---
title: "Duplexer for half-power airport surveillance radar"
subject: "RF and Microwave Engineering"
chapter: 4
key: "semester_7/rf_and_microwave_engineering:radar-duplexer-smatrix"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/radar-duplexer-smatrix
site: BCT Engineering
---

# Duplexer for half-power airport surveillance radar

RF and Microwave Engineering · Chapter 4: RF and Microwave Components and Devices

- Asked in: 2076 Bhadra, 10 marks
- Video: [Duplexer | Brach type | Radar Systems | Lec-64](https://www.youtube.com/watch?v=Yoy2s6tiDV4) — Education 4u
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/radar-duplexer-smatrix
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-4

## Question

A radar installation engineer is given a responsibility to install an Airport Surveillance Radar that requires half input power to the antenna than the transmitter that can deliver using a duplexer. Prepare its S-matrix.

## Short answer

Use a magic tee duplexer: transmitter at H-arm, antenna at port 1, matched load at port 2, receiver at E-arm. S = (1/sqrt(2))[[0,0,1,1],[0,0,1,-1],[1,1,0,0],[1,-1,0,0]]. Antenna gets half transmitter power and receiver is isolated.

## Answer

#### Reading of the problem
The radar must feed the antenna with **half** of the transmitter power, and the same antenna must be used for receiving (duplexing). Both needs are met by a **magic tee used as a duplexer**:
- Transmitter at the **H-arm (port 3)**.
- Antenna at collinear **port 1**; matched dummy load at collinear **port 2**.
- Receiver at the **E-arm (port 4)**.

```text
               RX (4, E-arm)
                  | |
 ANT (1) ========+   +======== (2) matched load
                  / /
                 / /
           TX (3, H-arm)
```

#### S-matrix of the duplexer (ideal magic tee)
```text
              | 0   0   1   1 |
 [S] = (1/√2) | 0   0   1  -1 |
              | 1   1   0   0 |
              | 1  -1   0   0 |
```

#### Transmit mode
TX sends a3 = a (power Pt = |a|²/2); antenna and load are matched, a1 = a2 = a4 = 0:
```text
 b1 = a3/√2   -> power to antenna = Pt/2
 b2 = a3/√2   -> power to dummy load = Pt/2
 b4 = 0       -> no power to receiver (S43 = 0)
 b3 = 0       -> no reflection to TX
```
So the antenna gets exactly **half the transmitter power** (−3 dB), and the receiver is **isolated** from the high-power transmitter.

#### Receive mode
The echo enters at port 1 (a1 = e):
```text
 b3 = e/√2  -> half to transmitter arm
 b4 = e/√2  -> half to receiver (E-arm)
 b2 = 0     -> nothing to the load
```
The receiver gets half the received echo power.

#### Remarks
- Total loss in the duplexer is 3 dB on transmit and 3 dB on receive, but the arrangement is **passive, reciprocal and needs no switching**, and the receiver is protected by the E–H isolation.
- If no power loss were acceptable, a ferrite **circulator** duplexer [[0,0,1],[1,0,0],[0,1,0]] would be used instead; the magic tee version fits this requirement because the antenna needs only half the power.
- Practical systems add a TR (transmit-receive) limiter before the receiver for extra protection against leakage from imperfect isolation.

## Check yourself

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

1. In the magic tee duplexer, the transmitter is connected to
   A. the E-arm, receiver on the H-arm
   B. port 1, antenna on the E-arm
   C. port 2, load on port 1
   D. the H-arm (port 3), with antenna on port 1, dummy load on port 2 and receiver on the E-arm (port 4)
   Answer: D. TX in the H-arm splits equally to ports 1 and 2 and none to the E-arm. (Key point: Port arrangement)

2. The antenna receives
   A. half the transmitter power (-3 dB), the other half going to the load
   B. the full transmitter power
   C. a quarter of the power
   D. twice the transmitter power
   Answer: A. b1 = a3/sqrt2 gives Pt/2. (Key point: Transmit power)

3. During transmit, the receiver is protected because
   A. the load absorbs everything
   B. S43 = 0 (E-H isolation)
   C. the receiver is shorted
   D. the antenna reflects all power
   Answer: B. No transmitter power appears at the E-arm. (Key point: Receiver protection)

4. The echo entering the antenna port goes
   A. all to the receiver
   B. all to the dummy load
   C. half to the receiver at E-arm and half back toward the transmitter arm
   D. all to the transmitter
   Answer: C. b3 = b4 = e/sqrt2 and b2 = 0. (Key point: Receive mode)

## Related in RF and Microwave Engineering

- [Power measurement tool for airport surveillance radar](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Aairport-radar-power-measurement) — RF and Microwave Engineering, chapter 8
- [Combining two identical radar transmitters](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Atwo-radar-transmitters-combining) — RF and Microwave Engineering, chapter 4, asked 2×

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