Electrical Engineering · updated 23 Oct 2026

Control Systems MCQ – 33 previous year questions with solutions

Every Control Systems question asked in GPSC DEE Electrical, GPSC AE Electrical, Vidyut Sahayak JE — arranged topic by topic, with the exam and paper shown on each question.

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Time Response MCQs 6

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Time Response › Second-Order Systems (2)

Q1Moderate
Consider the following statements regarding Second Order Systems:
1. Damping ratio determines the nature of system response such as overdamped, underdamped, or critically damped.
2. Underdamped systems show oscillatory response with decaying amplitude.
3. Critically damped system has fastest response without oscillations.
4. Overdamped system oscillates indefinitely with constant amplitude.
Which of the above statements is/are correct?
View answer & solution
Correct answer: A. 1, 2 and 3 only
Quick explanation

Second-order dynamics are governed by two independent energy-storage states and are described by natural frequency and damping ratio. An underdamped response oscillates with exponentially decaying amplitude, critical damping gives the fastest non-oscillatory response, and overdamping gives a slower non-oscillatory response. Key relation: Characteristic form: s²+2ζω_n s+ω_n²=0.

Exam focus: ζ1 overdamped. Common trap: Overdamped responses do not sustain oscillations.

Characteristic form: s²+2ζω_n s+ω_n²=0.
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Q2Hard
The open loop transfer function for a unit feedback system is G(s) = K/s(sT + 1). By what factor the amplitude gain K should be reduced so that the peak overshoot becomes 25% from 75%?
View answer & solution
Correct answer: C. 20
Quick explanation

Percent overshoot fixes the damping ratio through M_p=e^{−πζ/√(1−ζ²)}. For G(s)=K/[s(sT+1)], the closed-loop damping ratio varies inversely with √K. Converting 75% and 25% overshoot to ζ values and squaring their ratio gives the required gain reduction factor, matching 20.

Mp = exp[−πζ/√(1−ζ²)] ζ = −ln(Mp)/√[π²+ln²(Mp)] Knew/Kold=(ζold/ζnew)²
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Time Response › Closed-Loop Step Response (1)

Q3Hard
What is the response of the feedback function with open loop transfer function G(s) = 4/s(s + 5) and input is input step?
View answer & solution
Correct answer: A. 1 − 1/3(4e^(−t) − e^(−4t))
Quick explanation

For unity feedback, first form T(s)=G(s)/(1+G(s)), then multiply by the step input 1/s and perform partial fractions. For G(s)=4/[s(s+5)], the closed-loop denominator is s²+5s+4=(s+1)(s+4), giving the response 1−(1/3)(4e⁻ᵗ−e⁻⁴ᵗ).

T(s)=4/[s²+5s+4] =4/[(s+1)(s+4)] C(s)=T(s)/s =4/[s(s+1)(s+4)] c(t)=1−(4/3)e⁻ᵗ+(1/3)e⁻⁴ᵗ =1−(1/3)(4e⁻ᵗ−e⁻⁴ᵗ)
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Time Response › First-Order System (1)

Q4Easy
A first-order system has a time constant of 2 seconds and is subjected to a unit step input. The system gain is unity, then
1. The output reaches approximately 63 percent of its final value at 2 seconds.
2. The steady-state value of the output is equal to 1.
3. The response is exponential in nature and does not exhibit oscillations.
4. The output reaches 95 percent of final value in about 2 seconds.
Which of the above statements is/are correct?
View answer & solution
Correct answer: B. 1, 2 and 3 only
Quick explanation

A standard first-order system has one real pole and an exponential step response controlled by a time constant τ. For τ=2 s, y(t)=1−e^(−t/2). At t=2 s, y≈0.632. Key relation: G(s)=1/(τs+1); y(t)=1−e^(−t/τ).

Exam focus: 63% at τ, ~95% at 3τ, ~99% at 5τ. Common trap: Do not confuse one time constant with 95% settling.

G(s)=1/(τs+1); y(t)=1−e^(−t/τ).
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Time Response › Second-Order System (1)

Q5Hard
A second-order system has damping ratio 0.5 and natural frequency 4 rad/s, then
1. The system is underdamped and exhibits oscillatory response.
2. The damped natural frequency is less than the natural frequency.
3. The peak overshoot is zero for this system.
4. The settling time decreases as damping ratio increases (within underdamped region).
Which of the above statements is/are correct?
View answer & solution
Correct answer: C. 1, 2 and 4 only
Quick explanation

Second-order dynamics are governed by two independent energy-storage states and are described by natural frequency and damping ratio. With ζ=0.5 and ωn=4 rad/s, ωd=ωn√(1−ζ²)≈3.46 rad/s, so the response is underdamped and has nonzero overshoot. Key relation: Characteristic form: s²+2ζω_n s+ω_n²=0.

Exam focus: ζ1 overdamped. Common trap: Overdamped responses do not sustain oscillations.

Characteristic form: s²+2ζω_n s+ω_n²=0.
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Time Response › Time Response (1)

Q6Hard
The unit step response of a second order system is = 1 − e^(−5t) − 5t e^(−5t). Consider the following statements:
1. The under damped natural frequency is 5 rad/s.
2. The damping ratio is1.
3. The impulse response is 25t e^(−5t).
Which of the statements given above are correct?
View answer & solution
Correct answer: B. Only 2 and 3
Quick explanation

The response 1−e⁻⁵ᵗ−5te⁻⁵ᵗ is the standard unit-step response of a critically damped second-order system with repeated pole at −5. Hence ζ=1, and differentiating the step response gives impulse response h(t)=25te⁻⁵ᵗ. Statement 1 is false because there is no underdamped natural frequency.

c(t)=1−(1+5t)e⁻⁵ᵗ ζ = 1 ωn = 5 rad/s ωd = ωn√(1−ζ²)=0 h(t)=dc(t)/dt =25t e⁻⁵ᵗ
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Controllers & Feedback MCQs 5

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Controllers & Feedback › PI Controller (2)

Q7Moderate
Assertion (A): A PI controller increases the order of the system by units but decreases the steady state error.
Reason (R): A PI controller introduces a pole at either the origin or at a desired points on the negative real axis.
View answer & solution
Correct answer: C. (A) is true but (R) is false
Quick explanation

A PI controller adds an integrator pole at the origin and a finite zero, increasing system type and reducing/eliminating steady-state error for appropriate inputs. The assertion is therefore true. The reason is false because a PI controller's added pole is specifically at the origin; the finite negative-real-axis point is the controller zero, not an arbitrary added pole.

GPI(s)=Kp + Ki/s =Kp(s+Ki/Kp)/s Pole: s=0 Zero: s=−Ki/Kp
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Q8Moderate
Read the following statements in the context of a PI controller used to control a first order LTI system.
Statement 1: The order of overall closed loop system becomes higher than the open loop system.
Statement 2: Increasing integral gain makes the overall system to have higher damping.
Choose the correct options from the following:
View answer & solution
Correct answer: B. Statement 1 is correct but statement 2 is incorrect.
Quick explanation

A PI controller introduces an integrator 1/s, so a first-order plant normally produces a second-order closed-loop characteristic equation: Statement 1 is true. For a standard plant K/(τs+1) with fixed K_p, ζ = (1 + KK_p)/(2√(τKK_i)); increasing K_i reduces ζ, so Statement 2 is false.

C(s) = Kp + Ki/s = (Kps + Ki)/s G(s) = K/(τs + 1) Characteristic equation: τs² + (1 + KKp)s + KKi = 0 ζ = (1 + KKp)/(2√(τKKi)) G(s) = K/(τs + 1) C(s) = Kp + Ki/s C(s) = (Kps + Ki)/s 1 + K(Kps + Ki)/[s(τs + 1)] = 0 s(τs + 1) + K(Kps + Ki) = 0 τs² + (1 + KKp)s + KKi = 0 s² + [(1 + KKp)/τ]s + KKi/τ = 0 s² + 2ζωns + ωn² = 0 ωn = √(KKi/τ) ζ = (1 + KKp)/(2√(τKKi)) Ki ↑ ⇒ √Ki ↑ ⇒ ζ ↓ (K, τ, Kp fixed)
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Controllers & Feedback › Negative Feedback (1)

Q9Moderate
Read the following statements in the context of negative feedback system.
Statement 1: Negative feedback always makes the overall system response stable.
Statement 2: Negative feedback system may improve the system’s robustness towards parametric uncertainties.
Choose the correct option from the following:
View answer & solution
Correct answer: C. Statement 2 is correct and statement 1 is incorrect.
Quick explanation

Statement 1 is false because negative feedback does not always guarantee stable closed-loop poles. Statement 2 is true because negative feedback can reduce parameter sensitivity; for a standard loop, S = 1/[1 + G(s)H(s)]. Therefore option C is correct.

T(s) = G(s)/[1 + G(s)H(s)] Characteristic equation: 1 + G(s)H(s) = 0 Sensitivity: S = 1/[1 + G(s)H(s)] T(s) = G(s)/[1 + G(s)H(s)] Characteristic equation: 1 + G(s)H(s) = 0 S = 1/[1 + G(s)H(s)]
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Controllers & Feedback › PID Controller (1)

Q10Moderate
Consider the following statements regarding PID Controller Design:
1. Proportional control reduces rise time.
2. Integral control eliminates steady-state error.
3. Derivative control improves damping.
4. PID control guarantees stability always.
Which of the above statements is/are correct?
View answer & solution
Correct answer: C. 1, 2 and 3 only
Quick explanation

Proportional action generally speeds response but leaves finite error; integral action raises low-frequency gain and can eliminate step steady-state error; derivative action adds predictive damping and can reduce overshoot. None of these by itself guarantees stability for every plant/tuning. Key relation: u(t)=K_p e + K_i∫e dt + K_d de/dt.

Exam focus: I fixes offset; D improves damping; P sets overall response strength. Common trap: PID is not an unconditional stability guarantee.

u(t)=Kp e + Ki∫e dt + Kd de/dt.
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Controllers & Feedback › Role of Controller (1)

Q11Easy
What is the need for a controller?
View answer & solution
Correct answer: C. Modify control signals
Quick explanation

A controller processes the error/control information and modifies the control signal so the plant responds according to desired performance requirements. It may amplify, integrate, differentiate, shape or otherwise compensate the signal.

u(t)=Controller{e(t)} u=Kp e+Ki∫e dt+Kd de/dt
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Frequency Response MCQs 5

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Frequency Response › Stability Margins (2)

Q12Hard
A system has gain crossover frequency at which magnitude is unity and phase is –150 degrees, then
1. The phase margin of the system is 30 degrees.
2. The system is stable based on phase margin criterion.
3. Increasing gain may lead to instability.
4. Gain margin is zero at this operating point.
Which of the above statements is/are correct?
View answer & solution
Correct answer: D. 1, 2 and 3 only
Quick explanation

Gain margin and phase margin quantify distance from the critical feedback condition and are widely used as relative-stability measures; pole and zero locations shape the frequency response. At gain crossover, phase=−150°, so PM=180−150=30°. Gain margin must be evaluated at phase crossover, so it is not automatically zero here.

Exam focus: Positive phase margin usually indicates a stable negative-feedback loop with normal assumptions. Common trap: Frequency response is not independent of poles and zeros.

At gain crossover ωgc: PM=180°+∠G(jωgc).
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Q13Easy
Which of the following condition should be satisfied for ensuring stability of an LTI system?
View answer & solution
Correct answer: D. Both phase margin and gain margin should be positive.
Quick explanation

In the standard negative-feedback frequency-response stability test used in this question, both margins must be positive: GM > 0 dB and PM > 0°. Positive margins mean the loop has not yet reached the critical instability boundary.

Gain Margin: GM > 0 dB Phase Margin: PM > 0° GM > 0 dB PM > 0°
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Frequency Response › Bode Plot (1)

Q14Moderate
In a bode magnitude plot, which one of the following slopes would be exhibited at high frequencies by a 4^th order system having two zeros in left half of s-plane?
View answer & solution
Correct answer: B. −40 dB/decade
Quick explanation

Each first-order pole contributes −20 dB/decade and each first-order zero contributes +20 dB/decade after its corner frequency. With 4 poles and 2 zeros, the high-frequency slope is 20(2−4)=−40 dB/decade.

High-frequency slope = 20(Nz − Np) dB/decade Np = 4, Nz = 2 Slope = 20(2 − 4) = −40 dB/decade Final slope = 20(Nz − Np) dB/decade Pole contribution = 4(−20) = −80 dB/decade Zero contribution = 2(+20) = +40 dB/decade Net slope = −80 + 40 Net slope = −40 dB/decade Relative degree = Np − Nz = 4 − 2 = 2 Final slope = −20(2) = −40 dB/decade
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Frequency Response › Bode Plot and Stability Margins (1)

Q15Moderate
Consider the following statements regarding Frequency Response:
1. Frequency response describes system behavior as function of frequency.
2. Bode plot represents magnitude and phase variation with frequency.
3. Gain margin and phase margin indicate system stability.
4. Frequency response is independent of system poles.
Which of the above statements is/are correct?
View answer & solution
Correct answer: D. 1, 2 and 3 only
Quick explanation

Frequency response describes how a linear system's gain and phase vary with sinusoidal frequency. Gain margin and phase margin quantify distance from the critical feedback condition and are widely used as relative-stability measures; pole and zero locations shape the frequency response. Key relation: At gain crossover ω_gc: PM=180°+∠G(jω_gc).

Exam focus: Positive phase margin usually indicates a stable negative-feedback loop with normal assumptions. Common trap: Frequency response is not independent of poles and zeros.

At gain crossover ωgc: PM=180°+∠G(jωgc).
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Frequency Response › Gain Margin (1)

Q16Moderate
If the gain of an open-loop system is doubled, then the gain margin
View answer & solution
Correct answer: B. Decreases by factor of 2
Quick explanation

Linear gain margin is the reciprocal of the open-loop magnitude at the phase-crossover frequency: GM = 1/|L(jω_pc)|. Doubling the loop gain doubles that magnitude, so GM_new = GM_old/2. Thus it decreases by a factor of 2.

GM = 1/|L(jωpc)| Lnew(s) = 2Lold(s) GMnew = GMold/2 20log10(2) = 6.02 dB GM = 1/|L(jωpc)| Lnew(s) = 2Lold(s) GMnew = 1/[2|Lold(jωpc)|] GMnew = (1/2)GMold Gain ×2 ⇒ magnitude shift = +20log10(2) = +6.02 dB Therefore GM(dB) decreases by 6.02 dB
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State-Space Analysis MCQs 5

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State-Space Analysis › State Variables (2)

Q17Moderate
Consider the following statements regarding State Variable Analysis:
1. State variables define system dynamics completely.
2. State-space model can represent MIMO systems.
3. Transfer function can always be uniquely obtained from state model.
4. State model includes initial conditions.
Which of the above statements is/are correct?
View answer & solution
Correct answer: D. 1, 2 and 4 only
Quick explanation

A state-space model represents internal dynamics using a minimal set of state variables. It naturally handles MIMO systems and incorporates initial conditions. Key relation: ẋ=Ax+Bu; y=Cx+Du.

Exam focus: Eigenvalues of A are the state-model characteristic roots. Common trap: Output is not generally independent of state variables.

ẋ=Ax+Bu; y=Cx+Du.
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Q18Moderate
A system is represented in state-space form with two state variables and stable eigenvalues, then
1. The system order is two.
2. The system is stable since eigenvalues lie in left half-plane.
3. State variables completely describe system dynamics.
4. Output depends only on input and not on state variables.
Which of the above statements is/are correct?
View answer & solution
Correct answer: C. 1, 2 and 3 only
Quick explanation

A state-space model represents internal dynamics using a minimal set of state variables. Output generally depends on both state and direct input through y=Cx+Du. Key relation: ẋ=Ax+Bu; y=Cx+Du.

Exam focus: Eigenvalues of A are the state-model characteristic roots. Common trap: Output is not generally independent of state variables.

ẋ=Ax+Bu; y=Cx+Du.
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State-Space Analysis › Controllability and Observability (1)

Q19Moderate
Read the following statements:
Statement 1: A system is said to be completely controllable if it is possible to transfer the system state from an initial state X(t_0) to any desired state X(t) in a specified finite time by a control vector u(t).
Statement 2: A system is said to be completely observable, if every state X(t_0) can be completely identified by measurement of the outputs y(t) over the time interval 0 ≤ t ≤ ∞.
Choose the correct options from the following:
View answer & solution
Correct answer: A. Both statement 1 and statement 2 are correct.
Quick explanation

Statement 1 gives the standard meaning of complete controllability: the input can drive the state to any desired state in finite time. Statement 2 gives the observability idea: the initial state can be reconstructed from output measurements. For an n-state LTI model, the algebraic checks are rank(𝒞)=n and rank(𝒪)=n. Hence option A.

𝒞 = [ B AB A²B … An−1B ] Complete controllability: rank(𝒞) = n 𝒪 = [ C ; CA ; CA² ; … ; CAn−1 ] Complete observability: rank(𝒪) = n 𝒞 = [ B AB A²B … An−1B ] 𝒪 = [ C ; CA ; CA² ; … ; CAn−1 ] rank(𝒞) = n ⇒ completely controllable rank(𝒪) = n ⇒ completely observable
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State-Space Analysis › Stability (1)

Q20Moderate
Consider the following statements regarding Stability in State-Space Analysis:
1. Eigenvalues of the system matrix A define the characteristic roots that determine the system's stability.
2. A system is asymptotically stable if and only if, all eigenvalues of the system matrix A have strictly negative real parts (located in the Left-Half Plane).
3. The presence of at least one eigenvalue with a positive real part (located in the Right-Half Plane) implies that the system is unstable.
4. Stability is a property of the system itself and is determined by the internal system parameters and the A matrix.
Which of the above statements is/are correct?
View answer & solution
Correct answer: D. All of the above
Quick explanation

A state-space model represents internal dynamics using a minimal set of state variables. Stability of a continuous-time LTI model is tied to eigenvalues of A: strictly negative real parts imply asymptotic stability. Key relation: ẋ=Ax+Bu; y=Cx+Du.

Exam focus: Eigenvalues of A are the state-model characteristic roots. Common trap: Output is not generally independent of state variables.

ẋ=Ax+Bu; y=Cx+Du.
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State-Space Analysis › State-Space to Transfer Function (1)

Q21Hard
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?
View answer & solution
Correct answer: A. (8s + 1)/(s^2 + 8)
Quick explanation

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).

[Ẋ1 ; Ẋ2] = [−2 −3 ; 4 2][X1 ; X2] + [3 ; 5]U Y = [1 1][X1 ; X2], D = 0 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) A = [−2 −3 ; 4 2] B = [3 ; 5] C = [1 1] D = 0 G(s) = C(sI − A)−1B + D sI = [s 0 ; 0 s] sI − A = [s + 2 3 ; −4 s − 2] det(sI − A) = (s + 2)(s − 2) − (3)(−4) det(sI − A) = s² − 4 + 12 det(sI − A) = s² + 8 adj(sI − A) = [s − 2 −3 ; 4 s + 2] (sI − A)−1 = [s − 2 −3 ; 4 s + 2]/(s² + 8) [s − 2 −3 ; 4 s + 2][3 ; 5] = [3(s − 2) − 15 ; 12 + 5(s + 2)] = [3s − 21 ; 5s + 22] C·adj(sI − A)·B = [1 1][3s − 21 ; 5s + 22] = (3s − 21) + (5s + 22) = 8s + 1 G(s) = (8s + 1)/(s² + 8) + 0 Characteristic polynomial = det(sI − A) = s² + 8 All four options have denominator s² + 8
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Transfer Function & Modelling MCQs 4

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Transfer Function & Modelling › Transfer Function (2)

Q22Moderate
Consider the following statements regarding Transfer Function:
1. Transfer function is defined as ratio of Laplace transform of output to input under zero initial conditions.
2. It provides complete information about system internal states.
3. Poles of transfer function determine system stability.
4. Zeros influence transient and frequency response characteristics.
Which of the above statements is/are correct?
View answer & solution
Correct answer: D. 1, 3 and 4 only
Quick explanation

A transfer function is the Laplace-domain ratio of output to input for a linear time-invariant system under zero initial conditions. Its poles and zeros determine important input-output stability, transient and frequency-response features, but the transfer function does not uniquely reveal the internal state realization. Key relation: G(s)=Y(s)/U(s) with zero initial conditions.

Exam focus: Poles are denominator roots; zeros are numerator roots. Common trap: Transfer functions do not contain complete unique internal-state information.

G(s)=Y(s)/U(s) with zero initial conditions.
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Q23Moderate
An LTI system has a unit step response y(t) = (2 − e^−t)u(t). What is the transfer function of this system?
View answer & solution
Correct answer: C. (s + 2)/(s + 1)
Quick explanation

For a unit-step input, U(s) = 1/s and G(s) = Y(s)/U(s) = sY(s). Transforming the given response gives Y(s) = (s + 2)/[s(s + 1)]; therefore G(s) = (s + 2)/(s + 1).

Y(s) = 2/s − 1/(s + 1) Y(s) = (s + 2)/[s(s + 1)] U(s) = 1/s G(s) = Y(s)/U(s) = sY(s) G(s) = (s + 2)/(s + 1) G(s) = Y(s)/U(s) y(t) = 2u(t) − e−tu(t) Y(s) = 2/s − 1/(s + 1) Y(s) = [2(s + 1) − s]/[s(s + 1)] Y(s) = (s + 2)/[s(s + 1)] U(s) = 1/s G(s) = Y(s)/(1/s) G(s) = sY(s) G(s) = (s + 2)/(s + 1) Final step-response value = 2 G(0) = (0 + 2)/(0 + 1) = 2
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Transfer Function & Modelling › Denominator Polynomial (1)

Q24Easy
In a transfer function representation of a system, what does the denominator polynomial represent?
View answer & solution
Correct answer: D. Characteristics of the system
Quick explanation

D — Characteristics of the system. The denominator gives the characteristic equation; its roots determine the poles and natural response of the represented system.

G(s) = N(s)/D(s) D(s) = 0 gives the characteristic roots G(s) = Y(s)/U(s) = N(s)/D(s) Characteristic equation: D(s) = 0 s² + 3s + 2 = (s + 1)(s + 2) = 0 Poles: s = −1 and s = −2
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Transfer Function & Modelling › Time-Shifting Property (1)

Q25Moderate
Given the Laplace transform of a signal as Y(s) = 6e^−2s/(s + 4), what will be the initial value in time domain y(0)?
View answer & solution
Correct answer: C. 0
Quick explanation

The factor e^−2s is a 2 s transport delay. Therefore y(t)=6e^−4(t−2)u(t−2), which is zero for t<2 s. Hence y(0)=0.

ℒ−1{6/(s + 4)} = 6e−4tu(t) e−asF(s) ↔ f(t − a)u(t − a) y(t) = 6e−4(t−2)u(t − 2) y(0) = 0 e−asF(s) ↔ f(t − a)u(t − a) F(s) = 6/(s + 4) f(t) = 6e−4tu(t) a = 2 s Y(s) = e−2sF(s) y(t) = f(t − 2)u(t − 2) y(t) = 6e−4(t−2)u(t − 2) y(0) = 6e8·0 = 0 y(0+) = lims→∞ sY(s) = lims→∞ [6s e−2s/(s + 4)] = 0
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Steady-State Error MCQs 3

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Steady-State Error › System Type (2)

Q26Moderate
A unity feedback system has open-loop transfer function such that it is a type-1 system and subjected to a unit step input, then
1. The steady-state error for step input is zero.
2. The steady-state error for ramp input is finite.
3. Increasing system gain reduces steady-state error for ramp input.
4. The steady-state error for step input increases with increase in gain.
Which of the above statements is/are correct?
View answer & solution
Correct answer: A. 1, 2 and 3 only
Quick explanation

It determines the order of polynomial input that can be tracked with zero/finite/infinite steady-state error, provided the closed loop is stable. Static-error constants directly produce the truth set {1,2,3}. Key relation: Type 1: Kp=lim_{s→0}GH=∞ → ess,step=1/(1+Kp)=0; Kv=lim_{s→0}sGH finite → ess,ramp=1/Kv.

Exam focus: Type-1 mnemonic: step → zero error; ramp → finite error; parabolic → infinite error. Common trap: Ignoring the closed-loop stability condition when applying steady-state error formulas.

Type 1: Kp=lim_{s→0}GH=∞ → ess,step=1/(1+Kp)=0; Kv=lim_{s→0}sGH finite → ess,ramp=1/Kv.
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Q27Easy
For a unity feedback system, the open loop transfer function is G(s) = K(s + 2)/s² (s² + 7s + 12). What is the type of system?
View answer & solution
Correct answer: B. Two
Quick explanation

System type equals the number of pure integrators (poles at the origin) in the open-loop transfer function G(s)H(s). Here the denominator contains s², so there are two poles at the origin and the system is Type 2.

G(s)H(s)=K(s+2)/[s²(s+3)(s+4)] Number of poles at s=0 = 2
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Steady-State Error › Steady-State Error (1)

Q28Moderate
Consider the following statements regarding Steady-State Error:
1. Type of system determines steady-state error for standard inputs.
2. Increasing system gain reduces steady-state error.
3. Integral control eliminates steady-state error for step input.
4. Steady-state error is independent of system type.
Which of the above statements is/are correct?
View answer & solution
Correct answer: C. 1, 2 and 3 only
Quick explanation

It determines which standard polynomial inputs can be tracked with zero or finite steady-state error. Integral action increases low-frequency gain and improves steady-state accuracy, but it can also affect transient stability margins. Key relation: Kp=lim G(s), Kv=lim sG(s), Ka=lim s²G(s) as s→0 for unity feedback.

Exam focus: Type-0/1/2 classification comes from poles at the origin. Common trap: Do not infer stability from system type alone.

Kp=lim G(s), Kv=lim sG(s), Ka=lim s²G(s) as s→0 for unity feedback.
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Nonlinear Systems MCQs 2

Nonlinear Systems › Lyapunov Stability (1)

Q29Moderate
The method of investigating the stability using Liapunov function is _____________
View answer & solution
Correct answer: A. The direct method
Quick explanation

Investigating stability by constructing a Lyapunov function V(x) and examining its derivative along system trajectories is called Lyapunov's direct (second) method. It avoids explicitly solving the nonlinear differential equations.

V(x)>0 for x≠0, V(0)=0 V̇(x)<0 ⇒ asymptotic stability (standard local/global conditions apply)
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Nonlinear Systems › Nonlinear Analysis (1)

Q30Moderate
Consider the following statements regarding Non-linear Control Systems:
1. Superposition principle does not apply.
2. Linearization is used for analysis near operating point.
3. Non-linear systems are always unstable.
4. Phase plane analysis is used for non-linear systems.
Which of the above statements is/are correct?
View answer & solution
Correct answer: A. 1, 2 and 4 only
Quick explanation

Nonlinear systems do not obey superposition, but they can still be stable or unstable. Linearization around an operating point uses a local first-order approximation and is useful for small-signal analysis. Key relation: Local linearization uses the Jacobian evaluated at the operating point.

Exam focus: Nonlinear does not mean unstable; stability depends on the dynamics and operating point. Common trap: Do not apply global conclusions from a local linear model without checking validity range.

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Root Locus MCQs 2

Root Locus › Asymptotes (1)

Q31Moderate
The characteristic equation of a control system is given by s(s + 4)(s² + 2s + s) + k(s + 1) = 0. What are the angles of the asymptotes for the root loci?
View answer & solution
Correct answer: C. 60°, 180°, 300°
Quick explanation

Root-locus asymptote angles depend only on the excess number of open-loop poles over zeros: θ_k=(2k+1)180°/(n−m). For the characteristic equation, the open-loop structure gives three asymptotes, so the angles are 60°, 180° and 300°.

θk = (2k+1)180°/(n−m) k=0 → 60° k=1 → 180° k=2 → 300°
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Root Locus › Root Locus Rules (1)

Q32Moderate
Consider the following statements regarding Root Locus Plot:
1. Root locus starts at open-loop poles and ends at zeros.
2. Number of branches equals number of poles.
3. Root locus always lies in left half plane.
4. Breakaway points occur on real axis.
Which of the above statements is/are correct?
View answer & solution
Correct answer: A. 1, 2 and 4 only
Quick explanation

Branches start at open-loop poles and end at finite zeros or zeros at infinity; the number of branches equals the number of open-loop poles. Real-axis breakaway/break-in points can occur where multiple branches meet, and loci can enter the right-half plane. Key relation: Characteristic equation 1+KG(s)H(s)=0.

Exam focus: Start at poles, finish at zeros/infinity. Common trap: Root locus is not confined to the left-half plane.

Characteristic equation 1+KG(s)H(s)=0.
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Stability & Routh-Hurwitz MCQs 1

Stability & Routh-Hurwitz › Routh-Hurwitz Criterion (1)

Q33Moderate
Assertion (A): Routh criterion is in terms of array formulation which is more convenient to handle.
Reason (R): This method is used to investigate the method of stability of higher order systems.
View answer & solution
Correct answer: B. Both (A) and (R) are true and (R) is not the correct explanation of (A)
Quick explanation

Both statements are true: the Routh criterion converts a characteristic polynomial into an algebraic array and is especially convenient for determining the number of right-half-plane roots of higher-order systems without explicitly solving for all roots. However, the second statement is a use of the method rather than the precise reason the array formulation is convenient, so the final-key choice is B.

an s^n + a_{n−1}s^{n−1}+...+a0=0 = number of RHP roots.
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Control Systems MCQ FAQs

How many Control Systems MCQs are on SarkariResultOut?

33 previous year Control Systems questions across 9 topics, each with the exam it was asked in, the verified answer and a step-by-step solution.

Which exams asked these Control Systems questions?

GPSC DEE Electrical · DEE Master paper · 2026; GPSC AE Electrical · AE R&B · 2025 · Series M; GPSC DEE Electrical · DEE Electrical · 2024; Vidyut Sahayak JE · Second tier · 2024.

Can I download Control Systems MCQs as a PDF?

Yes. Tap “Download Control Systems PDF” — it is free after a quick sign-in and includes every answer and solution.