કોઈ સિસ્ટમનું સ્ટેટ-સ્પેસ મોડેલ છે:
[Ẋ_1 ; Ẋ_2] = [−2 −3 ; 4 2][X_1 ; X_2] + [3 ; 5]U
Y = [1 1][X_1 ; X_2], D = 0
આ સિસ્ટમનું ટ્રાન્સફર ફંક્શન શું થશે?
જવાબ અને પગલાવાર ઉકેલ
G(s)=C(sI−A)^−1B+D નો ઉપયોગ કરો. અહીં det(sI−A)=s^2+8. C·adj(sI−A)·B નો ગુણાકાર 8s+1 આપે છે; તેથી G(s)=(8s+1)/(s^2+8).
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
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.
Quick Trick
Denominator first: det(sI−A)=s²+8. Then compute only C·adj(sI−A)·B.
Why Other Options Are Wrong
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.
![Solution diagram – કોઈ સિસ્ટમનું સ્ટેટ-સ્પેસ મોડેલ છે: [Ẋ_1 ; Ẋ_2] = [−2 −3 ; 4 2][X_1 ; X_2] +…](https://sarkariresultout.com/wp-content/uploads/2026/09/sro-pyq-f64811998b09b02bb1e3eba4.webp)
A certain system has a state space model as
[Ẋ_1 ; Ẋ_2] = [−2 −3 ; 4 2][X_1 ; X_2] + [3 ; 5]U
Y = [1 1][X_1 ; X_2], D = 0
What will be the transfer function of this system?
Answer & step-by-step solution
Use G(s)=C(sI−A)^−1B+D. Here det(sI−A)=s^2+8. Multiplying C·adj(sI−A)·B gives 8s+1; therefore G(s)=(8s+1)/(s^2+8).
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.
Quick Trick
Denominator first: det(sI−A)=s²+8. Then compute only C·adj(sI−A)·B.
Why Other Options Are Wrong
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.
किसी सिस्टम का स्टेट-स्पेस मॉडल है:
[Ẋ_1 ; Ẋ_2] = [−2 −3 ; 4 2][X_1 ; X_2] + [3 ; 5]U
Y = [1 1][X_1 ; X_2], D = 0
इस सिस्टम का ट्रांसफर फंक्शन क्या होगा?
उत्तर और चरणबद्ध हल
G(s)=C(sI−A)^−1B+D का उपयोग करें। यहाँ det(sI−A)=s^2+8। C·adj(sI−A)·B का गुणन 8s+1 देता है; अतः G(s)=(8s+1)/(s^2+8)।
विस्तृत हल (मूल भाषा में)
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.
Quick Trick
Denominator first: det(sI−A)=s²+8. Then compute only C·adj(sI−A)·B.
Why Other Options Are Wrong
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.
More State-Space Analysis questions