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QUESTION: 1

Which of the following is not a Hurwitz polynomial?

Solution:

For option (a):

F_{1}(s) - (s + 1) (s^{2} + 2s + 3)

= s^{3} + 3s^{2} + 5s + 3

Since all coefficients of F_{1}(s) are positive and no term is missing, therefore it is a Hurwitz polynomial.

Now, F_{2}(s) - (s + 3) (s^{2} + s - 2)

= s^{3 }+ 4s^{2} + s - 6

Since F_{2}(s) has a negative coefficient, therefore

it is not a Hurwitz polynomial

Similarly, we can check for options (c) and (d).

QUESTION: 2

A Hurwitz polynomial has

Solution:

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

QUESTION: 3

The function represent a

Solution:

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.

QUESTION: 4

The circuit shown in the figure is a

Solution:

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

Thus, V_{0} = V_{i }

For s → ∞, L acts as

open circuit and C acts as short circuit.

∴ V_{0 }= 0

Hence, given circuit will act as a LPF.

QUESTION: 5

The driving point admittance of the network shown below is

Solution:

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

∴ Driving point admittance V(s)

QUESTION: 6

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

Solution:

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

QUESTION: 7

A function F(s) is defined as

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

Solution:

QUESTION: 8

What is the range of values of m in P(s), so that P(s) is a Hurwitz polynomial?

P(s) = (2s^{4} + s^{3} + ms^{2} + s + 2)

Solution:

Given P(s) - 2s^{4} + s^{3} + ms^{2} + s + 2

= (2s^{4} + ms^{2} + 2) + (s^{3} + 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.

QUESTION: 9

Consider the impedance functions:

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

Solution:

The pole-zero plots of F_{1}(s) and F_{2}(s) are shown below.

In Z_{1}(s): Poles and zeros do not alternate. Also, there is no zero-or a pole at origin.

In Z_{2}(s): Poles and zeros do not alternate. Hence, neither of Z_{1}(s) or Z_{2}(s) represents a LC immittance function.

QUESTION: 10

Driving point impedance is not realizable because the

Solution:

QUESTION: 11

Cauer and Foster forms of realizations are used only for

Solution:

QUESTION: 12

The driving point impedance function

can be realized as

Solution:

The continued fraction expansion is shown below

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

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:

Solution:

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.

Solution:

Foster’s type - I and type - II networks are not equivalent networks.

Hence, assertion is false.

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.

Solution:

### Synthesis of 1 port-network

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