00:01
Here to solve this problem first we draw the circuit in the s domain and inductor in s domain is represented by l into s and resistance in s domain is represented by r here the output voltage in the s domain is given by v out s and input voltage in the s domain is given by v in s now by using voltage division rule we obtain v out s equals to v in s into r over r plus l s.
00:45
And from this we obtain v out s over v in s equals to r over r plus l s.
00:57
Now here after taking l common from the denominator we obtain v out s over v in s equals to r over l over s plus r over l.
01:12
Here the ratio of r and l comes equal to, here r equals to 62 and l equals to 32 and l equals to 338 into 10 to the power negative 3.
01:26
Now from this we obtain r over l equals to 183 .431.
01:34
Now after putting the value of r over l here we obtain v out as over v in s equals to 183 .431 over s plus 183 .431 over s plus 183 .431.
02:04
Now next we assume that here this expression is equals to h s.
02:11
Now suppose this is the equation number 1.
02:14
Now from this equation we obtain h j omega equals to 183 .431 over j omega plus 183 .431 suppose here this is the equation number 2.
02:34
Now suppose we have a transfer function, hs equals to a over s plus b...