Solving for y parameters using a system of equations

Question

^{simulate this circuit – Schematic created using CircuitLab}

Question: Find the y parameters for the circuit shown below.

My attempt:

KCL

\$I_1 = I_3 + I_6 \$

\$I_4 = I_3 + I_5\$

\$I_5 = I_2 + I_6\$

EFC:

\$I_3 = \frac{V_1 - V_a}{20}\$

\$I_4 = \frac{V_a-0}{10}\$

\$I_5 = \frac{V_2 - V_a}{4}\$

\$I_6 = \frac{V_1 - V_2}{8}\$

Solving this systems for \$ V_1 \$ and \$ V_2\$ yields:

\$ V_1 = \frac{35}{2} + \frac{25}{2} \$

\$ V_2 = \frac{25 }{2} + \frac{27}{2} \$

Therefore our Z parameters are as follows:

\begin{bmatrix} \frac{35}{2} & \frac{25}{2} \\ \frac{25}{2} & \frac{27}{2} \end{bmatrix}

I know how to find the y parameters using delta Z, but I thought there was a way to solve for these items directly in the system of equations. Can I solve for the y parameters directly?

Answer 1

hint: \$ Y_{11} = \frac{I_1}{V_1} \$ when you short-circuit \$ V_2 \$.
This arises from \$ I_1 = V_1 Y_{11} + V_2 Y_{12} \$. The short-circuited \$ V_2 \$ forces the \$V_2 Y_{12} \$ term to zero.

The other equation that applies is \$ I_2 = V_1 Y_{21} + V_2 Y_{22} \$ .

Somehow, you just memorize these (they're burned into my non-volatile-memory). The other aid to memory is visualization of a current source in parallel with a conductance: one for input port and another for output port. Another aid to sanity-check your memory is to verify units. For Y-parameters, units must be in terms of conductance and/or susceptance....like a current divided by a voltage.

^{simulate this circuit – Schematic created using CircuitLab}

You seem to be more familiar with Z-parameters. Here, units are resistance, and instead of current sources, you have two voltage sources, with series-resistors. Of course, you could solve the Z-parameter equations, then transform the four parameters into Y-parameters, but that's a bit round-about.