Question

Provided is a channel of 10 MHz bandwidth to be shared by 5 users. It is decided the channel is going to be shared by the users using FDM. A guard band of 1 MHz between each channel is allocated to avoid interference. Each user is given an equal share of bandwidth. Assume each user has 1 MB of data to transfer and can utilize the channel at full capacity.

Calculate the time required to transfer all data when

(i) The SNR of each channel is 10 dB

(ii) The SNR of four channels is 10 dB and one channel is 5 dB

(iii)  FDM multiplexing is not used, and each user transmits data one at a time at SNR of 10 dB.

Provide suitable figures and description to explain your answer.

Answer 1

Given 10M Hz banduwilth shared by 5 Users with IMH2 gaurd band to Separate susers = 4 MH2 as guardeband is used is between So the users has 10-4 = GMHz each of total Bandwidth then each user has - GMHz Bandwidth 7 G = 1.2 MHz is band walth available to each user. and IMB data is transfered by each user and as it is FOM all the data is transferred sonul turnausly so C) SNR lo dB the time is required by Roxo Di Re - Roe data rate can be used on channel 0² amount of clata = IMB = 8 Mb [ B-86] L RG 2 Bowl log. [lt SMR] bps SNR = 10 dB or = 10 [ lod8 10 log EN SENR=10] [B-W2 1.2MM2] - Rg = 102 lag, [1 + 1o] Mbps = 102 log, C1] Mbps 24:151 Mbps So Time Required is - RBX D D & RB. - 40101 Mb 8 Mb + 4.101 Mbps 2 1.92 Sec di If SNR of 4 Channel is lod B and one is SdB. then RB is decided by Lower SNR for whole transmites so SAR 5dB a lo log. SNRJ SNR = 105 - 3:16

- RR z Bowl laga [I+SNR z 1. 2M log [ 1+ 3.162 z 2.4688 Mbps ] - So Time Required is z D £ RB 3:2 8 Mb 3·24 Sec & 2.4685 Mbps

all Now if PDM is not used and data is send one at a time and each user has : IMB data to traguit than total data transnit is. 5X IMB.2.5 MB D=5MB e 5X8 Mb an D (z 40 Mb

and each now ciser 2 B. W of lo.MHz as whale available to so the new Roz Bow [lege LIAS NR]] z 10 MHz ls. [1+1o ] 2 [o sa (1 Mbps z 34.894 Mbps D - Ro = 40 34.599

- 1.2MHylkinnot us, GR US₂ - USE G Biz GB₂ US, Glos Usa GB4 US, 10 MHz User Band of frequency & 102 MH2 Gaurd Band frequency & I MHz

Note : Ro= C= B. W loga [It sure is the Sharon Hartley capacity theorey where SNR is in Ratio ] be bits Byte z 8 bits B² Bytes

The shanon Hartley theorem gives the maximum data rate we can transmit over channel using a bandwidth at a required signal to noise ratio (SNR)

The figure in answer shows the bandwidth distribution when the FDM is used in which user sends data at same time so data to be sent is 1MB only

But when FDM is not used the bandwidth available to each user is 10Mhz but the data to be transmit becomes 1+1+1+1+1=5MB