Electrical Engineering › Control Systems

Steady-State Error MCQ [Free PDF] – 3 Objective Questions with Answers

Every Steady-State Error question asked in GPSC DEE Electrical, GPSC AE Electrical — solve, check instantly, then read the formula and exam tip.

3 PYQs2 subtopics2 examsUpdated 23 Oct 2026
Topic practice0/3✓ 0✗ 0

System Type (2)

Q1Moderate
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.
Verified · SRO editorial deskSolution page → Report
Q2Easy
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
Verified · SRO editorial deskSolution page → Report

Steady-State Error (1)

Q3Moderate
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.
Verified · SRO editorial deskSolution page → Report