In the first Cauer LC network, the first element is a series indu
A. Pole at ω = ∞
B. Zero at ω = ∞
C. Pole at ω = 0
D. Zero at ω = 0
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
Explanation:
Cauer form 1
The LC structure is shown below along with the system equation
- Consider a driving point function having a pole at infinity.
- This implies that the degree of the numerator is greater than that of the denominator.
- We always remove pole at infinity by inverting the remainder and dividing.
- That means an LC driving point function can be synthesised by the continued fraction expansion.
If Z(s) is the function to be synthesised, then the continued fraction expansion is as follows
\(Z\left( s \right) = {L_1}s + \frac{1}{{{C_2}s + \frac{1}{{{L_2}s + \cdots }}}}\)
Therefore, in the first Cauer network shown in Figure, the inductors are connected in series and the capacitors are connected in the shunt.
- If the driving point function, Z(s) has zero at infinite, that is,
- This implies the degree of its numerator is less than that of its denominator, the driving point function is inverted.
- In this case, the continued fraction will give a capacitive admittance as the first element, and a series inductance
This structure is known as Cauer 2 form
If Z(s) is the function to be synthesised, then the continued fraction expansion is Figure
\(Z\left( s \right) = \frac{1}{{{C_1}s}} + \frac{1}{{\frac{1}{{{L_1}s}} + \frac{1}{{\frac{1}{{{C_2}s}} + \frac{1}{{\frac{1}{{{L_2}s + \cdots }}}}}}}}\)
