00:01
Hello everyone, in this question it is given that g of x is equal to 1 by a or the modulus of x minus x naught is less than or equal to a by 2 and 0 modulus of x minus x naught is greater than a by 2.
00:19
So we have to find the fourier transform.
00:21
So the fourier transform is given by g bar of k equal to minus infinity to infinity e to the power minus ikx g of x dx.
00:35
Therefore minus infinity to x naught minus a by 2 e to the power minus ikx dot 0 dx plus x naught minus a by 2 into x naught plus a by 2 e to the power minus ikx 1 by a dx plus integral of x naught plus a by 2 infinity e to the power minus ikx dot 0 dx.
01:06
So this will give you 0 plus 1 by a integral of x naught minus a by 2 x naught plus a by 2 e to the power minus ikx dx minus plus 0.
01:23
So this will give you 1 by a minus 1 by ik e to the power minus iks limit x naught minus a by 2 x naught plus a by 2.
01:38
So which is equal to minus 1 by ik a e to the power minus ik x naught plus a by 2 minus e to the power minus ik x naught minus a by 2.
01:56
So which is equal to minus 1 by ik a e to the power minus ik x naught minus ik a by 2 minus e to the power minus ik x naught plus ik a by 2.
02:13
So this will give you minus minus 1 by ik a e to the power minus ik x naught 2 e to the power minus ik a by 2 minus e to the power minus ik x naught e to the power ik a by 2.
02:37
So which is equal to 1 by ik a e to the power minus ik x naught e to the power ik a by 2 minus e to the power minus ik a by 2.
02:55
So which is equal to 1 by ik a e to the power minus ik x naught 2 i sin k a by 2.
03:07
So which is equal to 2 by k a e to the power minus ik x naught sin k a by 2.
03:17
So this is the required answer.
03:20
Then the second part is that is this is this part is only of about g bar of x next to that is g bar of k...