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Two magic tees connected at E-arms

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.

Explain the properties of two Magic Tees if one connects their E-arms and derive its S-matrix.

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.
        (3) H                   H (7)
         |                       |
 (1) ===[A]=== (2)       (5) ===[B]=== (6)
         |                       |
         +------ E-E link -------+

Equations of each tee

 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

 a8 = b4 = (a1 - a2)/√2
 a4 = b8 = (a5 - a6)/√2

Substituting:

 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)

        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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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