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Combining two identical radar transmitters
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.
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.
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).
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
| 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):
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.
✓ Answer checked by Claude Opus 5.5 (high effort)
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