PYQs / GATE EE / 2015 / Set 1 / Q62 GATE EE 2015 Set 1 — Question 62 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 Hard Signal Flow & Mason's Gain System Modelling Control Systems Controllability & Observability State-Space Analysis
Control Systems → State-Space Analysis → Controllability & Observability
Last updated 5 September 2026
Question In the signal flow diagram given in the figure,
u 1 u_1 u 1 and
u 2 u_2 u 2 are possible inputs whereas
y 1 y_1 y 1 and
y 2 y_2 y 2 are possible outputs. When would the SISO system derived from this diagram be controllable and observable?
Correct answer (B) When u₂ is the only input and y₁ is the only output.
Solution Step 1: Write the state model from the diagram
Taking the outputs of the two
1 / s 1/s 1/ s blocks as the states
x 1 x_1 x 1 and
x 2 x_2 x 2 , the branch gains give
x ˙ 1 = 5 x 1 − 2 x 2 + u 1 \dot{x}_1 = 5x_1 - 2x_2 + u_1 x ˙ 1 = 5 x 1 − 2 x 2 + u 1 x ˙ 2 = 2 x 1 + x 2 + u 1 + u 2 \dot{x}_2 = 2x_1 + x_2 + u_1 + u_2 x ˙ 2 = 2 x 1 + x 2 + u 1 + u 2 y 1 = x 1 , y 2 = x 1 − x 2 y_1 = x_1, \qquad y_2 = x_1 - x_2 y 1 = x 1 , y 2 = x 1 − x 2 so
A = [ 5 − 2 2 1 ] , B u 1 = [ 1 1 ] , B u 2 = [ 0 1 ] , C y 1 = [ 1 0 ] , C y 2 = [ 1 − 1 ] A = \begin{bmatrix} 5 & -2 \\ 2 & 1 \end{bmatrix}, \quad B_{u_1} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \quad B_{u_2} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \quad C_{y_1} = \begin{bmatrix} 1 & 0 \end{bmatrix}, \quad C_{y_2} = \begin{bmatrix} 1 & -1 \end{bmatrix} A = [ 5 2 − 2 1 ] , B u 1 = [ 1 1 ] , B u 2 = [ 0 1 ] , C y 1 = [ 1 0 ] , C y 2 = [ 1 − 1 ] Step 2: Controllability of each single input
For
u 1 u_1 u 1 alone:
A B u 1 = [ 5 − 2 2 + 1 ] = [ 3 3 ] , [ B A B ] = [ 1 3 1 3 ] , det = 0 A B_{u_1} = \begin{bmatrix} 5-2 \\ 2+1 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \end{bmatrix}, \qquad [\,B \;\; AB\,] = \begin{bmatrix} 1 & 3 \\ 1 & 3 \end{bmatrix}, \qquad \det = 0 A B u 1 = [ 5 − 2 2 + 1 ] = [ 3 3 ] , [ B A B ] = [ 1 1 3 3 ] , det = 0 The two columns are parallel, so the system is
not controllable from
u 1 u_1 u 1 .
For
u 2 u_2 u 2 alone:
A B u 2 = [ − 2 1 ] , [ B A B ] = [ 0 − 2 1 1 ] , det = 2 ≠ 0 A B_{u_2} = \begin{bmatrix} -2 \\ 1 \end{bmatrix}, \qquad [\,B \;\; AB\,] = \begin{bmatrix} 0 & -2 \\ 1 & 1 \end{bmatrix}, \qquad \det = 2 \neq 0 A B u 2 = [ − 2 1 ] , [ B A B ] = [ 0 1 − 2 1 ] , det = 2 = 0 So the system
is controllable from
u 2 u_2 u 2 .
Step 3: Observability of each single output
For
y 1 y_1 y 1 :
C y 1 A = [ 5 − 2 ] , [ C C A ] = [ 1 0 5 − 2 ] , det = − 2 ≠ 0 C_{y_1}A = \begin{bmatrix} 5 & -2 \end{bmatrix}, \qquad \begin{bmatrix} C \\ CA \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 5 & -2 \end{bmatrix}, \qquad \det = -2 \neq 0 C y 1 A = [ 5 − 2 ] , [ C C A ] = [ 1 5 0 − 2 ] , det = − 2 = 0 So the system
is observable from
y 1 y_1 y 1 .
For
y 2 y_2 y 2 :
C y 2 A = [ 5 + ( − 1 ) ( 2 ) ( − 2 ) + ( − 1 ) ( 1 ) ] = [ 3 − 3 ] = 3 C y 2 C_{y_2}A = \begin{bmatrix} 5+(-1)(2) & \;(-2)+(-1)(1) \end{bmatrix} = \begin{bmatrix} 3 & -3 \end{bmatrix} = 3\,C_{y_2} C y 2 A = [ 5 + ( − 1 ) ( 2 ) ( − 2 ) + ( − 1 ) ( 1 ) ] = [ 3 − 3 ] = 3 C y 2 [ C C A ] = [ 1 − 1 3 − 3 ] , det = 0 \begin{bmatrix} C \\ CA \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 3 & -3 \end{bmatrix}, \qquad \det = 0 [ C C A ] = [ 1 3 − 1 − 3 ] , det = 0 So the system is
not observable from
y 2 y_2 y 2 (
C y 2 C_{y_2} C y 2 is a left eigenvector of
A A A , both eigenvalues being
3 3 3 ).
Step 4: Combine
Only the pairing input
u 2 u_2 u 2 with output
y 1 y_1 y 1 makes the resulting SISO system both controllable and observable.
Answer: (B) When
u 2 u_2 u 2 is the only input and
y 1 y_1 y 1 is the only output.
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Correct answer (B) When u₂ is the only input and y₁ is the only output.
Solution Step 1: Write the state model from the diagram
Taking the outputs of the two
1 / s 1/s 1/ s blocks as the states
x 1 x_1 x 1 and
x 2 x_2 x 2 , the branch gains give
x ˙ 1 = 5 x 1 − 2 x 2 + u 1 \dot{x}_1 = 5x_1 - 2x_2 + u_1 x ˙ 1 = 5 x 1 − 2 x 2 + u 1 x ˙ 2 = 2 x 1 + x 2 + u 1 + u 2 \dot{x}_2 = 2x_1 + x_2 + u_1 + u_2 x ˙ 2 = 2 x 1 + x 2 + u 1 + u 2 y 1 = x 1 , y 2 = x 1 − x 2 y_1 = x_1, \qquad y_2 = x_1 - x_2 y 1 = x 1 , y 2 = x 1 − x 2 so
A = [ 5 − 2 2 1 ] , B u 1 = [ 1 1 ] , B u 2 = [ 0 1 ] , C y 1 = [ 1 0 ] , C y 2 = [ 1 − 1 ] A = \begin{bmatrix} 5 & -2 \\ 2 & 1 \end{bmatrix}, \quad B_{u_1} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \quad B_{u_2} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \quad C_{y_1} = \begin{bmatrix} 1 & 0 \end{bmatrix}, \quad C_{y_2} = \begin{bmatrix} 1 & -1 \end{bmatrix} A = [ 5 2 − 2 1 ] , B u 1 = [ 1 1 ] , B u 2 = [ 0 1 ] , C y 1 = [ 1 0 ] , C y 2 = [ 1 − 1 ] Step 2: Controllability of each single input
For
u 1 u_1 u 1 alone:
A B u 1 = [ 5 − 2 2 + 1 ] = [ 3 3 ] , [ B A B ] = [ 1 3 1 3 ] , det = 0 A B_{u_1} = \begin{bmatrix} 5-2 \\ 2+1 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \end{bmatrix}, \qquad [\,B \;\; AB\,] = \begin{bmatrix} 1 & 3 \\ 1 & 3 \end{bmatrix}, \qquad \det = 0 A B u 1 = [ 5 − 2 2 + 1 ] = [ 3 3 ] , [ B A B ] = [ 1 1 3 3 ] , det = 0 The two columns are parallel, so the system is
not controllable from
u 1 u_1 u 1 .
For
u 2 u_2 u 2 alone:
A B u 2 = [ − 2 1 ] , [ B A B ] = [ 0 − 2 1 1 ] , det = 2 ≠ 0 A B_{u_2} = \begin{bmatrix} -2 \\ 1 \end{bmatrix}, \qquad [\,B \;\; AB\,] = \begin{bmatrix} 0 & -2 \\ 1 & 1 \end{bmatrix}, \qquad \det = 2 \neq 0 A B u 2 = [ − 2 1 ] , [ B A B ] = [ 0 1 − 2 1 ] , det = 2 = 0 So the system
is controllable from
u 2 u_2 u 2 .
Step 3: Observability of each single output
For
y 1 y_1 y 1 :
C y 1 A = [ 5 − 2 ] , [ C C A ] = [ 1 0 5 − 2 ] , det = − 2 ≠ 0 C_{y_1}A = \begin{bmatrix} 5 & -2 \end{bmatrix}, \qquad \begin{bmatrix} C \\ CA \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 5 & -2 \end{bmatrix}, \qquad \det = -2 \neq 0 C y 1 A = [ 5 − 2 ] , [ C C A ] = [ 1 5 0 − 2 ] , det = − 2 = 0 So the system
is observable from
y 1 y_1 y 1 .
For
y 2 y_2 y 2 :
C y 2 A = [ 5 + ( − 1 ) ( 2 ) ( − 2 ) + ( − 1 ) ( 1 ) ] = [ 3 − 3 ] = 3 C y 2 C_{y_2}A = \begin{bmatrix} 5+(-1)(2) & \;(-2)+(-1)(1) \end{bmatrix} = \begin{bmatrix} 3 & -3 \end{bmatrix} = 3\,C_{y_2} C y 2 A = [ 5 + ( − 1 ) ( 2 ) ( − 2 ) + ( − 1 ) ( 1 ) ] = [ 3 − 3 ] = 3 C y 2 [ C C A ] = [ 1 − 1 3 − 3 ] , det = 0 \begin{bmatrix} C \\ CA \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 3 & -3 \end{bmatrix}, \qquad \det = 0 [ C C A ] = [ 1 3 − 1 − 3 ] , det = 0 So the system is
not observable from
y 2 y_2 y 2 (
C y 2 C_{y_2} C y 2 is a left eigenvector of
A A A , both eigenvalues being
3 3 3 ).
Step 4: Combine
Only the pairing input
u 2 u_2 u 2 with output
y 1 y_1 y 1 makes the resulting SISO system both controllable and observable.
Answer: (B) When
u 2 u_2 u 2 is the only input and
y 1 y_1 y 1 is the only output.
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