Question

For the circuit given below, the value of the Transmission parameter B and D are __________

(a) D = 0.5385 + j0.6923 Ω, B = -6.923 + j25.385 Ω

(b) D = 0.6923 + j0.5385 Ω, B = 6.923 + j25.385 Ω

(c) D = -0.6923 + j0.5385 Ω, B= 25.385 + j6.923 Ω

(d) D = -0.5385 + j0.6923 Ω, B = -6.923 + j25.385 Ω

I had been asked this question in my homework.

The doubt is from Advanced Problems Involving Parameters topic in section Two-Port Networks of Network Theory

Answer 1

The correct choice is (a) D = 0.5385 + j0.6923 Ω, B = -6.923 + j25.385 Ω

Easiest explanation: Z1 = \(\frac{(-j15)(-j10)}{-j15-j10-j20}\) = j10

Z2 = \(\frac{(-j10)(-j20)}{-j15}\) = \(-\frac{j40}{3}\)

Z3 = \(\frac{(j15)(-j20)}{-j15}\) = j20

-I2 = \(\frac{20-j40/3}{20-\frac{j40}{3}+j20}I_1 = \frac{3-j2}{3+j}\) I1

∴ D = \(\frac{-I_1}{I_2} = \frac{3+j}{3-j2}\) = 0.5385 + j0.6923 Ω

V1 = [j10 + 2(9+j7)] I1

= jI1 (24 – j18)

So, B = –\(\frac{V_1}{I_2}  = \frac{-jI_1 (24-j18)}{-\frac{3-j2}{3+j} I_1}\)

= \(\frac{6}{13}\)(-15+j55)

∴ B = -6.923 + j25.385 Ω.