24
Answersi cant understand is there another method for solving it
over 11 years ago
Krishna sa...
Answers
using patial fraction method
2=A(1+s)+Bs
if s=0, then A=2
if s=-1,thenB=-2
2/(s(1 + s))=(2/s)+(-2/(1+s))
taking LI
2-2e^-t
t -> oo
2-0=2
over 7 years ago
Deepa
above trasfer function is strictly proper.so we are using direct intial value theorem
over 8 years ago
Rekha
By the final value theorem
lim t->infinity i(t)=lim s->0 S I(S)
so in the above problem
S I(S)= 2/(1+s)
then put t=0
answer is 2
over 10 years ago
J.praveenr...
apply final value theorem.........lim(s-.0)[s*f(s)]......then put s=0 in 2/(s+1)=2
over 10 years ago
Prasanna
use intital value theorem for t--infinity.. s.F(S) for s tends to 0.
=2/s(s+1)*s for s---> 0
=2/(s+1) for s__>0
=2/1
=2
about 11 years ago
Swapna Kat...
Limit s tends to zero (s.I(s)) is equal to i(t) at infinity.. apply dis u will get 2 as answer
about 11 years ago
Manoj bn
Final Value theorem Lt s->0 s*I(s) = Lt t->`oo` I(t)
Multiply s with I(s) and then substitute s with 0 and proceed you will get 2
over 11 years ago
Suraj para...
first take partial fraction which will result in 2/s -2/(s+1). taking inverse laplace will result in 2-2e^-t. Put t='00' which will give 2
over 11 years ago
Ankit chab...
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