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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)
    4.50
  • b)
    2.81
  • c)
    5.25
  • d)
    3.89
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Consider a causal second-order system with the transfer function with...
∴ A(s + 1)2 + Bs(s + 1) + Cs = 1
∴ As2 + 2As + A + Bs2 + Bs + Cs = 1
∴ A + B = 0
∴ 2A + B + C = 0
∴ A = 1
So B = –1
And C = –1
At t → ∞ stedy state will occur
∴ C(∞) = 1
Now we are asked to find time at which 94% of the steady state value reached.
∴ C(t) = 1 – e–t – te–t = 0.94
∴ e–t + te–t = 0.06
∴ e–t (1 + t) = 0.06
Now from the given options try all option you will get t = 4.50 sec.
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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, isa)4.50b)2.81c)5.25d)3.89Correct answer is option 'A'. Can you explain this answer?
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