- This topic has 17 replies, 1 voice, and was last updated 10 years, 8 months ago by Pan.
6th January 2010 at 06:07 #60601MKTGuest
I just want to know that:
Can we send data = more than the capacity of E1.
Imean if E1= 2Mbits/sec
then is it possible to send more than 2Mbps over a E1.
Regards6th January 2010 at 07:04 #60602MKTGuest
My question is somehow different in a sense that I want to know is it possible to send more than 2Mbps over an E1. As you know the source Node-B has to send user traffic over E1’s. Here we are using ATM over E1. I want to know is it possible that when 2Mbps delivered at the destination actually conveys more than 2Mbps. As you can see that 6Mbps ceiling for a cell doesn’t make a big sense coz it will be reached with 2-3 UE’s alone.6th January 2010 at 14:07 #60603PixGuest
1 E1 can carry up to 2Mb/s.
That’s the way it is, and it cannot be exceeded.
This is why the transmission networks are moving to IP transport, rather than E1.8th January 2010 at 04:43 #60604MKTGuest
Same Time slot can carry ( or can convey to be precise)THREE TIMES MORE in EDGE in comparison to GSM. Like that I asked this question, it might be possible to send more information or bits or bits/sec by using certain number of bits or bits/sec to convey or understand more than. Lets forget it.8th January 2010 at 09:07 #60605AliAsgherGuest
Lets say for EDGE we have got a different coding scheme which enables more bitrate giving us more throughput. The analogy in case of E1 will be routers (designed by Cisco) which actually compress/decompress the data coming in/out of the transmission. And hence they may reduce the usage of E1 to half or so.
We have got quite some of these routers in our network and they are doing a good job. 🙂8th January 2010 at 10:17 #60606PixGuest
Ho yeah, that’s correct. You can optimize the E1 capacity, by using special “traffic aggregator” equipments.8th January 2010 at 11:19 #60607BijoyGuest
Another option will be HDLC:-)
Bijoy8th January 2010 at 16:28 #60608PanGuest
Bijoy, HDSL May Be? 🙂 It is not pure E1. It have line coding differ from one, standardizided for E1.
MKT, E1 is only E1 and nothing more. It have bit rate = 2048 kbit/s from point of view of transmitter and line code HDB3 according to G.703. I.e. you have a “pipe” with throughput 2048 kbit/s.
But you may have a Source (or sources) which generates some bit stream with speed bit rate > 2048 kbit/s and you should negotiate your bit stream with throughput of your bottleneck “pipe”. You do this by means of any well-known methods of sourse coding (or compression) dependently from source characteristics (for example Shennon or Haffman, Lempel-Ziv, JPEG, MPEG or others..)
B.R.9th January 2010 at 06:31 #60609BijoyGuest
I thought we are talking about increasing capacity over physical E1(media),not about some ANSI or ETSI line coding(E1/T1)
Bijoy9th January 2010 at 07:08 #60610MKTGuest
Thanks for reply,
The question asked by me is not totally baseless.
Just look at the scenario.
A Bit must have a certain duration.
The minimum the duration of bit the more will be the number of bits per second transmitted.
For the case of E1,
we have a time slot ( usual with TDMA) and there we have certain bits fitted in it.
Now it is not possible to alter the standard itself, i mean the source and destination nodes cannot change the standard itself. Both the nodes have studied the same school of standardization and thus they have no choice other than to obey the rules of G.703.
But the case takes a twist when
both of them, i mean nodes agreed to interpret the sequence of bits being delivered to them by E1 in some other way.
for example they might start doing like this
we will not do anything until our tray is filled with 20 bits.
These bits are not the bits but actually a code that only source and destination knows.
For every 20 bits received the destination node will look the sequence
this sequence will be compared with a coding table.
and it might be possible that these 20bits can correspond to another sequence of 40 bits
Thus it might be possible that
even though you have received 20 bits in one second ,
but you have understood 40 bits and that too in one second.
What do you say?
As i am not well versed with transmission standards but
when i hear things like
ATM over SDH
IP over SDH
that, just confuses me why we are riding a transmission technique over other if it is not providing any gain in
what you call
bits per second OR
Information at a rapid rate!!!9th January 2010 at 09:15 #60611kamalGuest
One n d same..
It is called source coding which can vary according to devices which we are using..
Line coding remain same in case of E1 or T1..9th January 2010 at 16:42 #60612PanGuest
Dear, MKT! You are the second Shennon and trying to invent bicycle:) Try to search the Shennon theorem and an optimum source coding theory, etc..
B.R.10th January 2010 at 17:10 #60613MKTGuest
Thanks for complement(?)
Its is almost 13 years when i studied all these shannon, nyquist etc theorems. At that time( college) the only objective was to mug up the things and your engineering exam clear.
No matter what is the meaning.
But, now after 13 years i don’t have to give any exam but what i do sometimes is to pick some topic for discussion and go with it like a student.
Tell me an honest answer…all of you.
would the world gave the same respect and honor to Shannon if somebody today came and reveals to the world that a similar document do exists in the ancient era before christ but shannon was not aware of that.
will the world take away the respect from the shannon today?
Anyway don’t think that i am trying to compare myself with shannon, but it do happens with me many times.
may be i was lab assistant with these dignitaries is previous births:)10th January 2010 at 17:18 #60614MKTGuest
Could you please tell me how does
Nyquist and Shannon theorem differs with each other?
It appears to me as if both of them are doing the same work.
Means representing your information with something else.
Is that what you call coding.
Then Nyquist is representing the information by X bits
and Shannon is by Y bits.
Both focuses on sending something less that full.
where is the difference actually?
MKT11th January 2010 at 04:54 #60615PanGuest
Dear, MKT, You are bit confused with two fundamental theorems: Shannon’s theorem about the maximum limit of channel throughput and the sampling theorem (by Kotelnikov or Nyquist or Shannon or Whittaker or by anybody else?:)