A second order system has closed loop poles located as $s = - 3 \pm j 4.$ The time t at which the maximum value of the step response occurs (in seconds, rounded off to two decimal places) is

0.78

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Solution

The correct option is A 0.78
Given,
$P o l e s = - 3 \pm j 4 - ξ ω_{n} \pm ω_{n} \sqrt{1 - ξ^{2}}$

$S o, ω_{n} \sqrt{1 - ξ^{2}} = 4$

$a n d s i n c e t_{p} = \frac{π}{ω_{n} \sqrt{1 - ξ^{2}}} = - 0.78$

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A second order system has closed loop poles located as s=−3±j4.The time t at which the maximum value of the step response occurs (in seconds, rounded off to two decimal places) is 0.78

The correct option is A 0.78Given, Poles=−3±j4−ξωn±ωn√1−ξ2 So, ωn√1−ξ2=4 and since tp=πωn√1−ξ2=−0.78

A second order system has closed loop poles located as $s = - 3 \pm j 4.$ The time t at which the maximum value of the step response occurs (in seconds, rounded off to two decimal places) is

0.78

The correct option is A 0.78
Given,
$P o l e s = - 3 \pm j 4 - ξ ω_{n} \pm ω_{n} \sqrt{1 - ξ^{2}}$

$S o, ω_{n} \sqrt{1 - ξ^{2}} = 4$

$a n d s i n c e t_{p} = \frac{π}{ω_{n} \sqrt{1 - ξ^{2}}} = - 0.78$