6.02 Short-Circuit Admittance (Y) Parameters | 6.04 Hybrid (h) & Inverse Hybrid (g) Parameter Formulations
6.03 Transmission (ABCD) Parameters & Cascaded Networks
Core Idea
Transmission (ABCD) Parameters model two-port networks by directly expressing the sending-end variables () as linear combinations of the receiving-end variables (). 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 and 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 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 () to the receiving-end voltage and current () through two coupled linear algebraic equations:
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}
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 |
|---|---|---|---|---|
| $$A = \left. rac{V_1}{V_2} | ||||
| ight | _{I_2=0}$$ | Open-Circuit Voltage Ratio | Dimensionless | Reverse voltage gain of the network with Port 2 open-circuited. |
| $$B = \left. rac{V_1}{-I_2} | ||||
| ight | _{V_2=0}$$ | Short-Circuit Transfer Impedance | Ohms () | The ratio of input voltage to output short-circuit current. |
| $$C = \left. rac{I_1}{V_2} | ||||
| ight | _{I_2=0}$$ | Open-Circuit Transfer Admittance | Siemens () | The ratio of input current to output voltage when Port 2 is open-circuited. |
| $$D = \left. rac{I_1}{-I_2} | ||||
| ight | _{V_2=0}$$ | 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:
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:
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 , and common shunt resistor .
R1 R2
o-----/\/\/\-------+-------/\/\/\-----o
+ | +
V1 _ _ R3 V2
- | -
o--------------------+--------------------o
Step 1: Write KVL Equations with Standard Currents ( entering)
Step 2: Introduce the Leaving Current Convention ()
Substitute (where is the current flowing out of Port 2):
Step 3: Isolate Sending-End Current () from Equation 2
From Equation 2, solve for :
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.*