Control system miscellaneous
- A linear second-order single-input continuous-time system is described by the following set of differential equations :
X1̇ (t) = – 2 X1 (t) + 4 X2 (t)
X2̇ (t) = 2 X1 (t) – X2 (t) + u (t)
where X1 (t) and X2 (t) are the state variables and u (t) is the control variable.
The system is
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In matrix form
Ẋ1(t) 
= -2 4 X1(t) + 0 u(t) Ẋ2(t) 2 -1 X2(t) 1
Test for controllability[B : AB] = 0 -2 4 0 1 2 -1 1 = 0 4 Rank is 2, hence controllable 1 -1
For stability, sI – A = 0∴ s + 2 4 = 0 2 s + 1
⇒ s2 + 3s - 4 = 0
Routh criterion is,s2 1 -4 s 3 s -4
Since sign changes, hence system is not stable.
Correct Option: B
In matrix form
Ẋ1(t) = -2 4 X1(t) + 0 u(t) Ẋ2(t) 2 -1 X2(t) 1
Test for controllability[B : AB] = 0 -2 4 0 1 2 -1 1 = 0 4 Rank is 2, hence controllable 1 -1
For stability, sI – A = 0∴ s + 2 4 = 0 2 s + 1
⇒ s2 + 3s - 4 = 0
Routh criterion is,s2 1 -4 s 3 s -4
Since sign changes, hence system is not stable.
- A system is described by the state equation
Ẋ = AX + BU.
The output is given by Y = CX,where, A = -4 -1 , B = 1 , C = [ 1 0 ] 3 -1 1
Transfer function G(s) of the system is
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Transfer function G
G(s) = C(sI – A)– 1 B(sI - A) = s 0 - -4 -1 = s + 4 1 0 s 3 -1 -3 s + 1 
Here , G(s) = [1 0] s + 1 -1 1 s2 + 5s + 7 +3 s + 4 1 = s s2 + 5s + 7
Correct Option: A
Transfer function G
G(s) = C(sI – A)– 1 B(sI - A) = s 0 - -4 -1 = s + 4 1 0 s 3 -1 -3 s + 1 Here , G(s) = [1 0] s + 1 -1 1 s2 + 5s + 7 +3 s + 4 1 = s s2 + 5s + 7
- A linear time-invariant system describd by the state variable model
X1̇ = -1 0 X1 + 0 u = [ 1 2 ] X1 X2̇ 0 -2 X2 1 X2
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The linear time-invariant system:
X1̇ = -1 0 X1 + 1 u = [ 1 2 ] X1 X2̇ 1 -2 X2 0 X2 A = -1 0 , B = 0 , C = [1 2] 1 -2 1 [B AB] = 0 0 = |B AB| = 0 1 -2
Hence not completely controllable[CT ATCT] = 1 1 2 -4
[CT ATCT] ≠ 0
Hence observable.
Correct Option: E
The linear time-invariant system:
X1̇ = -1 0 X1 + 1 u = [ 1 2 ] X1 X2̇ 1 -2 X2 0 X2 A = -1 0 , B = 0 , C = [1 2] 1 -2 1 [B AB] = 0 0 = |B AB| = 0 1 -2
Hence not completely controllable[CT ATCT] = 1 1 2 -4
[CT ATCT] ≠ 0
Hence observable.
- The maximum phase shift that can be provided by a lead compensator with transfer function
G(s) = 1 + 6s is 1 + 2s
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G(s) = 1 + 6s 1 + 2s
Phase shift, φ = tan– 16ω – tan– 12ωdφ = 6 - 2 dω 1 + 36ω2 1 + 4ω2 For maximum value of φ , dφ = 0 dω
∴ 3 (1+ 4ω2) = 1 + 36 ω2
or ω = 1/ √12Thus , φ = tan– 1 36 - tan– 1 1 = 30° 12 √3
Correct Option: B
G(s) = 1 + 6s 1 + 2s
Phase shift, φ = tan– 16ω – tan– 12ωdφ = 6 - 2 dω 1 + 36ω2 1 + 4ω2 For maximum value of φ , dφ = 0 dω
∴ 3 (1+ 4ω2) = 1 + 36 ω2
or ω = 1/ √12Thus , φ = tan– 1 36 - tan– 1 1 = 30° 12 √3
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The system Ẋ = 2 3 X + 1 u is 0 5 0
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Ẋ = 2 3 X + 1 u 0 5 0 B = 1 , A = 2 3 0 0 5 AB = 2 3 1 = 2 0 5 0 0 ⇒ [B : AB] = 1 2 0 0
This is not a 2 × 2 matrix, hence system is uncontrollableNow [sI - A] = s 0 - 2 3 0 s 0 5 = s - 2 -3 0 s - 5
= (s – 2) (s – 5)
Two roots on positive half of the y-axis, hence the system is unstable .
Correct Option: B
Ẋ = 2 3 X + 1 u 0 5 0 B = 1 , A = 2 3 0 0 5 AB = 2 3 1 = 2 0 5 0 0 ⇒ [B : AB] = 1 2 0 0
This is not a 2 × 2 matrix, hence system is uncontrollableNow [sI - A] = s 0 - 2 3 0 s 0 5 = s - 2 -3 0 s - 5
= (s – 2) (s – 5)
Two roots on positive half of the y-axis, hence the system is unstable .
