Test: Network Synthesis - 1


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Test: Network Synthesis - 1 - Question 1

 Which of the following is a positive real function?

Detailed Solution for Test: Network Synthesis - 1 - Question 1

Hence, notaPRF.

has a zero (s = 1) in RH s-plane.

Hence, not a PRF.

  has a pole and a zero in RH s-piane. 
Hence, not a PRF.
Option (b) is a PRF as all poles and zeros lie in LH s-plane.

Test: Network Synthesis - 1 - Question 2

A network function can be completely specified by:

Test: Network Synthesis - 1 - Question 3

Which of the following property does not relates to LC impedance or admittance functions?

Test: Network Synthesis - 1 - Question 4

If the impulse response is realisable by delaying it appropriately and is bounded for bounded excitation, then the system is said to be

Detailed Solution for Test: Network Synthesis - 1 - Question 4

Since system is bounded for bounded excitation therefore, it will be stable.
Since the system is delayed therefore, output may exist before the input is applied.
Hence, the system will be non-causal.

Test: Network Synthesis - 1 - Question 5

A network-A has zeros only in the left-half of the s-plane while the network-B has zeros only in the right-half of the s-plane. The network-A and Network-B will be respectively called

Test: Network Synthesis - 1 - Question 6

An L-C admittance or impedance function

Test: Network Synthesis - 1 - Question 7

 If F1(s) and F2(s) are positive real functions, then which of the following are positive real function?

Test: Network Synthesis - 1 - Question 8

For the pole-zero plot shown below, the network function is

Detailed Solution for Test: Network Synthesis - 1 - Question 8

Here, zeros are at: s = 0, 0,-3

Poles are at s = - 1,

or, poles are at (s + 2)2 + 1 - 0 and at s = -1 or, s2 + 4s + 5 = 0 and at s = -1

Test: Network Synthesis - 1 - Question 9

A stable system must have

Test: Network Synthesis - 1 - Question 10

Which of the following is not a positive real function?

Detailed Solution for Test: Network Synthesis - 1 - Question 10

Here, a = 1 , b = 0, c = 2

Thus, a, b, c ≥ 0 but b < a

For PRF, b ≥ a therefore, option (d) is not a PRF.

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