If L-ft- 2s-1s2-2s-5 then f0- and f- are given by

F(0^+)= lim_s→∞sF(s) = 2s^2+2ss^2+2s+5=2 f(∞)= lim_s→ 0 sF(s) = 2s^2+2ss^2+2s+5=0 ...

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Control Systems

Basics of Time Response Analysis

If L[ft]= 2s+...

Question

If $L [f (t)] = \frac{2 (s + 1)}{s^{2} + 2 s + 5}$ then $f (0^{+})$ and $f (\infty)$ are given by

0, 2

respectively

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2, 0

respectively

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0, 1

respectively

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2 / 5, 0

respectively

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Solution

The correct option is B $2, 0$ respectively
$f (0^{+}) = lim s \to \infty s F (s) = \frac{2 s^{2} + 2 s}{s^{2} + 2 s + 5} = 2$

$f (\infty) = lim s \to 0 s F (s) = \frac{2 s^{2} + 2 s}{s^{2} + 2 s + 5} = 0$

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If L[f(t)]=2(s+1)s2+2s+5 then f(0+) and f(∞) are given by

The correct option is B 2,0 respectivelyf(0+)=lims→∞sF(s)=2s2+2ss2+2s+5=2 f(∞)=lims→0sF(s)=2s2+2ss2+2s+5=0

If $L [f (t)] = \frac{2 (s + 1)}{s^{2} + 2 s + 5}$ then $f (0^{+})$ and $f (\infty)$ are given by

The correct option is B $2, 0$ respectively
$f (0^{+}) = lim s \to \infty s F (s) = \frac{2 s^{2} + 2 s}{s^{2} + 2 s + 5} = 2$

$f (\infty) = lim s \to 0 s F (s) = \frac{2 s^{2} + 2 s}{s^{2} + 2 s + 5} = 0$