Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Topicwise Question Bank for Electrical Engineering  >  Test: Network Synthesis- 2 - Electrical Engineering (EE) MCQ

Test: Network Synthesis- 2 - Electrical Engineering (EE) MCQ


Test Description

15 Questions MCQ Test Topicwise Question Bank for Electrical Engineering - Test: Network Synthesis- 2

Test: Network Synthesis- 2 for Electrical Engineering (EE) 2024 is part of Topicwise Question Bank for Electrical Engineering preparation. The Test: Network Synthesis- 2 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Network Synthesis- 2 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Network Synthesis- 2 below.
Solutions of Test: Network Synthesis- 2 questions in English are available as part of our Topicwise Question Bank for Electrical Engineering for Electrical Engineering (EE) & Test: Network Synthesis- 2 solutions in Hindi for Topicwise Question Bank for Electrical Engineering course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Network Synthesis- 2 | 15 questions in 45 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Topicwise Question Bank for Electrical Engineering for Electrical Engineering (EE) Exam | Download free PDF with solutions
Test: Network Synthesis- 2 - Question 1

Which of the following is not a Hurwitz polynomial?​

Detailed Solution for Test: Network Synthesis- 2 - Question 1

For option (a):
F1(s) - (s + 1) (s2 + 2s + 3)
= s3 + 3s2 + 5s + 3
Since all coefficients of F1(s) are positive and no term is missing, therefore it is a Hurwitz polynomial.
Now, F2(s) - (s + 3) (s2 + s - 2)
= s3 + 4s2 + s - 6
Since F2(s) has a negative coefficient, therefore
it is not a Hurwitz polynomial    
Similarly, we can check for options (c) and (d).

Test: Network Synthesis- 2 - Question 2

A Hurwitz polynomial has

Detailed Solution for Test: Network Synthesis- 2 - Question 2

A Hurwitz polynomial must have no poles and zeros in the RH s-plane.

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Network Synthesis- 2 - Question 3

The function   represent a 

Detailed Solution for Test: Network Synthesis- 2 - Question 3

The pole-zero plot of Z(s) is shown below.

Here, singularity nearest to origin is a zero while near infinity is a pole. Also, pole-zero alternates on negative real axis.
Therefore, Z(s) will represent a R-L impedance function.

Test: Network Synthesis- 2 - Question 4

The circuit shown in the figure is a

Detailed Solution for Test: Network Synthesis- 2 - Question 4

For s → 0, L acts as short circuit and C acts as open circuit.

Thus, V0 = V
For s → ∞, L acts as
open circuit and C acts as short circuit.
∴ V= 0
Hence, given circuit will act as a LPF.

Test: Network Synthesis- 2 - Question 5

The driving point admittance of the network shown below is

Detailed Solution for Test: Network Synthesis- 2 - Question 5

Converting the given network in Laplace domain, the driving point impedance is



∴ Driving point admittance V(s)

Test: Network Synthesis- 2 - Question 6

 Which of the following is a PRF (positive real function)?

Detailed Solution for Test: Network Synthesis- 2 - Question 6

 is not a PRF since all the coefficients of numerator polynomial are not positive and there is a zero in RH s-plane.
  is not a PRF due to the same reason (zeros are at s = ± 1).
  is a PRF since pole is at s = and zero at s = - 2 .
Also, all coefficients of N' and Dl are positive.
And and  

Test: Network Synthesis- 2 - Question 7

A function F(s) is defined as 

The necessary condition required for F(s) to be a positive real function is

Test: Network Synthesis- 2 - Question 8

What is the range of values of m in P(s), so that P(s) is a Hurwitz polynomial?
P(s) = (2s4 + s3 + ms2 + s + 2)

Detailed Solution for Test: Network Synthesis- 2 - Question 8

Given P(s) - 2s4 + s3 + ms2 + s + 2
= (2s4 + ms2 + 2) + (s3 + s) = M(s) + N(s)
Continued fraction expansion of M(s)/N(s) is obtained as follows:

The quotients of the continued fraction expansion would be positive only when m > 2 and 

Hence for m > 4, P(s) will be Hurwitz.

Test: Network Synthesis- 2 - Question 9

Consider the impedance functions:

Out of the above impedance function which are driving point LC immittances of LC network? 

Detailed Solution for Test: Network Synthesis- 2 - Question 9

The pole-zero plots of F1(s) and F2(s) are shown below.

In Z1(s): Poles and zeros do not alternate. Also, there is no zero-or a pole at origin.
In Z2(s): Poles and zeros do not alternate. Hence, neither of Z1(s) or Z2(s) represents a LC immittance function.

Test: Network Synthesis- 2 - Question 10

Driving point impedance is not realizable because the

Test: Network Synthesis- 2 - Question 11

Cauer and Foster forms of realizations are used only for

Test: Network Synthesis- 2 - Question 12

The driving point impedance function

can be realized as

Detailed Solution for Test: Network Synthesis- 2 - Question 12

The continued fraction expansion is shown below

Hence, Z(s) can be realized as a L-C network.

Test: Network Synthesis- 2 - Question 13

Match List - I (Network functions) with List - II (Type of functions) and select the correct answer using the codes given below the lists:
List - I

List - II
1. LC immittance function
2. RC admittance function
3. RL admittance function
4. RLC network function
Codes:

Test: Network Synthesis- 2 - Question 14

Assertion (A): Foster’s type - I and Foster’s type - II networks are equivalent networks.
Reason (R): Foster’s type - I and Foster’s type - II networks are dual of each other.

Detailed Solution for Test: Network Synthesis- 2 - Question 14

Foster’s type - I and type - II networks are not equivalent networks.
Hence, assertion is false.

Test: Network Synthesis- 2 - Question 15

Assertion (A): The realization of one port LC networks can be done in two. configurations (commonly known as Cauer-I and Cauer-II forms) using the continued fraction expansion of the driving point impedance.
Reason (R): The basic form of the Cauer realization being a ladder type network.

212 tests
Information about Test: Network Synthesis- 2 Page
In this test you can find the Exam questions for Test: Network Synthesis- 2 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Network Synthesis- 2, EduRev gives you an ample number of Online tests for practice

Top Courses for Electrical Engineering (EE)

Download as PDF

Top Courses for Electrical Engineering (EE)