---
title: "Transmission parameters in terms of Y-parameters"
subject: "Electrical Circuit Theory"
chapter: 7
key: "semester_3/electrical_circuit_theory:transmission-line-parameters-in-terms-of"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/transmission-line-parameters-in-terms-of
site: BCT Engineering
---

# Transmission parameters in terms of Y-parameters

Electrical Circuit Theory · Chapter 7: Two-port Parameter of Networks

- Asked in: 2070 Asar (old course), 6 marks
- Video: [Transmission or ABCD parameters  in terms  of Y parameters](https://www.youtube.com/watch?v=B0d2cp4zTBc) — Engg-Course-Made-Easy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/transmission-line-parameters-in-terms-of
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-7

## Question

Express transmission line parameters in terms of Y-parameter.

## Short answer

A = −y22/y21, B = −1/y21, C = −Δy/y21, D = −y11/y21, where Δy = y11 y22 − y12 y21.

## Answer

The transmission (ABCD) parameters are defined by V1 = A V2 − B I2 and I1 = C V2 − D I2. The Y-parameters are defined by I1 = y11 V1 + y12 V2 and I2 = y21 V1 + y22 V2. ("Transmission line parameters" here means the transmission parameters of the two-port.)

#### Derivation

**Step 1.** Solve the second Y equation for V1:

```text
I2 = y21 V1 + y22 V2
V1 = −(y22/y21) V2 + (1/y21) I2
```

Comparing with V1 = A V2 − B I2:

```text
A = −y22/y21          B = −1/y21
```

**Step 2.** Substitute V1 into the first Y equation:

```text
I1 = y11 V1 + y12 V2
   = y11[−(y22/y21) V2 + (1/y21) I2] + y12 V2
   = [ y12 − y11 y22/y21 ] V2 + (y11/y21) I2
   = −[(y11 y22 − y12 y21)/y21] V2 + (y11/y21) I2
```

Comparing with I1 = C V2 − D I2, and writing Δy = y11 y22 − y12 y21:

```text
C = −Δy/y21           D = −y11/y21
```

#### Result

```text
A = −y22/y21      B = −1/y21
C = −Δy/y21       D = −y11/y21        with Δy = y11 y22 − y12 y21
```

**Checks.** (1) AD − BC = [y11y22 − Δy]/y21² = y12 y21/y21² = y12/y21, which equals 1 exactly when y12 = y21 (reciprocal). (2) For a series resistor R (y11 = y22 = 1/R, y12 = y21 = −1/R): A = 1, B = R, C = 0, D = 1, as expected.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the derivation started?
   A. Solve the second Y equation I2 = y21V1 + y22V2 for V1
   B. Set V1 = 0
   C. Solve the first equation for V2
   D. Take the inverse of Y
   Answer: A. Solving for V1 gives V1 = -(y22/y21)V2 + (1/y21)I2, matched to V1 = AV2 - BI2. (Key point: First step)

2. What are A and B?
   A. A = -y22/y21, B = -1/y21
   B. A = y22/y21, B = 1/y21
   C. A = -y11/y21, B = -1/y21
   D. A = -1/y21, B = -y22/y21
   Answer: A. Comparing coefficients with V1 = AV2 - BI2. (Key point: A and B)

3. What are C and D, with delta y = y11y22 - y12y21?
   A. C = -delta y/y21, D = -y11/y21
   B. C = delta y/y21, D = y11/y21
   C. C = -y11/y21, D = -delta y/y21
   D. C = -1/y21, D = -y22/y21
   Answer: A. Substituting V1 into I1 and comparing with I1 = CV2 - DI2. (Key point: C and D)

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