GATE Electrical Engineering (EE) 2018 Solved Paper

Question : 48 of 65

Marks: +1, -0

Consider a system governed by the following equations:

\;{d x_{1}(t)}/{d t}=x_{2}(t)-x_{1}(t)

\;{d x_{2}(t)}/{d t}=x_{1}(t)-x_{2}(t)

The initial conditions are such that

x_{1}(0)< x_{2}(0)<∞ .

Let

x_{1 f}=\lim ↙{t → ∞} x_{1}(t)

and

x_{2 f}=

\lim ↙{t → ∞} x_{2}(t)

.
Which one of the following is true?

Solution:

[\table {x}_{1} ; {x}_{2}]=[\table -1 , 1 ; 1 , -1][\table x_{1} ; x_{2}]

A=[\table -1 , 1 ; 1 , -1]

(s I-A)=[\table s+1 , -1 ; 1 , s+1]

[(s I-A]^{-1}=\;{1}/{s^{2}+2 s}[\table s+1 , -1 ; 1 , s+1]

[S I-A]^{-1}=[\table \;{s+1}/{s(A+2)} , \;{1}/{s(s+2)} ; \;{1}/{s(s+2)} , \;{s+1}/{s(s+2)}]

=[\table \;{1}/{2 s}+\;{1}/{2(s+2)} , \;{1}/{2 s}-\;{1}/{2(s+2)} ; \;{1}/{2s}-\;{1}/{2(s+2)} , \;{1}/{2 s}+\;{1}/{2(s+2)}]

ϕ(t)=\;{1}/{2}[\table 1+e^{-1 t} , 1-e^{-2 t} ; 1-e^{-2 t} , 1+e^{-2 t}]

x(t)=ϕ(t) x(0)=\;{1}/{2}[\table 1+e^{-2 t} , 1-e^{-2 t} ; 1-e^{-2 t} , 1+e^{-2 t}][\table x_{1}(0) ; x_{2}(0)]

=\;{1}/{2}[\table x_{1}(0)+x_{1}(0) e^{-2 t}+x_2(0)-x_2(0) e^{-2 t} ; x_{1}(0)-x_{1}(0) e^{-2 t}+x_{2}(0)+x_{2}(4 e^{-2 t}]]

x_{f}=\lim ↙{t → ∞} x(t)=\;{1}/{2}[\table x_{1}(0)+x_{2}(0) ; x_{1}(0)+x_{2}(0)]

Hence,

\;\; x_{1 f}=x_{2 f}<∞

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