| Author |
Message |
xischaune
Joined: 08 Mar 2006
Posts: 122
Helped: 11
|
 06 Jul 2006 4:40 No of CDMA channels??? |
|
|
|
|
Hi all...
If 10MHz bandwidth is available per direction in a cellular system....
Is it possible to find out the number of channels (or equivalently the number of users in the cell assuming a single-cell system)
The cellular system is CDMA requiring 10kbps user signals with Eb=N0 of 6dB
Will the result change if we take Eb/N0=6dB.
I am confused... i think it must be Eb/N0 and not Eb=N0
Thanx in advance....
|
|
| Back to top |
|
 |
manojkrshukla
Joined: 06 Jun 2006
Posts: 46
Helped: 5
|
| 06 Jul 2006 6:19 Re: No of CDMA channels??? |
|
|
|
|
You should take Eb/N0.
Another point is that if you know the bandwidth occupied in one channel (forword or reverse) i.e. B, then the bandwidth for the both channel will be 2B.
You should also add the bandwidth for control channels engaged for base station i.e C.
So, available banwidth for users=total bandwidth(W)-C
Now, no. of possible users=available bandwidth for users/2B
|
|
| Back to top |
|
|
xischaune
Joined: 08 Mar 2006
Posts: 122
Helped: 11
|
| 06 Jul 2006 6:22 Re: No of CDMA channels??? |
|
|
|
|
Thanx...
But from the given information...specially the 10kbps figure...is it possible to calculate the channel bandwidth per user and then ignoring control channels C, come out with the total number of possible users???
|
|
| Back to top |
|
|
janath
Joined: 21 May 2006
Posts: 33
Helped: 4
|
| 06 Jul 2006 17:48 Re: No of CDMA channels??? |
|
|
|
|
| xischaune wrote: |
Hi all...
If 10MHz bandwidth is available per direction in a cellular system....
Is it possible to find out the number of channels (or equivalently the number of users in the cell assuming a single-cell system)
The cellular system is CDMA requiring 10kbps user signals with Eb=N0 of 6dB
Will the result change if we take Eb/N0=6dB.
I am confused... i think it must be Eb/N0 and not Eb=N0
Thanx in advance.... |
If the channel has 10MHz BW, and the users are seperated by spreading codes and if single cell is assumed, the interference is mainly due to the cross correlation among the users. If perfect orthogonallity of sequences assumed, number of users would be
10M/10k=1000; so using 1024 spreading code will be sufficient and maximum there can be 1024 users.
But if sequence are like gold codes ; or random codes then you can find average interference per user for an N length code.
Then you can find N for given Eb/N0 ; in most of the system you can just ignore AWGN for simplicity when interference is dominanet
|
|
| Back to top |
|
|
