Steady State Error MCQ Quiz - Objective Question with Answer for Steady State Error - Download Free PDF
Last updated on Mar 21, 2025
Latest Steady State Error MCQ Objective Questions
Steady State Error Question 1:
For a feedback control system of type 2, the steady state error for a ramp input is:
Answer (Detailed Solution Below)
Steady State Error Question 1 Detailed Solution
Explanation:
Feedback Control System of Type 2
Definition: A feedback control system of type 2 is a system where the open-loop transfer function has two integrators. This means that the system has two poles at the origin in the s-plane. The presence of these two integrators allows the system to handle certain types of input signals with zero steady-state error.
Steady-State Error: The steady-state error of a control system is the difference between the desired output and the actual output as time approaches infinity. This error is a critical parameter in evaluating the performance of control systems.
Ramp Input: A ramp input is a type of input signal that increases linearly with time. Mathematically, it can be represented as R(t) = At, where A is a constant. When analyzing control systems, it is crucial to understand how the system responds to different types of inputs, including step, ramp, and parabolic inputs.
Type 2 System and Ramp Input: For a feedback control system of type 2, the steady-state error for a ramp input can be determined using the final value theorem and the concept of system type. The system type is determined by the number of integrators in the open-loop transfer function.
Let's denote the open-loop transfer function as G(s)H(s). For a type 2 system:
G(s)H(s) = K / s2 * G'(s)
where K is a gain constant and G'(s) is a polynomial that does not have poles at the origin.
When a ramp input R(t) = At is applied, the Laplace transform of the ramp input is R(s) = A / s2.
The steady-state error for a ramp input can be found using the final value theorem:
e_ss = lim (s → 0) s * E(s)
where E(s) is the Laplace transform of the error signal. The error signal E(s) is given by:
E(s) = R(s) / (1 + G(s)H(s))
Substitute R(s) and G(s)H(s):
E(s) = (A / s2) / (1 + K / s2 * G'(s))
For a type 2 system, the dominant term in the denominator for small values of s is K / s2. Therefore:
E(s) ≈ (A / s2) / (K / s2) = A / K
Using the final value theorem:
e_ss = lim (s → 0) s * (A / K) = 0
Conclusion: The steady-state error for a ramp input in a type 2 feedback control system is zero. Hence, the correct answer is option 2: Zero.
Additional Information
To further understand the analysis, let's evaluate the other options:
Option 1: Infinite
This option suggests that the steady-state error for a ramp input is infinite. This would be true for a type 0 system (a system with no integrators in the open-loop transfer function). However, for a type 2 system, the steady-state error is zero, so this option is incorrect.
Option 3: Unity
This option suggests that the steady-state error for a ramp input is unity. This would be true for a type 1 system (a system with one integrator in the open-loop transfer function). However, for a type 2 system, the steady-state error is zero, so this option is incorrect.
Option 4: More than unity but not infinite
This option suggests that the steady-state error for a ramp input is more than unity but not infinite. This is not accurate for a type 2 system, where the steady-state error is zero for a ramp input. Therefore, this option is incorrect.
Conclusion:
Understanding the relationship between the type of a feedback control system and its steady-state error for different types of inputs is essential for evaluating system performance. For a type 2 system, the steady-state error for a ramp input is zero, making option 2 the correct choice. This analysis helps in designing and tuning control systems to achieve desired performance criteria based on the type of input signals they are expected to handle.
Steady State Error Question 2:
Consider a system shown below -
If the system is disturbed so that C(0) = 1, then C(t) for a unit step input will be -
Answer (Detailed Solution Below)
Steady State Error Question 2 Detailed Solution
Concept
A transfer function is defined as the ratio of the Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.
\(TF = \frac{L[Output]}{L[Input]}\)
\(TF = \frac{{C\left( s \right)}}{{U\left( s \right)}}\)
Calculation
Given, \(U(s) = {1\over s}\)
C(0) = 1
\(\frac{{C\left( s \right)}}{{U\left( s \right)}}={2\over s}\)
\(C(s)={2\over s}\times U(s)\)
\(C(s)={2\over s}\times {1\over s}\)
\(C(s)={2\over s^2}\)
If the initial condition is considered, then the inverse Laplace of C(s) is:
\(C(t)=C(0)+{2t}\)
\(C(t)=1+{2t}\)
Steady State Error Question 3:
State whether the given statements with reference to the non-linear control systems are true or false.
1. Non-linear system can be analysed with the Lyapunov stability criteria.
2. Non-linear system does not fulfil superposition principle.
Answer (Detailed Solution Below)
Steady State Error Question 3 Detailed Solution
The correct answer is "option 2"
Concept:
- Nonlinear control theory covers a wider class of systems that do not obey the superposition principle.
- It applies to more real-world systems, because all real control systems are nonlinear.
- These systems are often governed by nonlinear differential equations.
- Nonlinear systems are the ones that don’t stick to simple, predictable patterns throughout their whole operation.
- Nonlinear Differential Equations: Methods to solve and understand the systems governed by nonlinear differential equations.
- Application:
- Nonlinear Control Strategies:
- Developing control techniques specifically tailored for nonlinear systems, such as sliding mode control, adaptive control, and back stepping control.
- Nonlinear Stability Analysis:
- Evaluating stability in systems that does not adhere to linear stability concepts using Lyapunov stability theory & other nonlinear stability analysis methods.
- Nonlinear Control Strategies:
Steady State Error Question 4:
Steady state error is the value of error signal when:
Answer (Detailed Solution Below)
Steady State Error Question 4 Detailed Solution
Concept:
Steady-state error is defined as the difference between the desired value and the actual value of a system output in the limit as time goes to infinity (i.e. when the response of the control system has reached steady-state).
- Steady-state error is a property of the input/output response for a linear system.
- A good control system will have a low steady-state error.
For example, the above figure is the time response of the First order Transfer Function against step Input. It can be seen that the steady-state error is zero.
IMPORTANT POINTS:
Steady-state error is highest if the input is parabolic, is generally lower for ramp input, and is even lower for a step input.
It should be noted that steady-state error depends on input, while stability does not depend on input.
Steady State Error Question 5:
A closed loop system is shown in the figure where k > 0 and α > 0. The steady state error due to a ramp input (R(s) = α/s2) is given by
Answer (Detailed Solution Below)
Steady State Error Question 5 Detailed Solution
Concept:
The steady-state error for a system is defined as:
\({e_{ss}} = \mathop {\lim }\limits_{s \to 0} \frac{{sR\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}\)
R(s) = Input
G(s) = open loop transfer function
H(s) = feedback gain = 1 for unity feedback system
Calculation:
Given r(t) = α tu(t)
Applying Laplace Transform
R(s) = α /s2
From the figure G(s)H(s) =\(\frac{k}{S(S+2)}\)
Steady-state error for a unit ramp input
\({e_{ss}} = \frac{\alpha}{{{K_V}}}\)
Where
KV = Velocity constant
\({K_V} = \mathop {\lim }\limits_{s \to 0} s\;G\left( s \right)H\left( s \right)\)
\({K_V} = \mathop {\lim }\limits_{s \to 0} s\frac{{k}}{{s(s + 2})}\) =\(\frac{k}{2}\)
Putting the value of KV we get
\({e_{ss}} = \frac{2\alpha}{{{K}}}\)
Hence the correct answer is 1
Top Steady State Error MCQ Objective Questions
The steady-state error due to unit step input to a type-1 system is:
Answer (Detailed Solution Below)
Steady State Error Question 6 Detailed Solution
Download Solution PDFConcept:
KP = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)
Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)
Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)
Steady-state error for different inputs is given by
Input |
Type -0 |
Type - 1 |
Type -2 |
Unit step |
\(\frac{1}{{1 + {K_p}}}\) |
0 |
0 |
Unit ramp |
∞ |
\(\frac{1}{{{K_v}}}\) |
0 |
Unit parabolic |
∞ |
∞ |
\(\frac{1}{{{K_a}}}\) |
From the above table, it is clear that for type – 1 system, a system shows zero steady-state error for step-input.
Consider a unity feedback system with forward transfer function given by
\(G\left( S \right) = \frac{1}{{\left( {s + 1} \right)\left( {s + 2} \right)}}\)
The steady-state error in the output of the system for a unit-step input is _________ (up to 2 decimal places).Answer (Detailed Solution Below) 0.65 - 0.69
Steady State Error Question 7 Detailed Solution
Download Solution PDF\(G\left( s \right) = \frac{1}{{\left( {s + 1} \right)\left( {s + 2} \right)}}\)
Steady-state error for the Unit step input is,
\({E_{ss}} = \frac{A}{{1 + {k_p}}}\)
\({k_p} = \mathop {\lim }\limits_{s \to 0} G(s)~H(s)\)
\( {k_p} = \mathop {\lim }\limits_{s \to 0} \frac{1}{(s+1)(s+2)}\)
= \(\frac{1}{2}\)
\({E_{ss}} = \frac{1}{{1 + \frac{1}{2}}} = \frac{2}{3} = 0.67\)
The D.C. gain and steady state error for step input for \(G(s)=\frac{s+1}{s^2+s+1}\) are:
Answer (Detailed Solution Below)
Steady State Error Question 8 Detailed Solution
Download Solution PDFConcept:
DC gain:
The DC gain is the ratio of the magnitude of the steady-state step response to the magnitude of step input.
DC Gain of a system is the gain at the steady-state which is at t tending to infinity i.e., s tending to zero.
DC gain is nothing but the error coefficients.
For type 0 system: \({K_P} = \mathop {\lim }\limits_{s \to 0} G\left( s \right)\)
For type 1 system: \({K_v} = \mathop {\lim }\limits_{s \to 0} sG\left( s \right)\)
For type 2 system: \({K_a} = \mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)\)
Steady state error for different inputs is given by
Input |
Type -0 |
Type - 1 |
Type -2 |
Unit step |
\(\frac{1}{{1 + {K_p}}}\) |
0 |
0 |
Unit ramp |
∞ |
\(\frac{1}{{{K_v}}}\) |
0 |
Unit parabolic |
∞ |
∞ |
\(\frac{1}{{{K_a}}}\) |
Calculation:
Given:
\(G(s)=\frac{s+1}{s^2+s+1}\)
It is a type 0 system, so:
\({K_P} = \mathop {\lim }\limits_{s \to 0} G\left( s \right)\)
Kp = 1 = DC Gain
Steady-state error for unit step input is given as:
\(e_{ss}=\frac{1}{1+K_p}\)
ess = 0.5
For a feedback control system of type-2, the steady-state error for a ramp input is
Answer (Detailed Solution Below)
Steady State Error Question 9 Detailed Solution
Download Solution PDFConcept:
KP = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)
Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)
Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)
Steady state error for different inputs is given by
Input |
Type -0 |
Type - 1 |
Type -2 |
Unit step |
\(\frac{1}{{1 + {K_p}}}\) |
0 |
0 |
Unit ramp |
∞ |
\(\frac{1}{{{K_v}}}\) |
0 |
Unit parabolic |
∞ |
∞ |
\(\frac{1}{{{K_a}}}\) |
From the above table, it is clear that for type – 2 system, a system shows zero steady-state error for Ramp-input.
The steady state error of a stable ‘type 1’ unity feedback system for a unit step function is equal to:
Answer (Detailed Solution Below)
Steady State Error Question 10 Detailed Solution
Download Solution PDFConcept:
Kp = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)
Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)
Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)
Steady state error for different inputs is given by
Input |
Type -0 |
Type - 1 |
Type -2 |
Unit step |
\(\frac{1}{{1 + {K_p}}}\) |
0 |
0 |
Unit ramp |
∞ |
\(\frac{1}{{{K_v}}}\) |
0 |
Unit parabolic |
∞ |
∞ |
\(\frac{1}{{{K_a}}}\) |
From the above table, it is clear that for type – 1 system, a system shows zero steady-state error for step-input, finite steady-state error for Ramp-input and \(\infty \) steady-state error for parabolic-input.
The steady-state error for a system is 0.1. Steady-state error for the previously mentioned system being closed loop with a unity negative feedback and pulse input for 1 sec having a magnitude of 10 is?
Answer (Detailed Solution Below)
Steady State Error Question 11 Detailed Solution
Download Solution PDFConcept:
The steady-state error for a system is defined as:
\({e_{ss}} = \mathop {\lim }\limits_{s \to 0} \frac{{sR\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}\)
R(s) = Input
G(s) = open loop transfer function
H(s) = feedback gain = 1 for unity feedback system
Analysis:
For a unit step input, we have:
\(R\left( s \right) = \frac{1}{s}\)
The steady-state error becomes:
\({e_{ss}} = \mathop {\lim }\limits_{s \to 0} \frac{{s\left( {\frac{1}{s}} \right)}}{{1 + G\left( s \right)}} = 0.1\)
1 + G(0) = 10
G(0) = 9
Again we have the input as:
r(t) = 10 [u(t) – u(t - 1)]
\(R\left( s \right) = 10\left[ {\frac{1}{s} - \frac{1}{s}{e^{ - s}}} \right]\)
= \(10\left( {\frac{{1 - {e^{ - s}}}}{s}} \right)\)
∴ The steady-state error for the input will be:
\(e_{ss}' = \mathop {\lim }\limits_{s \to 0} \frac{{s \times 10\frac{{\left( {1 - {e^{ - s}}} \right)}}{s}}}{{1 + G\left( s \right)}}\)
= \(\frac{{10\left( {1 - {e^0}} \right)}}{{10}}\)
\(e_{ss}' = 0\)
Important Points
The shifting in input signal does not affect the steady-state error.
A unity feedback system is given by
\(G\left( s \right) = \frac{{10\left( {s + 2} \right)}}{{{s^2}\left( {s + 5} \right)}}\)
For input, r(t) = 1 + 2t, t > 0 the steady state error e(t) is:Answer (Detailed Solution Below)
Steady State Error Question 12 Detailed Solution
Download Solution PDFConcept:
KP = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)
Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)
Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)
Steady state error for different inputs is given by
Input |
Type -0 |
Type - 1 |
Type -2 |
Unit step |
\(\frac{1}{{1 + {K_p}}}\) |
0 |
0 |
Unit ramp |
∞ |
\(\frac{1}{{{K_v}}}\) |
0 |
Unit parabolic |
∞ |
∞ |
\(\frac{1}{{{K_a}}}\) |
From the above table, it is clear that for type – 1 system, a system shows zero steady-state error for step-input, finite steady-state error for Ramp-input ∞ , and steady-state error for parabolic-input.
Application:
Given
\(G\left( s \right) = \frac{{10\left( {s + 2} \right)}}{{{s^2}\left( {s + 5} \right)}}\)
r(t) = u(t) + 2t u(t)
r(t) = r1(t) + r2 (t)
It is a type 2 system. Therefore, the steady-state error for both step and ramp input is zero.
Note:
If the type of the system increases, the error decreases, and accuracy increases but the stability decreases.
With reference to the error analysis of the control systems, the term 'acceleration error constant' stands for:
Answer (Detailed Solution Below)
Steady State Error Question 13 Detailed Solution
Download Solution PDFAcceleration error constant is a measure of the steady-state error of the system when the input is parabolic function.
Positional error constant is a measure of the steady-state error of the system when the input is unit step function.
Velocity error constant is a measure of the steady-state error of the system when the input is ramp function.
Key Points
Type of input |
Input: r(t) |
ess |
Error constant |
Step |
A.u(t) |
\(\frac{A}{{1 + {k_p}}}\) |
Position error constant \({k_p} = \mathop {\lim }\limits_{s \to 0} G\left( s \right)\)
|
Ramp |
A.t.u(t) |
\(\frac{A}{{{k_v}}}\) |
Velocity error constant \({k_v} = \mathop {\lim }\limits_{s \to 0} {s}G\left( s \right)\)
|
Parabolic |
A.u(t).t2/2 |
\(\frac{A}{{{k_a}}}\) |
Acceleration error constant \({k_a} = \mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)\) |
In the given, figure, plant \({G_p}(s) = \frac{{2.2}}{{(1 + 0.1s)(1 + 0.4s)(1 + 1.2s)}}\) and compensator \({G_c}(s) = K\left( {\frac{{1 + {T_1}s}}{{1 + {T_2}s}}} \right)\). The external disturbance input is D(s). It is desired that when the disturbance is a unit step, the steady-state error should not exceed 0.1 unit. The minimum value of K is ______. (Round off to 2 decimal places.)
Answer (Detailed Solution Below) 9.5 - 9.6
Steady State Error Question 14 Detailed Solution
Download Solution PDFGiven,
\({G_p}(s) = \frac{{2.2}}{{(1 + 0.1s)(1 + 0.4s)(1 + 1.2s)}}\) ...(1)
\({G_c}(s) = K\left( {\frac{{1 + {T_1}s}}{{1 + {T_2}s}}} \right)\) ....(2)
Maximum steady state error due to external input = 0.1
For step external disturbance input D(s) = 1/s .....(3)
Assume R(s) = 0,
⇒ C(s) = - E(s)
Then the output expression is given as
C(s) = [E(s)Gc(s) + D(s)]Gp(s)
- E(s) = E(s)Gc(s)Gp(s) + D(s)Gp(s)
Error function \(E(s)=\frac{{-D(s)G_p(s)}}{{1+G_c(s)G_p(s)}}\) ....(4)
Steady state error is given by
\(e_{ss}=\mathop {\lim }\limits_{s \to 0} s.E(s)\)
From equations (1), (2), (3)&(4)
\(|e_{ss}|=\mathop {\lim }\limits_{s \to 0} {\frac {{s.\frac {{1}}{{s}}G_p(s)}}{{1+G_c(s)G_p(s)}}}\)
\(e_{ss}≥\frac{{2.2}}{{1+2.2K}}\)
\(\frac{{2.2}}{{1+2.2K}}\le0.1\)
K ≥ 9.54
Minimum value of K is 9.54
The steady state error for a ramp input for a unity feedback system with open loop transfer function k/s(s + 1) is:
(Given k = 2)Answer (Detailed Solution Below)
Steady State Error Question 15 Detailed Solution
Download Solution PDFConcept:
KP = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)
Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)
Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)
Steady state error for different inputs is given by
Input |
Type - 0 |
Type - 1 |
Type - 2 |
Unit step |
\(\frac{1}{{1 + {K_p}}}\) |
0 |
0 |
Unit ramp |
∞ |
\(\frac{1}{{{K_v}}}\) |
0 |
Unit parabolic |
∞ |
∞ |
\(\frac{1}{{{K_a}}}\) |
From the above table, it is clear that for type – 1 system, a system shows zero steady-state error for step-input, finite steady-state error for Ramp-input, and ∞ steady-state error for parabolic-input.
Calculation:
\(G\left( s \right) = \frac{k}{{s\left( {s + 1} \right)}} = \frac{2}{{s\left( {s + 1} \right)}}\)
Velocity error coefficient will be:
\({K_v} = \mathop {lim}\limits_{s \to 0} s\frac{2}{{s\left( {s + 1} \right)}} = 2\)
Steady-state error, ess = 1/2 = 0.5