00:01
In this question, it is given c of s divided by r of s is equal to 1 divided by s plus 2.
00:08
And we have been given that r of t is equal to u of t and this is unit step function.
00:21
So, if we take the laplace of this r of t, so that we will get r of s as 1 divided by s.
00:34
So, from here, we can write c of s is equal to r of s divided by s plus 2.
00:41
And from here, we can write r s as 1 divided by s divided by s plus 2.
00:46
So, this c of s will be equal to 1 divided by h times s plus.
00:53
Now, we need to find out t equal to 0.
00:56
So, we will apply initial value theorem, then we can write this as limit s tends to infinity of s times c of s.
01:12
So, this will be limit s tends to infinity of s times 1 divided by s times of s plus 2...