The transfer function of s system is $G (s) = \frac{100}{(s + 1) (s + 100)} .$ For a unit -step input to the system the approximate settling time for $2 %$ criterion is

100 s e c

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4 s e c

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1 s e c

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0.01 s e c

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Solution

The correct option is B $4 s e c$
$G (s) = \frac{100}{(s + 1) (s + 100)}$
Taking dominant pole consideration,
pole at $s = - 100$ is not taken
$∴ G (s) = \frac{100}{(s + 1)}$
Now it is 1st order system
$∴ t_{s} = 4 T = 4 \times 1 = 4 s$

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The transfer function of s system is G(s)=100(s+1)(s+100). For a unit -step input to the system the approximate settling time for 2% criterion is

The correct option is B 4 secG(s)=100(s+1)(s+100) Taking dominant pole consideration, pole at s=−100 is not taken ∴G(s)=100(s+1) Now it is 1st order system ∴ts=4T=4×1=4 s

The transfer function of s system is $G (s) = \frac{100}{(s + 1) (s + 100)} .$ For a unit -step input to the system the approximate settling time for $2 %$ criterion is

The correct option is B $4 s e c$
$G (s) = \frac{100}{(s + 1) (s + 100)}$
Taking dominant pole consideration,
pole at $s = - 100$ is not taken
$∴ G (s) = \frac{100}{(s + 1)}$
Now it is 1st order system
$∴ t_{s} = 4 T = 4 \times 1 = 4 s$