Question 1 |
Let X(s)=\frac{3s+5}{s^{2}+10s+21} be the Laplace Transform of a signal x(t). Then, x(0^+) is
0 | |
3 | |
5 | |
21 |
Question 1 Explanation:
Given, X(s)=\left [ \frac{3s+5}{s^2+10s+21} \right ]
Using initial value theorem,
\begin{aligned} x(0^+) &= \lim_{s \to \infty }[sX(s)]\\ x(0^+) &= \lim_{s \to \infty } \left [ \frac{s(3s+5)}{s^2+10+21} \right ] \\ &= \lim_{s \to \infty }\left [ \frac{3+\frac{5}{s}}{1+\frac{10}{s}+\frac{21}{s^2}} \right ]=3 \end{aligned}
Using initial value theorem,
\begin{aligned} x(0^+) &= \lim_{s \to \infty }[sX(s)]\\ x(0^+) &= \lim_{s \to \infty } \left [ \frac{s(3s+5)}{s^2+10+21} \right ] \\ &= \lim_{s \to \infty }\left [ \frac{3+\frac{5}{s}}{1+\frac{10}{s}+\frac{21}{s^2}} \right ]=3 \end{aligned}
There is 1 question to complete.