00:01
In this question we have to find the inverse laplace transformation using residue theorem.
00:06
The function given to us is l inverse of minus s square plus 1 divided by s cube plus 9 s.
00:18
Now here f bar of s is equals to 1 minus s squared divided by s cube minus 9.
00:32
Now, the poles of this function is s is equal to 0.
00:46
Because if we rewrite this expiration as 1 minus s squared divided by s multiplied with s square plus 9 and equate the denominator with equal to 0, then we'll get s is equal to 0.
01:00
So at s is equal to 0, the function is not defined.
01:05
Is s is equal to 0 and the order of the pole is 1 so the residues of e raised to the power st multiplied with f bar of s is equal to e raised to the power st multiplied with s t s times s square plus 9.
01:47
At the pole, s is equal to 0 is residue at s is equal to 0.
01:59
E raised to the power s t multiplied with f bar of s is equal to limit s tends to 0.
02:10
Derivative of n minus 1 divided by n s raised to the power s, and minus 1 multiplied with s minus si raised to the power n multiplied with f bar of s this is the formula so here limit s tends to 0 this will be d raised to the power 0 divided by ds raised to the power 0 why because the n is order of the pole and it is given to us as 1 so 1 minus 1 will be 0...