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The correct answer is (c) all even or ODD terms are missing

Best explanation: The denominator polynomial in a transfer function may not have any missing terms between the highest and the LOWEST DEGREE, UNLESS all even or odd terms are missing and the polynomial P(s) may have missing terms between the lowest and the highest degree.

The correct choice is (c) 0

To EXPLAIN: The DEGREE of NUMERATOR polynomial in a transfer function may be as small as zero, independent of the degree of the denominator polynomial and for the voltage transfer ratio and the current transfer ratio, the maximum degree of P(s) MUST be equal to the degree of Q(s).

Correct OPTION is (d) may be NEGATIVE

To explain I WOULD say: The coefficients in the numerator polynomial of the TRANSFER function may be negative and the COMPLEX or imaginary poles and zeros must occur in conjugate pairs.

Correct choice is (a) positive

For EXPLANATION I would say: The coefficients in the DENOMINATOR polynomial of the transfer FUNCTION must be positive but should not be negative and the coefficients in the polynomials P(s) and Q(s) of transfer function must be real.

Right ANSWER is (b) 1

The best explanation: The lowest degree in numerator polynomial and denominator polynomial in a driving POINT function may DIFFER by at most one and the coefficients in the polynomials P(s) and Q(s) of NETWORK function must be real and positive.

Correct choice is (C) 0 or 1

Best explanation: The degree of numerator polynomial and denominator polynomial in a driving POINT function may differ by ZERO or one. And the polynomials P(s) and Q(s) may not have any missing terms between the HIGHEST and lowest degrees unless all even or odd terms are missing.

Right choice is (c) zero or negative

Easiest explanation: The real parts of all poles and ZEROS in a DRIVING point function must be zero or negative but should not be positive and the COMPLEX or imaginary poles and zeros must occur in conjugate pairs.

Correct CHOICE is (d) less than or equal to

For explanation I WOULD say: In a transfer FUNCTION, the degree of numerator polynomial is less than or equal to than the degree of the denominator polynomial. And the degree of the numerator polynomial of Z21(s) or Y21(s) is less than or equal to the degree of the denominator polynomial PLUS one.

The correct OPTION is (a) real

The explanation is: The coefficients of P(s), the NUMERATOR polynomial and of Q(s), the denominator polynomial in a transfer function must be real. THEREFORE all POLES and zeros if complex must occur in CONJUGATE pairs.

Right ANSWER is (d) simple

Explanation: POLES or ZEROS lying on the jω axis must be simple because on jω axis the imaginary part of poles or zeros will be zero.

Correct option is (B) NEGATIVE or zero

Easy explanation: The REAL part of all zeros and poles must be negative or zero. But the poles or zeros should not be positive because if they are positive, then they will lie in the right-half of the s-plane.

Correct answer is (a) zero

Easiest explanation: The driving POINT admittance function Y(s) = I(s)/V(s). In the driving point admittance function, a pole of Y (s) MEANS a zero of V (S) i.e., the short CIRCUIT condition.

Correct ANSWER is (c) OPEN circuit

To elaborate: A zero of N(s) is a zero of V(s), it signifies a short circuit. SIMILARLY, a pole of Z(s) is a zero of I(s). The poles of driving point impedance are those frequencies corresponding to open circuit conditions.

The CORRECT ANSWER is (d) zero

Easy explanation: In the driving point admittance FUNCTION, a zero of Y (s) means a zero of I (S) i.e., the open circuit CONDITION as the driving point admittance function is the ratio of I(s) to V(s).

Right ANSWER is (d) SHORT CIRCUIT

Explanation: The zeros of driving point impedance are those FREQUENCIES corresponding to short circuit CONDITIONS as a pole of Z(s) is a zero of I(s) and zero of N(s) is a zero of V(s), it signifies a short circuit.

The correct option is (C) 0

To ELABORATE: A function N (S) is said to have a pole (or ZERO) at INFINITY, if the function N (1/S) has a pole (or zero) at S = infinity. A zero or pole is said to be of multiplicity ‘r’ if (S-Z)^r or (S-P)^r is a factor of P(s) or Q(s).

The CORRECT CHOICE is (b) equal to

For explanation I WOULD say: The number of zeros including zeros at infinity is equal to the number of poles including poles at infinity and it cannot be GREATER than or less than the number of poles including poles at infinity.

Correct ANSWER is (c) transform IMPEDANCE

Explanation: The driving POINT function is the RATIO of polynomials in s. Polynomials are obtained from the transform impedance of the elements and their COMBINATIONS and if the zeros and poles are not repeated then the poles or zeros are said to be distinct or simple.

The correct choice is (d) ∞

The EXPLANATION: The quantities P1, P2 … Pm are called POLES of N (S) if N (S) = ∞ at those points. The pole is that FINITE value of S for which N (S) becomes infinity.

Correct choice is (c) n-m

For explanation I would say: If the number of ZEROS (n) are greater than the number of poles (m), then there will be (n-m) number of zeros at s = ∞ and to OBTAIN (n-m) zeros at s = ∞ the condition is n>m.

Right option is (a) multiple, multiple

The best explanation: If there are repeated poles or zeros, then FUNCTION is said to be having multiple poles or multiple zeros and the network function is stable if the poles and zeros LIE within the LEFT half of the s-plane.

The correct answer is (c) SIMPLE, simple

Explanation: If the poles or zeros are not repeated, then the FUNCTION is said to be having simple poles or simple zeros and the NETWORK function is said to be stable when the real PARTS of the poles and zeros are NEGATIVE.

The correct option is (a) ∞

The explanation: The NETWORK function is COMPLETELY defined by its poles and ZEROS and the network function N (S) BECOMES infinite when s in the transfer function is equal to anyone of the poles.

The correct answer is (C) 0

The explanation is: The network function N (S) BECOMES zero when s in the TRANSFER function is equal to anyone of the zeros as the network function is completely defined by its poles and zeros.

The CORRECT option is (a) x

Best EXPLANATION: The roots of the equation Q (S) = 0 are poles of the transfer FUNCTION. The poles in the transfer function are denoted by ‘x’.

The correct answer is (B) H

The explanation: The scale factor is DENOTED by the LETTER ‘H’ and its VALUE is equal to the RATIO of ao to bo.

The CORRECT choice is (a) real and positive

To explain: The COEFFICIENTS of the polynomials P (S) and Q (S) in the NETWORK function N (S) are real and positive for passive network. On factorising the network function we obtain the poles and zeros.

Correct option is (c) 2(s+1)

Best EXPLANATION: The driving point IMPEDANCE Z11 (S) is Z11 (S)=V1(S)/I1(S). V1 (S) = 2 I1 (S) + 2 sI1 (S) => V1(S) = (2+2s)I1(S) => V1(S)/I1(S) = 2(s+1). On substituting Z11 (S) = 2(S+1).

Correct option is (b) 2 s

Easiest EXPLANATION: The transfer function Z21 (S) is Z21 (S) = V2(S)/I1(S). V2 (S) = I1 (S) X 2S. V2(S)/I1(S)=2s. On substituting Z21 (S) = 2s.

The correct answer is (d) s/(s+1)

For explanation: Applying Kirchhoff’s LAW V1 (S) = 2 I1 (S) + 2 sI1 (S) V2 (S) = I1 (S) X 2s HENCE G21 (S) = V2(s)/V1(s) = 2 s/(2+2 s)=s/(s+1).

Correct option is (B) (s^2+2s+1)/s

Easy EXPLANATION: Applying Kirchoff’s law at port 1, Z(S)=V(S)/I(S), where V(s) is applied at port 1 and I(s) is current flowinmg through the NETWORK. Then Z(S)=V(S)/I(S) = 2+S+1/S = (s^2+2s+1)/s.

The CORRECT answer is (d) Transfer impedance

For explanation I WOULD say: Transfer impedance is the RATIO of voltage TRANSFORM at first port to the current transform at the second port and is DENOTED by Z(s). Z21(s) = V2(s)/I1(s) Z12(s) = V1(s)/I2(s).

The CORRECT option is (c) CURRENT TRANSFER RATIO

Explanation: Current transfer ratio is the ratio of the current transform at ONE port to current transform at other port and is denoted by α(s). α12(s) = I1(s)/I2(s) α21(s) = I2(s)/I1(s).

Right option is (a) VOLTAGE transfer ratio

For explanation I would say: Voltage transfer ratio is the ratio of voltage transform at FIRST port to the voltage transform at the SECOND port and is denoted by G(s). G21 = V2(s)/V1(s) G12 = V1(s)/V2(s).

The correct ANSWER is (d) (4s^2+19s+4)/(6s+8)

Easiest explanation: The TERM ADMITTANCE is defined as the inverse of the term IMPEDANCE. As the impedance isZ2(s) = 1/2s+1/(s+4)=(3s+4)/2s(s+4), the admittance PARALLEL combination of the last elements is Y2(s) = 1/2+2s(s+4)/(3s+4)=(4s^2+19s+4)/(6s+8).

Correct ANSWER is (c) (3s+4)/2s(s+4)

To explain: We got the IMPEDANCE of last two elements in PARALLEL combination as Z1(s) = 1/(s+4) and now the impedance of CAPACITOR is 1/2s. So the series combination of the last elements is Z2(s) = 1/2s+1/(s+4) = (3s+4)/2s(s+4).

The correct CHOICE is (a) 1/(s+4)

Easiest explanation: The impedance of RESISTOR is 4Ω and the impedance of CAPACITOR is s. So the impedance of the last two elements in the PARALLEL combination is Z1(s) = 1/(s+4).

Right answer is (c) R+LS+1/CS

The explanation: The value of the total IMPEDANCE after replacing the inductor and capacitor is Z (s) = R+LS+1/CS. By knowing the V(s) and Z(s) we can calculate I(s) as I(s) is GIVEN as the total transform voltage in the CIRCUIT divided by the total transform impedance.

Correct option is (a) Ls, L V0

For EXPLANATION I would say: The inductor has an initial current I0. It is represented by a transform impedance Ls in SERIES with a voltage SOURCE L V0.