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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 
  • a)
    5.25
  • b)
    2.81
  • c)
    4.50
  • d)
    3.89
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Consider a causal second-order system with the transfer functionwith a...
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 = 94/100 x 1
= 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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