PYQs / GATE EC / 2014 / Set 3 / Q57 GATE EC 2014 Set 3 — Question 57 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +2 / -0.67 Medium Solution of State Equations State-Space Analysis Control Systems
Control Systems → State-Space Analysis → Solution of State Equations
Last updated 5 September 2026
Question The state equation of a second-order linear system is given by
x ˙ ( t ) = A x ( t ) , x ( 0 ) = x 0 \dot{\mathbf{x}}(t) = \mathbf{Ax}(t), \quad \mathbf{x}(0) = \mathbf{x}_0 x ˙ ( t ) = Ax ( t ) , x ( 0 ) = x 0 For
x 0 = [ 1 − 1 ] \mathbf{x}_0 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} x 0 = [ 1 − 1 ] ,
x ( t ) = [ e − t − e − t ] \mathbf{x}(t) = \begin{bmatrix} e^{-t} \\ -e^{-t} \end{bmatrix} x ( t ) = [ e − t − e − t ] and for
x 0 = [ 0 1 ] \mathbf{x}_0 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} x 0 = [ 0 1 ] ,
x ( t ) = [ e − t − e − 2 t − e − t + 2 e − 2 t ] \mathbf{x}(t) = \begin{bmatrix} e^{-t} - e^{-2t} \\ -e^{-t} + 2e^{-2t} \end{bmatrix} x ( t ) = [ e − t − e − 2 t − e − t + 2 e − 2 t ] .
When
x 0 = [ 3 5 ] \mathbf{x}_0 = \begin{bmatrix} 3 \\ 5 \end{bmatrix} x 0 = [ 3 5 ] ,
x ( t ) \mathbf{x}(t) x ( t ) is
Correct answer (B) bmatrix 11e^(-t) - 8e^(-2t) \ -11e^(-t) + 16e^(-2t) bmatrix
Solution Since the system is linear, the principle of superposition applies. We can express the initial state
x 0 = [ 3 5 ] \mathbf{x}_0 = \begin{bmatrix} 3 \\ 5 \end{bmatrix} x 0 = [ 3 5 ] as a linear combination of the given initial states
v 1 = [ 1 − 1 ] \mathbf{v}_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} v 1 = [ 1 − 1 ] and
v 2 = [ 0 1 ] \mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} v 2 = [ 0 1 ] :
[ 3 5 ] = c 1 [ 1 − 1 ] + c 2 [ 0 1 ] \begin{bmatrix} 3 \\ 5 \end{bmatrix} = c_1 \begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 1 \end{bmatrix} [ 3 5 ] = c 1 [ 1 − 1 ] + c 2 [ 0 1 ] From the first row:
c 1 = 3 c_1 = 3 c 1 = 3 .
From the second row:
− c 1 + c 2 = 5 ⟹ − 3 + c 2 = 5 ⟹ c 2 = 8 -c_1 + c_2 = 5 \implies -3 + c_2 = 5 \implies c_2 = 8 − c 1 + c 2 = 5 ⟹ − 3 + c 2 = 5 ⟹ c 2 = 8 .
Thus, the response
x ( t ) \mathbf{x}(t) x ( t ) for
x 0 = [ 3 5 ] \mathbf{x}_0 = \begin{bmatrix} 3 \\ 5 \end{bmatrix} x 0 = [ 3 5 ] is:
x ( t ) = 3 x 1 ( t ) + 8 x 2 ( t ) \mathbf{x}(t) = 3 \mathbf{x}_1(t) + 8 \mathbf{x}_2(t) x ( t ) = 3 x 1 ( t ) + 8 x 2 ( t ) x ( t ) = 3 [ e − t − e − t ] + 8 [ e − t − e − 2 t − e − t + 2 e − 2 t ] \mathbf{x}(t) = 3 \begin{bmatrix} e^{-t} \\ -e^{-t} \end{bmatrix} + 8 \begin{bmatrix} e^{-t} - e^{-2t} \\ -e^{-t} + 2e^{-2t} \end{bmatrix} x ( t ) = 3 [ e − t − e − t ] + 8 [ e − t − e − 2 t − e − t + 2 e − 2 t ] x ( t ) = [ 3 e − t + 8 e − t − 8 e − 2 t − 3 e − t − 8 e − t + 16 e − 2 t ] = [ 11 e − t − 8 e − 2 t − 11 e − t + 16 e − 2 t ] \mathbf{x}(t) = \begin{bmatrix} 3e^{-t} + 8e^{-t} - 8e^{-2t} \\ -3e^{-t} - 8e^{-t} + 16e^{-2t} \end{bmatrix} = \begin{bmatrix} 11e^{-t} - 8e^{-2t} \\ -11e^{-t} + 16e^{-2t} \end{bmatrix} x ( t ) = [ 3 e − t + 8 e − t − 8 e − 2 t − 3 e − t − 8 e − t + 16 e − 2 t ] = [ 11 e − t − 8 e − 2 t − 11 e − t + 16 e − 2 t ] This matches option (B).
Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker. Review answer and solution without JavaScript Interactive answer checking needs JavaScript. The published solution is available below.
Correct answer (B) bmatrix 11e^(-t) - 8e^(-2t) \ -11e^(-t) + 16e^(-2t) bmatrix
Solution Since the system is linear, the principle of superposition applies. We can express the initial state
x 0 = [ 3 5 ] \mathbf{x}_0 = \begin{bmatrix} 3 \\ 5 \end{bmatrix} x 0 = [ 3 5 ] as a linear combination of the given initial states
v 1 = [ 1 − 1 ] \mathbf{v}_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} v 1 = [ 1 − 1 ] and
v 2 = [ 0 1 ] \mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} v 2 = [ 0 1 ] :
[ 3 5 ] = c 1 [ 1 − 1 ] + c 2 [ 0 1 ] \begin{bmatrix} 3 \\ 5 \end{bmatrix} = c_1 \begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 1 \end{bmatrix} [ 3 5 ] = c 1 [ 1 − 1 ] + c 2 [ 0 1 ] From the first row:
c 1 = 3 c_1 = 3 c 1 = 3 .
From the second row:
− c 1 + c 2 = 5 ⟹ − 3 + c 2 = 5 ⟹ c 2 = 8 -c_1 + c_2 = 5 \implies -3 + c_2 = 5 \implies c_2 = 8 − c 1 + c 2 = 5 ⟹ − 3 + c 2 = 5 ⟹ c 2 = 8 .
Thus, the response
x ( t ) \mathbf{x}(t) x ( t ) for
x 0 = [ 3 5 ] \mathbf{x}_0 = \begin{bmatrix} 3 \\ 5 \end{bmatrix} x 0 = [ 3 5 ] is:
x ( t ) = 3 x 1 ( t ) + 8 x 2 ( t ) \mathbf{x}(t) = 3 \mathbf{x}_1(t) + 8 \mathbf{x}_2(t) x ( t ) = 3 x 1 ( t ) + 8 x 2 ( t ) x ( t ) = 3 [ e − t − e − t ] + 8 [ e − t − e − 2 t − e − t + 2 e − 2 t ] \mathbf{x}(t) = 3 \begin{bmatrix} e^{-t} \\ -e^{-t} \end{bmatrix} + 8 \begin{bmatrix} e^{-t} - e^{-2t} \\ -e^{-t} + 2e^{-2t} \end{bmatrix} x ( t ) = 3 [ e − t − e − t ] + 8 [ e − t − e − 2 t − e − t + 2 e − 2 t ] x ( t ) = [ 3 e − t + 8 e − t − 8 e − 2 t − 3 e − t − 8 e − t + 16 e − 2 t ] = [ 11 e − t − 8 e − 2 t − 11 e − t + 16 e − 2 t ] \mathbf{x}(t) = \begin{bmatrix} 3e^{-t} + 8e^{-t} - 8e^{-2t} \\ -3e^{-t} - 8e^{-t} + 16e^{-2t} \end{bmatrix} = \begin{bmatrix} 11e^{-t} - 8e^{-2t} \\ -11e^{-t} + 16e^{-2t} \end{bmatrix} x ( t ) = [ 3 e − t + 8 e − t − 8 e − 2 t − 3 e − t − 8 e − t + 16 e − 2 t ] = [ 11 e − t − 8 e − 2 t − 11 e − t + 16 e − 2 t ] This matches option (B).
Understand the concept, then try another question Revisit Control Systems with concept notes, common mistakes and an original worked example before your next attempt.
More questions on State-Space Analysis ← Q56 Full paper Q58 → 