--- title: "Passive device to split power into half" subject: "RF and Microwave Engineering" chapter: 4 key: "semester_7/rf_and_microwave_engineering:power-splitter-device" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/power-splitter-device site: BCT Engineering --- # Passive device to split power into half RF and Microwave Engineering · Chapter 4: RF and Microwave Components and Devices - Asked in: 2073 Magh, 8 marks - 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/power-splitter-device - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-4 ## Question Choose a suitable passive microwave device to split power into half and explain its properties. ## Short answer A magic tee fed at the H-arm splits input power equally and in phase between its two collinear arms with 3 dB split, high isolation (S12=0), and all ports matched. Feeding the E-arm instead yields equal anti-phase halves. It can also act as a power combiner. ## Answer #### Choice: magic tee (hybrid tee) fed at the H-arm To split power into two equal halves, a **magic tee** is a good choice: power fed into its **H-arm (port 3)** divides equally and in phase between the collinear arms 1 and 2, while all ports remain matched and the outputs are isolated. (A simple H-plane tee also splits power equally, but its output arms are not matched or isolated; in microstrip circuits the equivalent device is the Wilkinson divider.) ```text E-arm (4) | | -----------+ +----------- (1) ===========+ +=========== (2) -----------+ +----------- / / / / H-arm (3) ``` #### S-matrix ```text | 0 0 1 1 | [S] = (1/√2) | 0 0 1 -1 | | 1 1 0 0 | | 1 -1 0 0 | ``` #### Power split With input a3 = a (power P = |a|²/2) and matched outputs: ```text b1 = a/√2 -> P1 = |a|²/4 = P/2 (−3 dB) b2 = a/√2 -> P2 = P/2 (−3 dB, same phase) b4 = 0 -> no power to E-arm b3 = 0 -> input matched, no reflection ``` **Result: each output gets exactly half the input power, in phase.** If anti-phase halves are needed, the input is applied at the E-arm instead (b1 = a/√2, b2 = −a/√2). #### Properties of the magic tee 1. **Equal split:** |S13| = |S23| = |S14| = |S24| = 1/√2 (3 dB). 2. **Phase:** H-arm gives in-phase outputs; E-arm gives 180° out-of-phase outputs. 3. **Isolation:** E and H arms are isolated (S34 = 0); collinear arms are isolated (S12 = 0), so a mismatch on one output does not disturb the other output. 4. **Matching:** all ports matched (Sᵢᵢ = 0) when E and H arms are matched with posts or irises. 5. **Lossless and reciprocal:** [S] is unitary and symmetric. If the two loads are unequal, the difference between their reflections goes to the E-arm, which can be terminated in a matched load. 6. **Reverse use:** equal in-phase inputs at 1 and 2 combine at the H-arm, so the same device works as a power combiner. #### Comparison with an H-plane tee | Feature | H-plane tee | Magic tee | |---|---|---| | Split ratio | Equal (3 dB) | Equal (3 dB) | | Ports | 3 | 4 | | Output isolation | No | Yes (S12 = 0) | | All ports matched | Not possible | Yes | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which port of a magic tee should be fed to split power equally and in phase between the collinear arms? A. E-arm (port 4) B. H-arm (port 3) C. Port 1 D. Port 2 Answer: B. Feeding the H-arm gives b1 = b2 = a/√2 in phase and nothing to the E-arm. (Key point: Device choice) 2. With input power P at the H-arm of a matched magic tee, what power appears at each collinear arm? A. P B. P/4 C. P/2 (−3 dB) D. P/√2 Answer: C. Each output amplitude is a/√2, so the power is |a|²/4 = P/2, i.e. 3 dB below the input. (Key point: Power split result) 3. What does S12 = 0 and S34 = 0 mean for a magic tee? A. Collinear arms are isolated from each other and the E and H arms are isolated B. All ports are shorted C. Both outputs get no power D. The tee is lossy Answer: A. Zero S-parameters mean no direct coupling, so a mismatch at one output does not disturb the other. (Key point: Isolation and matching) 4. What advantage does the magic tee have over a simple H-plane tee for splitting power? A. It splits unequally B. It has only three ports C. It has output isolation and all ports matched D. It needs no waveguide Answer: C. Both give a 3 dB split, but only the magic tee has isolated outputs and matched ports. (Key point: Magic tee versus H-plane tee) ## Related in RF and Microwave Engineering - [Device to measure 7.5 mW microwave power](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Ameasure-7-5mw-power) — RF and Microwave Engineering, chapter 8 - [Measurement parameters and low power device](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Alow-power-measurement-device) — RF and Microwave Engineering, chapter 8 - [Identify device from S-matrix with ±S14 entries](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Aidentify-device-s-matrix-hybrid) — RF and Microwave Engineering, chapter 4, asked 2× - [Identify device from numeric 4×4 S-matrix](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Aidentify-device-s-matrix-numeric) — RF and Microwave Engineering, chapter 4 - [Duplexer for half-power airport surveillance radar](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Aradar-duplexer-smatrix) — RF and Microwave Engineering, chapter 4 - [Identify device from S-matrix with S22 term](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Aidentify-device-s-matrix-s22) — RF and Microwave Engineering, chapter 4 ## Related in other subjects - [Series RLC resonance, half power, bandwidth](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aseries-resonance-half-power-bandwidth) — Electrical Circuit Theory, chapter 1, asked 3× - [Element values for changing half-power frequency](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Adenormalize-half-power-frequency-change) — Filter Design, chapter 1, asked 3× - [Series RLC resonance: f0, bandwidth, half power, Q](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aseries-resonance-derive-f0-bw-q) — Electrical Circuit Theory, chapter 1 - [Half power points, bandwidth and effect of Q](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aseries-resonance-bandwidth-q-effect) — Electrical Circuit Theory, chapter 1