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

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).

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

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.

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

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

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



∴ Driving point admittance V(s)

 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  

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.

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.

The continued fraction expansion is shown below

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

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