For an L section, shown in the figure, find the iterative impedance at ports 1 and 2, respectively:

Concept:

For the given T network Iterative impedance can be calculated by terminating the port with the input impedance of the other respective port.

Calculations:

Given that

1) Iterative impedance of port 1 is calculated by

Z_1(in) = R₁ + (R₂ || Z1(in))

Where,

R₁ = 200 Ω , R₂ = 400 Ω

Z1(in) = 200 + (400 || Z1(in))

\(Z_{1(in)} = 200 + \frac{400\times Z_{1(in)}}{400+Z_{1(in)}}\)

on rearranging the above equation we get

Z²1(in) - 200 Z1(in) - 80000 = 0

by solving from the above equation we get

Z_{1 (in)}  = 400 Ω,  - 200 Ω

resistance must be positive so 400 Ω will be the answer.

1) Iterative impedance of port 2 is calculated by

Z₂(in) = (R1 + Z2(in)) || R₂

Where,

R1 = 200 Ω , R2 = 400 Ω

Z2(in) = (200 + Z2(in)) || 400

\(Z_{2(in)} = \frac{(200+Z_{2(in)}) \times 400}{(200+Z_{2(in)})+400}\)

on rearranging the above equation we get

Z21(in) + 200 Z1(in) - 80000 = 0

by solving from the above equation we get

Z1 (in)  = - 400 Ω,  200 Ω

resistance must be positive so 200 Ω will be the answer.

Hence from the above answer option (4) is correct