
@FrankThomas OP asked for an explanation of why splitting a Cat5/RJ-45 cable won't work. Instead of explaining why not, Wes claimed (incorrectly) that it's technically possible. –  Spiff Jun 6 at 21:21
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I was not falsely claiming it was possible. It totally IS possible. You can tap the pairs on an RJ-45 cable the same way you can an analog RJ-11 phone cord. See this link RJ-45 splitter. Maybe my language was a bit harsh, but I can't think of a single, logical reason to do so when you've already got a 4-port switch to work with. –  Wes Sayeed Jun 6 at 22:09

If one were to split a cable such that two device's receive inputs got data from the third device's transmitter, and the first two devices' transmitters fed the third devices' receiver, then data transmitted by the third device might be received by the first two, and it's even possible that the third device might hear data transmitted by one of the first two, but reliability in either case would be poor.

Imagine a cable as a Slinky brand spring toy which is hanging vertically and floating at the bottom. If one briefly jostles the top of the spring, a wave will travel down the spring to the bottom, whereupon it will be reflected back up. Fastening the bottom end to the floor won't solve the problem. It will reverse the polarity of the reflected wave, but the reflection will still be there. The only way to avoid a reflection at the bottom of the spring would be to have enough give to prevent a like-phase reflection, but not so much give as to cause an anti-phase reflection.

Internet cables operate much the same way--a device sends out pulses and expects that the other device will have just enough "give" to absorb them cleanly. Anyplace the characteristics of a cable change will cause reflections and other such unwanted effects unless proper measures are taken to prevent them. If packets are sufficiently short, and code waits long enough before sending a packet than any reflections which were propagating through the cable have died down enough, it may be possible for some data to be sent through the cable. Since Ethernet communications generally don't include such delays, however, communications are apt to be unreliable. It's possible that a device might transmit e.g. the first ten packets of data it wants to send, resulting in the first two being received and the rest being garbled by the first; the receiver might, upon receiving the second packet, hold off on acknowledging it until it decides no more data is coming immediately (conveniently not making that determination until after the noise has died down). Upon getting the acknowledgment from the second packet, the transmitter would send the third through twelfth packets (again, with only two of the ten making it), the receiver would acknowledge the fourth, etc. Data might make it through, but slowly at best.

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+1 for the slinky analogy to explain impedance mismatches! –  Oliver Charlesworth Jun 7 at 19:39

If your network is 100BASE-TX with a Cat 5 cable you can split the cable, but you do so by using an adapter to use all four pairs in your existing long Cat 5 cable (usually only two of the four pairs are used).

So you'd need two very short patch cables at the router's end, two RJ45/Cat 5 LAN splitters, and some longer patch cables at the computers' end. In the diagram below, 'Y' signifies a splitter and an arrow signifies one Cat 5 cable.

.----------.                                         __  _ 
| Router   |                            .---------> [__]|=|
|          |       .----.               |           /::/|_|
|          |------>| Y  |             .----.
|          |       |    |------------>| Y  |
|          |------>|    |             '----'
|          |       '----'               |
'----------'                            |           __  _ 
                                        |          [__]|=|
                                        '--------->/::/|_|

The above solution won't require an extra power source, but it will require use of two ports on your router, but it would at least be tidier than two separate cables running in parallel.

See also the discussion on 100BASE-TX and its wiring in Fast Ethernet, subsection Copper.

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Keep in mind that an Ethernet network deals with the transmission of signals, which is a matter several orders of magnitude more finicky than the matter of transmission of electrical power.

The Wikipedia article 10BASE2 outlines the pros and cons of the 'old' Ethernet systems. While additional hosts could, in fact, be added without a hub, it was never quite as easy as 'splicing' another segment in.

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At least two misleading things here. There is no "breakdown voltage" involved (varying impedance doors not cause "spikes"), and delay has nothing to do with raw bit-rate. –  Oliver Charlesworth Jun 7 at 8:15
    
I guess I'll delete most of it, then... –  Nevin Williams Jun 7 at 16:29

I'm surprised I must disagree with Spiff -- in a sense it does work. We were hunting the cause of excessive packet errors in the factory. Among other things we found where some electrician had simply spliced a Y into a 100BASE-T network cable.

The two computers involved sometimes had network errors, but since this persisted for a long time while the users used a program that was on the network and all its data (except stuff written to the temp directory) was on the network I can conclusively say it's possible.

The switches are the traffic lights of the network -- without them packets run into each other badly. Normally the network protocol makes up for the lost packets, though.

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I suspect that the network interface cards on either end had something to do with that. Some of them are more tolerant/smarter than others when it comes to dealing with wiring faults. Back in the days of coaxial Ethernet, I had one particular card that somehow managed to work without a properly installed terminator. –  Andon M. Coleman Jun 8 at 1:32
    
@AndonM.Coleman Agreed. The cards involved must have been able to tolerate it. At this was done at least 10 years ago I would think that a feature that was on some plain vanilla cards back then would be in all cards by now. –  Loren Pechtel Jun 8 at 3:33

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