Question

Find initial value of X(s) = 2s/[s2+25+2]​

Answer 1

Given :

X(s) = 2s / [s²+25+2]

To find :

Initial value

Solution :

$\lim_{s \to \infty} sX(s)\\\\ \lim_{s \to \infty} s*\frac{2s}{s^{2} +25+2} \\\\ \lim_{s \to \infty}\frac{ 2s^{2} }{s^{2} +25+2}$

Divide numerator and denominator by s²

$\lim_{s \to \infty} \frac{1}{1+ 27/s^{2} }$

Any no. divided by infinity is equal to 0.

Therefore,

$= \frac{1}{1+27/ \infty}\\\\= \frac{1}{1 + 0} \\\\=1$

Initial value is 1