00:01
The filter has 3 db cutoff frequency at 1 kilohertz.
00:09
If its input is connected to 120 millivolt variable frequency signal, then find the output voltage at 200 hertz, b, 2 kilohertz and c, 10 kilohertz.
00:29
Observe that the output is taken across the resistance in the circuit since there is no feedback loop connected to the ideal of m using the voltage division method to find the expression v0s equals r by r plus 1 by sc into vis.
00:49
This can be written as src by src plus 1 into vis.
00:56
Replacing s with j omega we get v0 j omega equals j omega r c by j omega r c plus 1 into the input voltage this is the output function of a high pass filter simplifying the expression further v0 j omega equals 1 by 1 minus j by omega r c multiplied by v i j omega here the product rc is the time constant, toe equals rc, and the cutoff frequency is 1 by time constant, that is 1 by rc.
01:40
Replacing the expression of the cutoff frequency in the equation, we have v0jf equals vijf by 1 minus i, 1 minus j, f0 by f.
02:02
Then, substituting f -0 as 1 kilohertz, v0 j -f will be equals to v -i -jf by 1 -j -f by 1 -1 ,000 by f.
02:22
The magnitude of the output voltage in terms of j -f will be equals to magnitude of v -i -j -f by square root of 1 plus 1 ,000 -square...