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Electrical Engineering › Control Systems

Transfer Function & Modelling MCQ [Free PDF] – 4 Objective Questions with Answers

Every Transfer Function & Modelling question asked in GPSC DEE Electrical, Vidyut Sahayak JE — solve, check instantly, then read the formula and exam tip.

4 PYQs3 subtopics2 examsUpdated 23 Oct 2026
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Transfer Function (2)

Q1Moderate
Asked in: GPSC DEE Electrical · DEE Master paper · 2026
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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Q2Moderate
Asked in: GPSC DEE Electrical · DEE Electrical · 2024
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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Denominator Polynomial (1)

Q3Easy
Asked in: Vidyut Sahayak JE · Second tier · 2024
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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Time-Shifting Property (1)

Q4Moderate
Asked in: GPSC DEE Electrical · DEE Electrical · 2024
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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