Chapter 5 • 8 Hours • Expected Marks: 8M
Always Tested in Q5 or Q6 (5–6 Marks)
Two-Port Parameters of Network
Terminal descriptions of multi-terminal networks, evaluation of Z, Y, ABCD, and h parameter sets, inter-parameter conversions, cascade matrices, and reciprocity/symmetry conditions.
5.1 & 5.2 Definitions & The 6 Two-Port Parameter Sets
Concept of Two-Port Network
A two-port network is a four-terminal black-box where terminals are paired into an input port (Port 1) and an output port (Port 2). The fundamental port condition mandates that the current entering the upper terminal must exactly equal the current leaving the lower terminal for each port: \[I_1 = I_1', \quad I_2 = I_2'\]
The 6 Standard Two-Port Parameter Sets
| Parameter Set | Independent Variables | Dependent Variables | Governing Linear Matrix Equations |
|---|---|---|---|
| Impedance \([Z]\) | \(I_1, I_2\) | \(V_1, V_2\) |
\(V_1 = Z_{11} I_1 + Z_{12} I_2\) \(V_2 = Z_{21} I_1 + Z_{22} I_2\) |
| Admittance \([Y]\) | \(V_1, V_2\) | \(I_1, I_2\) |
\(I_1 = Y_{11} V_1 + Y_{12} V_2\) \(I_2 = Y_{21} V_1 + Y_{22} V_2\) |
| Transmission \([T]\) or \([ABCD]\) | \(V_2, -I_2\) | \(V_1, I_1\) |
\(V_1 = A V_2 - B I_2\) \(I_1 = C V_2 - D I_2\) |
| Inverse Transmission \([T']\) | \(V_1, -I_1\) | \(V_2, I_2\) |
\(V_2 = A' V_1 - B' I_1\) \(I_2 = C' V_1 - D' I_1\) |
| Hybrid \([h]\) | \(I_1, V_2\) | \(V_1, I_2\) |
\(V_1 = h_{11} I_1 + h_{12} V_2\) \(I_2 = h_{21} I_1 + h_{22} V_2\) |
| Inverse Hybrid \([g]\) | \(V_1, I_2\) | \(I_1, V_2\) |
\(I_1 = g_{11} V_1 + g_{12} I_2\) \(V_2 = g_{21} V_1 + g_{22} I_2\) |
5.3 Transformation Between Parameter Sets
Any valid two-port parameter set can be directly converted into any other by equating equations algebraically.
\[\text{h-parameters in terms of Z: } h_{11} = \frac{\Delta Z}{Z_{22}}, \quad h_{12} = \frac{Z_{12}}{Z_{22}}, \quad h_{21} = -\frac{Z_{21}}{Z_{22}}, \quad h_{22} = \frac{1}{Z_{22}}\]
\[\text{Z-parameters in terms of ABCD: } Z_{11} = \frac{A}{C}, \quad Z_{12} = \frac{AD - BC}{C}, \quad Z_{21} = \frac{1}{C}, \quad Z_{22} = \frac{D}{C}\]
\[\text{ABCD in terms of Z: } A = \frac{Z_{11}}{Z_{21}}, \quad B = \frac{\Delta Z}{Z_{21}}, \quad C = \frac{1}{Z_{21}}, \quad D = \frac{Z_{22}}{Z_{21}}\]
5.4 & 5.5 Interconnections, Reciprocity & Symmetry
Two-Port Network Interconnections
| Interconnection Type | Port Connection Scheme | Governing Matrix Operation |
|---|---|---|
| Series-Series | Inputs in series, outputs in series | \([Z] = [Z_A] + [Z_B]\) |
| Parallel-Parallel | Inputs in parallel, outputs in parallel | \([Y] = [Y_A] + [Y_B]\) |
| Cascade (Tandem) | Output of Network A feeds input of Network B | \([T] = [T_A] \times [T_B]\) (Matrix Multiplication) |
Conditions for Reciprocity and Symmetry
| Parameter | Reciprocity Condition (Transfer Ratio Equality) | Symmetry Condition (Port Invariance) |
|---|---|---|
| Impedance \([Z]\) | \(Z_{12} = Z_{21}\) | \(Z_{11} = Z_{22}\) |
| Admittance \([Y]\) | \(Y_{12} = Y_{21}\) | \(Y_{11} = Y_{22}\) |
| Transmission \([ABCD]\) | \(AD - BC = 1\) | \(A = D\) |
| Hybrid \([h]\) | \(h_{12} = -h_{21}\) | \(\Delta h = h_{11} h_{22} - h_{12} h_{21} = 1\) |
Master Worked Example (Two-Port with Dependent Source)
ABCD Parameters of Dependent Source Network
PYQ 2081 Ashwin Q6
Problem Statement:
Find the ABCD parameters for the two-port network: A dependent current source of \(0.5 I_x\) is in parallel with Port 1. A series resistor of \(8\,\Omega\) connects to Port 2. A shunt resistor of \(5\,\Omega\) is connected across Port 2, carrying downward current \(I_x\).
Calculations:
By inspection, across Port 2:
\[V_2 = 5 I_x \implies I_x = \frac{V_2}{5} = 0.2 V_2\]
Dependent current source value:
\[0.5 I_x = 0.5(0.2 V_2) = 0.1 V_2\]
1. Finding \(V_1\) in terms of \(V_2\) and \(I_2\):
Current flowing through the series \(8\,\Omega\) resistor from Port 1 towards Port 2:
At the node above the \(5\,\Omega\) resistor:
\[I_{\text{series}} + I_2 = I_x \implies I_{\text{series}} = I_x - I_2 = 0.2 V_2 - I_2\]
KVL from Port 1 to Port 2:
\[V_1 = 8 I_{\text{series}} + V_2 = 8(0.2 V_2 - I_2) + V_2 = 1.6 V_2 - 8 I_2 + V_2 = 2.6 V_2 - 8 I_2\]
Comparing with \(V_1 = A V_2 - B I_2\):
\[A = 2.6, \quad B = 8\,\Omega\]
2. Finding \(I_1\) in terms of \(V_2\) and \(I_2\):
At the input node (Port 1):
\[I_1 = 0.5 I_x + I_{\text{series}} = 0.1 V_2 + (0.2 V_2 - I_2) = 0.3 V_2 - I_2\]
Comparing with \(I_1 = C V_2 - D I_2\):
\[C = 0.3\text{ S}, \quad D = 1\]
Transmission Matrix:
\[\begin{bmatrix} A & B \\ C & D \end{bmatrix} = \begin{bmatrix} 2.6 & 8\,\Omega \\ 0.3\text{ S} & 1 \end{bmatrix}\]
Check Reciprocity: \(AD - BC = (2.6 \times 1) - (8 \times 0.3) = 2.6 - 2.4 = 0.2 \ne 1\). The network is NON-RECIPROCAL due to the dependent source!