--- title: "Two magic tees connected at E-arms" subject: "RF and Microwave Engineering" chapter: 4 key: "semester_7/rf_and_microwave_engineering:two-magic-tees-e-arms" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/two-magic-tees-e-arms site: BCT Engineering --- # Two magic tees connected at E-arms RF and Microwave Engineering · Chapter 4: RF and Microwave Components and Devices - Asked in: 2077 Chaitra, 10 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/two-magic-tees-e-arms - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-4 ## Question Explain the properties of two Magic Tees if one connects their E-arms and derive its S-matrix. ## Short answer Two magic tees joined at their E-arms form a six-port network. Input at port 1 sends half power to H-arm 3 and quarter powers to ports 5 and 6 in anti-phase; H-arms are isolated. The S-matrix is symmetric, matched and lossless. ## Answer Two identical ideal magic tees are joined by connecting their **E-arms** with a matched waveguide section (assumed of zero electrical length; a real length only adds a phase factor). The result is a six-port junction. **Port numbering** - Tee A: 1, 2 (collinear), 3 (H-arm), 4 (E-arm) - Tee B: 5, 6 (collinear), 7 (H-arm), 8 (E-arm) - Connection: port 4 joined to port 8, so a8 = b4 and a4 = b8. ```text (3) H H (7) | | (1) ===[A]=== (2) (5) ===[B]=== (6) | | +------ E-E link -------+ ``` #### Equations of each tee ```text Tee A: b1 = (a3 + a4)/√2 b2 = (a3 - a4)/√2 b3 = (a1 + a2)/√2 b4 = (a1 - a2)/√2 Tee B: b5 = (a7 + a8)/√2 b6 = (a7 - a8)/√2 b7 = (a5 + a6)/√2 b8 = (a5 - a6)/√2 ``` #### Applying the connection ```text a8 = b4 = (a1 - a2)/√2 a4 = b8 = (a5 - a6)/√2 ``` Substituting: ```text b1 = a3/√2 + (a5 - a6)/2 b2 = a3/√2 - (a5 - a6)/2 b3 = (a1 + a2)/√2 b5 = a7/√2 + (a1 - a2)/2 b6 = a7/√2 - (a1 - a2)/2 b7 = (a5 + a6)/√2 ``` #### S-matrix of the combination (port order 1, 2, 3, 5, 6, 7) ```text 1 2 3 5 6 7 1 | 0 0 1/√2 1/2 -1/2 0 | 2 | 0 0 1/√2 -1/2 1/2 0 | 3 | 1/√2 1/√2 0 0 0 0 | 5 | 1/2 -1/2 0 0 0 1/√2 | 6 | -1/2 1/2 0 0 0 1/√2 | 7 | 0 0 0 1/√2 1/√2 0 | ``` Check: the matrix is symmetric (reciprocal) and each row has |S|² sum = 1, rows mutually orthogonal (lossless). Diagonal is zero, so every port is matched. #### Properties 1. **Input at port 1:** half the power goes to H-arm 3 (b3 = a1/√2); the other half travels through the E-link and appears at ports 5 and 6 as **equal, 180° out-of-phase** signals (b5 = a1/2, b6 = −a1/2), each carrying a quarter of the power. Port 2 and port 7 get nothing. 2. **Only the difference (anti-phase) component** of signals at ports 1, 2 is passed to tee B; the sum component exits at A's H-arm. Likewise for tee B. 3. **H-arms 3 and 7 are isolated** from each other and from the far tee's collinear arms (S35 = S37 = 0). 4. **Input at H-arm 3** divides in phase between ports 1 and 2 only (normal magic tee behaviour); input at H-arm 7 divides in phase between 5 and 6. 5. **Anti-phase equal inputs at 1 and 2** (a1 = −a2 = a): b5 = a, b6 = −a, so all the power is transferred to tee B's collinear arms, again in anti-phase. In-phase inputs stay in tee A (all exit at port 3). 6. The network is matched, lossless and reciprocal. #### Applications - Transferring differential (Δ) signals between two hybrid circuits, e.g. monopulse comparators. - Balanced duplexers and balanced mixers using two hybrids. - Bridge-type measurement circuits for comparing two loads. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In this arrangement, which ports are joined? A. H-arm 3 to H-arm 7 B. Port 1 to port 5 C. Port 4 to port 7 D. E-arm 4 of tee A to E-arm 8 of tee B, giving a six-port Answer: D. a8 = b4 and a4 = b8. (Key point: Connection) 2. Input at port 1 results in A. half power to H-arm 3, and a quarter each to ports 5 and 6 with 180° phase difference B. all power to port 3 C. equal in-phase output at 5 and 6 D. half to port 2 Answer: A. b3 = a1/sqrt2, b5 = a1/2, b6 = -a1/2. (Key point: Input at port 1) 3. Only which component is passed from tee A to tee B? A. The sum (in-phase) component B. The difference (anti-phase) component of ports 1 and 2 C. Both components D. Neither Answer: B. The E-arm carries the difference signal over the link. (Key point: Transferred component) 4. Equal in-phase inputs at ports 1 and 2 exit A. at ports 5 and 6 B. at port 7 C. entirely at H-arm port 3 D. are reflected Answer: C. The sum component stays in tee A. (Key point: In-phase inputs) ## Related in RF and Microwave Engineering - [Short note: Two magic tees connected in H-plane](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Atwo-magic-tees-h-plane-note) — RF and Microwave Engineering, chapter 4 - [Three-port model and magic tee with shorted arms](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Athree-port-magic-tee-shorted-arms) — RF and Microwave Engineering, chapter 4 ## Related in other subjects - [Derive beam element stiffness matrix](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Abeam-element-stiffness-derivation) — Finite Element Method, chapter 3