PYQs / GATE EC / 2024 / Set 1 / Q53 GATE EC 2024 Set 1 — Question 53 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MSQ +2 / -0 Medium Transfer Function from States State-Space Analysis Control Systems
Control Systems → State-Space Analysis → Transfer Function from States
Last updated 5 September 2026
Question Consider a system
S S S represented in state space as
d x d t = [ 0 − 2 1 − 3 ] x + [ 1 0 ] r , y = [ 2 − 5 ] x . \frac{dx}{dt} = \begin{bmatrix} 0 & -2 \\ 1 & -3 \end{bmatrix} x + \begin{bmatrix} 1 \\ 0 \end{bmatrix} r, \quad y = \begin{bmatrix} 2 & -5 \end{bmatrix} x. d t d x = [ 0 1 − 2 − 3 ] x + [ 1 0 ] r , y = [ 2 − 5 ] x . Which of the state space representations given below has/have the same transfer function as that of
S S S ?
Correct answer (A) (dx)/(dt) = bmatrix 0 & 1 \ -2 & -3 bmatrix x + bmatrix 0 \ 1 bmatrix r, y = bmatrix 1 & 2 bmatrix x; (C) (dx)/(dt) = bmatrix -1 & 0 \ 0 & -2 bmatrix x + bmatrix -1 \ 3 bmatrix r, y = bmatrix 1 & 1 bmatrix x
Solution The transfer function is given by
H ( s ) = C ( s I − A ) − 1 B H(s) = C(sI - A)^{-1}B H ( s ) = C ( s I − A ) − 1 B .
For system
S S S :
s I − A = [ s 2 − 1 s + 3 ] ⟹ ( s I − A ) − 1 = 1 s 2 + 3 s + 2 [ s + 3 − 2 1 s ] sI - A = \begin{bmatrix} s & 2 \\ -1 & s+3 \end{bmatrix} \implies (sI - A)^{-1} = \frac{1}{s^2 + 3s + 2} \begin{bmatrix} s+3 & -2 \\ 1 & s \end{bmatrix} s I − A = [ s − 1 2 s + 3 ] ⟹ ( s I − A ) − 1 = s 2 + 3 s + 2 1 [ s + 3 1 − 2 s ] H ( s ) = [ 2 − 5 ] 1 s 2 + 3 s + 2 [ s + 3 − 2 1 s ] [ 1 0 ] = 2 ( s + 3 ) − 5 s 2 + 3 s + 2 = 2 s + 1 s 2 + 3 s + 2 H(s) = \begin{bmatrix} 2 & -5 \end{bmatrix} \frac{1}{s^2 + 3s + 2} \begin{bmatrix} s+3 & -2 \\ 1 & s \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \frac{2(s+3) - 5}{s^2 + 3s + 2} = \frac{2s + 1}{s^2 + 3s + 2} H ( s ) = [ 2 − 5 ] s 2 + 3 s + 2 1 [ s + 3 1 − 2 s ] [ 1 0 ] = s 2 + 3 s + 2 2 ( s + 3 ) − 5 = s 2 + 3 s + 2 2 s + 1 .
Checking options:
(A)
H ( s ) = [ 1 2 ] 1 s 2 + 3 s + 2 [ s + 3 1 − 2 s ] [ 0 1 ] = 1 + 2 s s 2 + 3 s + 2 H(s) = \begin{bmatrix} 1 & 2 \end{bmatrix} \frac{1}{s^2 + 3s + 2} \begin{bmatrix} s+3 & 1 \\ -2 & s \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \frac{1 + 2s}{s^2 + 3s + 2} H ( s ) = [ 1 2 ] s 2 + 3 s + 2 1 [ s + 3 − 2 1 s ] [ 0 1 ] = s 2 + 3 s + 2 1 + 2 s . Correct.
(B)
H ( s ) = [ 0 2 ] 1 s 2 + 3 s + 2 [ s + 3 1 − 2 s ] [ 1 0 ] = − 4 s 2 + 3 s + 2 H(s) = \begin{bmatrix} 0 & 2 \end{bmatrix} \frac{1}{s^2 + 3s + 2} \begin{bmatrix} s+3 & 1 \\ -2 & s \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \frac{-4}{s^2 + 3s + 2} H ( s ) = [ 0 2 ] s 2 + 3 s + 2 1 [ s + 3 − 2 1 s ] [ 1 0 ] = s 2 + 3 s + 2 − 4 . Incorrect.
(C)
H ( s ) = [ 1 1 ] [ s + 1 0 0 s + 2 ] − 1 [ − 1 3 ] = − 1 s + 1 + 3 s + 2 = − ( s + 2 ) + 3 ( s + 1 ) ( s + 1 ) ( s + 2 ) = 2 s + 1 s 2 + 3 s + 2 H(s) = \begin{bmatrix} 1 & 1 \end{bmatrix} \begin{bmatrix} s+1 & 0 \\ 0 & s+2 \end{bmatrix}^{-1} \begin{bmatrix} -1 \\ 3 \end{bmatrix} = \frac{-1}{s+1} + \frac{3}{s+2} = \frac{-(s+2) + 3(s+1)}{(s+1)(s+2)} = \frac{2s+1}{s^2+3s+2} H ( s ) = [ 1 1 ] [ s + 1 0 0 s + 2 ] − 1 [ − 1 3 ] = s + 1 − 1 + s + 2 3 = ( s + 1 ) ( s + 2 ) − ( s + 2 ) + 3 ( s + 1 ) = s 2 + 3 s + 2 2 s + 1 . Correct.
(D)
H ( s ) = 1 s + 1 + 2 s + 2 = 3 s + 4 s 2 + 3 s + 2 H(s) = \frac{1}{s+1} + \frac{2}{s+2} = \frac{3s+4}{s^2+3s+2} H ( s ) = s + 1 1 + s + 2 2 = s 2 + 3 s + 2 3 s + 4 . Incorrect.
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Correct answer (A) (dx)/(dt) = bmatrix 0 & 1 \ -2 & -3 bmatrix x + bmatrix 0 \ 1 bmatrix r, y = bmatrix 1 & 2 bmatrix x; (C) (dx)/(dt) = bmatrix -1 & 0 \ 0 & -2 bmatrix x + bmatrix -1 \ 3 bmatrix r, y = bmatrix 1 & 1 bmatrix x
Solution The transfer function is given by
H ( s ) = C ( s I − A ) − 1 B H(s) = C(sI - A)^{-1}B H ( s ) = C ( s I − A ) − 1 B .
For system
S S S :
s I − A = [ s 2 − 1 s + 3 ] ⟹ ( s I − A ) − 1 = 1 s 2 + 3 s + 2 [ s + 3 − 2 1 s ] sI - A = \begin{bmatrix} s & 2 \\ -1 & s+3 \end{bmatrix} \implies (sI - A)^{-1} = \frac{1}{s^2 + 3s + 2} \begin{bmatrix} s+3 & -2 \\ 1 & s \end{bmatrix} s I − A = [ s − 1 2 s + 3 ] ⟹ ( s I − A ) − 1 = s 2 + 3 s + 2 1 [ s + 3 1 − 2 s ] H ( s ) = [ 2 − 5 ] 1 s 2 + 3 s + 2 [ s + 3 − 2 1 s ] [ 1 0 ] = 2 ( s + 3 ) − 5 s 2 + 3 s + 2 = 2 s + 1 s 2 + 3 s + 2 H(s) = \begin{bmatrix} 2 & -5 \end{bmatrix} \frac{1}{s^2 + 3s + 2} \begin{bmatrix} s+3 & -2 \\ 1 & s \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \frac{2(s+3) - 5}{s^2 + 3s + 2} = \frac{2s + 1}{s^2 + 3s + 2} H ( s ) = [ 2 − 5 ] s 2 + 3 s + 2 1 [ s + 3 1 − 2 s ] [ 1 0 ] = s 2 + 3 s + 2 2 ( s + 3 ) − 5 = s 2 + 3 s + 2 2 s + 1 .
Checking options:
(A)
H ( s ) = [ 1 2 ] 1 s 2 + 3 s + 2 [ s + 3 1 − 2 s ] [ 0 1 ] = 1 + 2 s s 2 + 3 s + 2 H(s) = \begin{bmatrix} 1 & 2 \end{bmatrix} \frac{1}{s^2 + 3s + 2} \begin{bmatrix} s+3 & 1 \\ -2 & s \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \frac{1 + 2s}{s^2 + 3s + 2} H ( s ) = [ 1 2 ] s 2 + 3 s + 2 1 [ s + 3 − 2 1 s ] [ 0 1 ] = s 2 + 3 s + 2 1 + 2 s . Correct.
(B)
H ( s ) = [ 0 2 ] 1 s 2 + 3 s + 2 [ s + 3 1 − 2 s ] [ 1 0 ] = − 4 s 2 + 3 s + 2 H(s) = \begin{bmatrix} 0 & 2 \end{bmatrix} \frac{1}{s^2 + 3s + 2} \begin{bmatrix} s+3 & 1 \\ -2 & s \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \frac{-4}{s^2 + 3s + 2} H ( s ) = [ 0 2 ] s 2 + 3 s + 2 1 [ s + 3 − 2 1 s ] [ 1 0 ] = s 2 + 3 s + 2 − 4 . Incorrect.
(C)
H ( s ) = [ 1 1 ] [ s + 1 0 0 s + 2 ] − 1 [ − 1 3 ] = − 1 s + 1 + 3 s + 2 = − ( s + 2 ) + 3 ( s + 1 ) ( s + 1 ) ( s + 2 ) = 2 s + 1 s 2 + 3 s + 2 H(s) = \begin{bmatrix} 1 & 1 \end{bmatrix} \begin{bmatrix} s+1 & 0 \\ 0 & s+2 \end{bmatrix}^{-1} \begin{bmatrix} -1 \\ 3 \end{bmatrix} = \frac{-1}{s+1} + \frac{3}{s+2} = \frac{-(s+2) + 3(s+1)}{(s+1)(s+2)} = \frac{2s+1}{s^2+3s+2} H ( s ) = [ 1 1 ] [ s + 1 0 0 s + 2 ] − 1 [ − 1 3 ] = s + 1 − 1 + s + 2 3 = ( s + 1 ) ( s + 2 ) − ( s + 2 ) + 3 ( s + 1 ) = s 2 + 3 s + 2 2 s + 1 . Correct.
(D)
H ( s ) = 1 s + 1 + 2 s + 2 = 3 s + 4 s 2 + 3 s + 2 H(s) = \frac{1}{s+1} + \frac{2}{s+2} = \frac{3s+4}{s^2+3s+2} H ( s ) = s + 1 1 + s + 2 2 = s 2 + 3 s + 2 3 s + 4 . Incorrect.
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