The state equation of a second-order linear system is given by-t - Axt - x0- x0For x0 - 1- -1- - xt - e-t- -e-t- -and for x0 - - 0- 1- -xt - - e-t e-2t- -e-t - 2e-2t- -When x0 - - 3- 5- - xt is

Given, ẋ(t) = Ax(t) , x(0) = x_0 nbsp; Taking the Laplace transform sX(s) x(0) = AX(s) [sI A] X(s) = x(0) X(s) = [sI A]^ 1 x(0) x(t) = L^ 1[sI A]^ 1 ...

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Control Systems

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The state equ...

Question

The state equation of a second-order linear system is given by,
$˙ x (t) = A x (t), x (0) = x_{0}$

$F o r x_{0} = [\begin{matrix} 1 - 1 \end{matrix}], x (t) = [\begin{matrix} e^{- t} - e^{- t} \end{matrix}] and for x_{0} = [\begin{matrix} 01 \end{matrix}]$

$x (t) = [\begin{matrix} e^{- t} & e^{- 2 t} - e^{- t} & + 2 e^{- 2 t} \end{matrix}]$

$W h e n x_{0} = [\begin{matrix} 35 \end{matrix}], x (t) i s$

[\begin{matrix} - 8 e^{- t} & + & 11 e^{- 2 t} 8 e^{- t} & - & 22 e^{- 2 t} \end{matrix}]

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[\begin{matrix} 11 e^{- t} & - & 8 e^{- 2 t} - 11 e^{- t} & + & 16 e^{- 2 t} \end{matrix}]

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[\begin{matrix} 3 e^{- t} & - & 5 e^{- 2 t} - 3 e^{- t} & + & 10 e^{- 2 t} \end{matrix}]

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[\begin{matrix} - 5 e^{- t} & + & 6 e^{- 2 t} - 5 e^{- t} & + & 6 e^{- 2 t} \end{matrix}]

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Solution

The correct option is B $[\begin{matrix} 11 e^{- t} & - & 8 e^{- 2 t} - 11 e^{- t} & + & 16 e^{- 2 t} \end{matrix}]$
Given,
$˙ x (t) = A x (t), x (0) = x_{0}$

Taking the Laplace transform
$s X (s) - x (0) = A X (s)$

$[s I - A] X (s) = x (0)$

$X (s) = [s I - A]^{- 1} x (0)$

$x (t) = L^{- 1} [s I - A]^{- 1} x (0) . . . (i)$

Conditions given are
$F o r x_{0} = [\begin{matrix} 1 - 1 \end{matrix}], x (t) = [\begin{matrix} e^{- t} - e^{- t} \end{matrix}]$

$F o r x_{0} = [\begin{matrix} 01 \end{matrix}], x (t) = [\begin{matrix} e^{- t} & - e^{- 2 t} - e^{- t} & + 2 e^{- 2 t} \end{matrix}]$

Using the linearity property in equation (i)
$K_{1} x_{1} (t) = L^{- 1} [s I - A]^{- 1} x_{1} (0) K_{1}$
$K_{2} x_{2} (t) = L^{- 1} [s I - A]^{- 1} x_{2} (0) K_{2}$

Using the linearity property as
$K_{1} x_{1} (t) + K_{2} x_{2} (t) = L^{- 1} [s I - A]^{- 1} [K_{1} x_{1} (0) + K_{2} x_{2} (0)] . . . (i i)$

$A l s o X_{3} (s) = [s I - A]^{- 1} x_{3} (0)$

$S o, K_{1} [\begin{matrix} 1 - 1 \end{matrix}] + K_{2} [\begin{matrix} 01 \end{matrix}] = [\begin{matrix} 35 \end{matrix}]$

$[\begin{matrix} K_{1} + 0 K_{2} - K_{1} + K_{2} \end{matrix}] + [\begin{matrix} 35 \end{matrix}]$

$\Rightarrow K_{1} = 3$

$K_{2} = 8$

So, from equation (ii), we get $x (t)$

$x (t) = K_{1} x_{1} (t) + K_{2} x_{2} (t)$

$= 3 [\begin{matrix} e^{- t} - e^{- t} \end{matrix}] + 8 [\begin{matrix} e^{- t} - e^{- 2 t} - e^{- t} + 2 e^{- 2 t} \end{matrix}]$

$= [\begin{matrix} 11 e^{- t} & - & 8 e^{- 2 t} - 11 e^{- t} & + & 16 e^{- 2 t} \end{matrix}]$

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The state equation of a second-order linear system is given by, ˙x(t)=Ax(t),x(0)=x0 For x0=[1−1],x(t)=[e−t−e−t]and for x0=[01] x(t)=[e−te−2t−e−t+ 2e−2t] When x0=[35],x(t) is