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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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