GATE Electrical Engineering (EE) 2021 Solved Papers

Question : 60 of 65

Marks: +1, -0

In the given figure, plant

G_{p}(s)=\;{2.2}/{(1+0.1 s)(1+0.4 s)(1+1.2 s)}

and compensator

G_{c}(s)=K(\;{1+T_{1} s}/{1+T_{2} s})

. The external disturbance input is

{D}( {s})

. It is desired that when the disturbance is a unit step, the steady state error should not exceed

0.1

unit. The minimum value of

{K}

is __________ (Round off to 2 decimal places.)

Solution:

E(s)=R(s)-C(s)

{E}( {s})= {R}( {s})- {G}_{ {p}}( {s})[ {D}( {s})+ {E}( {s}) {G}_{ {c}}( {s})]

{E}= {R}- {G}_{ {p}} {D}- {G}_{ {p}} {G}_{ {c}} {E}

{E}(1+ {G}_{ {p}} {G}_{ {c}})= {R}- {G}_{ {p}} {D}

Put

{R}( {s})=0

E=\;{-G_{p} D}/{1+G_{p} G_{c}}

e_{s s}=\lim ↙{s → 0} s E(s)

e_{s s}=\;{\lim ↙{s → 0} s × {-1}/{s} × {2.2}/{(1+0.1 s)(1+0.4 s)(1+1.2 s)}}/{1+\;{2.2}/{(1+0.1 s)(1+0.4 s)(1+1.2 s)} × k(\;{1+T_{1} s}/{1+T_{2} s})}=\;{-2.2}/{1+2.2 k}

If,

{e}_{ {ss}} ≤ 0.1

{K} ≥-10.45

{K}_{\min }=-10.45

but '

k

' for compensator cannot be negative.
so, let us take

| {e}_{ {ss}}| ≤ 0.1

\;{2.2}/{1+20.2 {k}} ≤ 0.1

{K} ≥ 9.45

{K}_{\min }=9.45

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