Consider a causal second-order system with the transfer function with a unit-step as an input. Let c(s) be the corresponding output. The time taken by the system output c(t) to reach 94% of its steady-state value , rounded off to two decimal places, is

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UPSC ESE (Prelims) Electronics and Telecommunication Engineering 19 Feb 2023 Official Paper

Answer 1

Concept:

A = 1

B = -1

C = -1

Applying inverse Laplace transform:

C(t) = 1 – e^-t – te^-t

at steady state i.e. at t → ∞

C(t) = 1

94% of steady state

= 0.94.

0.94 = 1 – e^-t – te^-t

Substitute all options

Let us substitute options

option (a) t= 5.25

1 – e^-5.25 – 5.25 e^-5.25

= 0.967

option (b)

t = 4.50

1 – 0.938

≈ 0.94

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