00:01
Now in this particular question first we have to calculate the bandwidth.
00:04
So bandwidth is calculated by r upon 2 pi l which is equals to 20 into 10 raised to the power 3 which is divided by 2 pi multiplied by 16 which is the value for l.
00:21
So according to it the bandwidth is equals to 198 .94 hertz.
00:29
So this is your bandwidth.
00:32
Next we have to find the q factor.
00:35
So the q factor will be equals to 1 upon r into under root l upon c.
00:43
So it will be equals to what? it will be 1 upon 20 into 10 raised to the power 3 multiplied by under root 16 upon 4 into 10 raised to the power minus 12.
00:56
So solving this we will get the value for q factor it will be equals to 100.
01:02
Next we have to find the value for lower cut off frequency.
01:06
So lower cut off frequency will be f1 is equals to fr minus the bandwidth by 2.
01:16
So fr will be equals to 1 upon 2 pi into under root l c.
01:24
So writing the value for fr it will be 1 divided by 2 pi into under root l we have 16 into 4 into 10 raised to the power minus 12.
01:36
So the value for fr it will be equals to 198...