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Determine initial value and final value of the following signal, $X(S)=\frac{1}{s(s+2)}$
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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} $

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