00:01
This question, the lower cutoff frequency, fl, is equal to 1000 hertz.
00:09
The upper cutoff frequency, that is fh, that is equal to 8 ,000 hertz.
00:19
The pass band gain of the filter is 10 decibels.
00:25
Now the cutoff frequency can be calculated by fl.
00:34
The cutoff frequency for hpf or the high pass filter is fl that is equal to 1 by 2 pi rh, so, if you can write, rh is equal to 1x2 pi 2 pi into 1000 into 50 into 10 to the power minus 9.
00:56
Since c equals 1550 nanopferreds, this is equal to 3 .2 kilo -ooms.
01:05
Now the cutoff frequency for the low pass filter is, for the low pass filter is 1 by 2 pi rl, cl so then we can write cl equals 1 by 2 pi rl f h this will be equal to 1 by 2 into 3 .14 into rl and that is 8 ,000 into 5 ,000 this is equal to 4 nanofarats the pass band gain pass band gain is a equals to 10 deep decibels so so we can write write a equals to a h into a l where a h is the pass band gain of high pass filter and a l is the pass burn gain of low pass filter now considering a h equals al that is equal to 10 we can write a h equal to a h equal to a h equal to 1 plus r f1 divided by r s 1 so let r f1 equals to 10 kilo -ooms so we can write r s 1 equals to r f1 divided by a h minus 1 this is equal to 10 ,000 divided by 9 that is equal to 1 .1 kilo -ooms now al equals 1 plus r f2 divided by r s 2 so we can write r s 2 equals r f 2 divided by al minus 1 so this is equal to 10 ,000 by 9 again and that is equal to 1 .1 kilo -ooms.
03:23
Now for the b subpart, this was the answer for the a subpart.
03:27
Now for the b subpart we have to draw the drag down or the circuit.
03:30
So this is the circuit...