00:01
So, in this question we have given with two signals that is xt is given us like this.
00:09
We can draw it.
00:10
So, the responses are here.
00:12
This is minus pi.
00:13
This is pi.
00:15
Again, it is getting repeated at 10 pi or the midpoint is 10 pi.
00:24
Here it is minus 10 pi.
00:27
So, we can find the period of this signal.
00:33
This is your xt.
00:34
So, period of this signal we can find out that will be t equal to 10 pi.
00:44
So, omega naught will be 2 pi upon t.
00:48
That will be 2 pi by 10 pi which comes out to be 1 by pi.
00:58
So, xt we can express in terms of its fourier coefficient that is n tending to minus infinity to infinity summation of cn e to the power j omega naught nt.
01:11
So, in that case cn can be or the fourier coefficient, fourier series coefficient can be calculated as xt e power minus j omega nt dt.
01:30
So, putting these values here we can say that since it is a square signal we can say that this will be xt will be 1 and its amplitude is 1 and we have our time period that is minus pi to pi and here it is 1.
01:51
So, e to the power minus j omega naught nt dt.
01:57
So, we know that integration of e power x is e power x.
02:03
So, here it will come out as e to the power minus j omega naught nt upon minus j omega naught n.
02:17
That is the coefficient of t in the expression.
02:21
So, here it is limits are minus pi to pi.
02:24
So, further solving it we can get the values like this.
02:28
So, we can get the values but here time period we can substitute is 1 by 10 pi and in the inside here values will come out to be e to the power minus j omega naught n pi minus e to the power minus j omega naught n minus pi.
02:51
So, this will become plus pi and in the denominator we have minus j omega naught n.
03:00
So, putting the values of omega naught and rewriting them we can get the rearranged value like this e to the power j n j e to the power j n pi by pi and here it will be minus e to the power minus j n pi by pi whole divided by this minus term actually get rearranged in the numerator.
03:35
So, we will get n pi by pi n by pi.
03:42
So, this pi get cancelled with 10 and here we get 2.
03:48
So, we can rewrite our expression as let's see what we can write.
03:57
So, we can rewrite it e to the power j n pi by pi minus e to the power minus j n pi by pi whole divided by j 2 we can write and outside we will be having 1 by n pi or 1 by pi n we can write.
04:26
So, this term is the general expansion of sin theta.
04:31
So, we can write directly that this is that your c n value will be that's implies c n or fourier series coefficient for x t will be 1 by pi n sin n pi by pi.
04:49
So, this is the fourier series coefficient of x t.
04:53
So, we can finally write the fourier series expansion as c 0 c 1 c 0 c 1 c 2 etcetera can be calculated with n equal to 0 1 2 putting the values we can calculate...