Steady State Error MCQ Quiz in मराठी - Objective Question with Answer for Steady State Error - मोफत PDF डाउनलोड करा
Last updated on Apr 16, 2025
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Steady State Error Question 1:
Consider the system shown below.The sensitivity of the steady state error to parameter α for a step input is
Answer (Detailed Solution Below)
Steady State Error Question 1 Detailed Solution
The Steady-state error is defined for the unity -ve feedback system only.
Thus for finding variation in ess due to variation in α we need to find first equivalent OLTF of unity -ve feedback system and then find ess.
Convert the above system into unity -ve feedback system then the open-loop gain of the new system is:
Where:
The new open-loop transfer function G(s) will be:
The given input is step input and position error constant is given by:
Sensitivity will now be:
Steady State Error Question 2:
The unit-step response of a unity feedback system with open loop transfer function G(s) = K/((s + 1) (s + 2)) is shown in the figure. The value of K2 + 2K +2 is ________
Answer (Detailed Solution Below)
Steady State Error Question 2 Detailed Solution
Concept:
Steady-state error:
Steady-state error is the deviation of the output of the control system from the desired response during the steady-state. It is denoted by ess and steady-state error usually find with the help of the final value theorem. Let's consider a unity feedback system as shown in the below figure,
E(s) = error signal
R(s) = reference input
C(s) = actual output
G(s) = forward path gain
E(s) = R(s) - C(s) = R(s) / (1 + G(s))
Steady-state error
Where e(t) = L-1 [E(s)]
Steady state error for step input:
For step input R(s) = A/s
Where
Application:
The open-loop transfer function of the system is given as:
From the response plot of the system, we have the steady-state value = 0.75
So, the steady-state error for unit step input will be:
ess = 1 – 0.75 = 0.25
Now, the steady-state error is defined as:
The given input is a unit step input, i.e.
From equations (1) and (2), we have:
⇒ 2 + K = 8
⇒ K = 6
⇒ K2 + 2K + 2 = 62 + 2 × 6 + 2 = 50
Steady State Error Question 3:
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 3 Detailed Solution
Concept:
KP = position error constant =
Kv = velocity error constant =
Ka = acceleration error constant =
Steady state error for different inputs is given by
Input |
Type - 0 |
Type - 1 |
Type - 2 |
Unit step |
|
0 |
0 |
Unit ramp |
∞ |
|
0 |
Unit parabolic |
∞ |
∞ |
|
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:
Velocity error coefficient will be:
Steady-state error, ess = 1/2 = 0.5
Steady State Error Question 4:
The closed loop transfer function of a system is
Answer (Detailed Solution Below) 0
Steady State Error Question 4 Detailed Solution
Concept:
KP = position error constant =
Kv = velocity error constant =
Ka = acceleration error constant =
Steady state error for different inputs is given by
Input |
Type -0 |
Type - 1 |
Type -2 |
Unit step |
|
0 |
0 |
Unit ramp |
∞ |
|
0 |
Unit parabolic |
∞ |
∞ |
|
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
Calculation:
The given closed-loop transfer function is
To find a steady state error, we require an open loop transfer function.
The open-loop transfer function
Error constant
Steady-state error
Steady State Error Question 5:
The steady-state error due to the ramp input for a type two system is equal to-
Answer (Detailed Solution Below)
Steady State Error Question 5 Detailed Solution
Concept:
KP (position error constant) is given as:
Kv (velocity error constant) is given as:
Ka (acceleration error constant) is given as:
Steady-state error for different inputs is given by
Input |
Type -0 |
Type - 1 |
Type -2 |
Unit step |
|
0 |
0 |
Unit ramp |
∞ |
|
0 |
Unit parabolic |
∞ |
∞ |
|
Observation:
From the above table, it is clear that for the type-2 system, a system shows zero steady-state error for step-input and ramp-input; and finite steady-state error for a parabolic input.
Steady State Error Question 6:
A unity feedback system has a forward path transfer function G(s) =
Answer (Detailed Solution Below)
Steady State Error Question 6 Detailed Solution
Concept:
The steady-state error for different inputs is given by:
Ka = position error constant =
Kv = velocity error constant =
Ka = acceleration error constant =
Steady-state error for different inputs is given by
Input |
Type -0 |
Type - 1 |
Type -2 |
Unit step |
|
0 |
0 |
Unit ramp |
∞ |
|
0 |
Unit parabolic |
∞ |
∞ |
|
Calculation:
Given,
Type = 2
Error =
Ka = 10
Error =
Error = 0.1
Steady State Error Question 7:
The steady state error of a unity feedback linear system for a unit step input is 0.1. The steady state error of the same system, for a pulse input r(t) having magnitude of 10 and a duration of one second as shown in the figure is
Answer (Detailed Solution Below)
Steady State Error Question 7 Detailed Solution
Concept:
The steady-state error for a system is defined as:
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:
The steady-state error becomes:
1 + G(0) = 10
G(0) = 9
Again we have the input as:
r(t) = 10 [u(t) – u(t - 1)]
∴ The steady-state error for the input will be:
Steady State Error Question 8:
When a unit ramp input is applied to the unity feedback system having closed loop transfer function
Answer (Detailed Solution Below)
Steady State Error Question 8 Detailed Solution
Closed loop transfer function is,
Open loop transfer function will be,
Steady state error for ramp input is given by
Velocity error coefficient,
Steady State Error Question 9:
A unity feedback system has G(s) =
Answer (Detailed Solution Below)
Steady State Error Question 9 Detailed Solution
Concept:
Steady-state error:
Type of the system |
Step input |
Ramp input |
Parabolic input |
0 |
A/(1+Kp) |
∞ |
∞ |
1 |
0 |
A/Kv |
∞ |
2 |
0 |
0 |
A/Ka |
3 |
0 |
0 |
0 |
Where,
Kp is Static position error constant
Kp
Kv is Static velocity error constant
Kv
Ka is static acceleration error
Ka
G(s) is a forward gain of a control system.
A is the magnitude of the input function.
Calculation:
Given G(s)
Kv
Kv = 20
ess = A/KV
ess = 5/20
ess = 0.25
The steady-state error of the system with a Ramp input of magnitude '5' is 0.25
Tips and tricks:
First of all, check the type of system given.
- For the type-0 system, a steady-state exists only for a step input. For Ramp and Parabolic error will be ∞
- For the type-1 system, a steady-state exists only for a ramp input. For step and parabolic error will be 0 and ∞ respectively
- For the type-2 system, a steady-state exists only for a parabolic input. For step and ramp, input error will be 0
Steady State Error Question 10:
Which one of the following coefficient is associated with Unit Ramp function?
Answer (Detailed Solution Below)
Steady State Error Question 10 Detailed Solution
Explanation:
Type
It is defined as the number of poles at the origin in the open-loop transfer function.
Error: It is the deviation of output from the input.
- If the actual output of an open-loop system during the steady-state deviates from the reference input then the system is said to have a steady-state error.
- It is the index of accuracy of the control system.
- It is also called a ‘static error’.
Steady-state error is defined as:
Error constant and ess
The below table gives all information about the ess and error constant.
Type of input |
Input: r(t) |
ess |
Error constant |
Step |
A.u(t) |
|
Position error constant
|
Ramp |
A.t.u(t) |
|
Velocity error constant
|
Parabolic |
A.u(t).t2/2 |
|
Acceleration error constant |
Hence Static velocity error coefficient is associated with the Unit Ramp function.