--- title: "Hybrid tee with terminated arms" subject: "RF and Microwave Engineering" chapter: 4 key: "semester_7/rf_and_microwave_engineering:hybrid-tee-terminated-arms" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/hybrid-tee-terminated-arms site: BCT Engineering --- # Hybrid tee with terminated arms RF and Microwave Engineering · Chapter 4: RF and Microwave Components and Devices - Asked in: 2082 Baisakh, 4+4+2 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/hybrid-tee-terminated-arms - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-4 ## Question Using proper S-Matrices, mention the properties of a Hybrid-Tee in the situation of (a) terminated E-arm, (b) terminated H-arm and (c) terminated co-planar arms. ## Short answer With E-arm matched, H-arm input divides equally in phase to collinear ports; with H-arm terminated, E-arm input divides equally 180 degrees out of phase. With collinear arms terminated, E and H arms are matched and isolated, enabling balanced mixers and bridge measurements. ## Answer For an ideal hybrid (magic) tee with ports 1 and 2 collinear, port 3 the H-arm and port 4 the E-arm: ```text | 0 0 1 1 | [S] = (1/√2) | 0 0 1 -1 | | 1 1 0 0 | | 1 -1 0 0 | ``` When a port is terminated in a **matched load**, no wave returns into it (aₖ = 0), so its column is dropped; its row tells how much power the load absorbs. The remaining matrix describes the reduced network. #### (a) E-arm (port 4) terminated in a matched load Remaining ports 1, 2, 3: ```text | 0 0 1 | [S] = (1/√2) | 0 0 1 | | 1 1 0 | ``` - Input at H-arm (3) divides equally and **in phase** to ports 1 and 2: the device works as an **H-plane (in-phase) power divider**. - Input at port 1: half the power goes to port 3, half (b4 = a1/√2) is absorbed in the E-arm load; port 2 receives nothing. - In-phase equal inputs at 1 and 2 combine fully at port 3 (power combiner); any imbalance goes to the load. #### (b) H-arm (port 3) terminated in a matched load Remaining ports 1, 2, 4: ```text | 0 0 1 | [S] = (1/√2) | 0 0 -1 | | 1 -1 0 | ``` - Input at E-arm (4) divides equally to ports 1 and 2 but **180° out of phase**: it works as an **E-plane (anti-phase) divider**. - Input at port 1: half goes to port 4, half is absorbed in the H-arm load; ports 1 and 2 remain isolated. - Anti-phase equal inputs at 1 and 2 combine fully at port 4. #### (c) Co-planar (collinear) arms 1 and 2 terminated in matched loads Remaining ports 3 and 4: ```text | S33 S34 | | 0 0 | [S] = | | = | | | S43 S44 | | 0 0 | ``` - Both E and H arms are **matched** (no reflection) and **completely isolated** from each other (S34 = 0). - Power fed into either the E-arm or the H-arm is split equally into the two loads and fully absorbed. - This isolation between E and H arms is the basis of the balanced mixer (LO at one arm, signal at the other) and of bridge measurements: with identical loads on 1 and 2, no signal passes from H-arm to E-arm; any mismatch difference between the two loads shows up at the E-arm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With the E-arm terminated in a matched load, the remaining 3-port acts mainly as A. an in-phase (H-plane) power divider/combiner using H-arm and collinear arms B. an anti-phase divider with 180° difference C. a circulator D. a fully isolating short circuit Answer: A. The H-arm splits equally and in phase to ports 1 and 2; imbalance goes to the E-arm load. (Key point: Terminated E-arm) 2. With the H-arm terminated, an input at E-arm (port 4) gives at ports 1 and 2 A. equal power in phase B. equal power with 180° phase difference C. all the power at port 1 D. no output because the H-arm absorbs everything Answer: B. The E-arm is the difference arm, so outputs are equal and anti-phase. (Key point: Terminated H-arm) 3. With both collinear arms 1 and 2 terminated in matched loads, what is S34 and what does it mean? A. S34 = 1, power passes from H to E arm B. S34 = 1/sqrt2, half passes between arms C. S34 = 0, E and H arms are isolated and matched; power fed to either is absorbed in the two loads D. S34 = -1, power is fully reflected Answer: C. E-H isolation holds, and power entering either arm splits equally into the two loads. (Key point: Terminated collinear arms) 4. For input at port 1 with E-arm terminated, where does the power go? A. All to port 2 B. Half to port 2 and half to port 3 C. All to port 3 D. Half to port 3 and half absorbed in the E-arm load; port 2 gets nothing Answer: D. Port 1 couples to the E and H arms equally and is isolated from port 2. 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