Question

Comprehension

The state variable formulation of a system is given as

\(\left[ \begin{matrix} x_1 \\\ x_2 \end{matrix} \right] = \left[ \begin{matrix} -2 && 0 \\\ 0 && -1 \end{matrix} \right]\left[ \begin{matrix} x_1 \\\ x_2 \end{matrix} \right]+\left[ \begin{matrix} 1 \\\ 1 \end{matrix} \right]u, x_1(0)=0, x_2(0)=0 \:an\:y = \left[ \begin{matrix} 1 &&0 \end{matrix} \right]\left[ \begin{matrix} x_1 \\\ x_2 \end{matrix} \right]\)

1. \(\frac{1}{2}-\frac{1}{2}e^{-2t}\)
2. \(1-\frac{1}{2}e^{-2t}-\frac{1}{2}e^{-t}\)
3. e^-2t - e^-t
4. 1 - e^-t
5.

Answer 1

Correct Answer - Option 1 : \(\frac{1}{2}-\frac{1}{2}e^{-2t}\)

Given state model is

\(\left[ \begin{matrix} x_1 \\\ x_2 \end{matrix} \right] = \left[ \begin{matrix} -2 && 0 \\\ 0 && -1 \end{matrix} \right]\left[ \begin{matrix} x_1 \\\ x_2 \end{matrix} \right]+\left[ \begin{matrix} 1 \\\ 1 \end{matrix} \right]u \)    ------- (1)

\(y = \left[ \begin{matrix} 1 &&0 \end{matrix} \right]\left[ \begin{matrix} x_1 \\\ x_2 \end{matrix} \right]\)       ------- (2)

From equation (1),

(dx₁/dt) = 2 x₁ + u

Taking Laplace transform, we get

s x₁(s) - x₁(0) = 2 x₁(s) + (1/s)

⇒ x₁ (s) = 1/(s × (s+2))

From equation (2),

y = x₁

Taking Laplace transform we get,

⇒ y(s) = x₁(s)

⇒ y(s) = 1/(s × (s+2))

⇒ y(s) = \(\frac{1}{2}[\frac{1}{s}-\frac{1}{s+2}]\)

Taking inverse Laplace transform, we get

y(t) = \(\frac{1}{2}-\frac{1}{2}e^{-2t}\)