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Test: Stability - Electrical Engineering (EE) MCQ


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15 Questions MCQ Test - Test: Stability

Test: Stability for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: Stability questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Stability MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Stability below.
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Test: Stability - Question 1

The forward-path transfer function of a ufb system is

 

For system to be stable, the range of K is

Detailed Solution for Test: Stability - Question 1



Routh table is as shown in fig. S.6.211


Test: Stability - Question 2

Which among these is a classification of power system stability?

Test: Stability - Question 3

The open-loop transfer function of a ufb system is


The closed loop system will be stable if the value of K is

Detailed Solution for Test: Stability - Question 3

Routh table is as shown in fig.


200K > 0 → K > 0, 30K2 - 140K > 0
satisfy this condition.

Test: Stability - Question 4

The closed loop transfer function for this system is

Detailed Solution for Test: Stability - Question 4

First combine the parallel loop K/s2
and 2/s giving
 Then apply feedback formula with  and and then multiply with s2.

Test: Stability - Question 5

The poles location for this system is shown in fig.The value of K is

Detailed Solution for Test: Stability - Question 5

Denominator = s3 + s2 + 2s + K Routh table is as shown in fig. 

Row of zeros when K = 2,
s2 + 2 = 0, ⇒ s = -1, j√2, - j√2

Test: Stability - Question 6

The forward path transfer of ufb system is

The system is

Test: Stability - Question 7

The forward-path transfer function of a ufb system is T(s) =  

The system is

Detailed Solution for Test: Stability - Question 7

Closed loop transfer function


Routh table is as shown in fig. S.6.2.28

2 RHP poles so unstable.

Test: Stability - Question 8

The open loop transfer function of a system is as

The range of K for stable system will be

Detailed Solution for Test: Stability - Question 8

The characteristic equation is 1 + G(s)H(s) = 0

⇒ s(s - 0.2)(s2 + s + 0.6)+K(s + 0.1) = 0
s4 +0.8 s3 +0.4s2 +(K - 0.12)s +0.1K = 0
Routh table is as shown in fig. S.62.29


K > 0, 055 -125K > 0 ⇒ K < 0.44 -125K2 +0.63K -0066 >0
(K - 0.149)(K - 0355) < 0, 0.149 < K < 0.355 

Test: Stability - Question 9

The open-loop transfer function of a ufb control system is given by

For the system to be stable the range of K is

Detailed Solution for Test: Stability - Question 9

Characteristic equation

s(sT1 + 1)(sT2 +1) + K = 0
T1T2s3 + (T1 + T2)s2 + s + K = 0
Routh table is as shown in fig S.6.2.30 


Test: Stability - Question 10

If the roots of the have negative real parts, then the response is ____________ 

Detailed Solution for Test: Stability - Question 10

If the roots of the have negative real parts then the response is bounded and eventually decreases to zero.

Test: Stability - Question 11

The closed loop transfer function of a system is

The number of poles in RHP and in LHP are

Detailed Solution for Test: Stability - Question 11

3 RHP, 2 LHP poles.

Test: Stability - Question 12

The closed loop transfer function of a system is

The number of poles in LHP, in RHP, and on jω - axis are

Detailed Solution for Test: Stability - Question 12


No sign change exist from the s4 row down to the s0 row.
Thus, the even polynomial does not have RHP poles. Therefore because of symmetry all four poles must be on jw -axis.

Test: Stability - Question 13

For the system shown in fig. the number of poles on RHP, LHP, and imaginary axis areS

Detailed Solution for Test: Stability - Question 13

 Closed loop transfer function


Routh table is as shown in fig. S.6.2.34

From s4 row down to s0 there is one sign change. So LHP–1 + 1= 2 pole. RHP–1 pole, jw - axis - 2 pole.

Test: Stability - Question 14

If a system is given unbounded input then the system is:

Detailed Solution for Test: Stability - Question 14

If the system is given with the unbounded input then nothing can be clarified for the stability of the system.

Test: Stability - Question 15

For the open loop system of fig. location of poles on RHP, LHP, and an jω - axis are

Detailed Solution for Test: Stability - Question 15

Routh table is as shown in fig

Them is two sign change from the s4 mw down to the s° row. So two roots are on RHS. Because of symmetry rest two roots must be in LHP. From s6 to s4 there is 1 sign change so 1 on RHP and 1 on LHP.
Total LHP 3 root, RHP 3 root. 

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