Test: Network Synthesis- 2


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

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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.

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.

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