A function, in the Laplace domain, is given by$F\left( s \right) = \frac{2}{s} - \frac{1}{{s + 3}}$Its value by final value theorem in 't' domain will be

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TuteeHUB · Accepted Answer

$\mathop {\lim }\limits_{t \to \infty } f\left( t \right) = 1$

TuteeHUB · Answer

$\mathop {\lim }\limits_{t \to \infty } f\left( t \right) = 3$

TuteeHUB · Answer

$\mathop {\lim }\limits_{t \to \infty } f\left( t \right) = 2$

TuteeHUB · Answer

$\mathop {\lim }\limits_{t \to \infty } f\left( t \right) = 4$

1.	A function, in the Laplace domain, is given by\(F\left( s \right) = \frac{2}{s} - \frac{1}{{s + 3}}\)Its value by final value theorem in 't' domain will be
A.	\(\mathop {\lim }\limits_{t \to \infty } f\left( t \right) = 3\)
B.	\(\mathop {\lim }\limits_{t \to \infty } f\left( t \right) = 2\)
C.	\(\mathop {\lim }\limits_{t \to \infty } f\left( t \right) = 1\)
D.	\(\mathop {\lim }\limits_{t \to \infty } f\left( t \right) = 4\)
Answer» C. \(\mathop {\lim }\limits_{t \to \infty } f\left( t \right) = 1\)

A function, in the Laplace domain, is given by\(F\left( s \right) = \frac{2}{s} - \frac{1}{{s + 3}}\)Its value by final value theorem in 't' domain will be

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