Q:

Two-port network Current Characterization

I was trying to solve this two-pole exercise in which I had to found the z-parameters but I haven't find any clue in the textbook. How should I solve it?

A:

The parameters \$Z_{11}\$ and \$Z_{21}\$ are obtained by applying a current \$I_1\$ to the left-hand side port (port 1), leaving port 2 (on the right-hand side) open (i.e. \$I_2=0\$), and calculating the resulting voltages \$U_1\$ and \$U_2\$ at ports 1 and 2, respectively. In this case, since there is no current flowing through \$R_b\$ (because \$I_2=0\$), the current through \$R_a\$ is the same as the current through \$R_c\$. Equating the voltage \$U_1=ri_c\$ with the voltage across \$R_a\$ and \$R_c\$ gives
$$ri_c=(R_a+R_c)i_c\tag{1}$$
Since \$r\neq R_a+R_c\$, the only solution of Eq. (1) is \$i_c=0\$. This means that \$U_1=ri_c=0\$, and \$U_2=R_ci_c=0\$, and, consequently, \$Z_{11}=Z_{21}=0\$.
In order to calculate \$Z_{12}\$ and \$Z_{22}\$ we need to apply a current \$I_2\$ to port 2 and leave port 1 open. We get
$$R_ci_c=R_a(I_2-i_c)+ri_c\tag{2}$$
Solving (2) for \$i_c\$ gives
$$i_c=\frac{R_aI_2}{R_a+R_c-r}\tag{3}$$
From (3) we can calculate \$U_1\$ and \$U_2\$:
$$U_1=ri_c=\frac{rR_aI_2}{R_a+R_c-r}\tag{4}$$
which results in
$$Z_{12}=\frac{rR_a}{R_a+R_c-r}=\frac{6\cdot 12}{18}\,\Omega=4\,\Omega\tag{5}$$
And
$$U_2=R_bI_2+R_ci_c=R_bI_2+\frac{R_aR_cI_2}{R_a+R_c-r}\tag{6}$$
from which we get
$$Z_{22}=R_b+\frac{R_aR_c}{R_a+R_c-r}=6\,\Omega+\frac{12\cdot 12}{18}\,\Omega=14\,\Omega\tag{7}$$

