Which of the following is not a Hurwitz polynomial?
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 has
A Hurwitz polynomial must have no poles and zeros in the RH s-plane.
The function represent a
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.
The circuit shown in the figure is a
For s → 0, L acts as short circuit and C acts as open circuit.
Thus, V0 = Vi
For s → ∞, L acts as
open circuit and C acts as short circuit.
∴ V0 = 0
Hence, given circuit will act as a LPF.
The driving point admittance of the network shown below is
Converting the given network in Laplace domain, the driving point impedance is
∴ Driving point admittance V(s)
Which of the following is a PRF (positive real function)?
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.
A function F(s) is defined as
The necessary condition required for F(s) to be a positive real function is
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)
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.
Consider the impedance functions:
Out of the above impedance function which are driving point LC immittances of LC network?
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.
Driving point impedance is not realizable because the
Cauer and Foster forms of realizations are used only for
The driving point impedance function
can be realized as
The continued fraction expansion is shown below
Hence, Z(s) can be realized as a L-C network.
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
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.
Foster’s type - I and type - II networks are not equivalent networks.
Hence, assertion is false.
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.