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SarkariResultOutBMS · 2024

Previous Year Question

A certain system has a state space model as

[Ẋ1 ; Ẋ2] = [−2 −3 ; 4 2][X1 ; X2] + [3 ; 5]U
Y = [1 1][X1 ; X2], D = 0

What will be the transfer function of this system?

  1. (8s + 1)/(s2 + 8)
  2. (s + 8)/(s2 + 8)
  3. (2s + 1)/(s2 + 8)
  4. (s + 2)/(s2 + 8)

Correct answer: A

Explanation

Use G(s)=C(sI−A)−1B+D. Here det(sI−A)=s2+8. Multiplying C·adj(sI−A)·B gives 8s+1; therefore G(s)=(8s+1)/(s2+8).

Formula

G(s) = C(sI − A)−1B + D
sI − A = [s + 2 3 ; −4 s − 2]
det(sI − A) = (s + 2)(s − 2) − 3(−4) = s² + 8
(sI − A)−1 = adj(sI − A)/(s² + 8)

Step-by-step solution

Transfer Function from State-Space Model explanatory diagram

Correct Answer

A — (8s + 1)/(s² + 8)

The denominator comes from det(sI−A)=s²+8 and the numerator from C·adj(sI−A)·B=8s+1.

What This Question Is Really Testing

For a state-space realization ẋ = Ax + Bu, y = Cx + Du, the transfer function under zero initial conditions is obtained directly from the resolvent matrix (sI−A)−1. This is a matrix calculation, so sign discipline is essential.

Concept Foundation

From the question:

A = [−2 −3 ; 4 2]
B = [3 ; 5]
C = [1 1]
D = 0

Step 1 — Write the Transfer-Function Formula

G(s) = C(sI − A)−1B + D

Step 2 — Form the Matrix sI − A

sI = [s 0 ; 0 s]
sI − A = [s + 2 3 ; −4 s − 2]

Step 3 — Calculate the Determinant

det(sI − A) = (s + 2)(s − 2) − (3)(−4)
det(sI − A) = s² − 4 + 12
det(sI − A) = s² + 8

Step 4 — Write the Adjugate and Inverse

adj(sI − A) = [s − 2 −3 ; 4 s + 2]
(sI − A)−1 = [s − 2 −3 ; 4 s + 2]/(s² + 8)

Step 5 — Multiply the Adjugate by B

[s − 2 −3 ; 4 s + 2][3 ; 5]
= [3(s − 2) − 15 ; 12 + 5(s + 2)]
= [3s − 21 ; 5s + 22]

Step 6 — Multiply by C and Include D

C·adj(sI − A)·B = [1 1][3s − 21 ; 5s + 22]
= (3s − 21) + (5s + 22)
= 8s + 1
G(s) = (8s + 1)/(s² + 8) + 0

Hence the correct option is A.

Why This Method Works

Taking the Laplace transform of ẋ = Ax + Bu with zero initial state gives (sI−A)X(s)=BU(s). Thus X(s)=(sI−A)−1BU(s), and substitution into the output equation produces the transfer-function formula.

Independent Verification / Cross-Check

Characteristic polynomial = det(sI − A) = s² + 8
All four options have denominator s² + 8

The determinant independently confirms the common denominator. Only the numerator calculation is then needed to distinguish the options.

Concept Extension

  • D=0 means there is no direct input-to-output feedthrough.
  • The denominator is tied to the state matrix A.
  • Numerator errors usually come from signs in sI−A or the adjugate.

Exam Strategy

For a 2×2 state-space MCQ: calculate det(sI−A) first. It checks the denominator and often eliminates work before you calculate the numerator.

Other options

All options share the denominator s²+8. Direct multiplication C·adj(sI−A)·B gives exactly 8s+1; the numerators in B, C and D do not match the state-space model.

Common mistake

Writing sI + A instead of sI − A, or changing the off-diagonal signs incorrectly while forming the adjugate.

Exam tip

For a 2×2 realization, write sI−A and its determinant explicitly before any multiplication.

Quick method

Denominator first: det(sI−A)=s²+8. Then compute only C·adj(sI−A)·B.