00:01
Determine the center frequency and bandwidth of the bandpass filters given in the referring to the first figure the current flows in this loop is i1 and in the second loop is i2 applying kirchof's voltage law to the mesh 1 we get minus vs vs is the voltage from the source plus 1 i1 plus 1 by s i1 minus i2 equals 0.
00:35
Simplification will give us si1 plus i1 minus i2 equals s vs.
00:42
The equation becomes 1 plus s i1 minus i2 equals s vs.
00:49
Let it be the first equation.
00:52
Then applying kvl to the second mash 1 by s i2 minus i1 plus i2 by s plus i2 by s plus 1 i2 equals 0.
01:05
On simplification we get the equation i 1 equals 2 plus s i2.
01:12
This is our second equation.
01:15
Substituting the value for i 1 into the first equation we get i2 equals s by s square plus 3 s plus 1 multiply vs and from the figure the output voltage is i 2 into 1 which is i2, substituting the value of i to here, v0 will be s by s squared plus 3s plus 1 into vs.
01:46
From here finding the ratio v0 by vs, s by s by s squared plus 3s plus 1...