GATE Electronics and Communications (EC) 2019 Solved Paper

Question : 51 of 65

Marks: +1, -0

Consider a causal second-order system with the transfer function

G(s)=\;{1}/{1+2 s+s^{2}}

with a unit-step

R(s)=\;{1}/{5}

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

{\lim}↙{t → ∞}c(t)

rounded off to two decimal places, is

Solution:

G(s)=\;{C(s)}/{R(s)}

∴ C(s)=G(s) ⋅ R(s)

=\;{1}/{s(s^{2}+2 s+1)}

∴ C(s)=\;{1}/{s(s+1)^{2}}

∴ C(s)=\;{A}/{s}+\;{B}/{(s+1)}+\;{C}/{(s+1)^{2}}

∴ {A}( {s}+1)^{2}+ {Bs}( {s}+1)+ {Cs}=1

∴ {AS}^{2}+2 {As}+ {A}+ {B}s^{2}+ {Bs}+ {Cs}=1

∴ {A}+ {B}=0

∴ 2 {A}+ {B}+ {C}=0

∴ {A}=1

B=-1

And

C=-1

∴ C(s)=\;{1}/{s}+\;{-1}/{s+1}+\;{-1}/{(s+1)^{2}}

∴ C(t)=(1-e^{-t}-t e^{-t}) u(t)

{t} → ∞

steady 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}-t e^{-t}=0.94

∴ e^{-t}+t e^{-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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