00:01
Design a series rlc type band pass filter with the cutoff frequencies 10 khz and 11 kilohertz, assuming that the capacitance is 80 picofarad find rl and q.
00:25
The frequencies omega 1 less than omega -0, that is resonant frequencies is less than omega -2.
00:34
Here, omega 1 is the lower cutoff frequency, omega 2 is the upper cutoff frequency, and omega not is the center frequency.
00:44
For the series rlc circuit, the transfer function h -oega equals the output voltage by input can be written as r by r plus j -omega -l plus 1 by j -o -mega -c.
00:59
Taking out common r -by -l, it will be j -o -mega -1 -l -c.
01:07
Minus omega square plus j into r by l omega.
01:14
Omega.
01:15
Calculating the bandwidth of the filter, which is the difference of lower and upper cutoff frequencies, that is 2 pi, f2 minus 2 pi, f1, which is equals to 2 pi into 11 into 10 raised to the power 3 minus 10 into 10 raised to the power 3.
01:35
2 pi into 10 raise to the power 3 is the bandwidth.
01:39
The relation between center frequency and the side band frequency omega 0 is omega 1 plus omega 2 by 2.
01:49
Rewriting the above equation, f0 equals f1 plus f2 by 2.
01:55
Substitute the value and calculate the center frequency, omega 0 equals 2 pi f0.
02:06
The f0 value here is 10 plus 11 10 plus 11 into 10 raised to the power 3 divided by 2.
02:17
Which is equal to 20, the center frequency is 10 .5 kilohertz.
02:30
Substitute here the f0 value...