Consider a state-variable model of a system \left[\begin{array}{c}˙{ {x\end{array}}_{1} ; ˙{ {x}}_{2} ]=[\table 0 , 1 ; -\alpha , -2 \beta ][\table {x}_{1} ; {x}_{2} ]+[\table 0 ; \alpha ] {r}} {y}=[\table 1 , 0 ][\table {x}_{1} ; {x}_{2} ] where y is the output, and r is the input. The damping ratio \zeta and the undamped natural frequency \omega_{ {n}}( {rad} {/} {\sec}) of the system are given by

Correct Answer is : ζ=βα,ωn=α

\zeta=\frac{\beta}\sqrt{\alpha}, \omega_{ {n}}=\sqrt{\alpha}

\zeta=\sqrt{\alpha ;} \omega_{ {n}}=\frac{\beta}\sqrt{\alpha}

\alpha=\sqrt{\beta ;} \omega_{ {n}}=\sqrt{\alpha}

\zeta=\frac\sqrt{\alpha}{\beta}, \omega_{ {n}}=\sqrt{\beta}

GATE Electrical Engineering (EE) 2019 Solved Paper

Question : 48 of 65

Marks: +1, -0

Consider a state-variable model of a system

{[\table ˙{ {x}}_{1} ; ˙{ {x}}_{2} ]=[\table 0 , 1 ; -α , -2 β ][\table {x}_{1} ; {x}_{2} ]+[\table 0 ; α ] {r}}

{y}=[\table 1 , 0 ][\table {x}_{1} ; {x}_{2} ]

where

y

is the output, and

r

is the input. The damping ratio

ζ

and the undamped natural frequency

ω_{ {n}}( {rad} {/} {sec})

of the system are given by

Solution:

Given data:

{TF}= {C}[ {SI}- {A}]^{-1} {B}+ {D}

∵D=0

{TF}= {C} \;{ {Adj}[ {SI}- {A}]}/{[ {SI}- {A}]} {B}

[ {SI}- {A}]=[\table {S} , -1 ; α , {S}+2 β]

{TF}=\;{[\table 1 , 0][\table {S}+2 β , 1 ; -α , {S}][\table 0 ; α]}/{ {S}( {S}+2 β)+α}=\;{[\table 1 , 0][\table α ; {s} α]}/{ {s}^{2}+2 β {s}+α}

{TF}=\;{α}/{ {s}^{2}+2 β {s}+α}

ω_{ {n}}=√{α} {rad} / {sec}

2 ξ ω_{n}=2 β

β=ξ ω_{n} ⇒ ξ=\;{β}/{ω_{n}}=\;{β}/{√{α}}

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