GATE 2020 IN Q.50: solution and answer

GATE 2020 · INQ.50 · 2 marks · NAT

Consider the following state variable equations:

$\overset{x}{˙}_{1} (t) = x_{2} (t)$
$\overset{x}{˙}_{2} (t) = - 6 x_{1} (t) - 5 x_{2} (t)$

The initial conditions are $x_{1} (0) = 0$ and $x_{2} (0) = 1$ . At $t = 1$ second, the value of $x_{2} (1)$ (rounded off to two decimal places) is ____

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Answer and solution

ANSWER

$x_{2} (1) = 3 e^{- 3} - 2 e^{- 2} \approx - 0.12$ .

Official answer: -0.13 to -0.11Source: Official answer key, GATE 2020 IN (IIT Delhi)

Write the system as $\overset{x}{˙} = A x$ with $A = [0 - 6 1 - 5]$ and $x (0) = [01]$ . There is no input, so we want the zero-input (free) response.
Take the Laplace transform of $\overset{x}{˙} = A x$ : $s X (s) - x (0) = A X (s)$ , so $X (s) = (s I - A)^{- 1} x (0)$ , where $s I - A = [s 6 - 1 s + 5]$ .
Find the resolvent matrix. The determinant is $s (s + 5) + 6 = s^{2} + 5 s + 6 = (s + 2) (s + 3)$ , so $(s I - A)^{- 1} = \frac{1}{( s + 2 ) ( s + 3 )} [s + 5 - 6 1 s]$
Because $x (0) = [01]^{T}$ , only the second column of the resolvent matters: $X_{1} (s) = \frac{1}{( s + 2 ) ( s + 3 )}$ and $X_{2} (s) = \frac{s}{( s + 2 ) ( s + 3 )}$ .
Split $X_{2} (s)$ into partial fractions using residues. At $s = - 2$ : $\frac{s}{s + 3}_{s = - 2} = \frac{- 2}{1} = - 2$ . At $s = - 3$ : $\frac{s}{s + 2}_{s = - 3} = \frac{- 3}{- 1} = 3$ . So $X_{2} (s) = \frac{- 2}{s + 2} + \frac{3}{s + 3}$ .
Take the inverse Laplace transform: $x_{2} (t) = 3 e^{- 3 t} - 2 e^{- 2 t}$ . Checks: $x_{2} (0) = 3 - 2 = 1$ , which is correct. Also $x_{1} (t) = e^{- 2 t} - e^{- 3 t}$ gives $x_{1} (0) = 0$ , and differentiating it gives $\overset{x}{˙}_{1} = - 2 e^{- 2 t} + 3 e^{- 3 t} = x_{2}$ , which agrees with the first state equation.
Put in $t = 1 s$ : $x_{2} (1) = 3 e^{- 3} - 2 e^{- 2} = 3 (0.04979) - 2 (0.13534) = 0.14936 - 0.27067 = - 0.1213$
Rounded to two decimal places, $x_{2} (1) \approx - 0.12$ .

KEY IDEA

With no input, the state response is $X (s) = (s I - A)^{- 1} x (0)$ , or equivalently $x (t) = e^{A t} x (0)$ . Since $x (0) = [01]^{T}$ , $x (t)$ is just the second column of the state-transition matrix $Φ (t)$ . The eigenvalues $- 2$ and $- 3$ (the roots of $s^{2} + 5 s + 6$ ) tell you in advance which exponentials will appear.

WATCH OUT FOR

The common mistakes are these: picking the first column of $(s I - A)^{- 1}$ instead of the second, which gives $x_{2} = - 6/ ((s + 2) (s + 3))$ and $x_{2} (1) \approx - 0.51$ ; getting a sign wrong in the adjugate (the off-diagonal entries must become $+ 1$ and $- 6$ ); and reporting $x_{1} (1) = e^{- 2} - e^{- 3} \approx 0.09$ when the question asks for $x_{2} (1)$ . Also note that $x_{2} (1)$ comes out negative even though $x_{2} (0) = 1$ . The fast $e^{- 3 t}$ term dies away before the slower $- 2 e^{- 2 t}$ term does, so the answer really is negative. Do not drop the sign.

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GATE 2020 IN Q.50: solution and answer

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