--- title: "Combining two identical radar transmitters" subject: "RF and Microwave Engineering" chapter: 4 key: "semester_7/rf_and_microwave_engineering:two-radar-transmitters-combining" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/two-radar-transmitters-combining site: BCT Engineering --- # Combining two identical radar transmitters RF and Microwave Engineering · Chapter 4: RF and Microwave Components and Devices - Asked in: 2074 Magh, 6 marks; 2073 Bhadra, 8 marks (asked 2 times) - 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-radar-transmitters-combining - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-4 ## Question Suppose there are two identical radar transmitters and few passive devices in equipment stock. A particular application requires twice more input power to an antenna than either transmitter can deliver. As a RF engineer, give your appropriate solution for the above problem with necessary figures, mathematics (S-matrix) and sufficient explanation. ## Short answer Connect two phase-locked transmitters to the collinear arms of a magic tee, antenna to the H-arm, and a matched load on the E-arm. Equal in-phase inputs a1=a2=a yield antenna signal b3=√2·a, delivering 2P to antenna. The E-arm load absorbs imbalance, and S12=0 isolates the transmitters. ## Answer #### Solution: combine the two transmitters with a magic tee The required antenna power is 2P while each transmitter gives P. Since the transmitters are identical, their outputs can be added coherently in a **magic tee (hybrid tee)**: feed the two transmitters (locked to the same phase, e.g. driven by a common master oscillator) into the **collinear arms 1 and 2**, connect the antenna to the **H-arm (port 3)** and put a **matched load on the E-arm (port 4)**. ```text matched load | E-arm (4) | | TX-A ===(1)========+ +========(2)=== TX-B P, 0° / / P, 0° / / H-arm (3) | Antenna (receives 2P) ``` #### S-matrix of the magic tee ```text | b1 | | 0 0 1 1 | | a1 | | b2 | = (1/√2) | 0 0 1 -1 | | a2 | | b3 | | 1 1 0 0 | | a3 | | b4 | | 1 -1 0 0 | | a4 | ``` #### Mathematics Each transmitter delivers power P, so its wave amplitude is a = √(2P) (using P = |a|²/2). With equal, in-phase inputs a1 = a2 = a and a3 = a4 = 0 (antenna and load matched): ```text b3 = (a1 + a2)/√2 = 2a/√2 = √2·a b4 = (a1 - a2)/√2 = 0 b1 = b2 = 0 (no reflection to the transmitters) Power at antenna = |b3|²/2 = 2|a|²/2 = 2P Power in load = 0 ``` **Result: the antenna receives 2P, twice the power of either transmitter, with no power wasted.** #### Explanation and practical points - The H-arm acts as a **sum port** and the E-arm as a **difference port**. In-phase inputs add at the H-arm; any imbalance (amplitude or phase error) appears at the E-arm and is safely absorbed by the matched load instead of reflecting into the transmitters. - If the phase difference is θ, the antenna power is 2P·cos²(θ/2) and the load power 2P·sin²(θ/2), so phase locking is essential. - Because S12 = 0, the two transmitters are **isolated** from each other, so one cannot load or damage the other. - Ports are matched (Sᵢᵢ = 0), so the transmitters see no reflected power and the VSWR stays low. - Alternatively, the transmitters can be fed into the E and H arms (a3 = a4 = a); then b1 = √2·a and b2 = 0, so the combined 2P appears at collinear arm 1. Either arrangement needs the two inputs to be phase-locked; feeding the collinear arms keeps the H-arm as the single sum output, with the E-arm load absorbing any imbalance. - Isolators (circulators with a load) may be added at each transmitter output for further protection. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. To get 2P from two identical transmitters of power P, where is the antenna connected on the magic tee? A. H-arm (port 3), with the transmitters on collinear arms 1 and 2 B. E-arm (port 4), with the transmitters on collinear arms 1 and 2 C. Collinear arm 1, with transmitters on the E and H arms D. Collinear arm 2, with both transmitters on the H-arm Answer: A. In-phase inputs on arms 1 and 2 add at the H-arm (sum port), so the antenna goes there. (Key point: Port choice) 2. Why is a matched load placed on the E-arm? A. It makes the antenna power twice as large B. Any amplitude or phase imbalance appears at the E-arm and is absorbed there instead of reflecting into the transmitters C. It supplies the phase reference for the transmitters D. It reflects power back to improve the VSWR Answer: B. The E-arm is the difference port; with perfect in-phase equal inputs it gets zero, and errors are safely absorbed. (Key point: Role of the E-arm load) 3. With a1 = a2 = a (power P each), what power reaches the antenna? A. P, because the power splits equally between antenna and load B. 4P, because amplitudes add and powers add again C. 2P, because b3 = sqrt(2)·a gives |b3|^2/2 = 2P D. P/2, because of the 3 dB loss in the tee Answer: C. Amplitudes add: b3 = (a1+a2)/sqrt2 = sqrt2·a, so power is 2|a|^2/2 = 2P. (Key point: Power result) 4. If the two transmitters differ in phase by θ, the antenna power is A. 2P sin^2(θ/2), with the rest in the load B. 2P independent of θ C. P cos θ, with no power in the load D. 2P cos^2(θ/2), with the rest 2P sin^2(θ/2) going to the load Answer: D. Sum port amplitude depends on cos(θ/2), so phase locking is essential for full 2P. 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