Question

bandwidth- 5 MHz

transmission power-100mW

noise power -10mW

1) whats the signal to noise ratio (SNR)

2) if u send a file of 700 Mbits on this channel. what is the max data rate of this channel and how long will it take to transmit the file

3) if we need a max data rate of 30Mbps, how much is the required channel bandwidth?( both noise power and transmission power dont change)

Answer 1

Given :

1. bandwidth- 5 MHz

2.transmission power-100mW

3.noise power -10mW

solution :

A) whats the signal to noise ratio (SNR) -

SNR is calculated as following

SNR= Average Signal Power / Average Noise Power

Therefore here,

SNR = 100mW / 10mW

SNR = 10

signal to noise ratio (SNR) = 10

B) if u send a file of 700 Mbits on this channel. what is the max data rate of this channel and how long will it take to transmit the file

Capacity to max data rate of this channel is calculated as following

Max data rate of this channel = Bandwidth × log₂( 1+SNR )

Therefore here,

Max data rate of this channel = 5  × log₂( 1+ 10 )

Max data rate of this channel = 5 × 3.4594

Max data rate of this channel = 17.297 Mbps

now Time Requried for file of 700 Mbits on this channel = 700 / 17.297

Time Requried for file of 700 Mbits on this channel = 40.46 seconds

C)if we need a max data rate of 30Mbps, how much is the required channel bandwidth?( both noise power and transmission power dont change)

Here we can calculate required channel bandwidth need a max data rate of 30Mbps is as following

Max data rate of this channel = Bandwidth × log₂( 1+SNR )

we have , SNR = 10 , Max data rate = 30Mbps

therfore ,

Bandwidth =  Max data rate / log₂( 1+SNR )

Bandwidth = 30 / log₂( 1+ 10 )

Bandwidth = 30 / 3.4594

Bandwidth = 8.6720 MHz

Hence if we need a max data rate of 30Mbps, required channel bandwidth is 8.6720 MHz