00:01
So, here in this question we have to show a periodic where the efficiency is given that is equals to 50 % duty cycle dc offshore set 1 water where xt is with the peak amplitude we have to determine the energy and the power of the xt if the signal xt is periodic with the period t then here we are considering p of x is equals to 1 divided by t integral is from 0 to 1 of x raise to the power of t multiply by the dt.
00:29
So, this is the term which we are given here.
00:31
So, xt from here is equals to 2a which is divided by t that is multiplied by the t where the value of t is greater than and equals to 0 and less than and equals to t divided by 2 which from here is equals to 0 where t divided by 2 is less than and equals to t which is for the less than and equals to t.
00:48
So, we have to determine the energy of xt.
00:50
So, e of x from here is equals to integral from minus infinity to plus infinity x of t to its whole square which is multiplied by the dt.
00:58
So, the value of ex become equals to integral that is from 0 to t divided by 2 4 of a raise to the power 2 which is divided by t square which is multiplied by t of dt plus integral that is from t divided by 2 to t of 0 which is multiplied by the dt simplifying the term we get the value of ex that become equals to a raise to the power 2 of t which is divided by 6 this is the value of ex.
01:22
Now, the energy of the signal for n period is given as ex which is equals to n multiplied by the a square of t which is divided by 6 as the signal is extended from minus infinity to plus infinity...