#### Problem:

(a)  Calculate the impedance Z of an infinite chain of elements with impedances Z1 and Z2, as shown in the top figure.
(b)  Calculate Z1 and Z2 for the specific case shown in the bottom figure.

Solution:

• Concepts:
• Reasoning:
We treat the circuit as an infinite ladder network with characteristic impedance Z.

• Details of the calculation:
(a)  An equivalent network with impedance Z is shown in the figure.

Z = Z1 + ZZ2/(Z + Z2),  Z2 - Z1Z - Z1Z2 = 0,  Z = Z1/2 + (Z12/4 + Z1 Z2)½.
(b)  For the network in the bottom figure
Z1 = iωL - i/(ωC) = (i/(ωC))(ω2LC - 1),  Z2 = (-iωL)/(ω2LC - 1).
Z12 = (-1/(ωC)2)(ω2LC - 1)2,  Z1 Z2 = L/C.

#### Problem:

Consider the "twin-T" band-pass filter as shown below.  Find the highest and lowest frequency passed by this filter if it is properly terminated.

Solution:

• Concepts:
• Reasoning:
In order to act as a filter, the circuit must be terminated with the characteristic impedance Z0 of the ladder network.  Z0 = (Z12/4 + Z1Z2)½ if the sections of the ladder look as shown below.

If Z02 > 0 the frequency is passed.
If Z02 < 0 the frequency is not passed.  (The circuit absorbs no power.)

• Details of the calculation:
We treat the circuit as an infinite ladder network with characteristic impedance Z0.
Z0 = (Z12/4 + Z1Z2)½.
Z1 = iωL + 1/(iωC),  Z2 = 1/(iωC').
Z12 = -ω2L2 - 1/(ω2C2) + 2L/C.
Z1Z2 = L/C' - 1/(ω2C'C).
Z02 = L/2C + L/C' - ω2L2/4 - 1/(4ω2C2) - 1/(ω2C'C)
= (LC'+2LC)/(2CC')  - ω2L2/4 - (¼ + C/C')/(ω2C2).
Set Z02 = 0, then ω2(LC'+2LC)/(2CC')  - ω4L2/4 - (¼ + C/C')/C2 = 0.
Quadratic equation in ω2:  ω2 = 1/(LC) +2/(LC') ± 2/(LC').
ωmin2 = 1/(LC),  ωmax2 = 1/LC + 4/(LC')
If ω ≈ 0, Z02 <  0 the frequency is not passed.
If ω --> ∞,Z02 < 0 the frequency is not passed.
Frequencies between ωmin2 = 1/(LC) and ωmax2 = 1/LC + 4/(LC') are not filtered out.