RF and Microwave Components and Devices

A rectangular waveguide is a hollow metal pipe of inner width a (x-direction) and height b (y-direction), a > b, with the wave travelling along z. In a TM (transverse magnetic) mode, Hz = 0 and Ez ≠ 0.

   y
   ^
 b |+-----------------+
   ||                 |   wave travels in +z
   ||    air (μ0, ε0)  |   (into the page)
   |+-----------------+
   0------------------> x
                      a

Wave equation and solution

For fields varying as e^(j(ωt − βz)), Ez satisfies

 ∂²Ez/∂x² + ∂²Ez/∂y² + kc² Ez = 0
 kc² = k² - β²,   k = ω√(με)

By separation of variables, Ez = X(x)Y(y) with X = A1 cos kx·x + A2 sin kx·x, Y = B1 cos ky·y + B2 sin ky·y.

Boundary conditions: tangential E is zero on the walls, so Ez = 0 at x = 0, a and y = 0, b. This gives A1 = B1 = 0, sin kx·a = 0 and sin ky·b = 0:

 kx = mπ/a,  ky = nπ/b,  m, n = 1, 2, 3 ...
 kc² = (mπ/a)² + (nπ/b)²

Field equations (TMmn)

 Ez = E0 sin(mπx/a) sin(nπy/b) e^(-jβz)
 Ex = -(jβ/kc²)(mπ/a) E0 cos(mπx/a) sin(nπy/b) e^(-jβz)
 Ey = -(jβ/kc²)(nπ/b) E0 sin(mπx/a) cos(nπy/b) e^(-jβz)
 Hx =  (jωε/kc²)(nπ/b) E0 sin(mπx/a) cos(nπy/b) e^(-jβz)
 Hy = -(jωε/kc²)(mπ/a) E0 cos(mπx/a) sin(nπy/b) e^(-jβz)
 Hz = 0

Since Ez contains sin(mπx/a)·sin(nπy/b), m = 0 or n = 0 gives zero field. So TM00, TM10, TM01 do not exist; the lowest TM mode is TM11.

Characteristic equations

• Propagation constant: β = √(ω²με − kc²); the mode propagates only when ω²με > kc².
• Cut-off frequency:

 fc = (1/(2√(με))) √((m/a)² + (n/b)²)
    = (c/2) √((m/a)² + (n/b)²)   (air)

• Cut-off wavelength: λc = 2/√((m/a)² + (n/b)²)
• Guide wavelength: λg = λ0/√(1 − (fc/f)²) (longer than free-space λ0)
• Phase velocity: vp = ω/β = c/√(1 − (fc/f)²) > c
• Group velocity: vg = c·√(1 − (fc/f)²) < c, with vp·vg = c²
• Wave impedance:

 Z_TM = Ex/Hy = β/(ωε) = η·√(1 - (fc/f)²)
 (η = 377 Ω in air)

Z_TM is always less than η and becomes zero at cut-off.

Example

For an X-band guide (a = 2.286 cm, b = 1.016 cm), the TM11 cut-off is fc = (3×10⁸/2)√((1/0.02286)² + (1/0.01016)²) ≈ 16.2 GHz, much higher than the TE10 cut-off (6.56 GHz). This is why TM modes are not used as the operating mode in standard guides.

In a rectangular waveguide (inner dimensions a × b, a > b, propagation along z), a TE (transverse electric) mode has Ez = 0 and Hz ≠ 0. All other field components are derived from Hz.

   y
 b +-----------------+
   |  air (μ, ε)     |   propagation along +z
   +-----------------+
   0                 a  -> x

Derivation

With fields varying as e^(j(ωt − βz)), Hz satisfies the reduced wave equation

 ∂²Hz/∂x² + ∂²Hz/∂y² + kc² Hz = 0,   kc² = ω²με - β²

Separation of variables, Hz = X(x)·Y(y):

 X = A1 cos kx·x + A2 sin kx·x
 Y = B1 cos ky·y + B2 sin ky·y,     kx² + ky² = kc²

The transverse fields come from Maxwell's curl equations:

 Ex = -(jωμ/kc²) ∂Hz/∂y      Hx = -(jβ/kc²) ∂Hz/∂x
 Ey =  (jωμ/kc²) ∂Hz/∂x      Hy = -(jβ/kc²) ∂Hz/∂y

Boundary conditions (tangential E = 0 on walls): Ex = 0 at y = 0, b → ∂Hz/∂y = 0 there; Ey = 0 at x = 0, a → ∂Hz/∂x = 0 there. This gives A2 = B2 = 0 and

 kx = mπ/a,  ky = nπ/b,  m, n = 0, 1, 2 ... (not both zero)

TEmn field equations

 Hz = H0 cos(mπx/a) cos(nπy/b) e^(-jβz)
 Hx =  (jβ/kc²)(mπ/a) H0 sin(mπx/a) cos(nπy/b) e^(-jβz)
 Hy =  (jβ/kc²)(nπ/b) H0 cos(mπx/a) sin(nπy/b) e^(-jβz)
 Ex =  (jωμ/kc²)(nπ/b) H0 cos(mπx/a) sin(nπy/b) e^(-jβz)
 Ey = -(jωμ/kc²)(mπ/a) H0 sin(mπx/a) cos(nπy/b) e^(-jβz)
 Ez = 0

Because Hz uses cosines, one index may be zero (TE10, TE01, TE20 exist), but TE00 gives no field.

Characteristic parameters

• Cut-off wave number: kc = √((mπ/a)² + (nπ/b)²)
• Cut-off frequency: fc = (c/2)√((m/a)² + (n/b)²) (air-filled)
• Cut-off wavelength: λc = 2/√((m/a)² + (n/b)²); for TE10, λc = 2a
• Phase constant: β = √(k² − kc²) = k√(1 − (fc/f)²)
• Guide wavelength: λg = λ0/√(1 − (fc/f)²)
• Phase velocity: vp = c/√(1 − (fc/f)²); group velocity: vg = c√(1 − (fc/f)²); vp·vg = c²
• Wave impedance:

 Z_TE = Ex/Hy = ωμ/β = η/√(1 - (fc/f)²)     (η = 377 Ω)

Z_TE is always greater than η.

Dominant TE10 mode

With m = 1, n = 0 (and a > b), TE10 has the lowest cut-off fc = c/2a:

 Hz = H0 cos(πx/a) e^(-jβz)
 Hx = (jβa/π) H0 sin(πx/a) e^(-jβz)
 Ey = -(jωμa/π) H0 sin(πx/a) e^(-jβz)

Only Ey, Hx and Hz exist; E is maximum at the centre (x = a/2) and zero at the side walls.

Example: for a = 2.286 cm at 10 GHz, fc = 6.56 GHz, λg = 3.98 cm and Z_TE ≈ 500 Ω.

A magic tee (hybrid tee) is a four-port waveguide junction made by combining an E-plane tee and an H-plane tee at the same point of a main rectangular waveguide. Ports 1 and 2 are the collinear arms, port 3 is the H-arm (sum arm) and port 4 is the E-arm (difference arm). Matching posts or irises are placed inside so that the E and H arms are matched.

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

S-matrix of an ideal (matched) magic tee:

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

Properties ("magic" behaviour):

1. H-arm input (port 3): power divides equally between ports 1 and 2 in phase; nothing goes to port 4.
2. E-arm input (port 4): power divides equally between ports 1 and 2 180° out of phase; nothing goes to port 3.
3. E and H arms are isolated (S34 = S43 = 0), and ports 1 and 2 are isolated (S12 = 0).
4. Input at port 1 (or 2): splits equally to ports 3 and 4; nothing goes to port 2.
5. Equal in-phase inputs at ports 1 and 2 add at port 3 (sum) and cancel at port 4 (difference).
6. When the E and H arms are matched, all four ports are automatically matched (S11 = S22 = S33 = S44 = 0).

Applications:

• Balanced mixers (LO and signal fed to E and H arms, diodes in arms 1 and 2).
• Duplexers in radar (isolating transmitter and receiver).
• Impedance bridges and measurement of unknown impedances.
• Power combining/dividing and monopulse radar sum/difference networks.

A circulator is a multi-port (usually 3- or 4-port), non-reciprocal microwave device in which power entering any port is transmitted only to the next port in a fixed rotation (1 → 2 → 3 → 1) and not to the others. Non-reciprocity is produced by a magnetized ferrite placed at the junction.

           port 2
             |
          +--+--+
          | /-> |
 port 1 --| | F | --- port 3
          | <-/ |
          +-----+
     F = biased ferrite disc
     1 -> 2 -> 3 -> 1

S-matrix of an ideal 3-port circulator:

       | 0  0  1 |
 [S] = | 1  0  0 |
       | 0  1  0 |

• All ports matched (Sᵢᵢ = 0), lossless (unitary) and non-reciprocal (S21 ≠ S12).
• This agrees with the rule that a matched, lossless 3-port must be non-reciprocal.

Working principle: In the Y-junction circulator, the ferrite disc biased by a dc magnetic field splits the field into two counter-rotating modes with different propagation constants (because the ferrite permeability differs for the two senses of rotation). Their dimensions are chosen so the modes add at the next port and cancel at the following port. A 4-port circulator can be made with two magic tees and a non-reciprocal (Faraday rotation) phase shifter.

Performance parameters: insertion loss (typ. 0.3–0.5 dB), isolation (20–30 dB), VSWR and bandwidth.

Applications:

• Duplexer in radar and transceivers: transmitter → antenna → receiver using one antenna.
• Isolator: a circulator with port 3 terminated in a matched load, used to protect sources from reflections.
• Reflection-type (negative resistance) amplifiers such as tunnel diode and parametric amplifiers.
• Separating incident and reflected signals in measurement systems.

A directional coupler is a four-port passive device that samples a known small fraction of the power travelling in one direction in the main line, while ignoring power travelling in the opposite direction. Port 1 is the input, port 2 the through (direct) port, port 3 the coupled port and port 4 the isolated port. A two-hole waveguide coupler is a common example.

 Input (1) ===================== Through (2)
                 o      o        main guide
             <-- λg/4 -->        (two holes)
 Isolated (4) ------------------ Coupled (3)
                                 auxiliary guide

Two-hole principle: waves coupled through the two holes, spaced λg/4 apart, travel the same distance to port 3 and add in phase, but differ by λg/2 (180°) on the way to port 4 and cancel. So power reaches port 3 only.

Performance parameters (P1 = input power, P2 through, P3 coupled, P4 isolated):

• Coupling factor: C = 10 log(P1/P3) dB (e.g. 3, 10, 20 dB)
• Directivity: D = 10 log(P3/P4) dB (ideally infinite, typically 30–40 dB)
• Isolation: I = 10 log(P1/P4) dB = C + D
• Insertion loss: IL = 10 log(P1/P2) dB

S-matrix of an ideal symmetric coupler (α² + β² = 1):

       | 0   α   jβ  0  |
 [S] = | α   0   0   jβ |
       | jβ  0   0   α  |
       | 0   jβ  α   0  |

All ports matched, ports 1 and 4 (and 2 and 3) isolated.

Types: two-hole and multi-hole waveguide couplers, Bethe-hole coupler, branch-line (90° hybrid) coupler, coupled-line microstrip coupler, rat-race (180° hybrid).

Applications: power monitoring and sampling, reflectometers for VSWR and reflection coefficient measurement, signal injection, power splitting/combining and balanced amplifiers.

Identification

       | S11   S12   S13   S14 |
 [S] = | S12   S22   S13  -S14 |
       | S13   S13   0     0   |
       | S14  -S14   0     0   |

• Port 3 receives equal signals in phase from ports 1 and 2 (S31 = S32 = S13): port 3 behaves as the H-arm.
• Port 4 receives equal signals 180° out of phase from ports 1 and 2 (S41 = −S42 = S14): port 4 behaves as the E-arm.
• S34 = S43 = 0: the E and H arms are isolated; S33 = S44 = 0: those arms are matched.
• The matrix is symmetric, so the device is reciprocal.

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

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

Finding the ideal values (lossless condition)

For a lossless junction, [S]*ᵀ[S] = I.

 Row 3: |S13|² + |S13|² = 1  ->  S13 = 1/√2
 Row 4: |S14|² + |S14|² = 1  ->  S14 = 1/√2
 Row 1: |S11|² + |S12|² + 1/2 + 1/2 = 1
        -> S11 = 0 and S12 = 0
 Row 2: |S12|² + |S22|² + 1/2 + 1/2 = 1  ->  S22 = 0

So matching the E and H arms automatically matches ports 1 and 2 and isolates them:

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

Properties

1. Input at H-arm (3): b1 = b2 = a3/√2 – equal in-phase split; b4 = 0.
2. Input at E-arm (4): b1 = a4/√2, b2 = −a4/√2 – equal split, 180° apart; b3 = 0.
3. Input at port 1: b3 = b4 = a1/√2; port 2 gets nothing (S21 = 0), so collinear arms are isolated.
4. Equal in-phase inputs at 1 and 2: b3 = √2·a (all power at H-arm), b4 = 0. Equal anti-phase inputs: all power at E-arm.
5. All ports matched, lossless and reciprocal.

Uses

Balanced mixers, radar duplexers, power combiners/dividers, impedance bridges and monopulse comparators.

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; then the output appears at one collinear arm. Using the collinear arms as inputs is simpler since both inputs must be in phase.
• Isolators (circulators with a load) may be added at each transmitter output for further protection.

A cavity resonator is a closed metallic enclosure (a section of waveguide shorted at both ends, or a circular/re-entrant cavity) that stores electromagnetic energy at certain resonant frequencies, like an LC tank circuit at microwave frequencies. Lumped L and C are unusable at microwave frequencies because of radiation and skin-effect losses, so cavities are used instead.

     +---------------------+
     |                     |  b
     |   standing wave     |
     +---------------------+
     <-------- d ---------->
   rectangular cavity a × b × d
   coupled by probe, loop or iris

Resonant frequency (rectangular cavity, TEmnp or TMmnp):

 f_r = (c/2) √((m/a)² + (n/b)² + (p/d)²)

The cavity length must be a whole number of half guide wavelengths: d = p·λg/2. For a circular cavity (TE011, TM010) the resonant frequency depends on the Bessel roots. Example: an a = 2.286 cm, b = 1.016 cm, d = 2 cm cavity in TE101 resonates at (3×10⁸/2)√((1/0.02286)² + (1/0.02)²) ≈ 9.96 GHz.

Quality factor:

• Q = 2π × (energy stored)/(energy lost per cycle) = ω0·W/P
• Unloaded Q (wall loss only) is very high, 10³–10⁴ (higher than lumped circuits); loaded Q includes the external circuit: 1/QL = 1/Q0 + 1/Qext.

Types: rectangular, cylindrical (circular), re-entrant (used in klystrons), coaxial and dielectric resonators.

Coupling methods: probe (electric), loop (magnetic) and aperture/iris coupling.

Tuning: a plunger changes d, or a screw/dielectric rod perturbs the field.

Applications: klystron and magnetron tuned circuits, cavity wavemeters (frequency measurement), filters, stable oscillators, and measurement of dielectric constant (cavity perturbation).

A circular waveguide is a hollow metal tube of inner radius a. Fields are written in cylindrical coordinates (ρ, φ, z) with propagation e^(−jβz). In a TE mode, Ez = 0 and Hz ≠ 0.

          .-----.
        /    ρ    \
       |  o---->   |   radius a
        \         /    wave along z
          '-----'

Wave equation and solution

 ∂²Hz/∂ρ² + (1/ρ)∂Hz/∂ρ + (1/ρ²)∂²Hz/∂φ² + kc² Hz = 0
 kc² = k² - β²

Separation of variables, Hz = R(ρ)·Φ(φ), gives Φ = A sin nφ + B cos nφ and Bessel's equation for R, whose finite solution at ρ = 0 is Jn(kc ρ):

 Hz = (A sin nφ + B cos nφ) Jn(kc ρ) e^(-jβz)

Boundary condition: Eφ = 0 at ρ = a. Since Eφ ∝ J'n(kc ρ):

 J'n(kc a) = 0  ->  kc = p'nm / a

where p'nm is the m-th root of J'n(x).

TEnm field equations

 Hz = (A sin nφ + B cos nφ) Jn(kcρ) e^(-jβz)
 Eρ = -(jωμn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Eφ =  (jωμ/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Hρ = -(jβ/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Hφ = -(jβn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Ez = 0

Characteristic equations

• Cut-off frequency: fc = p'nm·c/(2πa) (air-filled)
• Cut-off wavelength: λc = 2πa/p'nm
• Phase constant: β = √(k² − (p'nm/a)²)
• Guide wavelength: λg = λ0/√(1 − (fc/f)²)
• Phase and group velocity: vp = c/√(1 − (fc/f)²), vg = c√(1 − (fc/f)²)
• Wave impedance: Z_TE = ωμ/β = η/√(1 − (fc/f)²)

Roots of J'n(x):

Important modes

• TE11 has the smallest root, so it is the dominant mode of the circular waveguide: fc = 1.841c/(2πa), λc = 3.41a. Example: a = 1 cm gives fc = 8.79 GHz.
• TE01 has circularly symmetric fields and its wall loss falls as frequency rises, so it is used for low-loss long-distance waveguide links.

For an ideal hybrid (magic) tee with ports 1 and 2 collinear, port 3 the H-arm and port 4 the E-arm:

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

When a port is terminated in a matched load, no wave returns into it (aₖ = 0), so its column is dropped; its row tells how much power the load absorbs. The remaining matrix describes the reduced network.

(a) E-arm (port 4) terminated in a matched load

Remaining ports 1, 2, 3:

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

• Input at H-arm (3) divides equally and in phase to ports 1 and 2: the device works as an H-plane (in-phase) power divider.
• Input at port 1: half the power goes to port 3, half (b4 = a1/√2) is absorbed in the E-arm load; port 2 receives nothing.
• In-phase equal inputs at 1 and 2 combine fully at port 3 (power combiner); any imbalance goes to the load.

(b) H-arm (port 3) terminated in a matched load

Remaining ports 1, 2, 4:

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

• Input at E-arm (4) divides equally to ports 1 and 2 but 180° out of phase: it works as an E-plane (anti-phase) divider.
• Input at port 1: half goes to port 4, half is absorbed in the H-arm load; ports 1 and 2 remain isolated.
• Anti-phase equal inputs at 1 and 2 combine fully at port 4.

(c) Co-planar (collinear) arms 1 and 2 terminated in matched loads

Remaining ports 3 and 4:

       | S33  S34 |   | 0  0 |
 [S] = |          | = |      |
       | S43  S44 |   | 0  0 |

• Both E and H arms are matched (no reflection) and completely isolated from each other (S34 = 0).
• Power fed into either the E-arm or the H-arm is split equally into the two loads and fully absorbed.
• This isolation between E and H arms is the basis of the balanced mixer (LO at one arm, signal at the other) and of bridge measurements: with identical loads on 1 and 2, no signal passes from H-arm to E-arm; any mismatch difference between the two loads shows up at the E-arm.

Field equations of a circular waveguide in TE mode

For a circular guide of radius a (cylindrical coordinates ρ, φ, z) with Ez = 0, Hz satisfies Bessel's equation and the solution finite at the axis is

 Hz = (A sin nφ + B cos nφ) Jn(kcρ) e^(-jβz)

The transverse fields from Maxwell's equations are:

 Eρ = -(jωμn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Eφ =  (jωμ/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Hρ = -(jβ/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Hφ = -(jβn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Ez = 0

Boundary condition Eφ(ρ = a) = 0 gives J'n(kc a) = 0, so kc = p'nm/a.

Characteristic equations:

• fc = p'nm·c/(2πa), λc = 2πa/p'nm
• β = √(k² − kc²), λg = λ0/√(1 − (fc/f)²)
• Z_TE = η/√(1 − (fc/f)²)
• Dominant mode TE11: p'11 = 1.841, λc = 3.41a.

Dominant and degenerate modes of a rectangular waveguide

Dominant mode: the mode with the lowest cut-off frequency (longest cut-off wavelength). For a > b:

 fc,mn = (c/2) √((m/a)² + (n/b)²)
 TE10: fc = c/(2a)  (lowest)

TE10 is the dominant mode. Example: a = 2.286 cm, b = 1.016 cm gives fc,TE10 = 6.56 GHz, next mode TE20 at 13.12 GHz. So between 6.56 and 13.12 GHz only TE10 propagates, giving single-mode operation (X-band, 8.2–12.4 GHz).

Degenerate modes: different modes having the same cut-off frequency (same m, n but different field patterns). In a rectangular guide, TEmn and TMmn (m, n ≥ 1) always share the same fc, e.g. TE11 and TM11, TE21 and TM21. In a square guide (a = b), TE10 and TE01 are also degenerate.

Example: in the X-band guide above, TE11 and TM11 both have fc = (3×10⁸/2)√((1/0.02286)² + (1/0.01016)²) ≈ 16.15 GHz.

Degenerate modes are undesirable because energy can transfer between them at discontinuities, causing distortion; guides are therefore operated in the dominant-mode band.

Three-port network model and its parameters

A three-port network (e.g. an E-plane or H-plane tee, circulator or power divider) is described by its scattering matrix, relating outgoing waves b to incident waves a:

            port 3
              |
          +---+---+
 port 1 --|  [S]  |-- port 2
          +-------+

 | b1 |   | S11  S12  S13 | | a1 |
 | b2 | = | S21  S22  S23 | | a2 |
 | b3 |   | S31  S32  S33 | | a3 |

Characteristic parameters:

• Sᵢᵢ: reflection coefficient at port i (return loss = −20 log|Sᵢᵢ|).
• Sᵢⱼ: transmission from port j to i (insertion loss = −20 log|Sᵢⱼ|).
• Reciprocal: Sᵢⱼ = Sⱼᵢ. Lossless: [S]*ᵀ[S] = I. Matched: Sᵢᵢ = 0.
• A 3-port cannot be lossless, reciprocal and fully matched at the same time.

Example, H-plane tee: S = [[1/2, −1/2, 1/√2], [−1/2, 1/2, 1/√2], [1/√2, 1/√2, 0]].

Magic tee with both E and H arms shorted

Ideal magic tee (1, 2 collinear; 3 = H-arm; 4 = E-arm):

 b1 = (a3 + a4)/√2      b3 = (a1 + a2)/√2
 b2 = (a3 - a4)/√2      b4 = (a1 - a2)/√2

A short at the reference plane of an arm reflects the wave with Γ = −1, so a3 = −b3 and a4 = −b4:

 a3 = -(a1 + a2)/√2,   a4 = -(a1 - a2)/√2

 b1 = (a3 + a4)/√2 = -(2a1)/2 = -a1
 b2 = (a3 - a4)/√2 = -(2a2)/2 = -a2

So the reduced two-port (ports 1 and 2) has

       | -1   0 |
 [S] = |        |
       |  0  -1 |

Behaviour: each collinear arm behaves like a short circuit: all power entering port 1 (or 2) is reflected back with 180° phase, and nothing is transmitted to the other arm. The power entering port 1 splits into the E and H arms, is reflected by the shorts, and recombines entirely at port 1.

General case (shorts at distances l3 and l4): Γ3 = −e^(−j2βl3), Γ4 = −e^(−j2βl4). Then

 b1 = ((Γ3 + Γ4)/2)·a1 + ((Γ3 - Γ4)/2)·a2
 b2 = ((Γ3 - Γ4)/2)·a1 + ((Γ3 + Γ4)/2)·a2

If the shorts differ in position by λg/4 (Γ4 = −Γ3), then S11 = 0 and |S21| = 1: all power passes from port 1 to port 2. By moving the shorts, the tee can act as a variable reflector, switch or phase shifter; this property is used in magic-tee phase shifters and duplexers.

The dominant mode is the mode with the lowest cut-off frequency. Here the broad wall is taken as the dimension along which the index m is counted, i.e. the wider side is a (the usual convention a > b; the "b > a" in the question is read as the broad dimension being the one carrying index 1).

For an air-filled rectangular guide, the cut-off frequency of the TEmn mode is

 fc,mn = (c/2) √((m/a)² + (n/b)²)

For TE modes m and n can be 0, 1, 2 … but not both zero (TE00 has no field).

The candidates for the lowest cut-off are:

 TE10: fc = c/(2a)
 TE01: fc = c/(2b)
 TE11: fc = (c/2)√(1/a² + 1/b²)

Since a > b, 1/a < 1/b, so c/(2a) < c/(2b) < (c/2)√(1/a² + 1/b²). Any higher m or n only increases fc, and TM modes start from TM11, which also has a higher fc. Hence TE10 has the lowest cut-off frequency and is the dominant mode, with λc = 2a.

(If the indices were assigned the other way, i.e. the wider wall were along y, the same argument would make TE01 the dominant mode; physically it is always the mode with one half-wave across the wider wall.)

For each junction, port numbering is: ports 1 and 2 are the collinear arms and port 3 is the side arm. A "perfectly working" device means a lossless junction with its side arm matched (for tees) or all ports matched (for the duplexer).

1. E-plane tee (series tee)

The side arm is in the plane of the E-field (on the broad wall). A wave fed into port 3 splits equally into ports 1 and 2 with 180° phase difference.

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

Check: each row has |S|² sum = 1/4 + 1/4 + 1/2 = 1 and rows are orthogonal, so it is lossless; S33 = 0 (matched side arm); S13 = −S23 (anti-phase split). Uses: power division with phase reversal, balanced mixers, part of the magic tee, impedance tuners (with a shorted side arm).

2. H-plane tee (shunt tee)

The side arm is in the plane of the H-field (on the narrow wall). A wave fed into port 3 splits equally into ports 1 and 2 in phase.

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

Again lossless and reciprocal, S33 = 0, and S13 = S23 (in-phase split). Uses: in-phase power dividers and combiners, feeding antenna arrays, part of the magic tee.

3. Duplexer (3-port circulator type)

A duplexer lets one antenna be shared by a transmitter (port 1), antenna (port 2) and receiver (port 3): transmitter power goes only to the antenna, and the received echo goes only to the receiver. A perfect duplexer is an ideal circulator:

       | 0  0  1 |
 [S] = | 1  0  0 |       TX(1) -> ANT(2) -> RX(3)
       | 0  1  0 |

• Matched at all ports, lossless (unitary), and non-reciprocal (S21 = 1 but S12 = 0), made with a magnetized ferrite.
• Transmitter → antenna: b2 = a1. Antenna → receiver: b3 = a2. Transmitter → receiver: S31 = 0 (isolation protects the receiver).

   TX (1) ---->+-----+----> ANT (2)
               | (→) |
               +--+--+
                  |
                  v
               RX (3)

Uses: radar and communication transceivers using a single antenna, isolators (with port 3 terminated), reflection amplifiers.

Note: the two tees are reciprocal and cannot have all three ports matched; only the non-reciprocal circulator can be lossless and matched at all ports.

Dominant mode

The dominant mode of a waveguide is the mode with the lowest cut-off frequency (longest cut-off wavelength). It is the first mode to propagate as frequency rises, so a guide is normally operated where only this mode can travel.

Dominant modes of rectangular and circular waveguides

• Rectangular waveguide (a > b): TE10, with fc = c/(2a), λc = 2a.
• Circular waveguide (radius r): TE11, with fc = 1.841c/(2πr), λc = 3.41r.

Does a 2.254 cm wide guide pass 6 GHz?

For the dominant TE10 mode, with a = 2.254 cm = 0.02254 m:

 fc = c/(2a) = (3 × 10⁸)/(2 × 0.02254)
    = 6.655 × 10⁹ Hz = 6.655 GHz

The signal frequency f = 6 GHz is below the cut-off frequency (6 < 6.655 GHz). Equivalently, λ0 = 5 cm > λc = 2a = 4.508 cm.

Answer: No. The lowest cut-off of the guide is 6.655 GHz, so a 6 GHz signal is evanescent (attenuated) and does not propagate.

A magic tee is an E-plane tee and an H-plane tee joined at the same point. Ports 1 and 2 are collinear arms, port 3 is the H-arm and port 4 the E-arm.

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

Derivation of the S-matrix

1. H-arm property: input at port 3 splits equally and in phase to 1 and 2: S13 = S23.
2. E-arm property: input at port 4 splits equally and in anti-phase: S14 = −S24.
3. Isolation of E and H arms (by geometry/symmetry): S34 = S43 = 0.
4. Matching: E and H arms matched with posts: S33 = S44 = 0.
5. Reciprocity: Sᵢⱼ = Sⱼᵢ.

So far:

       | S11  S12  S13  S14 |
 [S] = | S12  S22  S13 -S14 |
       | S13  S13   0    0  |
       | S14 -S14   0    0  |

6. Lossless (unitary) condition:

 Row 3: 2|S13|² = 1           ->  S13 = 1/√2
 Row 4: 2|S14|² = 1           ->  S14 = 1/√2
 Row 1: |S11|² + |S12|² + 1 = 1 -> S11 = S12 = 0
 Row 2: |S12|² + |S22|² + 1 = 1 -> S22 = 0

Result:

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

All ports are matched, ports 1–2 and 3–4 are isolated, and power fed to any port divides equally between the two ports adjacent to it.

Part 1: Air-filled waveguide, a = 3 cm, f = 6 GHz, TE10 mode

Constants: c = 3 × 10⁸ m/s, η0 = 120π ≈ 377 Ω.

Cut-off wavelength:

 λc = 2a = 2 × 3 = 6 cm

Cut-off frequency:

 fc = c/λc = (3 × 10⁸)/(0.06) = 5 × 10⁹ Hz = 5 GHz

Since f = 6 GHz > 5 GHz, TE10 propagates.

Common factor:

 λ0 = c/f = 3 × 10⁸ / 6 × 10⁹ = 5 cm
 √(1 - (fc/f)²) = √(1 - (5/6)²) = √0.3056 = 0.5528

Guide (propagation) wavelength:

 λg = λ0/0.5528 = 5/0.5528 = 9.045 cm

Phase velocity:

 vp = c/0.5528 = 5.427 × 10⁸ m/s

Group velocity:

 vg = c × 0.5528 = 1.658 × 10⁸ m/s
 (check: vp·vg = 9 × 10¹⁶ = c²)

Characteristic (wave) impedance:

 Z_TE = η0/0.5528 = 377/0.5528 = 682 Ω

Answer: fc = 5 GHz, λc = 6 cm, vg = 1.658 × 10⁸ m/s, vp = 5.427 × 10⁸ m/s, λg = 9.045 cm, Z_TE ≈ 682 Ω.

Part 2: Four basic parameters of a directional coupler

Ports: 1 = input, 2 = through, 3 = coupled, 4 = isolated.

 (1) Input =================== Through (2)
             coupling region
 (4) Isolated ---------------- Coupled (3)

1. Coupling factor: C = 10 log(P1/P3) dB = −20 log|S31|. The fraction of input power sent to the coupled port (e.g. 10 dB coupler sends 10%).
2. Directivity: D = 10 log(P3/P4) dB = 20 log(|S31|/|S41|). How well the coupler separates forward and reverse waves; ideally infinite, typically 30–40 dB.
3. Isolation: I = 10 log(P1/P4) dB = −20 log|S41| = C + D. Power leaking to the isolated port.
4. Insertion loss: IL = 10 log(P1/P2) dB = −20 log|S21|. Power lost in the main line, including the coupled power.

An E-plane tee (series tee) is a waveguide tee junction in which the side arm (port 3) is attached to the broad wall of the main guide, so the side arm axis is parallel to the E-field of the TE10 mode. Ports 1 and 2 are the collinear arms.

          port 3 (E-arm)
             | |
   ----------+ +----------
 (1)                    (2)
   -----------------------

Behaviour:

• Input at port 3 divides equally into ports 1 and 2 180° out of phase (the E-field lines bend in opposite directions).
• Equal in-phase inputs at 1 and 2 cancel at port 3; anti-phase inputs add at port 3.
• It acts like a series connection of the side arm with the main line.

S-matrix (lossless, side arm matched):

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

Uses: power division with phase reversal, balanced mixers, impedance tuners (with a sliding short in port 3) and as part of a magic tee.

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 (sum of powers at 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.

When two magic tees are connected by joining their H-arms with a short piece of waveguide, a six-port network is obtained. Let tee A have ports 1, 2 (collinear), 3 (H) and 4 (E); tee B have ports 5, 6 (collinear), 7 (H) and 8 (E). Port 3 is joined to port 7, so a7 = b3 and a3 = b7 (zero-length link assumed).

        (4) E                   E (8)
         |                       |
 (1) ===[A]=== (2)       (5) ===[B]=== (6)
         |                       |
         +------ H-H link -------+

Using b3 = (a1 + a2)/√2 and b7 = (a5 + a6)/√2 for each tee:

 b1 = a4/√2 + (a5 + a6)/2      b5 =  a8/√2 + (a1 + a2)/2
 b2 = -a4/√2 + (a5 + a6)/2     b6 = -a8/√2 + (a1 + a2)/2
 b4 = (a1 - a2)/√2             b8 = (a5 - a6)/√2

Properties of the combination:

• Power fed at port 1 goes half to E-arm 4 and a quarter each to ports 5 and 6, in phase; nothing returns to port 1 or reaches port 2 directly.
• The sum (in-phase) part of the signals at tee A passes through the H-link to tee B, while the difference part comes out of A's own E-arm. So the pair separates common-mode and differential signals.
• The two E-arms (4 and 8) are completely isolated from each other.
• The network stays matched, lossless and reciprocal (the 6 × 6 S-matrix is symmetric and unitary).

Uses: sum/difference networks in monopulse radar, balanced duplexers and mixers, and power combining arrangements where in-phase signals must be transferred between two hybrid circuits.

Assumption: a standard WR-90 (X-band) air-filled waveguide, a = 2.286 cm (breadth), b = 1.016 cm (height), operating at f = 10 GHz in the dominant TE10 mode. Constants: c = 3 × 10⁸ m/s, η0 = 120π ≈ 377 Ω.

Cut-off frequency (TE10):

 fc = c/(2a) = (3 × 10⁸)/(2 × 0.02286) = 6.562 GHz

Next mode TE20 (13.12 GHz) and TE01 (c/2b = 14.76 GHz) are above 10 GHz, so only TE10 propagates.

Common factor:

 √(1 - (fc/f)²) = √(1 - (6.562/10)²)
               = √(1 - 0.4306) = 0.7546
 λ0 = c/f = 3 cm

Phase velocity:

 vp = c/0.7546 = 3.976 × 10⁸ m/s

Guide (propagation) wavelength:

 λg = λ0/0.7546 = 3/0.7546 = 3.976 cm

Characteristic (wave) impedance:

 Z_TE = η0/0.7546 = 377/0.7546 = 499.6 Ω

Answer: fc = 6.562 GHz, vp = 3.976 × 10⁸ m/s, λg = 3.976 cm, Z_TE ≈ 500 Ω (for a = 2.286 cm, b = 1.016 cm, f = 10 GHz). Group velocity for reference: vg = c × 0.7546 = 2.264 × 10⁸ m/s.

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.

Reading of the problem

The radar must feed the antenna with half of the transmitter power, and the same antenna must be used for receiving (duplexing). Both needs are met by a magic tee used as a duplexer:

• Transmitter at the H-arm (port 3).
• Antenna at collinear port 1; matched dummy load at collinear port 2.
• Receiver at the E-arm (port 4).

               RX (4, E-arm)
                  | |
 ANT (1) ========+   +======== (2) matched load
                  / /
                 / /
           TX (3, H-arm)

S-matrix of the duplexer (ideal magic tee)

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

Transmit mode

TX sends a3 = a (power Pt = |a|²/2); antenna and load are matched, a1 = a2 = a4 = 0:

 b1 = a3/√2   -> power to antenna = Pt/2
 b2 = a3/√2   -> power to dummy load = Pt/2
 b4 = 0       -> no power to receiver (S43 = 0)
 b3 = 0       -> no reflection to TX

So the antenna gets exactly half the transmitter power (−3 dB), and the receiver is isolated from the high-power transmitter.

Receive mode

The echo enters at port 1 (a1 = e):

 b3 = e/√2  -> half to transmitter arm
 b4 = e/√2  -> half to receiver (E-arm)
 b2 = 0     -> nothing to the load

The receiver gets half the received echo power.

Remarks

• Total loss in the duplexer is 3 dB on transmit and 3 dB on receive, but the arrangement is passive, reciprocal and needs no switching, and the receiver is protected by the E–H isolation.
• If no power loss were acceptable, a ferrite circulator duplexer [[0,0,1],[1,0,0],[0,1,0]] would be used instead; the magic tee version fits this requirement because the antenna needs only half the power.
• Practical systems add a TR (transmit-receive) limiter before the receiver for extra protection against leakage from imperfect isolation.

A three-port directional coupler is analysed with the 3 × 3 S-matrix, using port 1 = input, 2 = through, 3 = coupled.

Step 1 – can a lossless, reciprocal, fully matched 3-port exist? Suppose S11 = S22 = S33 = 0 and Sᵢⱼ = Sⱼᵢ:

       |  0   S12  S13 |
 [S] = | S12   0   S23 |
       | S13  S23   0  |

Unitary conditions:

 |S12|² + |S13|² = 1
 |S12|² + |S23|² = 1
 |S13|² + |S23|² = 1
 S13*·S23 = 0,  S23*·S12 = 0,  S12*·S13 = 0

The last three need at least two of S12, S13, S23 to be zero, which then breaks one of the first three. So no lossless, reciprocal, matched 3-port coupler exists.

Step 2 – practical 3-port coupler. It is a 4-port coupler whose isolated port is terminated internally in a matched load. With the ideal 4-port coupler (S12 = α, S13 = jβ, α² + β² = 1), deleting port 4:

       |  0    α    jβ |
 [S] = |  α    0    0  |
       | jβ    0    0  |

It is matched and reciprocal but lossy (power entering port 2 or 3 is partly absorbed in the internal load). Coupling C = −20 log β dB; e.g. β = 0.316 gives a 10 dB coupler.

Identification

       | S11    0    S13   S14 |
 [S] = |  0    S22  -S13   S14 |
       | S13  -S13    0     0  |
       | S14   S14    0     0  |

• Ports 1 and 2 are isolated (S12 = 0); ports 3 and 4 are isolated (S34 = 0) and matched (S33 = S44 = 0).
• Port 3 gets equal and opposite-phase signals from ports 1 and 2 (S31 = −S32): port 3 is the E-arm.
• Port 4 gets equal and in-phase signals from ports 1 and 2 (S41 = S42): port 4 is the H-arm.
• The matrix is symmetric, so the device is reciprocal.

The device is a magic tee (hybrid E-H tee) with ports 1 and 2 as the collinear arms.

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

Judging the condition (lossless junction)

For a lossless device [S]*ᵀ[S] = I:

 Row 3: |S13|² + |S13|² = 1  -> |S13| = 1/√2
 Row 4: |S14|² + |S14|² = 1  -> |S14| = 1/√2
 Row 1: |S11|² + |S13|² + |S14|² = 1
        |S11|² + 1/2 + 1/2 = 1  -> S11 = 0
 Row 2: similarly            -> S22 = 0

So the given form is consistent only when S11 = S22 = 0: matching the E and H arms automatically matches the collinear arms. The device is then the ideal magic tee:

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

If S11, S22 ≠ 0 in practice, the junction is lossy or imperfectly matched.

Characteristics

• H-arm input (4): equal in-phase split to ports 1 and 2; no output at E-arm.
• E-arm input (3): equal split to 1 and 2 with 180° phase difference; no output at H-arm.
• Input at port 1: half to each of E and H arms; port 2 isolated.
• Sum/difference action: in-phase inputs at 1 and 2 add at the H-arm; anti-phase inputs add at the E-arm.
• Matched, lossless, reciprocal. Used in balanced mixers, duplexers, impedance bridges and monopulse comparators.

TM field expressions for an air-filled circular waveguide

For a circular guide of radius a (coordinates ρ, φ, z), a TM mode has Hz = 0 and Ez ≠ 0, with fields varying as e^(j(ωt − βz)).

        .-------.
      /     ρ     \
     |   o----->   |   radius a, air (μ0, ε0)
      \           /    propagation along z
        '-------'

Step 1 – wave equation for Ez:

 ∂²Ez/∂ρ² + (1/ρ)∂Ez/∂ρ + (1/ρ²)∂²Ez/∂φ² + kc² Ez = 0
 kc² = k² - β²,  k = ω√(μ0ε0)

Step 2 – separation of variables: Ez = R(ρ)·Φ(φ)·e^(−jβz) gives

 Φ'' + n²Φ = 0          ->  Φ = A sin nφ + B cos nφ
 ρ²R'' + ρR' + (kc²ρ² - n²)R = 0   (Bessel's equation)
 R = C·Jn(kcρ) + D·Yn(kcρ)

Yn is infinite at ρ = 0, so D = 0:

 Ez = (A sin nφ + B cos nφ) Jn(kcρ) e^(-jβz)

Step 3 – boundary condition: tangential Ez = 0 at the wall ρ = a:

 Jn(kc a) = 0   ->   kc = pnm / a

where pnm is the m-th zero of Jn(x).

Step 4 – transverse fields from Maxwell's equations (with Hz = 0):

 Eρ = -(jβ/kc²) ∂Ez/∂ρ        Hρ =  (jωε/(kc²ρ)) ∂Ez/∂φ
 Eφ = -(jβ/(kc²ρ)) ∂Ez/∂φ     Hφ = -(jωε/kc²) ∂Ez/∂ρ

giving the TMnm field strengths:

 Ez = (A sin nφ + B cos nφ) Jn(kcρ) e^(-jβz)
 Eρ = -(jβ/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Eφ = -(jβn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Hρ =  (jωε0 n/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Hφ = -(jωε0/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Hz = 0

Characteristic relations (air):

 fc = pnm·c/(2πa),   λc = 2πa/pnm
 β = √(k² - (pnm/a)²)
 Z_TM = Eρ/Hφ = β/(ωε0) = η0 √(1 - (fc/f)²)

Dominant modes in TE and TM

• TM: the smallest zero of Jn is p01 = 2.405, so TM01 is the lowest TM mode (fc = 2.405c/(2πa), λc = 2.61a).
• TE: the smallest zero of J'n is p'11 = 1.841, so TE11 is the lowest TE mode (λc = 3.41a).
• Since 1.841 < 2.405, TE11 is the dominant mode of the circular waveguide; TM01 is the second mode. Example: a = 1 cm gives fc(TE11) = 8.79 GHz and fc(TM01) = 11.48 GHz.

Choice: magic tee (hybrid tee) fed at the H-arm

To split power into two equal halves, a magic tee is a good choice: power fed into its H-arm (port 3) divides equally and in phase between the collinear arms 1 and 2, while all ports remain matched and the outputs are isolated. (A simple H-plane tee also splits power equally, but its output arms are not matched or isolated; in microstrip circuits the equivalent device is the Wilkinson divider.)

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

S-matrix

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

Power split

With input a3 = a (power P = |a|²/2) and matched outputs:

 b1 = a/√2  -> P1 = |a|²/4 = P/2   (−3 dB)
 b2 = a/√2  -> P2 = P/2            (−3 dB, same phase)
 b4 = 0     -> no power to E-arm
 b3 = 0     -> input matched, no reflection

Result: each output gets exactly half the input power, in phase. If anti-phase halves are needed, the input is applied at the E-arm instead (b1 = a/√2, b2 = −a/√2).

Properties of the magic tee

1. Equal split: |S13| = |S23| = |S14| = |S24| = 1/√2 (3 dB).
2. Phase: H-arm gives in-phase outputs; E-arm gives 180° out-of-phase outputs.
3. Isolation: E and H arms are isolated (S34 = 0); collinear arms are isolated (S12 = 0), so a mismatch on one output does not disturb the other output.
4. Matching: all ports matched (Sᵢᵢ = 0) when E and H arms are matched with posts or irises.
5. Lossless and reciprocal: [S] is unitary and symmetric; reflections from unequal loads go to the E-arm, which can be terminated in a matched load.
6. Reverse use: equal in-phase inputs at 1 and 2 combine at the H-arm, so the same device works as a power combiner.

Comparison with an H-plane tee

Waveguide junctions

Waveguide junctions are passive devices where three or more waveguide sections meet, used to divide, combine or route microwave power. The common types are:

• E-plane tee: side arm on the broad wall; splits power into two equal parts 180° out of phase.
• H-plane tee: side arm on the narrow wall; splits power equally in phase.
• Magic tee (E-H tee): combination of E- and H-plane tees; a four-port hybrid.
• Hybrid ring (rat-race) and directional couplers.

Operation of magic tee using S-parameters

Ports 1 and 2 are collinear, port 3 the H-arm, port 4 the E-arm:

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

• Input at H-arm: b1 = b2 = a3/√2, b4 = 0 – equal in-phase split, nothing to E-arm.
• Input at E-arm: b1 = a4/√2, b2 = −a4/√2, b3 = 0 – equal anti-phase split.
• Input at port 1: b3 = b4 = a1/√2, b2 = 0 – collinear arms isolated.
• Equal in-phase inputs at 1 and 2: all power at H-arm (sum); anti-phase: all at E-arm (difference).
• All ports matched, lossless and reciprocal; used in mixers, duplexers and bridges.

The dominant mode of a waveguide is the propagating mode with the lowest cut-off frequency (longest cut-off wavelength). Below its cut-off no energy propagates; between its cut-off and that of the next higher mode, only the dominant mode can travel, which gives clean single-mode operation.

Rectangular waveguide (a > b):

 fc,mn = (c/2) √((m/a)² + (n/b)²)
 TE10: fc = c/(2a),  λc = 2a   (dominant)

TE10 has only Ey, Hx and Hz components; E is maximum at the centre of the broad wall and zero at the side walls. TM modes start at TM11, which is always higher.

Circular waveguide (radius r): dominant mode TE11 (p'11 = 1.841), fc = 1.841c/(2πr), λc = 3.41r. The lowest TM mode is TM01 (λc = 2.61r).

Why the dominant mode is used:

• Single-mode operation avoids mode conversion and distortion.
• Lowest attenuation for a given size (in rectangular guides) and simple field pattern for coupling with probes.
• Easy design of components (tees, couplers) for one known field pattern.

Example: WR-90 guide (a = 2.286 cm, b = 1.016 cm): fc(TE10) = 6.56 GHz, next mode TE20 at 13.12 GHz. It is used in the X-band (8.2–12.4 GHz), where only TE10 propagates.

Magic tee based on S-parameters

A magic tee joins an E-plane tee and an H-plane tee at one point: ports 1 and 2 collinear, port 3 the H-arm, port 4 the E-arm.

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

S-matrix (derived from symmetry, reciprocity, E–H isolation and losslessness):

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

• H-arm input → equal in-phase outputs at 1 and 2, nothing at E-arm.
• E-arm input → equal 180° out-of-phase outputs at 1 and 2, nothing at H-arm.
• Input at 1 → half to E-arm and half to H-arm; port 2 isolated.
• Equal in-phase inputs at 1 and 2 add at the H-arm (sum); anti-phase inputs add at the E-arm (difference).
• All ports matched, lossless, reciprocal. Uses: balanced mixers, duplexers, impedance bridges, monopulse radar.

Dominant vs degenerate modes

Dominant mode among the given TM modes for a = 3b

For TM modes the field Ez ∝ sin(mπx/a)·sin(nπy/b), so m = 0 or n = 0 gives zero field. Therefore TM01, TM10, TM02 and TM20 do not exist. The remaining candidates are TM11, TM21 and TM12:

 fc,mn = (c/2) √((m/a)² + (n/b)²),   a = 3b
       = (c/2b) √((m/3)² + n²)

 TM11: √(1/9 + 1)  = 1.054   -> fc = 1.054 · c/(2b)
 TM21: √(4/9 + 1)  = 1.202   -> fc = 1.202 · c/(2b)
 TM12: √(1/9 + 4)  = 2.028   -> fc = 2.028 · c/(2b)

Answer: TM11 has the lowest cut-off frequency, so it is the dominant mode among the listed TM modes (order: TM11 < TM21 < TM12; TM10, TM01, TM20, TM02 do not exist).

In a circular waveguide of radius a, both TE and TM modes are solutions of Bessel's equation, but they differ in which longitudinal field exists, which boundary condition applies and therefore which Bessel roots set the cut-off.

TE modes (Ez = 0)

 Hz = (A sin nφ + B cos nφ) Jn(kcρ) e^(-jβz)
 Eρ = -(jωμn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Eφ =  (jωμ/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Hρ = -(jβ/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Hφ = -(jβn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)

Boundary condition Eφ = 0 at ρ = a gives J'n(kc·a) = 0, kc = p'nm/a.

TM modes (Hz = 0)

 Ez = (A sin nφ + B cos nφ) Jn(kcρ) e^(-jβz)
 Eρ = -(jβ/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Eφ = -(jβn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Hρ =  (jωεn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Hφ = -(jωε/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)

Boundary condition Ez = 0 at ρ = a gives Jn(kc·a) = 0, kc = pnm/a.

Differences

Mode roots and degeneracy

TE01 and TM11 share the root 3.832, so they are degenerate (same cut-off). Also, each mode with n ≥ 1 has two polarizations (sin nφ and cos nφ), another form of degeneracy; this makes TE11 polarization sensitive to bends and ellipticity.

Example

For a = 1 cm (air): fc(TE11) = 1.841 × 3 × 10⁸/(2π × 0.01) = 8.79 GHz; fc(TM01) = 11.48 GHz. Between 8.79 and 11.48 GHz only TE11 propagates.

Field equations of a rectangular waveguide in TM mode

For a guide of inner dimensions a × b (a > b), TM modes have Hz = 0. Solving ∇t²Ez + kc²Ez = 0 with Ez = 0 on all four walls gives kx = mπ/a, ky = nπ/b (m, n ≥ 1):

 Ez = E0 sin(mπx/a) sin(nπy/b) e^(-jβz)
 Ex = -(jβ/kc²)(mπ/a) E0 cos(mπx/a) sin(nπy/b) e^(-jβz)
 Ey = -(jβ/kc²)(nπ/b) E0 sin(mπx/a) cos(nπy/b) e^(-jβz)
 Hx =  (jωε/kc²)(nπ/b) E0 sin(mπx/a) cos(nπy/b) e^(-jβz)
 Hy = -(jωε/kc²)(mπ/a) E0 cos(mπx/a) sin(nπy/b) e^(-jβz)
 Hz = 0

Related parameters:

• kc² = (mπ/a)² + (nπ/b)²; β = √(ω²με − kc²)
• Cut-off frequency: fc = (c/2)√((m/a)² + (n/b)²) (air)
• Cut-off wavelength: λc = 2/√((m/a)² + (n/b)²)
• Guide wavelength: λg = λ0/√(1 − (fc/f)²)
• Phase velocity vp = c/√(1 − (fc/f)²); group velocity vg = c√(1 − (fc/f)²)
• Wave impedance: Z_TM = η√(1 − (fc/f)²), always less than η
• Lowest TM mode: TM11 (TM10 and TM01 do not exist because Ez would vanish everywhere).

TE10 vs TE20

 TE10: fc = c/(2a),  λc = 2a
 TE20: fc = 2c/(2a) = c/a,  λc = a

TE10 is the dominant mode; the band between fc(TE10) and fc(TE20) (if a ≥ 2b) is the single-mode operating range of the guide.

Field components of a rectangular waveguide in TE mode

For a guide with inner dimensions a × b (a > b) and Ez = 0, the solution of ∇t²Hz + kc²Hz = 0 with ∂Hz/∂n = 0 on the walls gives (m, n = 0, 1, 2 …, not both zero):

 Hz = H0 cos(mπx/a) cos(nπy/b) e^(-jβz)
 Hx =  (jβ/kc²)(mπ/a) H0 sin(mπx/a) cos(nπy/b) e^(-jβz)
 Hy =  (jβ/kc²)(nπ/b) H0 cos(mπx/a) sin(nπy/b) e^(-jβz)
 Ex =  (jωμ/kc²)(nπ/b) H0 cos(mπx/a) sin(nπy/b) e^(-jβz)
 Ey = -(jωμ/kc²)(mπ/a) H0 sin(mπx/a) cos(nπy/b) e^(-jβz)
 Ez = 0

with

 kc² = (mπ/a)² + (nπ/b)²,   β = √(ω²με - kc²)
 fc = (c/2)√((m/a)² + (n/b)²)
 Z_TE = ωμ/β = η/√(1 - (fc/f)²)
 λg = λ0/√(1 - (fc/f)²)

For the dominant TE10 mode (m = 1, n = 0): only Ey, Hx, Hz exist, and fc = c/(2a).

Can 1 GHz propagate in TE10 with a = 5 cm?

 fc(TE10) = c/(2a) = (3 × 10⁸)/(2 × 0.05)
          = 3 × 10⁹ Hz = 3 GHz

Equivalently, λc = 2a = 10 cm while λ0 = c/f = 30 cm.

Since f = 1 GHz < fc = 3 GHz (λ0 > λc), β = √(ω²με − kc²) is imaginary: the field decays exponentially (evanescent) instead of propagating.

Answer: the TE10 cut-off is 3 GHz, so a 1 GHz signal cannot propagate in this waveguide.

A rectangular waveguide is a hollow metal pipe of inner width a and height b (a > b, often a ≈ 2b). Being a single conductor, it cannot carry TEM waves; it carries TE and TM modes above their cut-off frequencies, acting as a high-pass filter.

   y
 b +-----------------+
   |   E (TE10)  ↑   |   propagation along z
   +-----------------+
   0                 a  -> x

Modes of propagation

• TEmn (H modes): Ez = 0, Hz = H0 cos(mπx/a) cos(nπy/b). m, n = 0, 1, 2 … but not both zero. Lowest: TE10.
• TMmn (E modes): Hz = 0, Ez = E0 sin(mπx/a) sin(nπy/b). m, n ≥ 1. Lowest: TM11.
• m and n are the numbers of half-wave field variations along the broad (a) and narrow (b) walls.
• Dominant mode: TE10 (lowest fc). Degenerate modes: TEmn and TMmn with the same m, n share the same fc.

Critical parameters (air-filled)

Also 1/λ0² = 1/λg² + 1/λc² and vp·vg = c².

Example (WR-90, a = 2.286 cm, b = 1.016 cm, f = 10 GHz, TE10): fc = 6.562 GHz, λg = 3.976 cm, vp = 3.976 × 10⁸ m/s, vg = 2.264 × 10⁸ m/s, Z_TE ≈ 500 Ω. TE20 cuts off at 13.12 GHz, so only TE10 propagates.

A Gunn diode is a two-terminal microwave source made of a single piece of n-type GaAs (or InP). It has no p-n junction. It works on the transferred electron effect (Ridley–Watkins–Hilsum theory), which gives the bulk material a negative differential resistance.

Principle

• The GaAs conduction band has two valleys. In the lower valley electrons are light (m* ≈ 0.068 m₀) and fast (μ ≈ 8000 cm²/V·s). The upper valley is about 0.36 eV higher, and there electrons are heavy (m* ≈ 1.2 m₀) and slow (μ ≈ 180 cm²/V·s).
• At low fields, nearly all electrons are in the lower valley, so current rises with field.
• Above a threshold field of about 3.2–3.4 kV/cm, electrons gain energy and move to the upper valley. Average drift velocity then falls as the field rises, so dI/dV < 0.
• In this negative-resistance region a high-field domain forms near the cathode, drifts to the anode at about 10⁷ cm/s and disappears. Then a new domain forms. Each domain gives one current pulse, so the frequency is about f = v_d / L.

 I |     peak
   |    /\
   |   /  \___  NDR region
   |  /
   | /
   +-------------- E
       E_th ~3.2 kV/cm

Modes of operation: transit-time (Gunn) mode, delayed-domain mode, quenched-domain mode and LSA (limited space-charge accumulation) mode.

Applications: local oscillators in receivers, police and automotive radar, microwave links, and sweep generators. Gunn diodes give mW up to about 1 W in the 1–100 GHz range, with low noise and a low bias voltage.

MASER stands for Microwave Amplification by Stimulated Emission of Radiation. It is a very low-noise microwave amplifier or oscillator that uses stimulated emission from atoms or molecules. C. H. Townes built the first one in 1954, an ammonia maser at 23.87 GHz.

Principle

• An atom in an upper energy level E₂ can be pushed down to E₁ by an incoming photon of frequency f = (E₂ − E₁)/h. It then emits a second photon of the same frequency and phase. This is stimulated emission.
• Amplification needs population inversion, meaning more atoms in E₂ than in E₁. This is obtained by pumping.
• In the common three-level solid-state maser (ruby, Cr³⁺ in Al₂O₃), a pump signal at f₁₃ lifts atoms from level 1 to level 3. Level 3 then becomes more populated than level 2, so a signal at f₃₂ is amplified. The levels are tuned by a DC magnetic field (Zeeman splitting).

 E3 ------------  <- pump f13 raises atoms
        | signal f32 amplified
 E2 ------------
 E1 ------------

Features

• Extremely low noise temperature (a few kelvin), because the active medium is cooled to liquid-helium temperature.
• Narrow bandwidth and low output power. Needs cryogenic cooling and a magnetic field.

Applications: front-end amplifiers in radio telescopes and deep-space communication receivers, and atomic frequency standards (the hydrogen maser at 1.42 GHz).

A waveguide tee is a three-port junction made by joining a side arm to a main rectangular waveguide. In an E-plane tee the side arm lies in the plane of the E-field. In an H-plane tee it lies in the plane of the H-field.

 E-plane tee (series)     H-plane tee (shunt)
        port 3                   port 3
         ||                       ||
 ====[  top  ]====       ====[ side  ]====
 port 1      port 2      port 1      port 2
 arm on broad wall       arm on narrow wall

E-plane tee (series tee)

• The side arm is joined to the broad wall, parallel to the E-field of the main guide.
• A signal fed into port 3 splits equally into ports 1 and 2, but the two outputs are 180° out of phase.
• In-phase equal signals at ports 1 and 2 cancel at port 3. Out-of-phase signals add at port 3, so port 3 gives the difference (Δ) of the inputs.
• It acts like a series connection.

H-plane tee (shunt tee)

• The side arm is joined to the narrow wall, so its axis is parallel to the H-field plane.
• A signal fed into port 3 splits equally into ports 1 and 2 in phase.
• In-phase signals at ports 1 and 2 add at port 3, so port 3 gives the sum (Σ).
• It acts like a shunt (parallel) connection.

Combining the two gives the magic tee (hybrid tee), used in mixers, duplexers and impedance bridges.

Energy is fed into or taken out of a waveguide or cavity resonator with probe (electric) coupling or loop (magnetic) coupling. A coaxial line is usually used to launch the wanted mode.

Probe coupling

• The inner conductor of the coaxial line is extended into the guide as a short antenna (probe) of about λ/4 length. It goes through the broad wall.
• The probe is placed where the E-field is maximum, parallel to the E-lines. For TE₁₀ this is the centre of the broad wall, about λg/4 from the shorted end.
• It works like a monopole antenna. Coupling is varied by changing the probe depth or position.

Loop coupling

• The inner conductor forms a small loop whose end is joined to the guide wall.
• The loop is placed where the H-field is maximum (for example near the shorted end wall), with the plane of the loop perpendicular to the H-lines so that the most magnetic flux links it.
• Coupling is varied by rotating the loop: rotating it by 90° reduces coupling to almost zero.

 Probe (E-field)        Loop (H-field)
   coax                    coax
    ||                      ||
 ===||=========          ===||====|
    |  <- probe             (_)   | <- short
 E-lines ||||           loop near short wall

Both can also excite higher modes when placed at those modes' field maxima. A third method is aperture (slot) coupling through a hole in the common wall.