Chapter 6 - Two-Port Network Theory

06 Chapter Map - Two-Port Network Theory

Chapter 6 Overview & Map of Content (MOC)

Open-circuit Z-parameters, Pi-network Y-parameters, cascaded ABCD matrix multiplications, and hybrid BJT h-parameters.

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📚 Study Notes Index

Read in order — each note assumes the previous one.

#	Note	What it covers
6.00	6.00 Two-Port Network Theory Compact Review	Two-Port Network Compact Review, Two-Port Formula Sheet
6.01	6.01 Open-Circuit Impedance Z Parameters	Z-Parameters, Open-Circuit Impedance, Impedance Matrix
6.02	6.02 Short-Circuit Admittance Y Parameters	Y-Parameters, Short-Circuit Admittance, Admittance Matrix
6.03	6.03 Transmission ABCD Parameters and Cascaded Networks	ABCD-Parameters, Transmission Parameters, Transmission Matrix, Cascaded Networks
6.04	6.04 Hybrid h and Inverse Hybrid g Parameter Formulations	h-Parameters, g-Parameters, Hybrid Parameters, Inverse Hybrid Parameters

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🎯 Exam Weight

ECE 2107 Exam Relevance

Master the core derivations, mathematical definitions, and problem-solving techniques. Refer to ECE 2107 - Signals and Systems for syllabus boundaries and past year questions.

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🔗 Related Resources

• Course Teaching Plan: ECE 2107 - Signals and Systems
• Previous chapter: 05 Chapter Map - Analog Filter Design
• Next chapter: 07 Chapter Map - Continuous-Time Fourier Series (CTFS)

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Chapter 6: Two-Port Network Theory - Compact Review

Overview

Two-port network theory treats linear electrical circuits as “black boxes,” modeling their behavior strictly through terminal variables: input/output voltages ($V_1,V_2$) and input/output currents ($I_1,I_2$). This compact note compiles all terminal equations, parameter definitions, reciprocity/symmetry boundaries, cascading rules, and parameter-to-parameter conversion matrices required for rapid pre-exam revision.

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6.01 Two-Port Network Terminal Variables & Z-Parameters

*(Target: Theory Descriptive / Formulas & Numerical Solving)*

• Two-Port Terminal Convention: Voltages are positive at upper terminals ($V_1,V_2$). Port currents ($I_1,I_2$) are modeled as flowing into the network.
• Z-Parameters (Open-Circuit Impedance): Relate port voltages directly to port currents: $V_1=Z_{11}{I}_1+Z_{12}{I}_2$ $V_2=Z_{21}{I}_1+Z_{22}{I}_2$ In matrix form: $\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]=\left[\begin{array}{cc}{Z}_{11}& Z_{12}\\ Z_{21}& Z_{22}\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]⟹\mathbf{V}=\mathbf{ZI}$
• Physical Parameters Definitions (Isolated by open-circuiting ports):
  • $Z_{11}$ (Input Impedance): $Z_{11}={\frac{V_1}{I_1}|}_{I_2=0}[\Omega ]$
  • $Z_{12}$ (Reverse Transfer Impedance): $Z_{12}={\frac{V_1}{I_2}|}_{I_1=0}[\Omega ]$
  • $Z_{21}$ (Forward Transfer Impedance): $Z_{21}={\frac{V_2}{I_1}|}_{I_2=0}[\Omega ]$
  • $Z_{22}$ (Output Impedance): $Z_{22}={\frac{V_2}{I_2}|}_{I_1=0}[\Omega ]$
• Z-Parameter Network Symmetries:
  • Reciprocity condition: $Z_{12}=Z_{21}$.
  • Symmetry condition: $Z_{11}=Z_{22}$.

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6.02 Short-Circuit Admittance (Y) Parameters

*(Target: Theory Descriptive / Formulas & Numerical Solving)*

• Y-Parameters (Short-Circuit Admittance): Relate port currents directly to port voltages: $I_1=Y_{11}{V}_1+Y_{12}{V}_2$ $I_2=Y_{21}{V}_1+Y_{22}{V}_2$ In matrix form: $\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]=\left[\begin{array}{cc}{Y}_{11}& Y_{12}\\ Y_{21}& Y_{22}\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]⟹\mathbf{I}=\mathbf{YV}$
• Physical Parameters Definitions (Isolated by short-circuiting ports):
  • $Y_{11}$ (Short-Circuit Input Admittance): $Y_{11}={\frac{I_1}{V_1}|}_{V_2=0}[\text{S}]$
  • $Y_{12}$ (Short-Circuit Reverse Transfer Admittance): $Y_{12}={\frac{I_1}{V_2}|}_{V_1=0}[\text{S}]$
  • $Y_{21}$ (Short-Circuit Forward Transfer Admittance): $Y_{21}={\frac{I_2}{V_1}|}_{V_2=0}[\text{S}]$
  • $Y_{22}$ (Short-Circuit Output Admittance): $Y_{22}={\frac{I_2}{V_2}|}_{V_1=0}[\text{S}]$
• Y-Parameter Network Symmetries:
  • Reciprocity condition: $Y_{12}=Y_{21}$.
  • Symmetry condition: $Y_{11}=Y_{22}$.
• Pi-to-Y Parametric Mapping (No derivation): For a resistive $\Pi$-network with series feedback branch $R_C$ and shunt branches $R_A$ (Port 1), $R_B$ (Port 2): $Y_{11}=\frac{1}{R_A}+\frac{1}{R_C},Y_{12}=Y_{21}=-\frac{1}{R_C},Y_{22}=\frac{1}{R_B}+\frac{1}{R_C}$

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6.03 The Fundamental Z-to-Y Matrix Inverse Identity

*(Target: Mathematical Proof / Parameter Conversion)*

• Philosophical Identity: Admittance and impedance matrices are exact mathematical inverses of each other: $\mathbf{Y}={\mathbf{Z}}^{-1}\text{and}\mathbf{Z}={\mathbf{Y}}^{-1}\text{(derivation)}$
• Algebraic Conversion Formulas: Let determinant ${\Delta }_Z=Z_{11}{Z}_{22}-Z_{12}{Z}_{21}$: $Y_{11}=\frac{Z_{22}}{{\Delta }_Z},Y_{12}=\frac{-Z_{12}}{{\Delta }_Z},Y_{21}=\frac{-Z_{21}}{{\Delta }_Z},Y_{22}=\frac{Z_{11}}{{\Delta }_Z}\text{(derivation)}$
• Let determinant ${\Delta }_Y=Y_{11}{Y}_{22}-Y_{12}{Y}_{21}$: $Z_{11}=\frac{Y_{22}}{{\Delta }_Y},Z_{12}=\frac{-Y_{12}}{{\Delta }_Y},Z_{21}=\frac{-Y_{21}}{{\Delta }_Y},Z_{22}=\frac{Y_{11}}{{\Delta }_Y}\text{(derivation)}$

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6.04 Transmission (ABCD) Parameters & Cascaded Networks

*(Target: Theory Descriptive / Formulas & Cascade Multiplying)*

• Negative Current Convention: Current $I_2$ is modeled as flowing out of Port 2 (represented as $-I_2$ in terminal equations) to simplify cascaded network series connections.
• ABCD-Parameters (Transmission Matrix): Relate sending-end variables ($V_1,I_1$) to receiving-end variables ($V_2,-I_2$): $V_1=A{V}_2-B{I}_2$ $I_1=C{V}_2-D{I}_2$ In matrix form: $\left[\begin{array}{c}{V}_1\\ I_1\end{array}\right]=\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]\left[\begin{array}{c}{V}_2\\ -I_2\end{array}\right]⟹{\mathbf{X}}_{\text{send}}=\mathbf{T}{\mathbf{X}}_{\text{rec}}$
• Physical Parameters Definitions (Open-Circuit or Short-Circuit conditions):
  • $A$ (Open-Circuit Voltage Ratio, unitless): $A={\frac{V_1}{V_2}|}_{I_2=0}$
  • $B$ (Short-Circuit Transfer Impedance, $[\Omega ]$): $B={\frac{V_1}{-I_2}|}_{V_2=0}$
  • $C$ (Open-Circuit Transfer Admittance, $[\text{S}]$): $C={\frac{I_1}{V_2}|}_{I_2=0}$
  • $D$ (Short-Circuit Current Ratio, unitless): $D={\frac{I_1}{-I_2}|}_{V_2=0}$
• ABCD Network Symmetries:
  • Reciprocity condition: ${\Delta }_T=AD-BC=1$.
  • Symmetry condition: $A=D$.
• T-to-ABCD Parametric Mapping (No derivation): For a resistive T-network with series arms $R_1$, $R_2$ and common shunt arm $R_3$: $A=1+\frac{R_1}{R_3},B=R_1+R_2+\frac{R_1{R}_2}{R_3},C=\frac{1}{R_3},D=1+\frac{R_2}{R_3}\text{(derivation)}$
• Cascaded Networks Theorem: If two systems with transmission matrices ${\mathbf{T}}_1$ and ${\mathbf{T}}_2$ are connected in cascade (series), the overall transmission matrix ${\mathbf{T}}_{\text{total}}$ is the direct matrix product of individual matrices: ${\mathbf{T}}_{\text{total}}={\mathbf{T}}_1\cdot {\mathbf{T}}_2=\left[\begin{array}{cc}{A}_1& B_1\\ C_1& D_1\end{array}\right]\left[\begin{array}{cc}{A}_2& B_2\\ C_2& D_2\end{array}\right]\text{(derivation)}$

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6.05 Hybrid (h) & Inverse Hybrid (g) Parameter Formulations

*(Target: Theory Descriptive / Transistor Modeling Context)*

• h-Parameters (Hybrid): Combine series input voltage and parallel output current, selecting input current $I_1$ and output voltage $V_2$ as independent variables. Heavily used for Bipolar Junction Transistor (BJT) modeling: $V_1=h_{11}{I}_1+h_{12}{V}_2$ $I_2=h_{21}{I}_1+h_{22}{V}_2$ In matrix form: $\left[\begin{array}{c}{V}_1\\ I_2\end{array}\right]=\left[\begin{array}{cc}{h}_{11}& h_{12}\\ h_{21}& h_{22}\end{array}\right]\left[\begin{array}{c}{I}_1\\ V_2\end{array}\right]⟹{\mathbf{X}}_h=\mathbf{H}{\mathbf{U}}_h$
• Physical h-Parameter Definitions:
  • $h_{11}$ (Short-Circuit Input Impedance, $[\Omega ]$): $h_{11}={\frac{V_1}{I_1}|}_{V_2=0}$
  • $h_{12}$ (Open-Circuit Reverse Voltage Gain, unitless): $h_{12}={\frac{V_1}{V_2}|}_{I_1=0}$
  • $h_{21}$ (Short-Circuit Forward Current Gain, unitless): $h_{21}={\frac{I_2}{I_1}|}_{V_2=0}$
  • $h_{22}$ (Open-Circuit Output Admittance, $[\text{S}]$): $h_{22}={\frac{I_2}{V_2}|}_{I_1=0}$
• g-Parameters (Inverse Hybrid): The mathematical dual of h-parameters, selecting input voltage $V_1$ and output current $I_2$ as independent variables: $I_1=g_{11}{V}_1+g_{12}{I}_2$ $V_2=g_{21}{V}_1+g_{22}{I}_2$ In matrix form: $\left[\begin{array}{c}{I}_1\\ V_2\end{array}\right]=\left[\begin{array}{cc}{g}_{11}& g_{12}\\ g_{21}& g_{22}\end{array}\right]\left[\begin{array}{c}{V}_1\\ I_2\end{array}\right]⟹{\mathbf{X}}_g=\mathbf{G}{\mathbf{U}}_g$
• Physical g-Parameter Definitions:
  • $g_{11}$ (Open-Circuit Input Admittance, $[\text{S}]$): $g_{11}={\frac{I_1}{V_1}|}_{I_2=0}$
  • $g_{12}$ (Short-Circuit Reverse Current Gain, unitless): $g_{12}={\frac{I_1}{I_2}|}_{V_1=0}$
  • $g_{21}$ (Open-Circuit Forward Voltage Gain, unitless): $g_{21}={\frac{V_2}{V_1}|}_{I_2=0}$
  • $g_{22}$ (Short-Circuit Output Impedance, $[\Omega ]$): $g_{22}={\frac{V_2}{I_2}|}_{V_1=0}$
• g-to-h Matrix Inverse Identity: $\mathbf{G}={\mathbf{H}}^{-1}=\frac{1}{{\Delta }_h}\left[\begin{array}{cc}{h}_{22}& -h_{12}\\ -h_{21}& h_{11}\end{array}\right]\text{where }{\Delta }_h=h_{11}{h}_{22}-h_{12}{h}_{21}\text{(derivation)}$
• Hybrid Symmetries (Reciprocity & Symmetry Rules):
  • Reciprocity: $h_{12}=-h_{21}$ and $g_{12}=-g_{21}$.
  • Symmetry: ${\Delta }_h=h_{11}{h}_{22}-h_{12}{h}_{21}=1$ and ${\Delta }_g=g_{11}{g}_{22}-g_{12}{g}_{21}=1$.
• Transistor CE Equivalent Notations: Input resistance $h_{ie}=h_{11}$, reverse gain feedback $h_{re}=h_{12}$, active current amplification factor (beta) $h_{fe}=h_{21}$, and output channel admittance $h_{oe}=h_{22}$.

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6.06 Master Parameter Symmetries & Conversions

*(Target: Summary Comparisons & Conversion Matrices)*

Table 6.1 Two-Port Network Symmetries & Boundary Criteria

This comparison table compiles the terminal equations, independent variables, and identical symmetry constraints across all parameter models:

Parameter Set	Governing Matrix Equations	Independent Variables	Dependent Variables	Reciprocity Condition	Symmetry Condition
Impedance (Z)	$\mathbf{V}=\mathbf{ZI}$	$I_1,I_2$	$V_1,V_2$	$Z_{12}=Z_{21}$	$Z_{11}=Z_{22}$
Admittance (Y)	$\mathbf{I}=\mathbf{YV}$	$V_1,V_2$	$I_1,I_2$	$Y_{12}=Y_{21}$	$Y_{11}=Y_{22}$
Transmission (T)	${\mathbf{X}}_{\text{send}}=\mathbf{T}{\mathbf{X}}_{\text{rec}}$	$V_2,-I_2$	$V_1,I_1$	$AD-BC=1$	$A=D$
Hybrid (h)	${\mathbf{X}}_h=\mathbf{H}{\mathbf{U}}_h$	$I_1,V_2$	$V_1,I_2$	$h_{12}=-h_{21}$	${\Delta }_h=1$
Inverse Hybrid (g)	${\mathbf{X}}_g=\mathbf{G}{\mathbf{U}}_g$	$V_1,I_2$	$I_1,V_2$	$g_{12}=-g_{21}$	${\Delta }_g=1$

Table 6.2 Matrix Conversion Reference Matrix

Use this table to seamlessly convert any parameter matrix directly into another:

Matrix	Z-Parameters	Y-Parameters	ABCD-Parameters	h-Parameters
$[\mathbf{Z}]$	$\left[\begin{array}{cc}{Z}_{11}& Z_{12}\\ Z_{21}& Z_{22}\end{array}\right]$	$\frac{1}{{\Delta }_Y}\left[\begin{array}{cc}{Y}_{22}& -Y_{12}\\ -Y_{21}& Y_{11}\end{array}\right]$	$\left[\begin{array}{cc}\frac{A}{C}& \frac{{\Delta }_T}{C}\\ \frac{1}{C}& \frac{D}{C}\end{array}\right]$	$\left[\begin{array}{cc}\frac{{\Delta }_h}{h_{22}}& \frac{h_{12}}{h_{22}}\\ -\frac{h_{21}}{h_{22}}& \frac{1}{h_{22}}\end{array}\right]$
$[\mathbf{Y}]$	$\frac{1}{{\Delta }_Z}\left[\begin{array}{cc}{Z}_{22}& -Z_{12}\\ -Z_{21}& Z_{11}\end{array}\right]$	$\left[\begin{array}{cc}{Y}_{11}& Y_{12}\\ Y_{21}& Y_{22}\end{array}\right]$	$\left[\begin{array}{cc}\frac{D}{B}& -\frac{{\Delta }_T}{B}\\ -\frac{1}{B}& \frac{A}{B}\end{array}\right]$	$\left[\begin{array}{cc}\frac{1}{h_{11}}& -\frac{h_{12}}{h_{11}}\\ \frac{h_{21}}{h_{11}}& \frac{{\Delta }_h}{h_{11}}\end{array}\right]$
$[\mathbf{T}]$	$\left[\begin{array}{cc}\frac{Z_{11}}{Z_{21}}& \frac{{\Delta }_Z}{Z_{21}}\\ \frac{1}{Z_{21}}& \frac{Z_{22}}{Z_{21}}\end{array}\right]$	$\left[\begin{array}{cc}-\frac{Y_{22}}{Y_{21}}& -\frac{1}{Y_{21}}\\ -\frac{{\Delta }_Y}{Y_{21}}& -\frac{Y_{11}}{Y_{21}}\end{array}\right]$	$\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]$	$\left[\begin{array}{cc}-\frac{{\Delta }_h}{h_{21}}& -\frac{h_{11}}{h_{21}}\\ -\frac{h_{22}}{h_{21}}& -\frac{1}{h_{21}}\end{array}\right]$
$[\mathbf{H}]$	$\left[\begin{array}{cc}\frac{{\Delta }_Z}{Z_{22}}& \frac{Z_{12}}{Z_{22}}\\ -\frac{Z_{21}}{Z_{22}}& \frac{1}{Z_{22}}\end{array}\right]$	$\left[\begin{array}{cc}\frac{1}{Y_{11}}& -\frac{Y_{12}}{Y_{11}}\\ \frac{Y_{21}}{Y_{11}}& \frac{{\Delta }_Y}{Y_{11}}\end{array}\right]$	$\left[\begin{array}{cc}\frac{B}{D}& \frac{{\Delta }_T}{D}\\ -\frac{1}{D}& \frac{C}{D}\end{array}\right]$	$\left[\begin{array}{cc}{h}_{11}& h_{12}\\ h_{21}& h_{22}\end{array}\right]$

Determinants: ${\Delta }_Z=Z_{11}{Z}_{22}-Z_{12}{Z}_{21}$, ${\Delta }_Y=Y_{11}{Y}_{22}-Y_{12}{Y}_{21}$, ${\Delta }_T=AD-BC$, ${\Delta }_h=h_{11}{h}_{22}-h_{12}{h}_{21}$.

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A. Common Mistakes That Cost Marks

Exam Pitfalls & Marks-Savers Checklist

1. The Negative Output Current Trap ($I_2^{\prime }$ vs. $I_2$): Standard two-port directions specify current entering both Port 1 and Port 2. However, ABCD parameters use $-I_2$ (leaving Port 2). Always check if your nodal loop currents have a negative sign on Port 2 terms when mapping to transmission forms.
2. Feedback Element Admittance Minus Sign: In resistive Pi-networks, the transfer admittance terms $Y_{12}$ and $Y_{21}$ are negative (i.e., $Y_{12}=Y_{21}=-1/R_C$). Forgetting the negative sign on transfer elements is a 3-mark penalty trap.
3. The Reciprocal Hybrid Sign Swap: Remember that in hybrid reciprocity, $h_{12}=-h_{21}$ (NOT $h_{12}=h_{21}$). A positive sign here indicates a non-reciprocal system (like an active transistor amplifier).

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B. PYQ Bank — Verbatim Questions & Answer Plans

1. Network Theory Classification [KUET 2025, 2022 - 4 to 5 Marks]

• Question: What is network theory? Classify the two-port network.
• Answer Plan:
  1. Define Network Theory (Section 6.00 Overview: study of analyzing, modeling, and solving electrical circuits using mathematical methods).
  2. Define Two-Port Network (Section 6.01: access terminals with input Port 1 and output Port 2).
  3. List and classify parameter matrix models: Open-circuit Impedance (Z), Short-circuit Admittance (Y), Hybrid (h), Inverse Hybrid (g), and Transmission (ABCD) parameters, mapping each to its specific application use-case.

2. General T-Network Parameter Extraction [KUET 2024, 2022 - 8 to 9 Marks]

• Question: Calculate the Z and Y parameters for a resistive T-network with resistor values of $$\begin{gathered} R_1=24 \\ ,\Omega \end{gathered}$$, $$\begin{gathered} R_2=8 \\ ,\Omega \end{gathered}$$, and $$\begin{gathered} R_3=8 \\ ,\Omega \end{gathered}$$.
• Answer Plan:
  1. Apply loop equations (KVL) around Port 1 and Port 2 to find Z-parameters: $V_1=32I_1+8I_2$, $V_2=8I_1+16I_2$.
  2. Write final Impedance matrix: $\mathbf{Z}=\left[\begin{array}{cc}32& 8\\ 8& 16\end{array}\right]\Omega$.
  3. Apply matrix inversion $[Y]=[Z{]}^{-1}$ to derive Y-parameters: ${\Delta }_Z=32\times 16-8\times 8=448⟹\mathbf{Y}=\frac{1}{448}\left[\begin{array}{cc}16& -8\\ -8& 32\end{array}\right]=\left[\begin{array}{cc}\frac{1}{28}& -\frac{1}{56}\\ -\frac{1}{56}& \frac{1}{14}\end{array}\right]\text{S}$

3. Cascaded Transmission ABCD Calculation [KUET 2025, 2024 - 12 Marks]

• Question: Obtain ABCD parameters for the network with series arms $$\begin{gathered} R_1=1 \\ ,\Omega \end{gathered}$$, $$\begin{gathered} R_2=2 \\ ,\Omega \end{gathered}$$, and shunt $$\begin{gathered} R_3=5 \\ ,\Omega \end{gathered}$$.
• Answer Plan:
  1. Substitute values into standard T-to-ABCD formulas:
     • $A=1+R_1/R_3=1+1/5=1.2$
     • $$\begin{gathered} B=R_1+R_2+R_1{R}_2/R_3=1+2+2/5=3.4 \\ ,\Omega \end{gathered}$$
     • $C=1/R_3=1/5=0.2\text{ S}$
     • $D=1+R_2/R_3=1+2/5=1.4$
  2. Assemble standard Transmission matrix: $\mathbf{T}=\left[\begin{array}{cc}1.2& 3.4\\ 0.2& 1.4\end{array}\right]$.

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C. Self-Check Before Moving On

• Can you write down all 5 sets of parameter matrix equations from memory?
• Do you know the exact reciprocity and symmetry conditions for Z, Y, ABCD, and h parameters?
• Can you convert a Z-parameter matrix to a Y-parameter matrix using the matrix inversion identity?
• Do you understand why ABCD parameters use a negative current convention for Port 2?

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Source: (k.Deergha Rao) signals and systems.pdf, 04 Network Theory.pdf, Rabiul sir class note.pdf

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6.00 Chapter Map - Two-Port Networks | 6.02 Short-Circuit Admittance (Y) Parameters

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6.01 Open-Circuit Impedance (Z) Parameters

Core Idea

Two-port network theory simplifies complex electrical networks by treating them as a “black box” characterized entirely by their terminal behaviors {the voltages and currents measurable at the external ports}. The Open-Circuit Impedance (Z) Parameters model this black box by expressing the port voltages ($V_1,V_2$) as linear combinations of the port currents ($I_1,I_2$). These parameters are calculated by selectively open-circuiting one port at a time, establishing a robust matrix representation of linear networks.

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1. Defining Two-Port Terminal Variables

An electrical network is classified as a two-port network if it possesses exactly two pairs of access terminals: Port 1 (usually designated as the input port) and Port 2 (usually designated as the output port).

To mathematically analyze these networks as black boxes, we enforce the Two-Port Terminal Convention:

• All port voltages are defined with positive polarity at the upper terminals ($V_1,V_2$).
• Crucial Port Current Direction Rule: Both terminal currents $I_1$ and $I_2$ must be modeled as flowing into their respective ports.

         I1 ---->                           <---- I2
        +--------o-------------------------o--------+
        |        |                         |        |
        |        |                         |        |
    V1  |      Port 1                    Port 2     |  V2
        |  (Input Terminal)          (Output Term.) |
        |        |                         |        |
        +--------o-------------------------o--------+

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2. The Core Z-Parameter Formulation

The Z-parameters relate the terminal voltages directly to the terminal currents through a system of two coupled linear algebraic equations:

$V_1=Z_{11}{I}_1+Z_{12}{I}_2$ $V_2=Z_{21}{I}_1+Z_{22}{I}_2$

Expressing these equations in standard compact vector-matrix form yields:

egin{bmatrix} V_1 \ V_2 \end{bmatrix} = egin{bmatrix} Z_{11} & Z_{12} \ Z_{21} & Z_{22} \end{bmatrix} egin{bmatrix} I_1 \ I_2 \end{bmatrix}

$\mathbf{V}=\mathbf{ZI}$

2.1 Individual Parameter Physical Definitions

By selectively forcing one of the port currents to zero, we isolate and calculate each parameter. Since $I=0$ corresponds to an open-circuit condition, these are termed the Open-Circuit Impedance Parameters:

Parameter	Mathematical Isolation	Physical Name	Description
$Z_{11}$	$$Z_{11} = \left. rac{V_1}{I_1}
ight	_{I_2=0}$$	Input Impedance	The impedance looking into Port 1 when Port 2 is open-circuited.
$Z_{12}$	$$Z_{12} = \left. rac{V_1}{I_2}
ight	_{I_1=0}$$	Reverse Transfer Impedance	The ratio of Port 1 voltage to Port 2 current when Port 1 is open-circuited.
$Z_{21}$	$$Z_{21} = \left. rac{V_2}{I_1}
ight	_{I_2=0}$$	Forward Transfer Impedance	The ratio of Port 2 voltage to Port 1 current when Port 2 is open-circuited.
$Z_{22}$	$$Z_{22} = \left. rac{V_2}{I_2}
ight	_{I_1=0}$$	Output Impedance	The impedance looking into Port 2 when Port 1 is open-circuited.

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3. Network Classification Criteria

3.1 Reciprocal Networks

A network is reciprocal if the transmission of a signal from Port 1 to Port 2 is identical to the transmission from Port 2 to Port 1. For any passive network containing only bilateral linear components {resistors, capacitors, and inductors, with no dependent sources}, the network is guaranteed to be reciprocal.

• Reciprocity Condition: $Z_{12}=Z_{21}$

3.2 Symmetrical Networks

A network is symmetrical if its electrical characteristics do not change when the input and output ports are physically swapped. This requires the network to have identical input and output terminal behaviors.

• Symmetry Condition: $Z_{11}=Z_{22}$

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4. High-Yield Worked Examples (The Exam Killers)

4.1 Example 1: The Resistive T-Network [PYQ 2022 Question 4b - 9 Marks]

Question: Find the Z-parameters for the resistive T-network shown in Fig. 4(b) with branch resistor values of $R_1=24\Omega$, $R_2=8\Omega$, and $R_3=8\Omega$.

              24 Ohm                  8 Ohm
        o-----/\/\/\/\-------+-------/\/\/\/\-----o
        +                    |                    +
     V1                     _ _  8 Ohm         V2
        -                    |                    -
        o--------------------+--------------------o

Step-by-Step Loop Equation Solution:

1. Define Loop Currents: Define loop current $I_1$ circulating clockwise in Loop 1 (left side) and loop current $I_2$ circulating counter-clockwise in Loop 2 (right side). Both currents flow into the upper terminals and meet in the common shunt branch $R_3=8\Omega$.

2. Write KVL around Loop 1: The voltage $V_1$ must equal the sum of the voltage drops across the series arm $R_1$ and the common shunt arm $R_3$: $V_1=I_1{R}_1+(I_1+I_2)R_3$ Substitute $R_1=24\Omega$ and $R_3=8\Omega$: $V_1=24I_1+8I_1+8I_2$ ${\mathbf{V}}_1=32{\mathbf{I}}_1+8{\mathbf{I}}_{2}ext---(Equation1)$

3. Write KVL around Loop 2: The voltage $V_2$ must equal the sum of the voltage drops across the series arm $R_2$ and the common shunt arm $R_3$: $V_2=I_2{R}_2+(I_1+I_2)R_3$ Substitute $R_2=8\Omega$ and $R_3=8\Omega$: $V_2=8I_2+8I_1+8I_2$ ${\mathbf{V}}_2=8{\mathbf{I}}_1+16{\mathbf{I}}_{2}ext---(Equation2)$

4. Compare with Standard Matrix Equations: Compare Equations 1 and 2 directly with the standard parameter definition equations:

   • $V_1=Z_{11}{I}_1+Z_{12}{I}_2⟹Z_{11}=32\Omega ,Z_{12}=8\Omega$
   • $V_2=Z_{21}{I}_1+Z_{22}{I}_2⟹Z_{21}=8\Omega ,Z_{22}=16\Omega$

5. Assemble the Final Impedance Matrix: \mathbf{Z} = egin{bmatrix} Z_{11} & Z_{12} \ Z_{21} & Z_{22} \end{bmatrix} = egin{bmatrix} 32 & 8 \ 8 & 16 \end{bmatrix} \Omega

6. Evaluate Reciprocity and Symmetry:

   • Reciprocity: Since $Z_{12}=Z_{21}=8\Omega$, the network is reciprocal.
   • Symmetry: Since $Z_{11}=32\Omega eq{Z}_{22}=16\Omega$, the network is asymmetrical.

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4.2 Example 2: Classroom Modified T-Network (The Scribbled Page Classic)

Question: Obtain the Z-parameters for the modified T-network where the branch values are scaled down to $R_1=2\Omega$, $R_2=4\Omega$, and $R_3=5\Omega$.

               2 Ohm                   4 Ohm
        o-----/\/\/\/\-------+-------/\/\/\/\-----o
        +                    |                    +
     V1                     _ _  5 Ohm         V2
        -                    |                    -
        o--------------------+--------------------o

Step-by-Step Loop Equation Solution:

1. Write KVL around Loop 1: $V_1=I_1{R}_1+(I_1+I_2)R_3=I_1(2)+5(I_1+I_2)$ ${\mathbf{V}}_1=7{\mathbf{I}}_1+5{\mathbf{I}}_2$

2. Write KVL around Loop 2: $V_2=I_2{R}_2+(I_1+I_2)R_3=I_2(4)+5(I_1+I_2)$ ${\mathbf{V}}_2=5{\mathbf{I}}_1+9{\mathbf{I}}_2$

3. Compare and Extract Parameters:

   • $Z_{11}=7\Omega$ (Input Impedance)
   • $Z_{12}=5\Omega$ (Reverse Transfer Impedance)
   • $Z_{21}=5\Omega$ (Forward Transfer Impedance)
   • $Z_{22}=9\Omega$ (Output Impedance)

4. Assemble the Final Matrix: \mathbf{Z} = egin{bmatrix} 7 & 5 \ 5 & 9 \end{bmatrix} \Omega This scaled network remains reciprocal ($Z_{12}=Z_{21}=5\Omega$) and asymmetrical ($Z_{11}eq{Z}_{22}$).

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4.3 Example 3: Challenge Classic — The Coupled Multi-Loop Lattice

Question: Calculate the Z-parameters for the bridged ladder network shown below, which contains an extra top-loop path.

                  Top Arm: 4 Ohm
            +-------/\/\/\/\-------+
            |                      |
            +--/\/\/\--+--/\/\/\---+
              1 Ohm    |    3 Ohm
                                              / 2 Ohm
                                              |
        o--------------+-----------o

Step-by-Step Nodal-Loop Combined Solution:

Let loop current $I_1$ enter Port 1, loop current $I_2$ enter Port 2, and let $I_3$ be the internal loop current circulating clockwise in the top loop.

1. Formulate Loop Equations:

   • Loop 1 (Port 1): $V_1=3I_1+2I_2-2I_{3}ext---(EquationA)$
   • Loop 2 (Port 2): $V_2=2I_1+5I_2+3I_{3}ext---(EquationB)$
   • Top Loop (Internal Loop 3): 8 I_3 - I_1 + 3 I_2 = 0 \implies I_3 = rac{1}{8} I_1 - rac{3}{8} I_2 \quad ext{--- (Equation C)}

2. Substitute Equation C back into Equations A and B to eliminate the internal state $I_3$:

   • Substitute in $V_1$:

ight)V_1 = 3 I_1 + 2 I_2 - rac{1}{4} I_1 + rac{3}{4} I_2V_1 = \left(3 - rac{1}{4} ight) I_1 + \left(2 + rac{3}{4} ight) I_2\mathbf{V_1 = rac{11}{4} I_1 + rac{11}{4} I_2} \implies Z_{11} = 2.75,\Omega, \ Z_{12} = 2.75,\Omega * **Substitute in $V_2$:** V_2 = 2 I_1 + 5 I_2 + 3\left(rac{1}{8} I_1 - rac{3}{8} I_2 ight)V_2 = 2 I_1 + 5 I_2 + rac{3}{8} I_1 - rac{9}{8} I_2V_2 = \left(2 + rac{3}{8} ight) I_1 + \left(5 - rac{9}{8} ight) I_2\mathbf{V_2 = rac{19}{8} I_1 + rac{31}{8} I_2} \implies Z_{21} = 2.375,\Omega, \ Z_{22} = 3.875,\Omega$$

3. Assemble the Final Matrix: \mathbf{Z} = egin{bmatrix} rac{11}{4} & rac{11}{4} \ rac{19}{8} & rac{31}{8} \end{bmatrix} = egin{bmatrix} 2.75 & 2.75 \ 2.375 & 3.875 \end{bmatrix} \Omega Note: Because this bridged network contains asymmetrical feedback paths, $Z_{12}eq{Z}_{21}$ ($2.75eq2.375$), making it non-reciprocal.

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5. Common Mistakes That Cost Marks

The Port Current Direction Convention Slip

Examiners love to test your attention to detail by drawing Port 2 with current $I_2$ flowing out of the port rather than into it. If you use KVL with an outward current, you must manually substitute $I_{2}o-I_2$ before extracting the Z-parameters. Forgetting this sign change negates the values of $Z_{12}$ and $Z_{22}$, leading to a zero-mark grading.

Dependent Loop/KCL Redundancy Trap

When solving multi-loop networks, do not write KCL at the reference ground node. This introduces redundant dependent variables that cannot be solved algebraically. Always choose independent nodal adder voltages or clockwise loop currents as your variables.

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6. PYQ Bank — Verbatim Questions & Answer Plans

6.1 PYQ 2022 Question 4b [9 Marks]

Question: Find the Z & Y parameters of two port network shown in Fig. 4(b).

• Answer Plan:
  1. Redraw Fig. 4(b) with labeled terminal variables ($V_1,I_1,V_2,I_2$) and clockwise loop currents.
  2. Write the KVL equations around Loop 1 and Loop 2 as shown in Section 4.1.
  3. Compare terms to isolate $Z_{11}=32\Omega ,Z_{12}=8\Omega ,Z_{21}=8\Omega ,Z_{22}=16\Omega$.
  4. State that Y-parameters can be found by inverting this Z-matrix ($[Y]=[Z{]}^{-1}$), which yields: [Y] = rac{1}{(32)(16) - 8^2} egin{bmatrix} 16 & -8 \ -8 & 32 \end{bmatrix} = egin{bmatrix} rac{1}{28} & -rac{1}{56} \ -rac{1}{56} & rac{1}{14} \end{bmatrix} \mho

6.2 Foundational Concept PYQ 2025/2022 [5 Marks]

Question: What is network theory? Classify the two-port parameters.

• Answer Plan:
  1. Define Network Theory as the study of analyzing, modeling, and solving electrical circuits using rigorous mathematical formulations.
  2. Define a Two-Port Network as a circuit block with two terminals on each side allowing energy to flow in and out.
  3. Classify and tabulate the five primary parameter sets (Z, Y, h, g, ABCD) as structured in the Chapter Map (MOC) Section 2.

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7. Self-Check Before Moving On

• Can you state the official terminal sign convention for currents ($I_1,I_2$) entering a two-port network? [9.01]
• Do you know how to write KVL loop equations for any T-network to extract Z-parameters? [9.02]
• Have you memorized the formal reciprocity and symmetry criteria for Z-parameters ($Z_{12}=Z_{21}$ and $Z_{11}=Z_{22}$)? [9.03]
• Can you explain why Z-parameters are called “open-circuit” parameters? [9.04]

---

Source: (k.Deergha Rao) signals and systems.pdf (Chapter 6), 04 Network Theory.pdf, Rabiul sir class note.pdf.

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6.01 Open-Circuit Impedance (Z) Parameters | 6.03 Transmission (ABCD) Parameters & Cascaded Networks

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6.02 Short-Circuit Admittance (Y) Parameters

Core Idea

Short-Circuit Admittance (Y) Parameters model the terminal behavior of a two-port network by expressing the input/output currents ($I_1,I_2$) as linear combinations of the terminal voltages ($V_1,V_2$). Unlike Z-parameters, which rely on open-circuit conditions, Y-parameters are calculated by selectively short-circuiting the ports ($V_1=0$ or $V_2=0$). This formulation is particularly optimized for analyzing parallel-connected networks and forms the mathematical basis for admittance matrix transformations.

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1. The Philosophy of Admittance Modeling

In circuit analysis, impedance represents a component’s opposition to current, whereas admittance represents the ease with which current flows through it {the mathematical reciprocal of impedance, measured in Siemens or Mhos}.

When analyzing multiple subnetworks connected in parallel, open-circuit parameters become algebraically complex to manipulate because the terminal voltages are identical across parallel nodes. By selecting the terminal voltages ($V_1,V_2$) as our independent variables, we can directly sum the terminal currents. This parallel-summation property makes Y-parameters the natural choice for multi-port power networks and transistor modeling.

---

2. The Core Y-Parameter Formulation

The Y-parameters relate the port currents ($I_1,I_2$) directly to the port voltages ($V_1,V_2$) through two coupled linear algebraic equations:

$I_1=Y_{11}{V}_1+Y_{12}{V}_2$ $I_2=Y_{21}{V}_1+Y_{22}{V}_2$

Expressing these equations in standard compact vector-matrix form yields:

egin{bmatrix} I_1 \ I_2 \end{bmatrix} = egin{bmatrix} Y_{11} & Y_{12} \ Y_{21} & Y_{22} \end{bmatrix} egin{bmatrix} V_1 \ V_2 \end{bmatrix}

$\mathbf{I}=\mathbf{YV}$

2.1 Individual Parameter Physical Definitions

By selectively forcing one of the port voltages to zero, we isolate and calculate each parameter. Since $V=0$ corresponds to a short-circuit condition, these are termed the Short-Circuit Admittance Parameters:

Parameter	Mathematical Isolation	Physical Name	Description
$Y_{11}$	$$Y_{11} = \left. rac{I_1}{V_1}
ight	_{V_2=0}$$	Short-Circuit Input Admittance at Port 1	The admittance looking into Port 1 when Port 2 is short-circuited.
$Y_{12}$	$$Y_{12} = \left. rac{I_1}{V_2}
ight	_{V_1=0}$$	Short-Circuit Reverse Transfer Admittance	The ratio of Port 1 current to Port 2 voltage when Port 1 is short-circuited.
$Y_{21}$	$$Y_{21} = \left. rac{I_2}{V_1}
ight	_{V_2=0}$$	Short-Circuit Forward Transfer Admittance	The ratio of Port 2 current to Port 1 voltage when Port 2 is short-circuited.
$Y_{22}$	$$Y_{22} = \left. rac{I_2}{V_2}
ight	_{V_1=0}$$	Short-Circuit Output Admittance at Port 2	The admittance looking into Port 2 when Port 1 is short-circuited.

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3. Network Classification Criteria

3.1 Reciprocal Networks

A network is reciprocal if the transfer admittances are equal, meaning the forward current induced by an input voltage matches the reverse current induced by the same voltage:

• Reciprocity Condition: $Y_{12}=Y_{21}$

3.2 Symmetrical Networks

A network is symmetrical if its input and output port admittance characteristics are identical, meaning the port terminals can be physically reversed in a circuit without changing the system response:

• Symmetry Condition: $Y_{11}=Y_{22}$

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4. The Fundamental Z-to-Y Matrix Inverse Identity

Since open-circuit Z-parameters relate voltages to currents ($\mathbf{V}=\mathbf{ZI}$), and short-circuit Y-parameters relate currents to voltages ($\mathbf{I}=\mathbf{YV}$), the admittance matrix is the exact mathematical inverse of the impedance matrix:

$\mathbf{Y}={\mathbf{Z}}^{-1}$

For any $2imes2$ two-port network matrix, this inverse relationship is expressed as:

egin{bmatrix} Y_{11} & Y_{12} \ Y_{21} & Y_{22} \end{bmatrix} = egin{bmatrix} Z_{11} & Z_{12} \ Z_{21} & Z_{22} \end{bmatrix}^{-1} = rac{1}{\Delta_Z} egin{bmatrix} Z_{22} & -Z_{12} \ -Z_{21} & Z_{11} \end{bmatrix}

where the determinant of the impedance matrix is ${\Delta }_Z=Z_{11}{Z}_{22}-Z_{12}{Z}_{21}$.

4.1 Parameter-by-Parameter Conversion Equations

\mathbf{Y_{11} = rac{Z_{22}}{\Delta_Z}}, \quad \mathbf{Y_{12} = rac{-Z_{12}}{\Delta_Z}}, \quad \mathbf{Y_{21} = rac{-Z_{21}}{\Delta_Z}}, \quad \mathbf{Y_{22} = rac{Z_{11}}{\Delta_Z}}

The Determinant Check

If a network is reciprocal, then $Z_{12}=Z_{21}⟹Y_{12}=Y_{21}$. If a network is symmetrical, then $Z_{11}=Z_{22}⟹Y_{11}=Y_{22}$. This serves as a rapid double-check during exam conditions.

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5. General Resistive $\Pi$-Network Derivation

The $\Pi$-network (or Pi-network) is the canonical circuit topology used to represent short-circuit Y-parameters, just as the T-network naturally maps to open-circuit Z-parameters.

                        feedback: Rc
                    +----/\/\/\/\----+
                    |                  |
           Port 1  _ _                _ _  Port 2
             V1    | | Ra             | | Rb    V2
                    |                  |
        o-----------+------------------+-----------o

To derive the parameters, we apply Kirchhoff’s Current Law (KCL) at the two terminal nodes:

Step 1: Write KCL at Node 1

The current entering Port 1 ($I_1$) splits into the shunt branch $R_A$ and the series feedback branch $R_C$: $I_1=I_{Ra}+I_{Rc}$ I_1 = rac{V_1}{R_A} + rac{V_1 - V_2}{R_C} Grouping voltage terms yields:

ight) V_1 - \left( rac{1}{R_C} ight) V_2 \quad ext{--- (Equation A)}$$ ### Step 2: Write KCL at Node 2 The current entering Port 2 ($I_2$) splits into the shunt branch $R_B$ and the series feedback branch $R_C$: $$I_2 = I_{Rb} - I_{Rc}$$ $$I_2 = rac{V_2}{R_B} + rac{V_2 - V_1}{R_C}$$ Grouping voltage terms yields: $$I_2 = - \left( rac{1}{R_C} ight) V_1 + \left( rac{1}{R_B} + rac{1}{R_C} ight) V_2 \quad ext{--- (Equation B)}$$ ### Step 3: Compare with Standard Y-Parameter Equations Comparing Equations A and B directly with the standard definition equations yields the general Pi-to-Y parameter mapping: $$\mathbf{Y_{11} = rac{1}{R_A} + rac{1}{R_C}}$$ $$\mathbf{Y_{12} = -rac{1}{R_C}}$$ $$\mathbf{Y_{21} = -rac{1}{R_C}}$$ $$\mathbf{Y_{22} = rac{1}{R_B} + rac{1}{R_C}}$$ --- ## 6. Comprehensive Worked Examples (The Exam Killers) ### 6.1 Example 1: The 13-Mark resistive $\Pi$-Network [PYQ 2025 Question 4c - 13 Marks] **Question:** For the resistive $\Pi$-network shown below with component values of $R_A = 2\,\Omega$, $R_B = 2\,\Omega$, and series feedback resistor $R_C = 6\,\Omega$, find the short-circuit Y-parameters. ``` Rc = 6 Ohm +----/\/\/\/\----+ | | Port 1 _ _ _ _ Port 2 V1 | | Ra = 2 Ohm | | Rb = 2 Ohm V2 | | o-----------+------------------+-----------o ``` #### Step-by-Step Nodal KCL Solution: 1. **Define Port Status for $Y_{11}$ and $Y_{21}$ ($V_2 = 0$):** Short-circuit Port 2, making $V_2 = 0$. This places the feedback resistor $R_C = 6\,\Omega$ directly in parallel with the Port 1 shunt resistor $R_A = 2\,\Omega$, while the shunt resistor $R_B = 2\,\Omega$ is shorted out and carries no current. * **Calculate $Y_{11}$:** The input current $I_1$ is: $$I_1 = rac{V_1}{R_A} + rac{V_1}{R_C} = V_1 \left( rac{1}{2} + rac{1}{6} ight) = V_1 \left( rac{3}{6} + rac{1}{6} ight) = rac{4}{6} V_1 = rac{2}{3} V_1$$ $$Y_{11} = \left. rac{I_1}{V_1} ight|_{V_2=0} = \mathbf{rac{2}{3} pprox 0.667 ext{ S}}$$ * **Calculate $Y_{21}$:** With $V_2 = 0$, the short-circuit current $I_2$ flows *into* Node 2 from the feedback arm. Because of the current direction convention, $I_2$ is negative relative to the current leaving Node 1: $$I_2 = rac{V_2 - V_1}{R_C} = rac{0 - V_1}{6} = -rac{1}{6} V_1$$ $$Y_{21} = \left. rac{I_2}{V_1} ight|_{V_2=0} = \mathbf{-rac{1}{6} pprox -0.167 ext{ S}}$$ 2. **Define Port Status for $Y_{22}$ and $Y_{12}$ ($V_1 = 0$):** Short-circuit Port 1, making $V_1 = 0$. This shorts out the shunt resistor $R_A = 2\,\Omega$ and puts $R_C = 6\,\Omega$ in parallel with $R_B = 2\,\Omega$. * **Calculate $Y_{22}$:** The output current $I_2$ is: $$I_2 = rac{V_2}{R_B} + rac{V_2}{R_C} = V_2 \left( rac{1}{2} + rac{1}{6} ight) = rac{2}{3} V_2$$ $$Y_{22} = \left. rac{I_2}{V_2} ight|_{V_1=0} = \mathbf{rac{2}{3} pprox 0.667 ext{ S}}$$ * **Calculate $Y_{12}$:** The feedback current entering Node 1 is: $$I_1 = rac{V_1 - V_2}{R_C} = rac{0 - V_2}{6} = -rac{1}{6} V_2$$ $$Y_{12} = \left. rac{I_1}{V_2} ight|_{V_1=0} = \mathbf{-rac{1}{6} pprox -0.167 ext{ S}}$$ 3. **Assemble the Admittance Matrix:** $$\mathbf{Y} = egin{bmatrix} Y_{11} & Y_{12} \ Y_{21} & Y_{22} \end{bmatrix} = egin{bmatrix} rac{2}{3} & -rac{1}{6} \ -rac{1}{6} & rac{2}{3} \end{bmatrix} ext{ S}$$ 4. **Evaluate Reciprocity and Symmetry:** * **Reciprocity:** Since $Y_{12} = Y_{21} = -rac{1}{6} ext{ S}$, the network is **reciprocal**. * **Symmetry:** Since $Y_{11} = Y_{22} = rac{2}{3} ext{ S}$, the network is **symmetrical**. --- ### 6.2 Example 2: Inversion of Exam Impedance Matrix [PYQ 2022 Question 4b - 9 Marks] **Question:** Obtain the Y-parameters for the 2022 T-network exam problem by directly inverting its impedance matrix $[Z] = egin{bmatrix} 32 & 8 \ 8 & 16 \end{bmatrix} \Omega$. #### Step-by-Step Matrix Inversion Solution: 1. **Calculate the Impedance Determinant ($\Delta_Z$):** $$\Delta_Z = Z_{11}Z_{22} - Z_{12}Z_{21}$$ $$\Delta_Z = (32)(16) - (8)(8) = 512 - 64 = \mathbf{448}$$ 2. **Apply the Inverse Matrix Formula:** $$\mathbf{Y} = \mathbf{Z}^{-1} = rac{1}{\Delta_Z} egin{bmatrix} Z_{22} & -Z_{12} \ -Z_{21} & Z_{11} \end{bmatrix}$$ $$\mathbf{Y} = rac{1}{448} egin{bmatrix} 16 & -8 \ -8 & 32 \end{bmatrix}$$ 3. **Evaluate Individual Admittance Coefficients:** * $$Y_{11} = rac{16}{448} = \mathbf{rac{1}{28} ext{ S}} pprox 0.0357 ext{ S}$$ * $$Y_{12} = rac{-8}{448} = \mathbf{-rac{1}{56} ext{ S}} pprox -0.0179 ext{ S}$$ * $$Y_{21} = rac{-8}{448} = \mathbf{-rac{1}{56} ext{ S}} pprox -0.0179 ext{ S}$$ * $$Y_{22} = rac{32}{448} = \mathbf{rac{1}{14} ext{ S}} pprox 0.0714 ext{ S}$$ 4. **Assemble the Final Admittance Matrix:** $$\mathbf{Y} = egin{bmatrix} rac{1}{28} & -rac{1}{56} \ -rac{1}{56} & rac{1}{14} \end{bmatrix} ext{ S}$$ --- ## 7. Common Mistakes That Cost Marks > [!danger] **The Negative Sign Omission on Transfer Admittances** > > The most frequent mark-loss in Chapter 6 examinations occurs when writing the transfer parameters $Y_{12}$ and $Y_{21}$ for a resistive Pi-network. Because port current $I_1$ flows **into** the upper node, KCL dictates that the current flowing from Node 1 to Node 2 is proportional to the difference $(V_1 - V_2)$. This introduces a negative sign on $V_2$ in the first port equation, making $Y_{12} = -1/R_C$. Writing these transfer terms as positive values will result in a **50% deduction** of the numerical question's marks. > [!warning] **The Inverse Unit Trap** > > Admittance is the reciprocal of impedance, meaning its unit is **Siemens (S)** or **Mhos ($\mho$)**. Writing the unit of Y-parameters as **Ohms ($\Omega$)** indicates a fundamental conceptual misunderstanding, resulting in a **1 to 2 mark penalty** on any numerical evaluation. --- ## 8. PYQ Bank — Verbatim Questions & Answer Plans ### 8.1 PYQ 2025 Question 4c [13 Marks] **Question:** For the following $\Pi$-network as shown in Fig. 4(c), find the Y-parameters of it. * **Answer Plan:** 1. Draw the schematic of the resistive Pi-network, labeling shunt resistors $R_A, R_B$ and feedback resistor $R_C$. 2. Write the Node 1 and Node 2 KCL algebraic equations as derived in **Section 5**. 3. Plug in the specific resistor values and solve the short-circuit cases as shown in **Section 6.1**. 4. Assemble the final matrix and state that the network is reciprocal and symmetrical. ### 8.2 PYQ 2022 Question 4b [9 Marks] **Question:** Find the Z & Y parameters of two port network shown in Fig. 4(b). * **Answer Plan:** 1. Obtain the Z-parameters first using loop analysis (as shown in **Note 6.01, Section 4.1**). 2. Apply the fundamental matrix inverse identity $\mathbf{Y} = \mathbf{Z}^{-1}$. 3. Compute $\Delta_Z = 448$ and substitute terms to derive the exact fractional admittances shown in **Section 6.2**. --- ## 9. Self-Check Before Moving On - [ ] Can you define Short-Circuit Admittance mathematically and physically? [10.01] - [ ] Do you know how to derive the Pi-network parameters using nodal KCL analysis? [10.02] - [ ] Have you memorized the warning to always include a negative sign on $Y_{12}$ and $Y_{21}$? [10.03] - [ ] Can you convert any Z-parameter matrix into Y-parameters using the determinant inverse formula? [10.04] --- *Source: (k.Deergha Rao) signals and systems.pdf (Chapter 6), 04 Network Theory.pdf, Rabiul sir class note.pdf.* --- [[6.02_Short-Circuit_Admittance_Y_Parameters|6.02 Short-Circuit Admittance (Y) Parameters]] | [[6.04_Hybrid_h_and_Inverse_Hybrid_g_Parameter_Formulations|6.04 Hybrid (h) & Inverse Hybrid (g) Parameter Formulations]] --- # 6.03 Transmission (ABCD) Parameters & Cascaded Networks > [!abstract] Core Idea > > **Transmission (ABCD) Parameters** model two-port networks by directly expressing the sending-end variables ($V_1, I_1$) as linear combinations of the receiving-end variables ($V_2, -I_2$). Unlike Z or Y parameters, ABCD parameters utilize a **negative current convention** for Port 2 current, modeling the current as flowing *out* of Port 2 rather than into it. This formulation is uniquely optimized for **cascaded networks**—where multiple two-port subsystems are connected in series—allowing the overall system matrix to be calculated via simple, sequential matrix multiplication. --- ## 1. The Philosophy of Transmission Modeling In telecommunication systems, power distribution networks, and cascading filter chains, we analyze the transmission of signals from an **input source** (sending-end) to an **output load** (receiving-end). ### 1.1 The Negative Output Current Convention The standard two-port current convention defines both currents $I_1$ and $I_2$ as flowing **into** their respective ports. However, when two networks are cascaded in series, the current flowing *out* of the first network is exactly the current flowing *into* the second network. To make cascading mathematically seamless, transmission parameters define Port 2 current as **leaving** the port. To preserve the standard terminal coordinate system, we introduce a negative sign on the Port 2 current term, using $-I_2$ in all formulation equations: ``` I1 ----> I2' = -I2 ----> +--------o-------------------------o--------+ | | | | | | | | V1 | Port 1 Port 2 | V2 | (Sending End) (Receiving End) | | | | | +--------o-------------------------o--------+ ``` --- ## 2. The Core ABCD Parameter Formulation The transmission parameters relate the sending-end voltage and current ($V_1, I_1$) to the receiving-end voltage and current ($V_2, -I_2$) through two coupled linear algebraic equations: $$V_1 = A V_2 - B I_2$$ $$I_1 = C V_2 - D I_2$$ Expressing these equations in standard compact **vector-matrix form** yields: $$egin{bmatrix} V_1 \ I_1 \end{bmatrix} = egin{bmatrix} A & B \ C & D \end{bmatrix} egin{bmatrix} V_2 \ -I_2 \end{bmatrix}$$ $$\mathbf{X}_{ ext{send}} = \mathbf{T}\mathbf{X}_{ ext{rec}}$$ ### 2.1 Individual Parameter Physical Definitions By selectively forcing either the receiving-end current to zero (open-circuit) or the receiving-end voltage to zero (short-circuit), we mathematically isolate and evaluate each parameter: | Parameter | Mathematical Isolation | Physical Name | Dimensions | Description | | :--- | :--- | :--- | :--- | :--- | | **$A$** | $$A = \left. rac{V_1}{V_2} ight|_{I_2=0}$$ | **Open-Circuit Voltage Ratio** | Dimensionless | Reverse voltage gain of the network with Port 2 open-circuited. | | **$B$** | $$B = \left. rac{V_1}{-I_2} ight|_{V_2=0}$$ | **Short-Circuit Transfer Impedance** | Ohms ($\Omega$) | The ratio of input voltage to output short-circuit current. | | **$C$** | $$C = \left. rac{I_1}{V_2} ight|_{I_2=0}$$ | **Open-Circuit Transfer Admittance** | Siemens ($\mho$) | The ratio of input current to output voltage when Port 2 is open-circuited. | | **$D$** | $$D = \left. rac{I_1}{-I_2} ight|_{V_2=0}$$ | **Short-Circuit Current Ratio** | Dimensionless | Reverse current gain of the network with Port 2 short-circuited. | --- ## 3. Network Classification Criteria ### 3.1 Reciprocal Networks A two-port network is **reciprocal** if the transmission of a signal from Port 1 to Port 2 is identical to transmission from Port 2 to Port 1. For ABCD parameters, this physical property is governed by the **matrix determinant rule**: * **Reciprocity Condition:** $$\Delta_T = AD - BC = 1$$ ### 3.2 Symmetrical Networks A two-port network is **symmetrical** if its input and output terminals can be physically swapped without altering the overall electrical characteristics of the system. For ABCD parameters, this requires: * **Symmetry Condition:** $$A = D$$ --- ## 4. Rigorous Derivation of T-Network ABCD Parameters Let us derive the general algebraic formulas for the ABCD parameters of a standard resistive T-network with series branch resistors $R_1$, $R_2$ and common shunt resistor $R_3$. ``` R1 R2 o-----/\/\/\-------+-------/\/\/\-----o + | + V1 _ _ R3 V2 - | - o--------------------+--------------------o ``` ### Step 1: Write KVL Equations with Standard Currents ($I_1, I_2$ entering) $$V_1 = (R_1 + R_3)I_1 + R_3 I_2$$ $$V_2 = R_3 I_1 + (R_2 + R_3)I_2$$ ### Step 2: Introduce the Leaving Current Convention ($I_2 = -I_2'$) Substitute $I_2 = -I_2'$ (where $I_2'$ is the current flowing *out* of Port 2): $$V_1 = (R_1 + R_3)I_1 - R_3 I_2' \quad ext{--- (Equation 1)}$$ $$V_2 = R_3 I_1 - (R_2 + R_3)I_2' \quad ext{--- (Equation 2)}$$ ### Step 3: Isolate Sending-End Current ($I_1$) from Equation 2 From Equation 2, solve for $I_1$: $$R_3 I_1 = V_2 + (R_2 + R_3)I_2'$$ $$I_1 = \left(rac{1}{R_3} ight) V_2 + \left(rac{R_2 + R_3}{R_3} ight) I_2'$$ $$\mathbf{I_1 = \left(rac{1}{R_3} ight) V_2 + \left(1 + rac{R_2}{R_3} ight) I_2'} \quad ext{--- (Isolated Current)}$$ ### Step 4: Substitute $I_1$ back into Equation 1 to Solve for $V_1$ $$V_1 = (R_1 + R_3)\left[ rac{1}{R_3} V_2 + \left(1 + rac{R_2}{R_3} ight) I_2' ight] - R_3 I_2'$$ $$V_1 = \left( rac{R_1 + R_3}{R_3} ight) V_2 + \left[ rac{(R_1 + R_3)(R_2 + R_3)}{R_3} - R_3 ight] I_2'$$ $$V_1 = \left( 1 + rac{R_1}{R_3} ight) V_2 + \left[ rac{R_1 R_2 + R_1 R_3 + R_2 R_3 + R_3^2 - R_3^2}{R_3} ight] I_2'$$ $$\mathbf{V_1 = \left(1 + rac{R_1}{R_3} ight) V_2 + \left(R_1 + R_2 + rac{R_1 R_2}{R_3} ight) I_2'} \quad ext{--- (Isolated Voltage)}$$ ### Step 5: Direct Parameter Comparison Comparing our isolated equations with $V_1 = A V_2 + B I_2'$ and $I_1 = C V_2 + D I_2'$ yields the **General T-Network Formulas**: $$\mathbf{A = 1 + rac{R_1}{R_3}}, \quad \mathbf{B = R_1 + R_2 + rac{R_1 R_2}{R_3}}, \quad \mathbf{C = rac{1}{R_3}}, \quad \mathbf{D = 1 + rac{R_2}{R_3}}$$ --- ## 5. Comprehensive Worked Examples (The Exam Killers) ### 5.1 Example 1: The asymmetrical T-Network [Rabiul Sir class Note / Exam standard] **Question:** Obtain the ABCD parameters for a resistive T-network with branch impedances $R_1 = 1\,\Omega$, $R_2 = 2\,\Omega$, and $R_3 = 5\,\Omega$. #### Method A: Direct Formula Substitution Using the formulas derived in **Section 4**: * $A = 1 + rac{R_1}{R_3} = 1 + rac{1}{5} = \mathbf{1.2} = \mathbf{rac{6}{5}}$ * $B = R_1 + R_2 + rac{R_1 R_2}{R_3} = 1 + 2 + rac{1 imes 2}{5} = 3 + 0.4 = \mathbf{3.4\,\Omega} = \mathbf{rac{17}{5}\,\Omega}$ * $C = rac{1}{R_3} = rac{1}{5} = \mathbf{0.2 ext{ S}} = \mathbf{rac{1}{5} ext{ S}}$ * $D = 1 + rac{R_2}{R_3} = 1 + rac{2}{5} = \mathbf{1.4} = \mathbf{rac{7}{5}}$ $$\mathbf{T} = egin{bmatrix} A & B \ C & D \end{bmatrix} = egin{bmatrix} 1.2 & 3.4 \ 0.2 & 1.4 \end{bmatrix} = egin{bmatrix} rac{6}{5} & rac{17}{5} \ rac{1}{5} & rac{7}{5} \end{bmatrix}$$ #### Method B: Boundary-Condition Analysis (Rigorously proving each parameter) This method is highly favored by examiners because it shows complete theoretical understanding of the boundary states. **1. Open-Circuit Port 2 ($I_2 = 0$):** * With Port 2 open-circuited, no current flows through $R_2 = 2\,\Omega$. * Therefore, the voltage across the shunt resistor $R_3 = 5\,\Omega$ is exactly the output voltage $V_2$: $$V_2 = I_1 R_3 = 5 I_1 \implies I_1 = 0.2 V_2 \implies C = rac{I_1}{V_2} = \mathbf{0.2 ext{ S}}$$ * The input voltage $V_1$ is the voltage drop across $R_1$ and $R_3$: $$V_1 = I_1 (R_1 + R_3) = I_1 (1 + 5) = 6 I_1$$ * Substitute $I_1 = 0.2 V_2$: $$V_1 = 6 (0.2 V_2) = 1.2 V_2 \implies A = rac{V_1}{V_2} = \mathbf{1.2}$$ **2. Short-Circuit Port 2 ($V_2 = 0$):** * With Port 2 short-circuited to ground, the output terminals are closed, forcing $V_2 = 0$. * The parallel combination of the shunt branch $R_3 = 5\,\Omega$ and output branch $R_2 = 2\,\Omega$ is: $$R_p = R_3 \parallel R_2 = rac{5 imes 2}{5 + 2} = rac{10}{7}\,\Omega$$ * The total equivalent input resistance of the network is: $$R_{ ext{eq}} = R_1 + R_p = 1 + rac{10}{7} = rac{17}{7}\,\Omega$$ * Therefore, the sending-end voltage is related to the input current by: $$V_1 = R_{ ext{eq}} I_1 = rac{17}{7} I_1 \implies I_1 = rac{7}{17} V_1$$ * By the **current division rule**, the current flowing out of Port 2 ($I_2' = -I_2$) is: $$I_2' = I_1 \left( rac{R_3}{R_3 + R_2} ight) = I_1 \left( rac{5}{5 + 2} ight) = rac{5}{7} I_1$$ * Substitute $I_1 = rac{7}{17} V_1$: $$I_2' = rac{5}{7} \left( rac{7}{17} V_1 ight) = rac{5}{17} V_1 \implies V_1 = rac{17}{5} I_2' \implies B = rac{V_1}{I_2'} = \mathbf{3.4\,\Omega}$$ * Now, isolate $I_1$ in terms of $I_2'$: $$I_2' = rac{5}{7} I_1 \implies I_1 = rac{7}{5} I_2' \implies D = rac{I_1}{I_2'} = \mathbf{1.4}$$ *Check Reciprocity:* $$AD - BC = (1.2)(1.4) - (3.4)(0.2) = 1.68 - 0.68 = \mathbf{1} \quad ext{(Reciprocal!)}$$ --- ### 5.2 Example 2: Symmetrical T-Network Realization [PYQ 2022 Question 4c] **Question:** Determine the ABCD matrix for the T-network shown in Fig. 4(c) with values $R_1 = 2\,\Omega$, $R_2 = 2\,\Omega$, and $R_3 = 5\,\Omega$. #### Step-by-Step Solution: 1. **Identify parameters:** $R_1 = 2\,\Omega, \ R_2 = 2\,\Omega, \ R_3 = 5\,\Omega$. 2. **Calculate $A$ (Open-Circuit voltage ratio):** $$A = 1 + rac{R_1}{R_3} = 1 + rac{2}{5} = \mathbf{1.4}$$ 3. **Calculate $B$ (Short-Circuit impedance):** $$B = R_1 + R_2 + rac{R_1 R_2}{R_3} = 2 + 2 + rac{2 imes 2}{5} = 4 + 0.8 = \mathbf{4.8\,\Omega}$$ 4. **Calculate $C$ (Open-Circuit admittance):** $$C = rac{1}{R_3} = rac{1}{5} = \mathbf{0.2 ext{ S}}$$ 5. **Calculate $D$ (Short-Circuit current ratio):** $$D = 1 + rac{R_2}{R_3} = 1 + rac{2}{5} = \mathbf{1.4}$$ $$\mathbf{T} = egin{bmatrix} A & B \ C & D \end{bmatrix} = egin{bmatrix} 1.4 & 4.8 \ 0.2 & 1.4 \end{bmatrix} = egin{bmatrix} rac{7}{5} & rac{24}{5} \ rac{1}{5} & rac{7}{5} \end{bmatrix}$$ *Symmetry & Reciprocity Checks:* * **Symmetry:** Since $A = D = 1.4$, the network is **symmetrical**. * **Reciprocity:** $$AD - BC = (1.4)(1.4) - (4.8)(0.2) = 1.96 - 0.96 = \mathbf{1} \quad ext{(Reciprocal!)}$$ --- ## 6. Cascaded Networks & Matrix Multiplication When two networks $N_a$ and $N_b$ are connected in **cascade** {series connection of two-port networks}, the receiving end of Network $A$ becomes the sending end of Network $B$: ``` I1 ----> Ia' = Ib ----> Ib' = -I2 ----> +--------o-------------o--------o-------------o--------+ | | | | | | V1 | Na | V_mid | Nb | V2 | | | | | | +--------o-------------o--------o-------------o--------+ ``` ### 6.1 Cascade Proof By definition: $$egin{bmatrix} V_1 \ I_1 \end{bmatrix} = \mathbf{T}_a egin{bmatrix} V_{ ext{mid}} \ I_a' \end{bmatrix} \quad ext{and} \quad egin{bmatrix} V_{ ext{mid}} \ I_b \end{bmatrix} = \mathbf{T}_b egin{bmatrix} V_2 \ -I_2 \end{bmatrix}$$ Since $I_a' = I_b$, we substitute the second matrix equation directly into the first: $$egin{bmatrix} V_1 \ I_1 \end{bmatrix} = \mathbf{T}_a \mathbf{T}_b egin{bmatrix} V_2 \ -I_2 \end{bmatrix}$$ $$\mathbf{T}_{ ext{total}} = \mathbf{T}_a \cdot \mathbf{T}_b = egin{bmatrix} A_a & B_a \ C_a & D_a \end{bmatrix} egin{bmatrix} A_b & B_b \ C_b & D_b \end{bmatrix}$$ > [!danger] **The Multiplication Order Constraint** > > Matrix multiplication is **non-commutative** ($\mathbf{T}_a \mathbf{T}_b eq \mathbf{T}_b \mathbf{T}_a$). You must always multiply the transmission matrices in the exact physical sequence of the signal path, from the sending-end to the receiving-end. Swapping the order results in a completely incorrect system model. --- ### 6.2 Worked Cascade Numerical **Question:** Calculate the overall ABCD transmission matrix for a cascade of two identical asymmetrical T-networks from **Section 5.1** ($R_1 = 1\,\Omega, R_2 = 2\,\Omega, R_3 = 5\,\Omega$). $$\mathbf{T}_a = \mathbf{T}_b = egin{bmatrix} 1.2 & 3.4 \ 0.2 & 1.4 \end{bmatrix}$$ #### Step-by-Step Multiplication: $$\mathbf{T}_{ ext{total}} = egin{bmatrix} 1.2 & 3.4 \ 0.2 & 1.4 \end{bmatrix} egin{bmatrix} 1.2 & 3.4 \ 0.2 & 1.4 \end{bmatrix} = egin{bmatrix} A_{ ext{total}} & B_{ ext{total}} \ C_{ ext{total}} & D_{ ext{total}} \end{bmatrix}$$ * **$A_{ ext{total}}$:** $$A_{ ext{total}} = (1.2)(1.2) + (3.4)(0.2) = 1.44 + 0.68 = \mathbf{2.12}$$ * **$B_{ ext{total}}$:** $$B_{ ext{total}} = (1.2)(3.4) + (3.4)(1.4) = 4.08 + 4.76 = \mathbf{8.84\,\Omega}$$ * **$C_{ ext{total}}$:** $$C_{ ext{total}} = (0.2)(1.2) + (1.4)(0.2) = 0.24 + 0.28 = \mathbf{0.52 ext{ S}}$$ * **$D_{ ext{total}}$:** $$D_{ ext{total}} = (0.2)(3.4) + (1.4)(1.4) = 0.68 + 1.96 = \mathbf{2.64}$$ $$\mathbf{T}_{ ext{total}} = egin{bmatrix} 2.12 & 8.84 \ 0.52 & 2.64 \end{bmatrix} = egin{bmatrix} rac{53}{25} & rac{221}{25} \ rac{13}{25} & rac{66}{25} \end{bmatrix}$$ *Check Reciprocity of Cascade:* $$\Delta_{T_{ ext{total}}} = (2.12)(2.64) - (8.84)(0.52) = 5.5968 - 4.5968 = \mathbf{1} \quad ext{(Perfectly Verified!)}$$ --- ## 7. Common Mistakes That Cost Marks > [!danger] **The Symmetrical Component Assumption** > > Students often assume that if a network is passive, it must be symmetrical ($A = D$). This is only true if the input and output branch values are identical ($R_1 = R_2$ in a T-network). Passive networks are always reciprocal ($AD - BC = 1$), but they are **not** always symmetrical. Always check $A = D$ explicitly before declaring symmetry. > [!warning] **The Minus Sign in Parameter Multiplication** > > When cascading, do not manually negate any elements of the calculated matrices. The negative sign on $I_2$ is already embedded in the mathematical definition of the ABCD parameters to make direct matrix multiplication valid. Direct, un-modified matrix multiplication yields the mathematically correct, flawless terminal results. --- ## 8. PYQ Bank — Verbatim Questions & Answer Plans ### 8.1 PYQ 2024 Question 4b [12 Marks] **Question:** Obtain the ABCD parameters for a resistive T-network with branch impedances $R_1 = 1\,\Omega$, $R_2 = 2\,\Omega$, and $R_3 = 5\,\Omega$. * **Answer Plan:** Redraw the T-network circuit diagram, apply either KVL loop equations (Method A in **Section 5.1**) or open/short boundary conditions (Method B in **Section 5.1**) to rigorously solve for $A = 1.2$, $B = 3.4\,\Omega$, $C = 0.2 ext{ S}$, and $D = 1.4$. Show the final matrix representation to lock in all 12 marks. ### 8.2 PYQ 2022 Question 4c [3 Marks] **Question:** Determine the ABCD matrix for the $2\,\Omega, 2\,\Omega, 5\,\Omega$ T-network shown in Fig. 4(c). * **Answer Plan:** Identify $R_1 = 2\,\Omega, R_2=2\,\Omega, R_3=5\,\Omega$. Use the quick companion formulas: $A = 1 + R_1/R_3 = 1.4$, $B = R_1 + R_2 + R_1R_2/R_3 = 4.8\,\Omega$, $C = 1/R_3 = 0.2 ext{ S}$, $D = 1 + R_2/R_3 = 1.4$ to quickly secure full marks. --- ## 9. Self-Check Before Moving On - [ ] Can you explain why the Port 2 current is defined as leaving the port in transmission parameters? - [ ] Do you know the exact matrix determinant check for reciprocity ($\Delta_T = 1$) and symmetry ($A = D$)? - [ ] Can you derive the ABCD parameters for any T-network using KVL/KCL algebra? - [ ] Have you memorized the warning that matrix multiplication is non-commutative when solving cascaded networks? --- *Source: (k.Deergha Rao) signals and systems.pdf (Chapter 6), 04 Network Theory.pdf, Rabiul sir class note.pdf.* --- [[6.03_Transmission_ABCD_Parameters_and_Cascaded_Networks|6.03 Transmission (ABCD) Parameters & Cascaded Networks]] | [[6.00_Chapter_Map_-_Two-Port_Networks|6.00 Chapter Map - Two-Port Networks]] --- # 6.04 Hybrid (h) & Inverse Hybrid (g) Parameter Formulations > [!abstract] Core Idea > > **Hybrid (h) and Inverse Hybrid (g) Parameters** represent two-port networks by combining open-circuit and short-circuit terminal conditions. Unlike Z-parameters (pure open-circuit) or Y-parameters (pure short-circuit), hybrid parameters mix voltage and current independent variables. This mathematical formulation is exceptionally optimized for **bipolar junction transistor (BJT) modeling** and electronic circuit simulations, where the input and output terminals exhibit extremely disparate impedance scales. --- ## 1. The Philosophy of Hybrid Modeling In electronics, different parts of a circuit require different types of physical measurements. For example, a transistor's input terminal typically has a low impedance (easy to short-circuit, difficult to open-circuit), while its output terminal has a high impedance (easy to open-circuit, difficult to short-circuit). By defining the input current ($I_1$) and output voltage ($V_2$) as our **independent variables**, we create a hybrid formulation that is both practically measurable in laboratories and computationally stable: * **Input port behaves like a series circuit** {represented as a voltage dependent on input current and output voltage}. * **Output port behaves like a parallel circuit** {represented as a current dependent on input current and output voltage}. --- ## 2. The Core h-Parameter Formulation The **h-parameters** relate terminal variables through the following system of coupled linear equations: $$V_1 = h_{11}I_1 + h_{12}V_2$$ $$I_2 = h_{21}I_1 + h_{22}V_2$$ Expressing these equations in standard compact **vector-matrix form** yields: $$egin{bmatrix} V_1 \ I_2 \end{bmatrix} = egin{bmatrix} h_{11} & h_{12} \ h_{21} & h_{22} \end{bmatrix} egin{bmatrix} I_1 \ V_2 \end{bmatrix}$$ $$\mathbf{X}_{h} = \mathbf{H}\mathbf{U}_{h}$$ ### 2.1 Individual Parameter Physical Definitions By selectively setting the independent variables $I_1 = 0$ (open-circuit) or $V_2 = 0$ (short-circuit), we isolate and calculate each parameter: | Parameter | Mathematical Isolation | Physical Name | Unit | Description | | :--- | :--- | :--- | :--- | :--- | | **$h_{11}$** | $$h_{11} = \left. rac{V_1}{I_1} ight|_{V_2=0}$$ | **Short-Circuit Input Impedance** | Ohms ($\Omega$) | The impedance looking into Port 1 when Port 2 is short-circuited. | | **$h_{12}$** | $$h_{12} = \left. rac{V_1}{V_2} ight|_{I_1=0}$$ | **Open-Circuit Reverse Voltage Gain** | Unitless | The voltage feedback ratio from output to input when Port 1 is open. | | **$h_{21}$** | $$h_{21} = \left. rac{I_2}{I_1} ight|_{V_2=0}$$ | **Short-Circuit Forward Current Gain** | Unitless | The forward current amplification factor when Port 2 is short-circuited. | | **$h_{22}$** | $$h_{22} = \left. rac{I_2}{V_2} ight|_{I_1=0}$$ | **Open-Circuit Output Admittance** | Siemens (S) | The admittance looking into Port 2 when Port 1 is open-circuited. | --- ## 3. The Dual: Inverse Hybrid (g) Parameters The **g-parameters** are the exact mathematical dual of the h-parameters, using input voltage ($V_1$) and output current ($I_2$) as the independent variables: $$I_1 = g_{11}V_1 + g_{12}I_2$$ $$V_2 = g_{21}V_1 + g_{22}I_2$$ Expressing these equations in standard compact **vector-matrix form** yields: $$egin{bmatrix} I_1 \ V_2 \end{bmatrix} = egin{bmatrix} g_{11} & g_{12} \ g_{21} & g_{22} \end{bmatrix} egin{bmatrix} V_1 \ I_2 \end{bmatrix}$$ $$\mathbf{X}_{g} = \mathbf{G}\mathbf{U}_{g}$$ ### 3.1 Individual Parameter Physical Definitions | Parameter | Mathematical Isolation | Physical Name | Unit | Description | | :--- | :--- | :--- | :--- | :--- | | **$g_{11}$** | $$g_{11} = \left. rac{I_1}{V_1} ight|_{I_2=0}$$ | **Open-Circuit Input Admittance** | Siemens (S) | Input admittance when Port 2 is open. | | **$g_{12}$** | $$g_{12} = \left. rac{I_1}{I_2} ight|_{V_1=0}$$ | **Short-Circuit Reverse Current Gain** | Unitless | Reverse current transfer ratio when Port 1 is shorted. | | **$g_{21}$** | $$g_{21} = \left. rac{V_2}{V_1} ight|_{I_2=0}$$ | **Open-Circuit Forward Voltage Gain** | Unitless | Forward voltage amplification factor when Port 2 is open. | | **$g_{22}$** | $$g_{22} = \left. rac{V_2}{I_2} ight|_{V_1=0}$$ | **Short-Circuit Output Impedance** | Ohms ($\Omega$) | Output impedance when Port 1 is short-circuited. | ### 3.2 The g-to-h Matrix Inverse Relationship Because g-parameters and h-parameters swap independent and dependent variables, the g-parameter matrix is the exact mathematical inverse of the h-parameter matrix: $$\mathbf{G} = \mathbf{H}^{-1}$$ $$egin{bmatrix} g_{11} & g_{12} \ g_{21} & g_{22} \end{bmatrix} = rac{1}{\Delta_h} egin{bmatrix} h_{22} & -h_{12} \ -h_{21} & h_{11} \end{bmatrix}$$ where the determinant of the hybrid matrix is $\Delta_h = h_{11}h_{22} - h_{12}h_{21}$. --- ## 4. Transistor Modeling Context (BJT CE Amplifier) In analog microelectronics, h-parameters are the standard representation for the Common-Emitter (CE) configuration of a Bipolar Junction Transistor (BJT). The parameters are rewritten with descriptive subscripts to represent their physical equivalents: * **$h_{ie} = h_{11}$** (Input Impedance with output shorted): Low resistance representing the forward-biased base-emitter junction. * **$h_{re} = h_{12}$** (Reverse Voltage gain with input open): Extremely small feedback ratio representing base-width modulation. * **$h_{fe} = h_{21}$** (Forward Current gain with output shorted): The transistor's active-region current amplification factor ($eta$). * **$h_{oe} = h_{22}$** (Output Admittance with input open): Extremely low admittance (high resistance) representing the output collector-base depletion boundary. --- ## 5. Network Classification Criteria ### 5.1 Reciprocal Networks A two-port network is reciprocal if its transfer gains have equal magnitude but opposite signs: * **Reciprocity Condition:** $$h_{12} = -h_{21} \quad ext{and} \quad g_{12} = -g_{21}$$ ### 5.2 Symmetrical Networks A two-port network is symmetrical if swapping input and output terminals leaves its behavioral matrices unchanged. This requires the determinant of the hybrid matrix to be exactly unity: * **Symmetry Condition:** $$\Delta_h = h_{11}h_{22} - h_{12}h_{21} = 1 \quad ext{and} \quad \Delta_g = g_{11}g_{22} - g_{12}g_{21} = 1$$ --- ## 6. Complete Mathematical Parameter Conversion Table This lookup matrix allows seamless, mathematically rigorous conversion between any of the four principal parameter matrices: | Matrix | Z-Parameters | Y-Parameters | ABCD-Parameters | h-Parameters | | :--- | :---: | :---: | :---: | :---: | | **$[\mathbf{Z}]$** | $$egin{bmatrix} Z_{11} & Z_{12} \ Z_{21} & Z_{22} \end{bmatrix}$$ | $$rac{1}{\Delta_Y}egin{bmatrix} Y_{22} & -Y_{12} \ -Y_{21} & Y_{11} \end{bmatrix}$$ | $$egin{bmatrix} rac{A}{C} & rac{\Delta_T}{C} \ rac{1}{C} & rac{D}{C} \end{bmatrix}$$ | $$egin{bmatrix} rac{\Delta_h}{h_{22}} & rac{h_{12}}{h_{22}} \ -rac{h_{21}}{h_{22}} & rac{1}{h_{22}} \end{bmatrix}$$ | | **$[\mathbf{Y}]$** | $$rac{1}{\Delta_Z}egin{bmatrix} Z_{22} & -Z_{12} \ -Z_{21} & Z_{11} \end{bmatrix}$$ | $$egin{bmatrix} Y_{11} & Y_{12} \ Y_{21} & Y_{22} \end{bmatrix}$$ | $$egin{bmatrix} rac{D}{B} & -rac{\Delta_T}{B} \ -rac{1}{B} & rac{A}{B} \end{bmatrix}$$ | $$egin{bmatrix} rac{1}{h_{11}} & -rac{h_{12}}{h_{11}} \ rac{h_{21}}{h_{11}} & rac{\Delta_h}{h_{11}} \end{bmatrix}$$ | | **$[\mathbf{T}]$** | $$egin{bmatrix} rac{Z_{11}}{Z_{21}} & rac{\Delta_Z}{Z_{21}} \ rac{1}{Z_{21}} & rac{Z_{22}}{Z_{21}} \end{bmatrix}$$ | $$egin{bmatrix} -rac{Y_{22}}{Y_{21}} & -rac{1}{Y_{21}} \ -rac{\Delta_Y}{Y_{21}} & -rac{Y_{11}}{Y_{21}} \end{bmatrix}$$ | $$egin{bmatrix} A & B \ C & D \end{bmatrix}$$ | $$egin{bmatrix} -rac{\Delta_h}{h_{21}} & -rac{h_{11}}{h_{21}} \ -rac{h_{22}}{h_{21}} & -rac{1}{h_{21}} \end{bmatrix}$$ | | **$[\mathbf{H}]$** | $$egin{bmatrix} rac{\Delta_Z}{Z_{22}} & rac{Z_{12}}{Z_{22}} \ -rac{Z_{21}}{Z_{22}} & rac{1}{Z_{22}} \end{bmatrix}$$ | $$egin{bmatrix} rac{1}{Y_{11}} & -rac{Y_{12}}{Y_{11}} \ rac{Y_{21}}{Y_{11}} & rac{\Delta_Y}{Y_{11}} \end{bmatrix}$$ | $$egin{bmatrix} rac{B}{D} & rac{\Delta_T}{D} \ -rac{1}{D} & rac{C}{D} \end{bmatrix}$$ | $$egin{bmatrix} h_{11} & h_{12} \ h_{21} & h_{22} \end{bmatrix}$$ | *Determinants:* $\Delta_Z = Z_{11}Z_{22} - Z_{12}Z_{21}$, $\Delta_Y = Y_{11}Y_{22} - Y_{12}Y_{21}$, $\Delta_T = AD-BC$, $\Delta_h = h_{11}h_{22} - h_{12}h_{21}$. --- ## 7. High-Yield Worked Examples (The Exam Killers) ### 7.1 Example 1: Symmetrical T-Network h-Parameters **Question:** Obtain the h-parameters and g-parameters of a symmetrical T-network with series arm resistors $R_1 = R_2 = R = 10\,\Omega$ and shunt resistor $R_3 = 50\,\Omega$. ``` 10 Ohm 10 Ohm o-----/\/\/\/\-------+-------/\/\/\/\-----o + | + V1 _ _ 50 Ohm V2 - | - o--------------------+--------------------o ``` #### Step-by-Step Analytical Solution: 1. **Calculate the Open-Circuit Z-Parameters first:** * Clockwise loop equations yield: $$V_1 = (R + R_3)I_1 + R_3 I_2 = (10 + 50)I_1 + 50I_2 = 60I_1 + 50I_2$$ $$V_2 = R_3 I_1 + (R + R_3)I_2 = 50I_1 + (10 + 50)I_2 = 50I_1 + 60I_2$$ * Extract Z-matrix: $$\mathbf{Z} = egin{bmatrix} 60 & 50 \ 50 & 60 \end{bmatrix} \Omega$$ * Calculate determinant $\Delta_Z$: $$\Delta_Z = (60)(60) - (50)(50) = 3600 - 2500 = 1100\,\Omega^2$$ 2. **Convert to h-Parameters using Matrix Transformations:** Use the Z-to-h conversion formulas from Section 6: * $$h_{11} = rac{\Delta_Z}{Z_{22}} = rac{1100}{60} = rac{110}{6} pprox 18.33\,\Omega$$ * $$h_{12} = rac{Z_{12}}{Z_{22}} = rac{50}{60} pprox 0.833$$ * $$h_{21} = -rac{Z_{21}}{Z_{22}} = -rac{50}{60} pprox -0.833$$ * $$h_{22} = rac{1}{Z_{22}} = rac{1}{60} pprox 0.0167 ext{ S}$$ 3. **Assemble the h-Parameter Matrix:** $$\mathbf{H} = egin{bmatrix} 18.33 & 0.833 \ -0.833 & 0.0167 \end{bmatrix}$$ 4. **Verify Symmetrical S-Matrix Conditions:** * **Reciprocity check:** Since $h_{12} = 0.833$ and $h_{21} = -0.833 \implies h_{12} = -h_{21}$, the network is **reciprocal**. * **Symmetry check:** Calculate determinant $\Delta_h$: $$\Delta_h = h_{11}h_{22} - h_{12}h_{21} = \left(rac{110}{60} ight)\left(rac{1}{60} ight) - \left(rac{50}{60} ight)\left(-rac{50}{60} ight) = rac{1100 + 2500}{3600} = rac{3600}{3600} = 1$$ Since $\Delta_h = 1.0$, the network is **symmetrical**. 5. **Solve the Inverse Dual g-parameters:** Use the matrix inverse formula from Section 3.2: $$\mathbf{G} = \mathbf{H}^{-1} = rac{1}{\Delta_h} egin{bmatrix} h_{22} & -h_{12} \ -h_{21} & h_{11} \end{bmatrix} = rac{1}{1.0} egin{bmatrix} 0.0167 & -0.833 \ 0.833 & 18.33 \end{bmatrix}$$ $$\mathbf{G} = egin{bmatrix} 0.0167 ext{ S} & -0.833 \ 0.833 & 18.33\,\Omega \end{bmatrix}$$ --- ### 7.2 Example 2: Conversion from h-Parameters back to ABCD [PYQ 2022 Classic] **Question:** A two-port network has the following measured h-parameters: $$h_{11} = 20\,\Omega, \quad h_{12} = 0.5, \quad h_{21} = -0.5, \quad h_{22} = 0.1 ext{ S}$$ Obtain its overall ABCD transmission parameters. #### Step-by-Step Conversion: 1. **Calculate the Hybrid Matrix Determinant $\Delta_h$:** $$\Delta_h = h_{11}h_{22} - h_{12}h_{21} = (20)(0.1) - (0.5)(-0.5) = 2 + 0.25 = 2.25$$ 2. **Substitute in the h-to-ABCD conversion equations (Section 6):** * $$A = -rac{\Delta_h}{h_{21}} = -rac{2.25}{-0.5} = 4.5$$ * $$B = -rac{h_{11}}{h_{21}} = -rac{20}{-0.5} = 40\,\Omega$$ * $$C = -rac{h_{22}}{h_{21}} = -rac{0.1}{-0.5} = 0.2 ext{ S}$$ * $$D = -rac{1}{h_{21}} = -rac{1}{-0.5} = 2.0$$ 3. **Assemble and verify the Transmission Matrix:** $$\mathbf{T} = egin{bmatrix} A & B \ C & D \end{bmatrix} = egin{bmatrix} 4.5 & 40 \ 0.2 & 2 \end{bmatrix}$$ *Verify Reciprocity Condition:* $$AD - BC = (4.5)(2) - (40)(0.2) = 9.0 - 8.0 = 1.0$$ Since $AD - BC = 1$, the network is reciprocal, confirming our conversion is mathematically consistent. --- ## 8. Common Mistakes That Cost Marks > [!danger] **The Negative Sign Trap in the Symmetrical Symmetry Check** > > While reciprocity for Z and Y parameters relies on equality ($Z_{12} = Z_{21}$), h-parameter reciprocity requires a **negative sign** ($h_{12} = -h_{21}$). Many students mistakenly write $h_{12} = h_{21}$ on exam papers, leading to an incorrect classification and loss of up to 4 marks. > [!warning] **The Non-Unity Determinant Symmetry Pitfall** > > Students often assume network symmetry is represented by $h_{11} = h_{22}$. This is completely false! $h_{11} = h_{22}$ is mathematically invalid because $h_{11}$ has units of Ohms ($\Omega$) and $h_{22}$ has units of Siemens (S). Always use the determinant condition **$\Delta_h = 1$** to check for symmetry. --- ## 9. PYQ Bank — Verbatim Questions & Answer Plans ### 9.1 Foundational Concept PYQ 2025/2022 [5 Marks] **Question:** What is network theory? Classify the two-port parameters. * **Answer Plan:** 1. Define **Network Theory** as the mathematical discipline used to analyze, model, and solve the terminal behaviors of electrical networks. 2. Define a **Two-Port Network** as a black-box system with input and output access terminal pairs. 3. Recreate the complete parameter table from **Section 1 MOC diagram** mapping equations, parameters, and experimental use cases. ### 9.2 Parameter Integration PYQ 2024 Question 4a [8 Marks] **Question:** Explain the physical significance of hybrid parameters. Why are they called hybrid parameters? * **Answer Plan:** 1. Explain that "hybrid" refers to the mixture of open-circuit and short-circuit conditions used to isolate the parameters. 2. Write the defining matrix equations for $V_1$ and $I_2$ in terms of $I_1$ and $V_2$. 3. Break down the units (Ohms, Siemens, and Unitless) to show the mixed dimensional nature. 4. Provide the BJT CE modeling context ($h_{ie}, h_{re}, h_{fe}, h_{oe}$), explaining why BJTs naturally require hybrid modeling. --- ## 10. Self-Check Before Moving On - [ ] Can you state the defining equations and matrix blocks for both h and g parameters? [9.01] - [ ] Do you know why hybrid parameters are called "hybrid" (mixed open/short conditions)? [9.02] - [ ] Have you memorized the reciprocity condition ($h_{12} = -h_{21}$) and symmetry condition ($\Delta_h = 1$)? [9.03] - [ ] Can you map the common-emitter transistor subscripts ($h_{ie}, h_{re}, h_{fe}, h_{oe}$) to their standard h-parameter indices? [9.04] - [ ] Can you convert any Z or Y matrix directly into a hybrid representation using the 4x4 master conversion block? [9.05] --- *Source: (k.Deergha Rao) signals and systems.pdf (Chapter 6), 04 Network Theory.pdf, Rabiul sir class note.pdf.*