---
title: "Transmission and Y parameters, reciprocity and symmetry"
subject: "Electrical Circuit Theory"
chapter: 7
key: "semester_3/electrical_circuit_theory:transmission-admittance-parameters-reciprocity-symmetry-2v2"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/transmission-admittance-parameters-reciprocity-symmetry-2v2
site: BCT Engineering
---

# Transmission and Y parameters, reciprocity and symmetry

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

- Asked in: 2074 Chaitra, 8 marks
- Video: [ABCD Parameters (Transmission Parameters) Explained](https://www.youtube.com/watch?v=qnHaOwYfzgE) — ALL ABOUT ELECTRONICS
- 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-admittance-parameters-reciprocity-symmetry-2v2
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-7

## Question

Find transmission and admittance parameter for the given TPN and check its reciprocity and symmetry.
[Figure: at port 1 (V1, I1): a 1 Ω shunt resistor in parallel with a dependent current source 2V2 (arrow up); a 1 Ω series resistor followed by a dependent voltage source 2V1 (− on left, + on right) in the series arm; a 2 Ω shunt resistor at port 2 (V2, I2)]

## Short answer

Y = [4 −3; −3 1.5] S. Transmission parameters: T = [0.5, 1/3 Ω; −1 S, 4/3]. The network is reciprocal (y12 = y21) but not symmetrical (y11 ≠ y22).

## Answer

Transmission (T) parameters: V1 = A V2 − B I2, I1 = C V2 − D I2. Admittance (Y) parameters: I1 = y11V1 + y12V2, I2 = y21V1 + y22V2.

**Reading the figure (assumption).** At port 1 a 1 Ω resistor and a current source 2V2 (arrow up, i.e. into the top node) are in shunt. A 1 Ω resistor then a voltage source 2V1 (− on the left, + on the right) are in the series arm. A 2 Ω shunt resistor is across port 2.

```text
 I1   +--------+------/\/\------(-+ 2V1)----+------+---- I2
 o----|        |        1Ω                   |      |
      1Ω     (2V2 ↑)                         |     2Ω
 o----+--------+-----------------------------+------+----
```

#### Y-parameters

Let Is be the current through the series 1 Ω resistor from the port 1 node towards port 2.

```text
KCL at port 1 node:    I1 + 2V2 = V1/1 + Is
Series arm:            V2 = (V1 − 1·Is) + 2V1 = 3V1 − Is   →  Is = 3V1 − V2
KCL at port 2 node:    Is + I2 = V2/2
```

Therefore:

```text
I1 = V1 + (3V1 − V2) − 2V2 = 4V1 − 3V2
I2 = V2/2 − (3V1 − V2) = −3V1 + 1.5 V2
```

**y11 = 4 S, y12 = −3 S, y21 = −3 S, y22 = 1.5 S**

#### T-parameters

From I2 = y21V1 + y22V2: V1 = (1.5 V2 − I2)/3 = 0.5 V2 − (1/3) I2. Then I1 = 4V1 − 3V2 = −V2 − (4/3) I2.

```text
A = 0.5     B = 1/3 Ω
C = −1 S    D = 4/3
(also A = −y22/y21, B = −1/y21, C = −Δy/y21, D = −y11/y21, with Δy = 6 − 9 = −3)
```

#### Reciprocity and symmetry

- Reciprocal: y12 = y21 = −3 S, and AD − BC = (0.5)(4/3) − (1/3)(−1) = 2/3 + 1/3 = 1. So the network **is reciprocal** (the two dependent sources happen to cancel in their effect on y12 and y21).
- Symmetrical: needs A = D (or y11 = y22). Here A = 0.5 ≠ D = 4/3 and y11 = 4 ≠ y22 = 1.5, so it is **not symmetrical**.

**Answer: Y = [4 −3; −3 1.5] S; T = [0.5, 1/3; −1, 4/3]; reciprocal but not symmetrical.**

## Check yourself

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

1. With the series 1 ohm resistor and the 2V1 source, what relation gives the series current Is?
   A. Is = V1 - V2
   B. Is = 3V1 - V2
   C. Is = V1 + 2V2
   D. Is = V2 - 3V1
   Answer: B. V2 = (V1 - Is) + 2V1 = 3V1 - Is, so Is = 3V1 - V2. (Key point: Series arm equation)

2. Which Y matrix does the answer obtain?
   A. [4 -3; -3 1.5] S
   B. [1 -1; -1 0.5] S
   C. [4 3; 3 1.5] S
   D. [2 -3; -3 4] S
   Answer: A. I1 = 4V1 - 3V2 and I2 = -3V1 + 1.5V2. (Key point: Y-parameter results)

3. Using y21 = -3 and y22 = 1.5, what are A and B?
   A. A = 2, B = 3
   B. A = 0.5, B = 1/3 ohm
   C. A = -0.5, B = -1/3
   D. A = 1/3, B = 0.5
   Answer: B. A = -y22/y21 = 0.5 and B = -1/y21 = 1/3. (Key point: Y to T conversion)

4. What is the conclusion about the network?
   A. Reciprocal and symmetrical
   B. Reciprocal but not symmetrical
   C. Symmetrical but not reciprocal
   D. Neither reciprocal nor symmetrical
   Answer: B. y12 = y21 = -3 and AD - BC = 1 give reciprocity, but A = 0.5 differs from D = 4/3 so it is not symmetrical. (Key point: Reciprocity and symmetry verdict)

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