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Identify device from numeric 4×4 S-matrix

Short answer

It is a magic tee (hybrid E-H tee) with ports 1 and 2 collinear, port 3 E-arm, port 4 H-arm, normalized by 1/sqrt(2). H-arm input splits equally in phase to ports 1 and 2; E-arm input splits equally anti-phase; collinear arms are isolated.

Identify and explain the properties of a microwave passive device having following S-Matrix with necessary diagram: [S] = [0 0 1 1; 0 0 −1 1; 1 −1 0 0; 1 1 0 0].

Identification

       | 0   0   1   1 |
 [S] = | 0   0  -1   1 |
       | 1  -1   0   0 |
       | 1   1   0   0 |
  • Ports 1 and 2 are isolated (S12 = 0) and ports 3 and 4 are isolated (S34 = 0); all ports are matched (Sᵢᵢ = 0); the matrix is symmetric (reciprocal).
  • Port 4 receives equal, in-phase signals from ports 1 and 2 (S41 = S42 = 1): port 4 is the H-arm (sum arm).
  • Port 3 receives equal, anti-phase signals from ports 1 and 2 (S31 = 1, S32 = −1): port 3 is the E-arm (difference arm).

These are the properties of a magic tee (hybrid E-H tee, 180° hybrid) with ports 1 and 2 as collinear arms.

Normalization: as written, each row has |S|² sum = 2. A passive lossless device needs a sum of 1, so the matrix is the magic-tee matrix with the common factor 1/√2 left out:

              | 0   0   1   1 |
 [S] = (1/√2) | 0   0  -1   1 |
              | 1  -1   0   0 |
              | 1   1   0   0 |

Diagram

              E-arm (3)
                | |
     -----------+ +-----------
 (1) ===========+   +=========== (2)
     -----------+ +-----------
               / /
              / /  H-arm (4)

Properties (with the 1/√2 factor)

  1. Input at H-arm (port 4): b1 = b2 = a4/√2, b3 = 0. Power divides equally and in phase between the collinear arms; the E-arm gets nothing.
  2. Input at E-arm (port 3): b1 = a3/√2, b2 = −a3/√2, b4 = 0. Equal split with 180° phase difference; the H-arm gets nothing.
  3. Input at port 1: b3 = b4 = a1/√2, b2 = 0. Half the power goes to each of the E and H arms; the opposite collinear arm is isolated.
  4. Equal in-phase inputs at 1 and 2 (a1 = a2 = a): b4 = √2·a (all the input power comes out of the H-arm), b3 = 0.
  5. Equal anti-phase inputs at 1 and 2 (a1 = −a2 = a): b3 = √2·a (all power at E-arm), b4 = 0.
  6. Matched and lossless: with E and H arms matched, all four ports are matched; [S]*ᵀ[S] = I (unitary).
  7. Reciprocal: Sᵢⱼ = Sⱼᵢ.

Applications

  • Balanced mixer: RF and LO fed to the E and H arms, mixer diodes in arms 1 and 2; LO noise cancels and the LO is isolated from the RF port.
  • Duplexer: transmitter and receiver at the isolated E and H arms, antenna at a collinear arm.
  • Power combiner/divider: combining two coherent sources (sum at H-arm).
  • Impedance bridge: with a known load on one collinear arm and an unknown on the other, the E-arm output is zero only when the loads are equal.
  • Monopulse radar: sum (Σ) and difference (Δ) signals from two antenna feeds.

✓ Answer checked by Claude Opus 5.5 (high effort)

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