00:01
Hello students, to solve this question as for the ideal differentiator 1 upon j omega now for part a, s is equal to 1 upon j omega square s.
00:16
Now we can write this as 1 upon omega square 6 omega square upon 1 plus omega square cube that equals to 6 upon 1 omega square whole cube.
00:29
Now the power, so the power spectrum of output is, so this will be 6 upon 1 plus omega square whole power cube.
00:45
Now for part b, so the power is equal to 1 upon 2 pi integration of minus infinity to infinity power spectral density d omega.
01:05
So from this we can also write 1 upon 2 pi minus infinity to infinity 6 upon 1 plus omega square whole cube d omega.
01:15
Now let omega is tan theta, so d omega is equal to sec square theta d theta where omega is infinity so theta will be pi by 2 and if minus infinity then theta is then theta will be minus pi by 2.
01:37
Now we can substitute the limits, so power will be equal to 1 by 2 pi integration of minus pi by 2 to pi by 2 6 upon 1 plus tan square theta whole cube sec square theta d theta.
01:55
Also further this can be simplified as 1 upon 2 pi integration of minus pi by 2 to pi by 2 6 upon sec square theta whole cube into sec square theta d theta.
02:12
So we get 1 upon 2 pi integration of minus pi by 2 to pi by 2 6 upon sec power 4 theta d theta.
02:24
Now further solving, on further solving we get the power and that is 3 by 2 integration of minus pi by 2 to pi by 2 cos square theta cos, so we can first write 6 cos power 4 theta d theta...