Test: Time Domain Analysis of Control Systems- 2
20 Questions MCQ Test GATE Electrical Engineering (EE) 2023 Mock Test Series | Test: Time Domain Analysis of Control Systems- 2
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The system shown in figure below has unit step input.
In order that the steady state error is 0.1, the value of K required should be
In order that the steady state error is 0.1, the value of K required should be
Since Input is unit step, therefore steady state error is
Here,
The steady-state error coefficients for a system are Kp = ∞ , Kv = finite constant and Ka = 0. The type of the system is
We have,
= finite constant
From above values of Kp, Kv and Ka we conclude that the type of the system should be one.
The steady-state error of a feedback control system with an acceleration input becomes finite in a
Steady state error with an acceleration input having an amplitude of A is given by
where,
Hence, if the type of the system = 2, then Ka = some non-zero value or finite value due to which we will get some finite vaiue of Ka.
Consider the unity feedback system shown below:
The settling time of the resulting second order system for 2% tolerance band will be
The characteristic equation of the given closed loop system is
Comparing above equation with
Thus, setting time for 2% tolerance band is
or, 
For a control system, the Laplace transform of error signal e(t) is given by 
. The steady state value of the error will be
Given,
(Using final value theorem)
or,
For a type one system, the steady-state error due to step input is equal to
where,
(Here, A = magnitude of step input)
Since system is type -1, therefore Kp = ∞
For an unity feedback control system with 
the value of K for damping ratio of 0.5 is
Characteristic equation is
or, K = 64
The damping ratio of a system having the characteristic equation s2 + 2s + 8 = 0 is
Given, s2 + 2s + 8 = 0
Here, ωn = √8 rad/s = 2√2 rad/sec
system is underdamped.
The closed-loop transfer function of a unity - feedback system is given by 
The steady state error to a unit ramp input is:
Given,
(Since H(s) = 1)
For a unit ramp input,
The peak overshoot of step-input response of an underdamped second-order system is an indication of
If damping of system increases, peak overshoot Mp decreases and vice-versa.
Consider the position control system shown below:
The value of K such that the steady state error is 20° for input θr = 300t rad/sec, is
Input = 300t rad/sec = ramp input of magnitude A = 300.
For ramp input, steady state error is
where,
Now, from given block diagram, we have:
So, value of gain is K = 42.97
Assertion (A): With the increase in bandwidth of the system the response of the system becomes fast.
Reason (R): Damping ratio of the system decreases with the increase in bandwidth.
When BW is increased, the system response becomes fast due to fait in rise time (tr).
With the increase in bandwidth of the system, damping ratio (ξ) decreases.
Thus, both assertion and reason are true but, reason is not the correct explanation of assertion.
For the stable system described by the block diagram shown below, Match List - I (Static error coefficients) with List - II (Values) and select the correct answer using the codes given below the lists:
List-I
A. Ka
B. Kv
C. Kp
List-II
1. ∞
2. 0
3. 2
4. -1
Codes:
We have,
Also,
and
The unit step response of a system is given by c(t) - 1 + 0.25 e-50t - 1.25 e-10t
The given system is
Given, 
Since, ξ > 1, therefore given system is overdamped.
Match List - I (Transfer function of systems) with List - II (Nature of damping) and select the correct answer using the codes given below the lists:
Codes:
Characteristic equations are:
• s2 + 8s + 12 = 0
• s2 + 8s + 16 = 0
Here, ωn = 4 and 2 ξωn = 8
or, ξ = 1 (critically damped)
• s2 + 8s + 20 = 0
Here, ωn = √20
and 2ξωn = 8
or, ξ = 0.894 (ξ < 1 ∴ underdamped)
• s2 + 4 = 0
or, ωn = 2
and ξ = 0 (∴ undamped)
The block diagram of an electronic pacemaker is given in figure below.
What is the value of K for which the steady-state error to a unit ramp input is 0.02?
Here,
As the input is a unit ramp function, therefore
Which one of the following equations gives the steady-state error for a unity feedback system excited by r(t) = 2 + 5t + 2t2 ?
Given, r(t) = 2 + 5t + 2t2
An unity feedback control system with closed loon transfer function is given by
The steady state error due a unit ramp input response is
The second order approximation using dominant pole concept for the transfer function
Given,
In time constant form,
Using dominant pole concept, the given transfer function can be approximated to
The unit step response of the system represented by the block diagram shown below is
From given block diagram, we have
∴ c(t) = (4e-t - 3e-2t - 1)
= Required step response
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