0
310views
Determine initial value and final value of the following signal, $X(S)=\frac{1}{s(s+2)}$
1 Answer
written 2.0 years ago by |
Solution:
Initial value:
$ x(0)=\underset{s \rightarrow \infty}{\mathrm{Lt}} S X(S)\\ $
$ =\operatorname{Lt}_{s \rightarrow \infty} s \frac{1}{s(s+2)}=\frac{1}{\infty}=0\\ $
Final value,
$ x(\infty)=\underset{s \rightarrow 0}{\operatorname{Lt}} S X(S)\\ $
$ =\operatorname{Lt}_{s \rightarrow 0} s \frac{1}{s(s+2)}=\frac{1}{2} $