Z-Parameters (Open-Circuit Impedance)
Z-Parameter Equations
V_1 = Z_{11}I_1 + Z_{12}I_2, \quad V_2 = Z_{21}I_1 + Z_{22}I_2
| Symbol | Description | Unit |
|---|---|---|
| Z11 | Input impedance (port 2 open) | Ω |
| Z12 | Reverse transfer impedance | Ω |
| Z21 | Forward transfer impedance | Ω |
| Z22 | Output impedance (port 1 open) | Ω |
Worked example
A T-network has Z1=4Ω (series arm 1), Z2=6Ω (series arm 2), Z3=3Ω (shunt arm). Find all Z-parameters.
Given: Z1=4Ω, Z2=6Ω, Z3=3Ω
- For T-network: Z11 = Z1 + Z3 = 4 + 3 = 7Ω
- Z22 = Z2 + Z3 = 6 + 3 = 9Ω
- Z12 = Z21 = Z3 = 3Ω (reciprocal network)
- Matrix: [V1; V2] = [7 3; 3 9][I1; I2]
Answer: Z11=7Ω, Z12=Z21=3Ω, Z22=9Ω
Z-Parameter Measurement
Z_{11} = \frac{V_1}{I_1}\bigg|_{I_2=0}, \quad Z_{21} = \frac{V_2}{I_1}\bigg|_{I_2=0}
| Symbol | Description | Unit |
|---|---|---|
| I2=0 | Port 2 open-circuited | — |
Worked example
With port 2 open, V1=10V causes I1=2A and V2=6V. Find Z11 and Z21.
Given: V1=10V, I1=2A, V2=6V, I2=0
- Z11 = V1/I1 = 10/2 = 5Ω
- Z21 = V2/I1 = 6/2 = 3Ω
Answer: Z11 = 5Ω, Z21 = 3Ω
Y-Parameters (Short-Circuit Admittance)
Y-Parameter Equations
I_1 = Y_{11}V_1 + Y_{12}V_2, \quad I_2 = Y_{21}V_1 + Y_{22}V_2
| Symbol | Description | Unit |
|---|---|---|
| Y11 | Input admittance (port 2 shorted) | S |
| Y12 | Reverse transfer admittance | S |
| Y21 | Forward transfer admittance | S |
| Y22 | Output admittance (port 1 shorted) | S |
Worked example
A π-network has Ya=0.2S (shunt at port 1), Yb=0.25S (shunt at port 2), Yc=0.1S (series arm). Find Y-parameters.
Given: Ya=0.2S, Yb=0.25S, Yc=0.1S
- For π-network: Y11 = Ya + Yc = 0.2 + 0.1 = 0.3S
- Y22 = Yb + Yc = 0.25 + 0.1 = 0.35S
- Y12 = Y21 = -Yc = -0.1S (reciprocal network)
Answer: Y11=0.3S, Y12=Y21=-0.1S, Y22=0.35S
Y from Z Conversion
[Y] = [Z]^{-1} = \frac{1}{\Delta Z}\begin{bmatrix} Z_{22} & -Z_{12} \\ -Z_{21} & Z_{11} \end{bmatrix}
| Symbol | Description | Unit |
|---|---|---|
| ΔZ | Determinant of Z-matrix = Z11·Z22 - Z12·Z21 | Ω² |
Worked example
Given Z11=7, Z12=Z21=3, Z22=9 (all Ω), find Y-parameters.
Given: Z11=7, Z12=3, Z21=3, Z22=9
- ΔZ = Z11*Z22 - Z12*Z21 = 7*9 - 3*3 = 63 - 9 = 54
- Y11 = Z22/ΔZ = 9/54 = 1/6 S
- Y12 = -Z12/ΔZ = -3/54 = -1/18 S
- Y21 = -Z21/ΔZ = -1/18 S
- Y22 = Z11/ΔZ = 7/54 S
Answer: Y11=1/6 S, Y12=Y21=-1/18 S, Y22=7/54 S
h-Parameters (Hybrid)
h-Parameter Equations
V_1 = h_{11}I_1 + h_{12}V_2, \quad I_2 = h_{21}I_1 + h_{22}V_2
| Symbol | Description | Unit |
|---|---|---|
| h11 | Input impedance (short-circuit), hi | Ω |
| h12 | Reverse voltage ratio (open-circuit), hr | — |
| h21 | Forward current gain (short-circuit), hf | — |
| h22 | Output admittance (open-circuit), ho | S |
Worked example
BJT has hie=1kΩ, hfe=100, hre=2×10⁻⁴, hoe=25μS. Find voltage gain with RL=4kΩ.
Given: hie=1000Ω, hfe=100, hoe=25e-6 S, RL=4000Ω
- Voltage gain Av = hfe * RL / hie (simplified, ignoring hoe*RL if small)
- Check: hoe*RL = 25e-6 * 4000 = 0.1 (small, can be ignored)
- Av ≈ -hfe * RL / hie = -100 * 4000 / 1000 = -400
- Negative sign indicates phase inversion in CE configuration
Answer: Av ≈ -400
h from Z Conversion
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}}
| Symbol | Description | Unit |
|---|---|---|
| ΔZ | Z-matrix determinant | Ω² |
Worked example
Convert Z-parameters (Z11=10, Z12=2, Z21=2, Z22=8 Ω) to h-parameters.
Given: Z11=10, Z12=2, Z21=2, Z22=8
- ΔZ = 10*8 - 2*2 = 80 - 4 = 76
- h11 = ΔZ/Z22 = 76/8 = 9.5Ω
- h12 = Z12/Z22 = 2/8 = 0.25
- h21 = -Z21/Z22 = -2/8 = -0.25
- h22 = 1/Z22 = 1/8 = 0.125S
Answer: h11=9.5Ω, h12=0.25, h21=-0.25, h22=0.125S
ABCD (Transmission) Parameters
ABCD Parameter Equations
V_1 = AV_2 - BI_2, \quad I_1 = CV_2 - DI_2
| Symbol | Description | Unit |
|---|---|---|
| A | Reverse voltage ratio (I2=0) | — |
| B | Transfer impedance (V2=0) | Ω |
| C | Transfer admittance (I2=0) | S |
| D | Reverse current ratio (V2=0) | — |
Worked example
A series impedance Z=jωL=j10Ω represents a transmission line section. Find its ABCD parameters.
Given: Series element Z = j10Ω
- For a series impedance Z: A=1, B=Z, C=0, D=1
- A = 1 (voltage ratio, no change in voltage)
- B = j10Ω
- C = 0 (no shunt path)
- D = 1
Answer: A=1, B=j10Ω, C=0, D=1
Cascade Connection (ABCD)
\begin{bmatrix}A&B\\C&D\end{bmatrix}_{total} = \begin{bmatrix}A_1&B_1\\C_1&D_1\end{bmatrix}\begin{bmatrix}A_2&B_2\\C_2&D_2\end{bmatrix}
| Symbol | Description | Unit |
|---|---|---|
| ABCD₁ | Parameters of first network | — |
| ABCD₂ | Parameters of second network | — |
Worked example
Two two-port networks in cascade: N1 has A1=2,B1=10,C1=0.1,D1=3; N2 has A2=1,B2=5,C2=0,D2=1. Find total ABCD.
Given: N1: [2,10;0.1,3], N2: [1,5;0,1]
- A = A1*A2 + B1*C2 = 2*1 + 10*0 = 2
- B = A1*B2 + B1*D2 = 2*5 + 10*1 = 20
- C = C1*A2 + D1*C2 = 0.1*1 + 3*0 = 0.1
- D = C1*B2 + D1*D2 = 0.1*5 + 3*1 = 3.5
Answer: Total: A=2, B=20, C=0.1, D=3.5
Reciprocity and Symmetry Conditions
\text{Reciprocal: } AD - BC = 1, \quad \text{Symmetric: } A = D
| Symbol | Description | Unit |
|---|---|---|
| AD-BC | Determinant of ABCD matrix | — |
Worked example
Check if a network with A=2, B=10, C=0.3, D=2 is reciprocal and symmetric.
Given: A=2, B=10, C=0.3, D=2
- Reciprocity: AD - BC = 2*2 - 10*0.3 = 4 - 3 = 1 ✓ (Reciprocal)
- Symmetry: A = D? 2 = 2 ✓ (Symmetric)
Answer: Network is both reciprocal and symmetric
Network Interconnections
Series-Series (Z-parameters add)
[Z]_{total} = [Z_1] + [Z_2]
| Symbol | Description | Unit |
|---|---|---|
| [Z1], [Z2] | Z-parameter matrices of individual networks | Ω |
Worked example
Two networks in series-series connection: Z1=[2 1;1 3], Z2=[4 2;2 5] (all Ω). Find total Z.
Given: Z1=[2,1;1,3], Z2=[4,2;2,5]
- Z_total = Z1 + Z2 (element-wise addition)
- Z11 = 2+4 = 6Ω
- Z12 = 1+2 = 3Ω
- Z21 = 1+2 = 3Ω
- Z22 = 3+5 = 8Ω
Answer: Z_total = [6 3; 3 8] Ω
Parallel-Parallel (Y-parameters add)
[Y]_{total} = [Y_1] + [Y_2]
| Symbol | Description | Unit |
|---|---|---|
| [Y1], [Y2] | Y-parameter matrices of individual networks | S |
Worked example
Two networks in parallel: Y1=[0.3 -0.1; -0.1 0.4] S, Y2=[0.2 -0.05; -0.05 0.3] S. Find total Y.
Given: Y1 and Y2 as given
- Y_total = Y1 + Y2
- Y11 = 0.3+0.2 = 0.5S
- Y12 = -0.1+(-0.05) = -0.15S
- Y21 = -0.15S
- Y22 = 0.4+0.3 = 0.7S
Answer: Y_total = [0.5 -0.15; -0.15 0.7] S
Image Parameters and Characteristic Impedance
Image Impedances
Z_{i1} = \sqrt{\frac{AB}{CD}}, \quad Z_{i2} = \sqrt{\frac{DB}{CA}}
| Symbol | Description | Unit |
|---|---|---|
| Z_i1 | Image impedance at port 1 | Ω |
| Z_i2 | Image impedance at port 2 | Ω |
Worked example
Find image impedances for a network with A=2, B=8Ω, C=0.5S, D=2.
Given: A=2, B=8, C=0.5, D=2
- Zi1 = sqrt(AB/CD) = sqrt(2*8 / 0.5*2) = sqrt(16/1) = sqrt(16) = 4Ω
- Zi2 = sqrt(DB/CA) = sqrt(2*8 / 0.5*2) = sqrt(16) = 4Ω
- Since A=D, Zi1 = Zi2 = 4Ω (symmetric network)
Answer: Z_i1 = Z_i2 = 4Ω
Propagation Function
e^\gamma = \sqrt{AD} + \sqrt{BC}, \quad \gamma = \alpha + j\beta
| Symbol | Description | Unit |
|---|---|---|
| γ | Propagation constant | Np/section |
| α | Attenuation constant | Np |
| β | Phase constant | rad |
Worked example
A network has A=D=1.5, B=5Ω, C=0.2S. Find e^γ.
Given: A=D=1.5, B=5, C=0.2
- e^γ = sqrt(AD) + sqrt(BC)
- sqrt(AD) = sqrt(1.5*1.5) = 1.5
- sqrt(BC) = sqrt(5*0.2) = sqrt(1) = 1
- e^γ = 1.5 + 1 = 2.5
- γ = ln(2.5) = 0.916 Np (real → purely attenuating)
Answer: e^γ = 2.5, γ = 0.916 Np
Quick reference
| Formula | Expression |
|---|---|
| Z-parameter equations | V1=Z11·I1+Z12·I2; V2=Z21·I1+Z22·I2 |
| Y-parameter equations | I1=Y11·V1+Y12·V2; I2=Y21·V1+Y22·V2 |
| h-parameter equations | V1=h11·I1+h12·V2; I2=h21·I1+h22·V2 |
| ABCD equations | V1=A·V2-B·I2; I1=C·V2-D·I2 |
| Z11 measurement | Z11 = V1/I1 at I2=0 |
| Y11 measurement | Y11 = I1/V1 at V2=0 |
| Y from Z | [Y] = [Z]^{-1} |
| Cascade (ABCD) | [ABCD]_total = [ABCD]_1 × [ABCD]_2 |
| Series connection | [Z]_total = [Z]_1 + [Z]_2 |
| Parallel connection | [Y]_total = [Y]_1 + [Y]_2 |
| Reciprocity (ABCD) | AD - BC = 1 |
| Symmetry | A = D |
| Image impedance | Z_i1 = sqrt(AB/CD) |
| Propagation function | e^γ = sqrt(AD) + sqrt(BC) |
| h from Z | h11=ΔZ/Z22; h21=-Z21/Z22 |
Exam tips
- GATE problems on two-port networks frequently ask for a specific parameter given a circuit topology — memorise the T-network Z-parameter and π-network Y-parameter formulas directly.
- The sign convention for ABCD parameters uses -I2 (current leaving port 2); forgetting this sign leads to incorrect cascade multiplication results.
- h-parameters are the standard model for BJT amplifiers — h21 (hfe) is the short-circuit current gain and h11 (hie) is the input impedance; always apply at port 2 short-circuited condition.
- When networks are cascaded, use ABCD parameter multiplication; for series connection use Z-parameter addition; for parallel connection use Y-parameter addition — mixing these up is a common mistake.
- For a reciprocal network, Z12=Z21 (or Y12=Y21, or AD-BC=1); for a symmetric reciprocal network additionally A=D and Z11=Z22 — check these conditions when asked to verify network properties.