00:01
Hello students, these are the given credentials of the question and here in this question we are asked to find out the ratio of received signal power and noise ratio, sorry noise power.
00:15
So let's first find out the received signal power that is eb.
00:24
We know that the equation for eb is equal to eb is equal to received power that is pr, pr into bit rate that is fb.
00:51
On calculation this becomes eb is equal to 10 into minus 11 divided by 1 into 10 raised to 6 implies eb is equal to that is the received signal power is equal to 10 raised to minus 17.
01:15
Further let's find the noise power that is no.
01:24
No is equal to boltzmann constant multiplied by the effective temperature into bandwidth.
01:36
Substituting the values we have no is equal to 1 .363806 into 10 raised to minus 23, the effective temperature is 296 and the bandwidth is 750 into 10 raised to 3 implies on calculation this becomes no is equal to 3 .267 becomes 3 point sorry becomes 3 .06 into 10 raised to minus 15.
02:28
Now let's form a ratio that is eb divided by n0 is equal to 10 raised to minus 17 divided by 3 into 3 .06 into 10 raised to minus 15 implies eb by n0 is equal to 3 .26 into 10 raised to minus 3.
03:15
Thus we can report this answer as eb by no is approximately equal to 3 .26 into 10 raised to minus 3...