00:01
Hello students, for part a, by using the equation v0 of s is equals to vi into s as per that 1 divided by 2s square plus s plus 2.
00:19
From this we can substitute the values 1 divided by 2c of s square plus s r divided by 2 plus 1 by c.
00:35
From this we will get the equation v0 of s divided by vi of s is equals to 1 by 2 divided by c of s square plus s r by 2 plus 1 by c.
00:58
Now we are going to compare this equation with v0 of s into vi of s is equals to d divided by c of s square plus r divided by c into s plus f by c.
01:20
From this we get the answer for the first question d is equals to 0 .5, e is equals to 2 and f is equals to 1.
01:33
For part b, we can use the transfer function as follow vi divided by vi of s is vc is equals to 1 by 2 of c divided by s square plus s r by 2 plus 1 by c.
02:01
Now we are going to compare this equation with the standard transform function as per that h of s is equals to 1 divided by s square plus 2 epsilon omega n into s plus omega n square.
02:20
As per that we will get omega n square is equals to 1 by c.
02:28
So omega n is equals to root 1 by c and the t value will be r to the root c divided by 4.
02:40
From this we can write s t by 2 is equals to negative 10 plus or minus j 10...