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Electrical Engg (EEE)
Control Systems

Electrical machines, power systems, circuits

14 Q 7 Topics Take Mock Test
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Topics in Electrical Engg (EEE)
In a Type-2 system with step, ramp, and parabolic inputs, the steady-state error with parabolic input A·t²/2 and loop gain Kv = 5 is:
A Zero
B A/Kv
C A/Ka where Ka is the acceleration error constant
D Infinity
Correct Answer:  C. A/Ka where Ka is the acceleration error constant
EXPLANATION

For a Type-2 system, parabolic error is ess = A/Ka where Ka = lim[s→0] s²G(s)H(s). Since Kv relates to ramp response, the parabolic steady-state error depends on Ka (acceleration constant).

Test
The sensitivity function S(s) in a feedback control system is defined as:
A T(s) = C(s)/R(s)
B S(s) = 1/[1+G(s)H(s)]
C S(s) = G(s)/[1+G(s)H(s)]
D S(s) = H(s)/[1+G(s)H(s)]
Correct Answer:  B. S(s) = 1/[1+G(s)H(s)]
EXPLANATION

Sensitivity S(s) = ∂Y/∂G ÷ Y/G = 1/[1+G(s)H(s)] for unity feedback systems. It measures effect of parameter variations.

Test
For a system with Gc(s)G(s)H(s) = 100/[s(s+5)], the phase at ω = 5 rad/s is approximately:
A -90°
B -135°
C -180°
D -225°
Correct Answer:  D. -225°
EXPLANATION

At ω=5: Phase = -90° (from 1/s) - arctan(5/1) - arctan(5/5) = -90° - 78.7° - 45° ≈ -213.7° ≈ -225°

Test
In a root locus plot, the asymptotes for a system with 5 poles and 2 zeros meet at a point called:
A Break-in point
B Centroid
C Pole-zero cancellation point
D Breakaway point
Correct Answer:  B. Centroid
EXPLANATION

Centroid (center of asymptotes) = [Σpoles - Σzeros]/[number of poles - number of zeros]. Number of asymptotes = 5-2 = 3.

Test
For the characteristic equation s⁴ + 8s³ + 24s² + 32s + 15 = 0, using Routh-Hurwitz criterion, the system is:
A Stable with 2 poles on RHS
B Marginally stable
C Unstable with 4 poles on RHS
D Stable with all poles on LHS
Correct Answer:  D. Stable with all poles on LHS
EXPLANATION

Routh table construction shows all positive elements in first column, indicating all poles in left half plane, making the system stable.

Test
For improving transient response with minimal steady-state error impact, a lead compensator should be designed to add phase lead at:
A Very low frequencies (
B The gain crossover frequency region
C Very high frequencies (>10ωgc)
D Both low and high frequencies equally
Correct Answer:  B. The gain crossover frequency region
EXPLANATION

Lead compensator adds phase lead at its designed center frequency ωm. For transient improvement, it should coincide with gain crossover frequency to increase phase margin

Test
For a system to be controllable using state feedback u = -Kx, which condition must be satisfied?
A Rank[B] = n
B Rank[A] = n
C Rank[B AB A²B ... Aⁿ⁻¹B] = n
D Rank[C] = n
Correct Answer:  C. Rank[B AB A²B ... Aⁿ⁻¹B] = n
EXPLANATION

Controllability requires the controllability matrix to have full rank n. This ensures all states can be moved from origin to any desired state

Test
A feedback system's sensitivity function S(s) = 1/(1+L(s)) where L(s) is loop gain. For large |L(s)|, what happens to |S(s)|?
A Becomes very large
B Becomes very small
C Remains constant
D Oscillates rapidly
Correct Answer:  B. Becomes very small
EXPLANATION

When |L(s)| >> 1, the sensitivity function |S(s)| ≈ 1/|L(s)| becomes very small. This shows that high loop gain reduces sensitivity to parameter variations.

Test
A negative feedback system's loop gain L(s) = G(s)H(s) has a pole-zero excess of 2. What can be concluded?
A The system must be unstable
B The system can be unstable or stable depending on gain
C The system is always stable
D The number of asymptotes is 2
Correct Answer:  B. The system can be unstable or stable depending on gain
EXPLANATION

Pole-zero excess affects the phase behavior at high frequencies. A system with excess of 2 can be either stable or unstable depending on the gain value and pole locations.

Test
Q.10 Hard Control Systems
For a second-order system with natural frequency ωn = 5 rad/s and ζ = 0.7, the peak time tp is approximately:
A 0.45 seconds
B 0.63 seconds
C 0.89 seconds
D 1.27 seconds
Correct Answer:  B. 0.63 seconds
EXPLANATION

tp = π/(ωn√(1-ζ²)) = π/(5√(1-0.49)) = π/(5×0.714) ≈ 0.88 seconds ≈ 0.89 seconds. Closest answer is B.

Test
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