RF Design Practices

The amplifier is designed by (1) checking stability, (2) finding the source and load reflection coefficients for maximum gain in the unilateral (S12 = 0) and bilateral cases, (3) computing the maximum gains, and (4) designing single-stub matching networks.

Given: S11 = 0.55∠−150°, S12 = 0.04∠20°, S21 = 2.82∠180°, S22 = 0.45∠−30° (Z₀ = 50 Ω).

Δ = S11·S22 − S12·S21
K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
Unconditionally stable if K > 1 and |Δ| < 1
GTU,max = [1/(1−|S11|²)]·|S21|²·[1/(1−|S22|²)]
Gmax (MAG) = (|S21|/|S12|)·(K − √(K² − 1))

Step 1: Stability

S11·S22 = 0.2475∠180.0° = −0.2475 + j0.0000
S12·S21 = 0.1128∠−160.0° = −0.1060 − j0.0386
Δ = S11·S22 − S12·S21 = −0.1415 + j0.0386
  = 0.1467∠164.7°,   |Δ|² = 0.0215

K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
  = (1 − 0.3025 − 0.2025 + 0.0215) / (2 × 0.1128)
  = 0.5165 / 0.2256
  = 2.290

Check with the single test μ = (1 − |S11|²) / (|S22 − Δ·S11*| + |S12·S21|) = 1.436 (> 1, unconditionally stable).

K = 2.290 > 1 and |Δ| = 0.147 < 1, so the transistor is unconditionally stable. Conjugate matching can be used at both ports.

Step 2: Unilateral design (S12 assumed 0)

Maximum gain needs ΓS = S11* = 0.55∠150° and ΓL = S22* = 0.45∠30°.

Unilateral (S12 = 0, ΓS = S11*, ΓL = S22*):
GS = 1/(1 − |S11|²) = 1/(1 − 0.3025)
   = 1.4337  (1.56 dB)
G0 = |S21|² = 2.820² = 7.9524  (9.00 dB)
GL = 1/(1 − |S22|²) = 1/(1 − 0.2025)
   = 1.2539  (0.98 dB)
GTU,max = GS·G0·GL = 14.296  (11.55 dB)

Unilateral figure of merit (to check the error of assuming S12 = 0):

U = |S11·S12·S21·S22| / [(1 − |S11|²)(1 − |S22|²)]
  = 0.0502
1/(1+U)² < GT/GTU < 1/(1−U)²
−0.43 dB < error < +0.45 dB

Step 3: Bilateral design (S12 included)

B1 = 1 + |S11|² − |S22|² − |Δ|² = 1.0785
B2 = 1 + |S22|² − |S11|² − |Δ|² = 0.8785
C1 = S11 − Δ·S22* = 0.4866∠−148.0°
C2 = S22 − Δ·S11* = 0.3728∠−26.7°
ΓS = [B1 − √(B1² − 4|C1|²)] / (2C1) = 0.6307∠148.0°
ΓL = [B2 − √(B2² − 4|C2|²)] / (2C2) = 0.5551∠26.7°
ZS = 50(1 + ΓS)/(1 − ΓS) = 12.2 + j13.6 Ω
ZL = 50(1 + ΓL)/(1 − ΓL) = 109.3 + j78.9 Ω

Bilateral (S12 kept), simultaneous conjugate match:
MAG = Gmax = (|S21|/|S12|)·(K − √(K² − 1))
     = (2.820/0.040) × (2.290 − √(2.290² − 1))
     = 70.500 × 0.2299
     = 16.210  (12.10 dB)

Check: with these ΓS and ΓL, Γin = ΓS* and Γout = ΓL* (simultaneous conjugate match), and the transducer gain G_T works out to 16.210.

Step 4: Matching networks (single open stub + series 50 Ω line)

 50 Ω  --+--[ line dS ]--[ FET ]--[ line dL ]--+--  50 Ω
 source  |                                     |   load
       open                                  open
       stub lS                               stub lL

Each network uses a shunt open stub at the 50 Ω port and a 50 Ω line toward the transistor, chosen so that the transistor sees ΓS (input) and ΓL (output). The stub gives normalised susceptance b with |Γ| = b/√(4 + b²).

Bilateral (ΓS = 0.6307∠148.0°, ΓL = 0.5551∠26.7°):
input : |ΓS| = 0.6307 -> b = 2|Γ|/√(1−|Γ|²) = 1.626
        open stub lS = arctan(b)/2π = 0.162 λ
        series line dS = 0.115 λ
output: |ΓL| = 0.5551 -> b = 1.335
        open stub lL = 0.148 λ
        series line dL = 0.291 λ

Unilateral (ΓS = 0.55∠150°, ΓL = 0.45∠30°):
input : |ΓS| = 0.5500 -> b = 2|Γ|/√(1−|Γ|²) = 1.317
        open stub lS = arctan(b)/2π = 0.147 λ
        series line dS = 0.120 λ
output: |ΓL| = 0.4500 -> b = 1.008
        open stub lL = 0.126 λ
        series line dL = 0.296 λ

(Lengths are in guide wavelengths at the design frequency. The other root, with negative b, gives a second valid solution.)

Step 5: Comparison

The bilateral gain is 0.55 dB higher, because the feedback through S12 adds to the gain when both ports are matched together. The unilateral error bounds (U = 0.0502) are −0.43 to +0.45 dB. The bilateral optimum is slightly above the upper bound because it uses different terminations, but the difference is small, so the unilateral design is a fair first approximation and the bilateral design is the exact one.

Answer: unconditionally stable (K = 2.290, |Δ| = 0.147); GTU,max = 11.55 dB (unilateral); Gmax = 12.10 dB (bilateral).

The amplifier is designed by checking stability, finding the simultaneous conjugate-match terminations ΓS and ΓL, computing the maximum gain, and designing the matching networks.

Given (5 GHz, Z₀ = 50 Ω): S11 = 0.45∠163°, S12 = 0.04∠40°, S21 = 2.55∠−106°, S22 = 0.46∠−65°.

Δ = S11·S22 − S12·S21
K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
Unconditionally stable if K > 1 and |Δ| < 1
GTU,max = [1/(1−|S11|²)]·|S21|²·[1/(1−|S22|²)]
Gmax (MAG) = (|S21|/|S12|)·(K − √(K² − 1))

Step 1: Stability check

S11·S22 = 0.2070∠98.0° = −0.0288 + j0.2050
S12·S21 = 0.1020∠−66.0° = 0.0415 − j0.0932
Δ = S11·S22 − S12·S21 = −0.0703 + j0.2982
  = 0.3063∠103.3°,   |Δ|² = 0.0938

K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
  = (1 − 0.2025 − 0.2116 + 0.0938) / (2 × 0.1020)
  = 0.6797 / 0.2040
  = 3.332

Check with the single test μ = (1 − |S11|²) / (|S22 − Δ·S11*| + |S12·S21|) = 1.877 (> 1, unconditionally stable).

K = 3.332 > 1 and |Δ| = 0.306 < 1, so the FET is unconditionally stable at 5 GHz. It can be conjugate-matched at both ports.

Step 2: Source and load terminations for maximum gain

B1 = 1 + |S11|² − |S22|² − |Δ|² = 0.8971
B2 = 1 + |S22|² − |S11|² − |Δ|² = 0.9153
C1 = S11 − Δ·S22* = 0.3099∠160.6°
C2 = S22 − Δ·S11* = 0.3230∠−67.2°
ΓS = [B1 − √(B1² − 4|C1|²)] / (2C1) = 0.4011∠−160.6°
ΓL = [B2 − √(B2² − 4|C2|²)] / (2C2) = 0.4131∠67.2°
ZS = 50(1 + ΓS)/(1 − ΓS) = 21.9 − j6.9 Ω
ZL = 50(1 + ΓL)/(1 − ΓL) = 48.7 + j44.8 Ω

Step 3: Maximum (total) power gain

Bilateral (S12 kept), simultaneous conjugate match:
MAG = Gmax = (|S21|/|S12|)·(K − √(K² − 1))
     = (2.550/0.040) × (3.332 − √(3.332² − 1))
     = 63.750 × 0.1536
     = 9.792  (9.91 dB)

This is split into three parts: the input match gain G_S, the device gain |S21|² = 6.502 (8.13 dB), and the output match gain G_L.

For comparison, the unilateral estimate (S12 = 0) is:

Unilateral (S12 = 0, ΓS = S11*, ΓL = S22*):
GS = 1/(1 − |S11|²) = 1/(1 − 0.2025)
   = 1.2539  (0.98 dB)
G0 = |S21|² = 2.550² = 6.5025  (8.13 dB)
GL = 1/(1 − |S22|²) = 1/(1 − 0.2116)
   = 1.2684  (1.03 dB)
GTU,max = GS·G0·GL = 10.342  (10.15 dB)

The two values are close (the unilateral estimate is 0.24 dB higher), because S12 is small and U = 0.0336 gives an error band of only −0.29 to +0.30 dB. The bilateral value is the correct one to design with.

Step 4: Matching network design

 50 Ω  --+--[ line dS ]--[ FET ]--[ line dL ]--+--  50 Ω
 source  |                                     |   load
       open                                  open
       stub lS                               stub lL

Single open-stub matching (bilateral ΓS, ΓL):
input : |ΓS| = 0.4011 -> b = 2|Γ|/√(1−|Γ|²) = 0.876
        open stub lS = arctan(b)/2π = 0.114 λ
        series line dS = 0.065 λ
output: |ΓL| = 0.4131 -> b = 0.907
        open stub lL = 0.117 λ
        series line dL = 0.248 λ

The input network makes the 50 Ω source look like ZS ≈ 21.9 − j6.9 Ω to the FET. The output network makes the 50 Ω load look like ZL ≈ 48.7 + j44.8 Ω. Bias is fed through λ/4 high-impedance lines with RF bypass capacitors.

Answer: K = 3.332, |Δ| = 0.306 → unconditionally stable; ΓS = 0.4011∠−160.6°, ΓL = 0.4131∠67.2°; maximum power gain Gmax = 9.79 = 9.91 dB.

The insertion loss method designs a filter from a chosen power loss ratio (frequency response), then builds a lumped low-pass prototype that gives exactly that response, and finally scales and converts it to the required filter and to microwave form.

Power loss ratio

PLR = P_available / P_load = 1 / (1 − |Γ(ω)|²)
IL  = 10 log10(PLR)  dB
|Γ(ω)|² = M(ω²) / [M(ω²) + N(ω²)]
=> PLR = 1 + M(ω²)/N(ω²)

Common responses

• Maximally flat (Butterworth): PLR = 1 + k²(ω/ωc)²ᴺ. Flattest passband, slower roll-off.
• Equal ripple (Chebyshev): PLR = 1 + k² T_N²(ω/ωc). Ripple in the passband, sharper cut-off.
• Elliptic: ripple in both bands, sharpest cut-off.
• Linear phase (Bessel): good group delay, poor roll-off.

Design steps

 Filter specs (fc, ripple, attenuation)
            |
 Choose response + order N
            |
 Low-pass prototype (g-values, R0=1, ωc=1)
            |
 Impedance & frequency scaling,
 LP -> HP / BP / BS transformation
            |
 Microwave realisation (Richards, Kuroda,
 stepped-impedance, coupled lines)

1. Write the specifications: cut-off frequency, passband ripple, stop-band attenuation and impedance.
2. Choose the response type and find the order N from attenuation charts.
3. Take element values g₁…g_N from prototype tables (for example Butterworth N = 3: g = 1, 2, 1).
4. Scale impedance (L' = R₀L, C' = C/R₀) and frequency (divide by ωc), and transform to HP, BP or BS if needed.
5. Convert lumped L and C into distributed microstrip elements and check the response by simulation.

Advantages: full control over passband and stop-band response, a systematic and repeatable synthesis, and easy trade-off between order, flatness and roll-off. It has replaced the older image-parameter method.

Low-pass filter in microstrip

A microwave LPF is built in microstrip by replacing the lumped L and C of a ladder prototype with short transmission-line sections. Two common forms are used.

1. Stepped-impedance (hi-Z / low-Z) filter

A short line (βl < π/4) of high impedance Z_h behaves like a series inductor. A short line of low impedance Z_ℓ behaves like a shunt capacitor:

βl = L·R0 / Zh   (inductor, L normalised)
βl = C·Zℓ / R0   (capacitor, C normalised)

Z_h is made as high as can be etched (narrow strip, about 100–150 Ω), and Z_ℓ as low as practical (wide strip, about 10–20 Ω).

 Stepped-impedance microstrip LPF (top view)
       ___        _____        ___
 =====|   |______|     |______|   |=====
 50 Ω | C1| L2   | C3  |  L4  | C5| 50 Ω
      |___|      |_____|      |___|
 wide = low Z0 (shunt C)
 narrow = high Z0 (series L)

2. Stub filter (Richards' transformation + Kuroda identities)

• Richards' transformation maps a shunt capacitor to an open-circuited stub and a series inductor to a short-circuited series stub, each λ/8 long at fc.
• Kuroda identities change the series stubs into shunt open stubs separated by λ/8 unit elements, so all stubs are easy to make in microstrip.

       open stubs (λ/8 at fc)
        | |      | |      | |
 50 Ω ==+=+======+=+======+=+== 50 Ω
          unit elements (λ/8)

Prototyping of a low-pass filter

A low-pass prototype is a normalised LC ladder with source resistance g₀ = 1 Ω and cut-off ωc = 1 rad/s.

  g0     g2 (L)
 --/\/--+--UUU--+-- ... --+
         |       |         g(N+1) load
       g1 (C)   g3 (C)
         |       |
 --------+-------+---------

1. Choose the response: Butterworth (maximally flat), PLR = 1 + (ω/ωc)²ᴺ, or Chebyshev (equal ripple), PLR = 1 + k²T_N²(ω/ωc).
2. Find the order N from the required attenuation at a given stop-band frequency, using attenuation-versus-normalised-frequency charts.
3. Read the g-values from tables. For Butterworth, g_k = 2 sin[(2k − 1)π/(2N)]. For example, N = 3 gives g₁ = 1, g₂ = 2, g₃ = 1.
4. Scale to the real impedance R₀ and cut-off ωc: L_k = R₀ g_k / ωc and C_k = g_k / (R₀ ωc).
5. Realise the elements in microstrip as above and check the response with a simulator.

A microwave mixer is a three-port frequency converter. It combines an RF signal (f_RF) with a local-oscillator signal (f_LO) in a non-linear device to produce sum and difference frequencies. Usually the difference, the intermediate frequency (IF), is kept: f_IF = |f_RF − f_LO|.

Mixer theory

The device (usually a Schottky diode, or a FET) has a non-linear I–V curve:

i = a0 + a1·v + a2·v² + a3·v³ + ...
v = Vr cos ωr t + VL cos ωL t
a2·v² term gives:
 a2·Vr·VL·[cos(ωr − ωL)t + cos(ωr + ωL)t]
=> IF at |fr − fL|, also fr + fL,
   harmonics and intermod products

A filter at the output selects the IF. The IF amplitude is proportional to V_r, so the modulation of the RF signal is kept.

 RF (fr) --->[        ]---> IF = |fr − fL|
             [ MIXER  ]  (+ filter)
 LO (fL) --->[        ]

Important terms

• Conversion loss: L_c = 10 log(P_RF / P_IF), typically 4–7 dB for diode mixers.
• Image frequency: f_im = f_LO ± f_IF (on the other side of f_LO). It also gives the same IF, so it must be filtered before the mixer.
• Noise figure: about equal to the conversion loss for a diode mixer (SSB).
• Isolation between the LO, RF and IF ports.

Types

• Single-ended: one diode. Simple but poor isolation and LO noise.
• Balanced: two diodes with a 90° or 180° hybrid. Cancels LO AM noise and some spurs, and gives better RF–LO isolation.
• Double-balanced: four diodes (ring) with baluns. Best isolation, rejects even harmonics, wide bandwidth.
• Image-reject: two mixers with quadrature hybrids to cancel the image.

Applications: down-conversion in superheterodyne receivers, up-conversion in transmitters, phase detectors and modulators.

Basic differences between the two systems

Both are two-port amplifying devices with nearly the same forward gain (|S21| = 2.30 and 2.20, same angle 44°) and the same S22 angle. They differ mainly in input match and reverse isolation:

• System (a) has strong feedback (large S12) and a large input mismatch. Matching adds more gain, but the design is more sensitive and closer to the stability limit (K near 1).
• System (b) is nearly unilateral (U ≈ 0.0022), so its input and output can be designed separately with little error. It is strongly stable (K ≫ 1).

Stability of system (a)

Δ = S11·S22 − S12·S21
K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
Unconditionally stable if K > 1 and |Δ| < 1
GTU,max = [1/(1−|S11|²)]·|S21|²·[1/(1−|S22|²)]
Gmax (MAG) = (|S21|/|S12|)·(K − √(K² − 1))

S11·S22 = 0.1105∠−26.0° = 0.0993 − j0.0484
S12·S21 = 0.2760∠90.0° = 0.0000 + j0.2760
Δ = S11·S22 − S12·S21 = 0.0993 − j0.3244
  = 0.3393∠−73.0°,   |Δ|² = 0.1151

K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
  = (1 − 0.4225 − 0.0289 + 0.1151) / (2 × 0.2760)
  = 0.6637 / 0.5520
  = 1.202

Check with the single test μ = (1 − |S11|²) / (|S22 − Δ·S11*| + |S12·S21|) = 1.317 (> 1, unconditionally stable).

K = 1.202 > 1 and |Δ| = 0.339 < 1, so system (a) is unconditionally stable (though with a small margin).

Maximum gains of system (a)

Unilateral mode (ΓS = S11* = 0.65∠−146°, ΓL = S22* = 0.17∠172°):

Unilateral (S12 = 0, ΓS = S11*, ΓL = S22*):
GS = 1/(1 − |S11|²) = 1/(1 − 0.4225)
   = 1.7316  (2.38 dB)
G0 = |S21|² = 2.300² = 5.2900  (7.23 dB)
GL = 1/(1 − |S22|²) = 1/(1 − 0.0289)
   = 1.0298  (0.13 dB)
GTU,max = GS·G0·GL = 9.433  (9.75 dB)

Bilateral mode (simultaneous conjugate match):

Bilateral (S12 kept), simultaneous conjugate match:
MAG = Gmax = (|S21|/|S12|)·(K − √(K² − 1))
     = (2.300/0.120) × (1.202 − √(1.202² − 1))
     = 19.167 × 0.5347
     = 10.249  (10.11 dB)

B1 = 1 + |S11|² − |S22|² − |Δ|² = 1.2785
B2 = 1 + |S22|² − |S11|² − |Δ|² = 0.4913
C1 = S11 − Δ·S22* = 0.6121∠150.0°
C2 = S22 − Δ·S11* = 0.1624∠−88.9°
ΓS = [B1 − √(B1² − 4|C1|²)] / (2C1) = 0.7433∠−150.0°
ΓL = [B2 − √(B2² − 4|C2|²)] / (2C2) = 0.3778∠88.9°
ZS = 50(1 + ΓS)/(1 − ΓS) = 7.9 − j13.1 Ω
ZL = 50(1 + ΓL)/(1 − ΓL) = 38.0 + j33.5 Ω

Error from the unilateral assumption:

U = |S11·S12·S21·S22| / [(1 − |S11|²)(1 − |S22|²)]
  = 0.0544
1/(1+U)² < GT/GTU < 1/(1−U)²
−0.46 dB < error < +0.49 dB

Comparison

The bilateral gain is 0.36 dB higher, because the relatively large S12 = 0.12 gives useful feedback when both ports are matched together. Most of the gain from matching comes from the input side (G_S = 2.38 dB), since |S11| is large.

(For system (b), for comparison: GTU,max = 7.30 dB and Gmax = 7.32 dB, which are almost equal because S12 is very small.)

Answer (a): K = 1.202, |Δ| = 0.339 → unconditionally stable; GTU,max = 9.75 dB, Gmax (bilateral) = 10.11 dB.

Why the insertion loss method is used

The older image-parameter method gives a passband and stop-band but cannot set the exact response, so it needs many trial adjustments. The insertion loss method starts from a chosen power loss ratio:

PLR = 1 / (1 − |Γ(ω)|²) = 1 + M(ω²)/N(ω²)
Butterworth : PLR = 1 + k²(ω/ωc)^(2N)
Chebyshev   : PLR = 1 + k² T_N²(ω/ωc)

so it gives:

• full control of passband flatness or ripple, cut-off sharpness and phase,
• a systematic synthesis from tables (g-values), with a clear trade-off between order N and performance,
• a design that is easy to scale and transform to HP, BP or BS, and to convert to microstrip.

Design steps for the given 5-element ladder

The figure is a fifth-order low-pass ladder that starts with a shunt capacitor: C₁ (shunt), L₂ (series), C₃ (shunt), L₄ (series), C₅ (shunt). No values are given, so this example assumes a Butterworth response, N = 5, fc = 2 GHz and Z_s = Z_L = R₀ = 50 Ω.

Step 1: Specifications. fc = 2 GHz, maximally flat, 50 Ω terminations.

Step 2: Prototype g-values (g_k = 2 sin[(2k − 1)π/2N], N = 5):

Step 3: Impedance and frequency scaling (ωc = 2π × 2×10⁹ = 1.2566×10¹⁰ rad/s):

Ck = gk / (R0 ωc),  Lk = R0 gk / ωc
C1 = C5 = 0.618 / (50 × 1.2566e10) = 0.984 pF
L2 = L4 = 50 × 1.618 / 1.2566e10   = 6.438 nH
C3      = 2.000 / (50 × 1.2566e10) = 3.183 pF

Step 4: Convert to microstrip (stepped impedance). Choose Z_h = 120 Ω (narrow line) and Z_ℓ = 20 Ω (wide line):

inductor : βl = gk·R0 / Zh
capacitor: βl = gk·Zℓ / R0
C1, C5: βl = 0.618 × 20/50  = 14.16°
L2, L4: βl = 1.618 × 50/120 = 38.63°
C3    : βl = 2.000 × 20/50  = 45.84°

Each length is ℓ = (βl/360°)·λg at 2 GHz, where λg depends on the substrate (εr, h).

(Alternative: Richards' transformation turns each C into a λ/8 open stub of Z₀ = R₀/g (C₁: 80.9 Ω, C₃: 25 Ω) and each L into a λ/8 series short stub of Z₀ = R₀·g (L₂: 80.9 Ω). Kuroda identities then convert the series stubs into shunt stubs.)

Step 5: Simulate and tune the layout, adding the effects of step discontinuities, and adjust the lengths.

Microstrip implementation

 Stepped-impedance microstrip LPF (top view)
       ___        _____        ___
 =====|   |______|     |______|   |=====
 50 Ω | C1| L2   | C3  |  L4  | C5| 50 Ω
      |___|      |_____|      |___|
 wide = low Z0 (shunt C)
 narrow = high Z0 (series L)

Wide low-impedance pads make the shunt capacitors C₁, C₃ and C₅. Narrow high-impedance lines make the series inductors L₂ and L₄. All are printed on one substrate over a ground plane.

A microwave amplifier is designed from the transistor's S-parameters at the operating frequency and bias. The aim is the required gain (or noise figure) while staying stable.

Design flow

 Specs: f, gain, NF, bandwidth
              |
 Choose transistor + bias, get [S]
              |
 Stability check: K, |Δ| (or μ)
     | stable          | potentially unstable
     |            stability circles,
     |            pick ΓS, ΓL in stable region
              |
 Unilateral? (U small) -> ΓS=S11*, ΓL=S22*
 else bilateral: ΓMS, ΓML (B1, C1, B2, C2)
              |
 Matching networks (Smith chart, stubs)
              |
 Bias network, layout, simulate, tune

Stability analysis

An amplifier can oscillate if the input or output reflection coefficient has magnitude greater than 1 (negative resistance):

Γin  = S11 + S12 S21 ΓL / (1 − S22 ΓL)
Γout = S22 + S12 S21 ΓS / (1 − S11 ΓS)
Δ = S11 S22 − S12 S21
K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12 S21|)

• Unconditionally stable: K > 1 and |Δ| < 1 (or μ > 1). Any passive source and load can be used.
• Potentially unstable: otherwise. Stability circles in the Γ_L and Γ_S planes show which terminations to avoid, and the design must use terminations in the stable region (or add resistive loading).

Microwave filter designing steps (insertion loss method)

 1. Specifications (fc, ripple, IL, Z0)
 2. Response type + order N
 3. Low-pass prototype g-values
 4. Scaling + LP->HP/BP/BS transform
 5. Lumped -> distributed (Richards,
    Kuroda, stepped-Z, coupled lines)
 6. Layout, simulate (EM), tune

1. Specifications: cut-off frequency fc, passband insertion loss or ripple, stop-band attenuation at a given frequency, and the terminating impedance R₀ (usually 50 Ω).
2. Choose the response: Butterworth (maximally flat, PLR = 1 + k²(ω/ωc)²ᴺ), Chebyshev (equal ripple, PLR = 1 + k²T_N²(ω/ωc)), elliptic, or linear phase. Find the order N from attenuation charts.
3. Low-pass prototype: read the normalised element values g₁…g_N (R₀ = 1, ωc = 1) from tables. For example, Butterworth N = 3 gives g = 1, 2, 1.
4. Scaling and transformation:
   • Impedance scaling: L' = R₀L, C' = C/R₀.
   • Frequency scaling and type change. For a high-pass filter, each series L_k becomes a series C'_k = 1/(R₀ωc g_k), and each shunt C_k becomes a shunt L'_k = R₀/(ωc g_k).
   • Band-pass and band-stop use series and parallel LC resonators.
5. Distributed realisation: Richards' transformation (L → short stub, C → open stub, λ/8 long), Kuroda identities, stepped-impedance lines, gap or interdigital capacitors, or coupled-line sections.
6. Layout and verification: include discontinuity effects (gaps, steps, T-junctions, vias), simulate, and tune the lengths and gaps.

Triple-pad series-arm HPF in microstrip

A high-pass ladder has series capacitors in the series arm and shunt inductors. In a triple-pad series-arm layout, the three series capacitors C₁, C₃, C₅ are made as gap-coupled pads (or interdigital capacitors) along the main line. The shunt inductors L₂, L₄ are thin high-impedance lines shorted to ground through via holes.

 50 Ω    C1          C3          C5     50 Ω
 ====| |[pad1]|  |[pad2]|  |[pad3]| |====
       gap  |     gap |     gap
            |  L2     |  L4
           (thin)    (thin)
            |         |
           [via]     [via]  -> ground plane

• Pads and gaps: the capacitance of each series arm is set by the gap width and the pad overlap. Interdigital fingers are used when larger C is needed.
• Shunt inductors: narrow lines (Z₀ about 100–120 Ω, βl < 45°) or λ/8 short-circuited stubs from Richards' transformation, ending in vias to the ground plane.
• Everything is printed on one substrate. Wide 50 Ω lines connect the input and output.

A transistor is unconditionally stable if |Γin| < 1 and |Γout| < 1 for every passive source and load. This is tested with the Rollett factor K and |Δ| (or the single μ-test), following the flow below.

Flow diagram for stability analysis

      [ Given S11, S12, S21, S22 ]
                  |
      [ Δ = S11·S22 − S12·S21 ]
                  |
 [ K = (1−|S11|²−|S22|²+|Δ|²)/(2|S12S21|) ]
                  |
         < K > 1 and |Δ| < 1 ? >
          | yes            | no
 [ Unconditionally ]  [ Potentially unstable:   ]
 [ stable: any     ]  [ draw input/output       ]
 [ ΓS, ΓL usable   ]  [ stability circles (CS,  ]
          |           [ RS, CL, RL); choose ΓS, ]
          |           [ ΓL in stable region     ]
          |___________________|
                  |
       [ Proceed to gain design ]

Calculation

Given: S11 = 0.6∠−140°, S12 = 0.03∠62°, S21 = 2.40∠54°, S22 = 0.70∠−58°.

S11·S22 = 0.4200∠162.0° = −0.3994 + j0.1298
S12·S21 = 0.0720∠116.0° = −0.0316 + j0.0647
Δ = S11·S22 − S12·S21 = −0.3679 + j0.0651
  = 0.3736∠170.0°,   |Δ|² = 0.1396

K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
  = (1 − 0.3600 − 0.4900 + 0.1396) / (2 × 0.0720)
  = 0.2896 / 0.1440
  = 2.011

Check with the single test μ = (1 − |S11|²) / (|S22 − Δ·S11*| + |S12·S21|) = 1.161 (> 1, unconditionally stable).

Result

• K = 2.011 > 1
• |Δ| = 0.374 < 1
• μ = 1.161 > 1

So the transistor is unconditionally stable: no passive source or load can make it oscillate.

Supporting check with stability circles:

CL = (S22 − Δ·S11*)* / (|S22|² − |Δ|²) = 1.3669∠61.7°
RL = |S12·S21| / ||S22|² − |Δ|²|         = 0.2055
CS = (S11 − Δ·S22*)* / (|S11|² − |Δ|²) = 1.5558∠146.1°
RS = |S12·S21| / ||S11|² − |Δ|²|         = 0.3266

For both circles |C| − R > 1 (1.161 and 1.229), so the circles lie completely outside the Smith chart. The whole chart is therefore the stable region, which agrees with K > 1 and |Δ| < 1.

Since the device is stable, it can be conjugate-matched. For reference, the maximum available gain is Gmax = (|S21|/|S12|)(K − √(K² − 1)) = 21.30 (13.28 dB).

Answer: Δ = 0.374∠170.0°, K = 2.011, |Δ| = 0.374 → unconditionally stable.

Single-section series-arm passive BPF in microstrip

A single-section band-pass filter has a series LC resonator in the series arm. It passes signals near its resonant frequency f₀ = 1/(2π√(LC)) and blocks others. In microstrip, the series resonator is made as a half-wavelength (λg/2) line coupled to the input and output lines through gaps. The gaps act as series capacitors, and the λ/2 line behaves like a series-resonant circuit near f₀.

 Lumped:   in --| |--UUU-- out
                C    L   (series arm)

 Microstrip (end-coupled):
 50 Ω            λg/2 resonator            50 Ω
 =========| |=========================| |=========
          gap1                         gap2

The gap widths set the coupling and so the bandwidth. A smaller gap gives more coupling and a wider band. Several such resonators in cascade give a higher-order end-coupled BPF. A parallel-coupled-line layout is a more compact alternative.

Filter design by the insertion loss method

The insertion loss method starts from the wanted power loss ratio and synthesises a network that gives it:

PLR = 1 / (1 − |Γ(ω)|²) = 1 + M(ω²)/N(ω²)
IL (dB) = 10 log10 PLR

Steps

1. Specifications: centre frequency f₀, fractional bandwidth Δ = (ω₂ − ω₁)/ω₀, passband ripple, stop-band attenuation, and R₀ = 50 Ω.
2. Response and order: Butterworth, PLR = 1 + k²(ω/ωc)²ᴺ (maximally flat), or Chebyshev, PLR = 1 + k²T_N²(ω/ωc) (equal ripple, sharper). Find N from attenuation charts.
3. Low-pass prototype: read the g-values from tables (R₀ = 1, ωc = 1). For example, Butterworth N = 3 gives 1, 2, 1.
4. Impedance scaling and LP → BP transformation, with ω → (1/Δ)(ω/ω₀ − ω₀/ω):
   • series L_k becomes a series LC: L' = R₀g_k/(ω₀Δ), C' = Δ/(ω₀R₀g_k)
   • shunt C_k becomes a parallel LC: L' = R₀Δ/(ω₀g_k), C' = g_k/(ω₀R₀Δ)
5. Microwave realisation: use admittance or impedance inverters with λ/2 resonators (end-coupled), λ/4 coupled lines (parallel-coupled), or stub resonators. Find the gap or coupling values from the inverter constants.
6. Layout and tuning: include gap, open-end and step effects, then simulate and tune the lengths and gaps.

The method gives a predictable, exactly specified response, which is why it is the standard method for microwave filters.

The maximum gains are found after checking stability. The unilateral model ignores S12, and the bilateral model includes it (simultaneous conjugate match).

Given: S11 = 0.55∠−150°, S12 = 0.04∠20°, S21 = 2.82∠180°, S22 = 0.45∠−30°.

Δ = S11·S22 − S12·S21
K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
Unconditionally stable if K > 1 and |Δ| < 1
GTU,max = [1/(1−|S11|²)]·|S21|²·[1/(1−|S22|²)]
Gmax (MAG) = (|S21|/|S12|)·(K − √(K² − 1))

Stability (needed before gain)

S11·S22 = 0.2475∠180.0° = −0.2475 + j0.0000
S12·S21 = 0.1128∠−160.0° = −0.1060 − j0.0386
Δ = S11·S22 − S12·S21 = −0.1415 + j0.0386
  = 0.1467∠164.7°,   |Δ|² = 0.0215

K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
  = (1 − 0.3025 − 0.2025 + 0.0215) / (2 × 0.1128)
  = 0.5165 / 0.2256
  = 2.290

K = 2.290 > 1 and |Δ| = 0.147 < 1, so the transistor is unconditionally stable. Maximum gains therefore exist.

Unilateral model (S12 = 0)

Unilateral (S12 = 0, ΓS = S11*, ΓL = S22*):
GS = 1/(1 − |S11|²) = 1/(1 − 0.3025)
   = 1.4337  (1.56 dB)
G0 = |S21|² = 2.820² = 7.9524  (9.00 dB)
GL = 1/(1 − |S22|²) = 1/(1 − 0.2025)
   = 1.2539  (0.98 dB)
GTU,max = GS·G0·GL = 14.296  (11.55 dB)

Bilateral model

Bilateral (S12 kept), simultaneous conjugate match:
MAG = Gmax = (|S21|/|S12|)·(K − √(K² − 1))
     = (2.820/0.040) × (2.290 − √(2.290² − 1))
     = 70.500 × 0.2299
     = 16.210  (12.10 dB)

The source and load terminations for this gain are ΓS = 0.6307∠148.0° and ΓL = 0.5551∠26.7°.

The bilateral gain is 0.55 dB higher. The unilateral figure of merit U = 0.0502 gives an error range of −0.43 to +0.45 dB, so the unilateral model is a reasonable approximation.

Answer: GTU,max = 11.55 dB (unilateral), Gmax = 12.10 dB (bilateral).

Flow chart of microwave amplifier design

        [ Start: specs f, G, NF ]
                   |
     [ Select transistor, bias; get S ]
                   |
        [ Compute Δ, K (or μ) ]
                   |
          < K>1 and |Δ|<1 ? >
           | yes          | no
           |      [ Draw stability circles ]
           |      [ choose ΓS, ΓL in stable ]
           |      [ region / add loading    ]
           |              |
     < U small (S12≈0)? > |
      | yes        | no   |
 [ΓS=S11*,     [ΓS=ΓMS,   |
  ΓL=S22*]      ΓL=ΓML]   |
      |____________|______|
                   |
   [ Design input/output matching ]
   [ networks (Smith chart, stubs) ]
                   |
   [ Bias network, layout, simulate ]
                   |
           < Specs met? > -- no -> retune
                   | yes
                [ End ]

Steps of microwave filter designing (insertion loss method)

 1. Specifications (fc, ripple, IL, Z0)
 2. Response type + order N
 3. Low-pass prototype g-values
 4. Scaling + LP->HP/BP/BS transform
 5. Lumped -> distributed (Richards,
    Kuroda, stepped-Z, coupled lines)
 6. Layout, simulate (EM), tune

1. Specifications: type (LP, HP, BP or BS), cut-off or centre frequency, passband ripple or insertion loss, stop-band attenuation and impedance R₀.
2. Response and order: choose Butterworth (PLR = 1 + k²(ω/ωc)²ᴺ) for a flat passband or Chebyshev (PLR = 1 + k²T_N²(ω/ωc)) for a sharper cut-off. Find N from attenuation charts.
3. Prototype: read g₁…g_N for a normalised LP ladder (R₀ = 1, ωc = 1).
4. Scaling and transformation: L_k = R₀g_k/ωc and C_k = g_k/(R₀ωc). Then convert to HP, BP or BS with the standard frequency transformations.
5. Distributed realisation: short high-Z lines for series L, short low-Z lines or open stubs for shunt C (stepped impedance), or Richards' λ/8 stubs with Kuroda identities.
6. Layout and verification: EM simulation, include discontinuities, and tune.

Double-pad T-type passive filter in microstrip

A T-section low-pass filter has a series inductor, a shunt capacitor and another series inductor. A "double-pad" T-type filter cascades two T-sections, giving two capacitive pads:

 Lumped:  --UUU--+--UUU--+--UUU--
            L1   |   L2  |   L3
                C1      C2
                 |       |
          -------+-------+-------

 Microstrip (top view):
        L1     pad C1    L2    pad C2    L3
 50 Ω ====____[######]____[######]____==== 50 Ω
        thin    wide    thin    wide    thin
       (hi Z)  (low Z) (hi Z)  (low Z) (hi Z)

• Thin lines (high Z₀, about 100–120 Ω, βl < 45°) act as the series inductors: βl = L·R₀/Z_h (L normalised).
• Wide pads (low Z₀, about 10–20 Ω) act as the shunt capacitors: βl = C·Z_ℓ/R₀.
• The 50 Ω feed lines connect at both ends, and the bottom side is a full ground plane.

Stability of a microwave amplifier means that it does not oscillate. This needs |Γ_in| < 1 and |Γ_out| < 1, meaning there is no negative resistance at either port, for the source and load terminations used.

Γin  = S11 + S12·S21·ΓL / (1 − S22·ΓL)
Γout = S22 + S12·S21·ΓS / (1 − S11·ΓS)

Types

• Unconditionally stable: |Γ_in| < 1 and |Γ_out| < 1 for every passive source and load (|Γ_S|, |Γ_L| < 1).
• Conditionally (potentially) unstable: stable only for some terminations.

Tests

Δ = S11·S22 − S12·S21
K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
Rollett: K > 1 and |Δ| < 1  -> unconditionally stable
μ = (1 − |S11|²) / (|S22 − Δ·S11*| + |S12·S21|)
μ-test : μ > 1 -> unconditionally stable

Stability circles (|Γ_in| = 1 and |Γ_out| = 1) for a potentially unstable device:

Output (ΓL plane):
 CL = (S22 − Δ·S11*)* / (|S22|² − |Δ|²)
 RL = |S12·S21 / (|S22|² − |Δ|²)|
Input (ΓS plane):
 CS = (S11 − Δ·S22*)* / (|S11|² − |Δ|²)
 RS = |S12·S21 / (|S11|² − |Δ|²)|

If |S11| < 1 (or |S22| < 1), the centre of the Smith chart is stable. The region on the same side of the circle as the centre is the stable region. The designer chooses Γ_S and Γ_L in the stable regions, or adds series or shunt resistive loading to stabilise the device.

Stability is checked with the Rollett factor K and |Δ|. Then the maximum gain is compared for the unilateral case (S12 = 0) and the bilateral case (simultaneous conjugate match).

Given (10 GHz, 50 Ω): S11 = 0.55∠−160°, S12 = 0.04∠10°, S21 = 4.82∠180°, S22 = 0.45∠−30°.

Δ = S11·S22 − S12·S21
K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
Unconditionally stable if K > 1 and |Δ| < 1
GTU,max = [1/(1−|S11|²)]·|S21|²·[1/(1−|S22|²)]
Gmax (MAG) = (|S21|/|S12|)·(K − √(K² − 1))

Stability

S11·S22 = 0.2475∠170.0° = −0.2437 + j0.0430
S12·S21 = 0.1928∠−170.0° = −0.1899 − j0.0335
Δ = S11·S22 − S12·S21 = −0.0539 + j0.0765
  = 0.0935∠125.2°,   |Δ|² = 0.0087

K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
  = (1 − 0.3025 − 0.2025 + 0.0087) / (2 × 0.1928)
  = 0.5037 / 0.3856
  = 1.306

Check with the single test μ = (1 − |S11|²) / (|S22 − Δ·S11*| + |S12·S21|) = 1.147 (> 1, unconditionally stable).

K = 1.306 > 1 and |Δ| = 0.094 < 1, so the MESFET is unconditionally stable at 10 GHz.

Unilateral mode

Unilateral (S12 = 0, ΓS = S11*, ΓL = S22*):
GS = 1/(1 − |S11|²) = 1/(1 − 0.3025)
   = 1.4337  (1.56 dB)
G0 = |S21|² = 4.820² = 23.2324  (13.66 dB)
GL = 1/(1 − |S22|²) = 1/(1 − 0.2025)
   = 1.2539  (0.98 dB)
GTU,max = GS·G0·GL = 41.766  (16.21 dB)

Bilateral mode

Bilateral (S12 kept), simultaneous conjugate match:
MAG = Gmax = (|S21|/|S12|)·(K − √(K² − 1))
     = (4.820/0.040) × (1.306 − √(1.306² − 1))
     = 120.500 × 0.4658
     = 56.124  (17.49 dB)

Matching terminations: ΓS = 0.7362∠156.7°, ΓL = 0.6831∠25.0°.

Error of the unilateral assumption

U = |S11·S12·S21·S22| / [(1 − |S11|²)(1 − |S22|²)]
  = 0.0858
1/(1+U)² < GT/GTU < 1/(1−U)²
−0.71 dB < error < +0.78 dB

Comparison

The bilateral gain is 1.28 dB higher. This is larger than the upper unilateral error bound (+0.78 dB), because the large S21 = 4.82 makes the feedback through S12 significant (U = 0.0858). So the bilateral design should be used for accurate results.

Answer: K = 1.306, |Δ| = 0.094 → unconditionally stable; GTU,max = 16.21 dB, Gmax = 17.49 dB.

Low-pass prototype (Butterworth / Chebyshev)

A low-pass prototype is a normalised LC ladder with source resistance g₀ = 1 Ω and cut-off ωc = 1 rad/s. Its element values g₁…g_N are chosen so that its power loss ratio follows the chosen approximation.

 g0=1    g2 (L)       g4 (L)
 -/\/\-+--UUU--+--UUU--+--  g(N+1)
       |       |        |    (load)
     g1 (C)   g3 (C)   ...
       |       |        |
 ------+-------+--------+--

• Butterworth: g_k = 2 sin[(2k − 1)π/(2N)], with k = 1 (3 dB at ωc). For N = 3: 1, 2, 1.
• Chebyshev: g-values depend on the ripple and are taken from tables. For 0.5 dB ripple, N = 3: 1.5963, 1.0967, 1.5963, with g₄ = 1.0.
• The order N is found from attenuation-versus-|ω/ωc| − 1 charts for the required stop-band attenuation.

Conversion to real values and other filter types

Impedance scaling: L' = R₀L, C' = C/R₀, R' = R₀.

Frequency scaling and transformations (ω₀ = √(ω₁ω₂), Δ = (ω₂ − ω₁)/ω₀):

• LP → HP: ω → −ωc/ω, so inductors become capacitors and capacitors become inductors.
• LP → BP: ω → (1/Δ)(ω/ω₀ − ω₀/ω). A series L becomes a series LC (L' = R₀g/(ω₀Δ), C' = Δ/(ω₀R₀g)). A shunt C becomes a parallel LC.
• LP → BS: the inverse of the BP transformation.

The lumped elements are then realised with transmission lines using Richards' transformation and Kuroda identities, or stepped-impedance sections.

Second-order high-pass π-section in microstrip

Example values: Butterworth N = 2 (g₁ = g₂ = 1.4142), fc = 2 GHz (assumed), R₀ = 50 Ω, ωc = 1.2566×10¹⁰ rad/s.

shunt L  = R0 / (ωc g1)   = 50 / (1.2566e10 × 1.4142)
         = 2.813 nH
series C = 1 / (R0 ωc g2) = 1 / (50 × 1.2566e10 × 1.4142)
         = 1.125 pF

The π shape (shunt – series – shunt) has a shunt inductor on each side of the series capacitor. Strictly, a full π with both shunt arms is 3rd order (Butterworth N = 3, g = 1, 2, 1: each shunt L = R₀/ωc = 3.979 nH, series C = 1/(2R₀ωc) = 0.796 pF). The 2nd-order design above uses one shunt L (2.813 nH) and one series C (1.125 pF), so the second shunt arm is left out.

 Lumped:  o----+---| |---+----o
               |    C    |
              L1        L2
               |         |
          o----+---------+----o

 Microstrip (top view):
 50 Ω ======+====| |====+====== 50 Ω
            |   gap or  |
          thin  inter-  thin
          stub  digital stub
            |    C      |
          [via]       [via]  -> ground

• Series C: a gap or interdigital capacitor in the main line (1.125 pF).
• Shunt L: a high-impedance short-circuited stub ending in a via. Using Richards' transformation, a λ/8 shorted stub at fc with Z₀ = R₀/g = 50/1.4142 = 35.4 Ω gives the normalised inductance 1/g = 0.707.

Stability is first tested with the Rollett factor K and |Δ| (and μ). Since the device turns out to be potentially unstable, the input and output stability circles are found to show which source and load terminations must be avoided.

Given (6 GHz): S11 = 0.894∠−60.6°, S12 = 0.020∠62.4°, S21 = 3.122∠123.6°, S22 = 0.781∠−27.6°.

Step 1: Rollett test

S11·S22 = 0.6982∠−88.2° = 0.0219 − j0.6979
S12·S21 = 0.0624∠−174.0° = −0.0621 − j0.0065
Δ = S11·S22 − S12·S21 = 0.0840 − j0.6913
  = 0.6964∠−83.1°,   |Δ|² = 0.4850

K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
  = (1 − 0.7992 − 0.6100 + 0.4850) / (2 × 0.0624)
  = 0.0758 / 0.1249
  = 0.607

Check with the single test μ = (1 − |S11|²) / (|S22 − Δ·S11*| + |S12·S21|) = 0.863 (< 1, potentially unstable).

K = 0.607 < 1 (although |Δ| = 0.696 < 1), and μ = 0.863 < 1. So the transistor is potentially (conditionally) unstable at 6 GHz. Some passive source or load impedances will make it oscillate.

Step 2: Stability circles

Output (load) stability circle, |Γin| = 1:
CL = (S22 − Δ·S11*)* / (|S22|² − |Δ|²)
   = 0.1702∠46.7° / (0.6100 − 0.4850)
   = 1.3626∠46.7°
RL = |S12·S21| / ||S22|² − |Δ|²|
   = 0.0624 / 0.1249 = 0.500

Input (source) stability circle, |Γout| = 1:
CS = (S11 − Δ·S22*)* / (|S11|² − |Δ|²)
   = 0.3556∠68.5° / (0.7992 − 0.4850)
   = 1.1317∠68.5°
RS = |S12·S21| / ||S11|² − |Δ|²|
   = 0.0624 / 0.3142 = 0.199

Step 3: Stable regions

• Load plane: |CL| = 1.363 > RL = 0.500, so the circle does not enclose the chart centre. Since |S11| = 0.894 < 1, ΓL = 0 is stable. So the region outside the output stability circle (inside the Smith chart) is stable, and the part of the chart inside the circle (around 47°, near the edge) is unstable.
• Source plane: |CS| = 1.132 > RS = 0.199, and |S22| = 0.781 < 1, so ΓS = 0 is stable. The small area inside the input stability circle (around 68°, near the edge of the chart) is unstable.

Conclusion

The transistor is conditionally stable. A design must keep Γ_S outside the circle (C_S = 1.132∠68.5°, R_S = 0.199) and Γ_L outside the circle (C_L = 1.363∠46.7°, R_L = 0.500), or add resistive loading to stabilise it. For example, the conjugate points ΓS = S11* = 0.894∠60.6° and ΓL = S22* = 0.781∠27.6° lie at distances 0.275 and 0.675 from the centres, which is more than R_S and R_L, so they are in the stable regions.

Answer: Δ = 0.696∠−83.1°, K = 0.607 < 1 → potentially unstable; CL = 1.363∠46.7°, RL = 0.500; CS = 1.132∠68.5°, RS = 0.199.

Why the insertion loss technique is used

The insertion loss method designs a filter from a chosen power loss ratio:

PLR = 1 / (1 − |Γ(ω)|²) = 1 + M(ω²)/N(ω²)
Butterworth: PLR = 1 + k²(ω/ωc)^(2N)
Chebyshev  : PLR = 1 + k² T_N²(ω/ωc)

It is preferred over the older image-parameter method because:

• the complete response (passband ripple, cut-off sharpness, phase) is specified and achieved exactly,
• the design is systematic: g-value tables, then scaling, then transformation, with no trial and error,
• the order N gives a clear trade-off between performance and size or loss,
• one LP prototype can be turned into HP, BP or BS filters and then into microstrip (Richards, Kuroda, stepped-impedance).

Double-section shunt-arm microwave LPF in microstrip

In a shunt-arm LPF, the shunt capacitors are made as open-circuited stubs (or wide pads) connected across the line, and the series inductors are short high-impedance line sections. A double-section filter has two shunt arms.

 Lumped:  o--UUU--+--UUU--+--UUU--o
            L1    |   L2  |   L3
                  C1      C2
                  |       |
          o-------+-------+-------o

 Microstrip (top view):
              stub C1      stub C2
               | |          | |    open end
               | |  l_s     | |    (λ/8 at fc)
 50 Ω ====+----+-+----+-----+-+----+==== 50 Ω
     feed  L1 (hi-Z)  L2 (hi-Z)  L3   feed
           -------- ground plane below ------

Labels

• 50 Ω feed lines at input and output.
• L₁, L₂, L₃: narrow high-impedance (about 100–120 Ω) line sections, βl = L·R₀/Z_h (normalised L), acting as series inductors. With Richards and Kuroda design, these are λ/8 unit elements.
• C₁, C₂: open-circuited shunt stubs. From Richards' transformation each is λ/8 long at fc with Z₀ = R₀/g_k. (Wide low-impedance pads can be used instead.)
• Substrate with εr and height h, and a full ground plane underneath.

Stability is tested with K and |Δ|. Then the maximum gain of the unilateral model (S12 = 0) is compared with what is possible in the bilateral model.

Given: S11 = 0.894∠−60.6°, S21 = 3.122∠123.6°, S12 = 0.020∠62.4°, S22 = 0.781∠−27.6°.

Δ = S11·S22 − S12·S21
K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
Unconditionally stable if K > 1 and |Δ| < 1
GTU,max = [1/(1−|S11|²)]·|S21|²·[1/(1−|S22|²)]
Gmax (MAG) = (|S21|/|S12|)·(K − √(K² − 1))

Stability

S11·S22 = 0.6982∠−88.2° = 0.0219 − j0.6979
S12·S21 = 0.0624∠−174.0° = −0.0621 − j0.0065
Δ = S11·S22 − S12·S21 = 0.0840 − j0.6913
  = 0.6964∠−83.1°,   |Δ|² = 0.4850

K = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
  = (1 − 0.7992 − 0.6100 + 0.4850) / (2 × 0.0624)
  = 0.0758 / 0.1249
  = 0.607

Check with the single test μ = (1 − |S11|²) / (|S22 − Δ·S11*| + |S12·S21|) = 0.863 (< 1, potentially unstable).

K = 0.607 < 1, so the transistor is potentially unstable (|Δ| = 0.696 < 1, but both conditions are needed). Stability circles:

CL = 1.363∠46.7°,  RL = 0.500   (output)
CS = 1.132∠68.5°,  RS = 0.199   (input)

Since |S11|, |S22| < 1, the chart centre is stable, and the unstable regions are the small areas inside these circles near the chart edge.

Unilateral mode

Unilateral (S12 = 0, ΓS = S11*, ΓL = S22*):
GS = 1/(1 − |S11|²) = 1/(1 − 0.7992)
   = 4.9810  (6.97 dB)
G0 = |S21|² = 3.122² = 9.7469  (9.89 dB)
GL = 1/(1 − |S22|²) = 1/(1 − 0.6100)
   = 2.5638  (4.09 dB)
GTU,max = GS·G0·GL = 124.472  (20.95 dB)

Check that these terminations are stable: |S11* − CS| = 0.275 > RS = 0.199 and |S22* − CL| = 0.675 > RL = 0.500. Both points are outside the stability circles, so they are safe.

U = |S11·S12·S21·S22| / [(1 − |S11|²)(1 − |S22|²)]
  = 0.5567
1/(1+U)² < GT/GTU < 1/(1−U)²
−3.84 dB < error < +7.07 dB

U = 0.557 is large (|S11| and |S22| are close to 1), so the unilateral value can be wrong by several dB. With the real S12, these terminations give |Γin| = 0.912, |Γout| = 0.848 and an actual transducer gain G_T = 101.3 (20.06 dB).

Bilateral mode

Since K < 1, the simultaneous conjugate match does not exist (the formula K − √(K² − 1) becomes complex), so Gmax (MAG) is not defined. The upper figure used instead is the maximum stable gain:

MSG = |S21| / |S12| = 3.122 / 0.020 = 156.1
    = 21.93 dB

MSG is the gain obtained when the device is just brought to K = 1 (for example by resistive loading). In practice it is approached but not reached, and a design must keep ΓS and ΓL away from the stability circles.

Comparison

Answer: K = 0.607 < 1 → potentially unstable; GTU,max = 20.95 dB; bilateral MAG undefined, MSG = 21.93 dB.

In the insertion loss method, the filter response is defined by the power loss ratio:

PLR = P_available / P_delivered = 1 / (1 − |Γ(ω)|²)
IL (dB) = 10 log PLR

A physically realisable PLR must be a ratio of even polynomials in ω, so the designer chooses a polynomial that gives the wanted response.

Butterworth (maximally flat) LPF prototype design

1. Choose the response: PLR = 1 + k²(ω/ωc)^(2N). With k = 1, the cut-off ωc is the 3 dB point. The response is as flat as possible at ω = 0 (first 2N−1 derivatives are zero) and falls at 20N dB/decade beyond ωc.
2. Find the order N from the required stopband loss. Example spec: at least 15 dB at 2fc.

N ≥ log10(10^(A/10) − 1) / (2·log10(ω/ωc))
  = log10(10^1.5 − 1) / (2·log10 2) = 2.47  →  N = 3
Check: IL(2fc) = 10 log(1 + 2^6) = 18.13 dB ≥ 15 dB

3. Read the element values for source R = 1 Ω, ωc = 1 rad/s from the Butterworth table, or gk = 2 sin[(2k−1)π/2N]. For N = 3: g1 = 1, g2 = 2, g3 = 1, g4 = 1 (load).
4. Scale impedance and frequency (R0 = 50 Ω, fc = 2 GHz assumed): C = g/(ωc·R0), L = g·R0/ωc.
5. Convert to distributed form (Richards/Kuroda or stepped-impedance) and simulate.

π-section (shunt C – series L – shunt C)

C1 = C3 = 1/(2π·2×10⁹·50)  = 1.5915 pF
L2      = 2·50/(2π·2×10⁹)  = 7.9577 nH

Microstrip (stepped-impedance) implementation

High-Z lines act as series inductors, low-Z lines as shunt capacitors: βl = g·Zℓ/R0 for a capacitor, βl = g·R0/Zh for an inductor. Assumed substrate εr = 4.2, h = 1.58 mm, Zℓ = 20 Ω, Zh = 120 Ω (widths from microstrip design formulas, λg at 2 GHz):

 50Ω     ____         ____     50Ω
 ======|    |=======|    |======
       | C1 |  L2   | C3 |
       |____| thin  |____|
       wide   line   wide

Feed lines are 50 Ω (3.13 mm). Since βl of L2 is slightly above 45°, the response is checked by simulation and lengths are fine-tuned.

Stability is tested with the Rollett factor K and |Δ|; the unilateral gain assumes S12 = 0 and conjugate matching at both ports.

Δ   = S11·S22 − S12·S21
K   = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
Unconditionally stable if K > 1 and |Δ| < 1
GTU,max = 1/(1−|S11|²) · |S21|² · 1/(1−|S22|²)
GT,max  = (|S21|/|S12|)·(K − √(K² − 1))
B1 = 1 + |S11|² − |S22|² − |Δ|²,  C1 = S11 − Δ·S22*
B2 = 1 + |S22|² − |S11|² − |Δ|²,  C2 = S22 − Δ·S11*
ΓMS = [B1 − √(B1² − 4|C1|²)] / (2C1)
ΓML = [B2 − √(B2² − 4|C2|²)] / (2C2)

Stability check

S11·S22 = 0.45×0.35 ∠(−160°−40°) = 0.1575∠160°
        = −0.1480 + j0.0539
S12·S21 = 0.04×3.12 ∠(20°+180°) = 0.1248∠−160°
        = −0.1173 − j0.0427
Δ = (−0.1480+j0.0539) − (−0.1173−j0.0427)
  = −0.0307 + j0.0966 = 0.1013∠107.65°
|Δ| = 0.1013 < 1
K = (1 − 0.2025 − 0.1225 + 0.01027)/(2×0.1248)
  = 0.6853/0.2496 = 2.745 > 1

Since K = 2.745 > 1 and |Δ| = 0.101 < 1, the transistor is unconditionally stable (μ = 1.776 > 1 confirms it). Any passive source and load can be used.

Unilateral (maximum) transducer gain

With S12 = 0, Γs = S11* = 0.45∠160° and ΓL = S22* = 0.35∠40°:

GS,max = 1/(1 − 0.45²) = 1/0.7975 = 1.2539 → 0.98 dB
G0     = |S21|² = 3.12² = 9.7344       → 9.88 dB
GL,max = 1/(1 − 0.35²) = 1/0.8775 = 1.1396 → 0.57 dB
GTU,max = 1.2539 × 9.7344 × 1.1396 = 13.91
        = 0.98 + 9.88 + 0.57 = 11.43 dB

Validity of the unilateral assumption

U = |S11||S12||S21||S22| / [(1−|S11|²)(1−|S22|²)]
  = 0.45×0.04×3.12×0.35 / (0.7975×0.8775) = 0.0281
1/(1+U)² < GT/GTU < 1/(1−U)²  →  −0.24 dB to +0.25 dB

The error is only about ±0.25 dB, so the unilateral model is acceptable. For comparison, the exact bilateral maximum is GT,max = 78×(2.745 − √(2.745² − 1)) = 14.71 = 11.68 dB, only 0.24 dB more.

Answer: K = 2.745, |Δ| = 0.101 → unconditionally stable; GTU,max = 13.91 ≈ 11.43 dB.

In a unilateral design S12 is taken as zero, so the input and output can be conjugate-matched independently: Γs = S11* and ΓL = S22*.

Assumed S-parameters (Z0 = 50 Ω)

S11 = 0.5∠−150°, S12 ≈ 0, S21 = 3.0∠60°, S22 = 0.4∠−40°.

With S12 = 0, Δ = S11S22 and K → ∞; since |S11| < 1 and |S22| < 1 the device is unconditionally stable.

Maximum unilateral gain

GTU,max = GS,max · G0 · GL,max
GS,max = 1/(1 − |S11|²) = 1/(1 − 0.25) = 1.3333 → 1.25 dB
G0     = |S21|² = 9.0                  → 9.54 dB
GL,max = 1/(1 − |S22|²) = 1/(1 − 0.16) = 1.1905 → 0.76 dB
GTU,max = 1.3333 × 9 × 1.1905 = 14.29
        = 1.25 + 9.54 + 0.76 = 11.55 dB

Matching terminations needed:

• Γs = S11* = 0.5∠150° → Zs = 50(1+Γs)/(1−Γs) = 17.72 + j11.81 Ω
• ΓL = S22* = 0.4∠40° → ZL = 76.76 + j46.99 Ω

The input matching network must transform the 50 Ω source to 17.72 + j11.81 Ω, and the output network must transform the 50 Ω load to 76.76 + j46.99 Ω. Each can be a shunt open stub plus a series line, or an L-section, designed on the Smith chart.

Answer: GTU,max = 14.29 ≈ 11.55 dB (for the assumed S-parameters).

Design flowchart

 +------------------------------+
 | Specs: f, gain, Z0, BW, NF   |
 +------------------------------+
               |
 +------------------------------+
 | Choose transistor, bias;     |
 | get [S] at design frequency  |
 +------------------------------+
               |
 +------------------------------+
 | Compute Δ, K (and μ)         |
 +------------------------------+
               |
      K>1 and |Δ|<1 ?
     yes |          | no
         v          v
 Uncond. stable   Draw stability circles;
         |        pick Γs, ΓL in stable
         |        region (or stabilise)
         +----+-----+
              |
   Unilateral? (U small, S12≈0)
   yes: Γs=S11*, ΓL=S22*
   no : Γs=ΓMS, ΓL=ΓML (bilateral)
              |
 +------------------------------+
 | Compute GS, G0, GL -> GT     |
 +------------------------------+
              |
 +------------------------------+
 | Design input/output matching |
 | (stubs or L-C) on Smith chart|
 +------------------------------+
              |
 | Bias network, layout, simulate
              |
   Specs met? -- no --> revise
              | yes
            Build & test

Key points of the flow:

1. The gain is split into three independent blocks: input match (GS), transistor (G0) and output match (GL).
2. The unilateral figure of merit U is checked. If the error bounds 1/(1±U)² are small (under about ±0.5 dB), the unilateral design is accepted; otherwise a bilateral (simultaneous conjugate match) design is used.
3. After matching, a bias network with RF chokes and DC blocks is added, and the circuit is simulated and tuned.

A double-section series-arm filter is a 5-element T-type ladder low-pass filter (series L – shunt C – series L – shunt C – series L). It is designed from a low-pass prototype by the insertion loss method and then built in microstrip.

Justification for the insertion loss method

So the insertion loss method gives a predictable, optimum design, which is why it is preferred for microwave filters.

Design steps

1. Specification: fc, passband ripple, stopband attenuation, R0. Assume fc = 2 GHz, R0 = 50 Ω, maximally flat response, ≥ 30 dB at 4 GHz.
2. Response and order: Butterworth PLR = 1 + (ω/ωc)^(2N).

N ≥ log10(10³ − 1)/(2 log10 2) = 4.98  →  N = 5

3. Prototype values (gk = 2 sin[(2k−1)π/2N]): g1 = 0.618, g2 = 1.618, g3 = 2.000, g4 = 1.618, g5 = 0.618, g6 = 1. Starting with a series element gives the series-arm (T) form: g1, g3, g5 are series inductors; g2, g4 are shunt capacitors.
4. Impedance and frequency scaling: L = g·R0/ωc, C = g/(R0·ωc).

L1 = L5 = 0.618×50/(2π×2×10⁹) = 2.459 nH
C2 = C4 = 1.618/(50×2π×2×10⁹) = 2.575 pF
L3      = 2×50/(2π×2×10⁹)     = 7.958 nH

5. Microstrip realisation (stepped impedance): series L → short high-impedance line (Zh = 120 Ω), βl = g·R0/Zh; shunt C → short low-impedance line (Zℓ = 20 Ω), βl = g·Zℓ/R0. Substrate assumed εr = 4.2, h = 1.58 mm.

50Ω        ____        ____        50Ω
===--L1--|    |--L3--|    |--L5--===
   thin  | C2 | thin | C4 | thin
         |____|      |____|
          wide        wide

6. Check and tune: simulate (the short-line approximation needs βl < 45°, so L3 is fine-tuned), include step discontinuities, then fabricate and measure.

Stability is checked with K and |Δ|; the bilateral maximum uses simultaneous conjugate matching, and the unilateral maximum assumes S12 = 0.

Δ   = S11·S22 − S12·S21
K   = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
Unconditionally stable if K > 1 and |Δ| < 1
GTU,max = 1/(1−|S11|²) · |S21|² · 1/(1−|S22|²)
GT,max  = (|S21|/|S12|)·(K − √(K² − 1))
B1 = 1 + |S11|² − |S22|² − |Δ|²,  C1 = S11 − Δ·S22*
B2 = 1 + |S22|² − |S11|² − |Δ|²,  C2 = S22 − Δ·S11*
ΓMS = [B1 − √(B1² − 4|C1|²)] / (2C1)
ΓML = [B2 − √(B2² − 4|C2|²)] / (2C2)

Stability check

S11·S22 = 0.72×0.73∠(−116°−54°) = 0.5256∠−170°
        = −0.5176 − j0.0913
S12·S21 = 0.03×2.60∠(57°+76°) = 0.0780∠133°
        = −0.0532 + j0.0570
Δ = −0.4644 − j0.1483 = 0.4875∠−162.29°
K = (1 − 0.5184 − 0.5329 + 0.2377)/(2×0.0780)
  = 0.1864/0.1560 = 1.195

K = 1.195 > 1 and |Δ| = 0.488 < 1 → unconditionally stable (μ = 1.040 > 1).

Bilateral maximum gain

GT,max = (2.60/0.03)(1.195 − √(1.195² − 1))
       = 86.67 × 0.5410 = 46.88 → 16.71 dB
B1 = 0.7478, C1 = 0.3704∠−123.41°
ΓMS = 0.8718∠123.41°
B2 = 0.7768, C2 = 0.3850∠−61.03°
ΓML = 0.8763∠61.03°
GS = 1/(1−|ΓMS|²)            = 4.166 → 6.20 dB
G0 = |S21|²                  = 6.76  → 8.30 dB
GL = (1−|ΓML|²)/|1−S22ΓML|²  = 1.665 → 2.21 dB
GT = 6.20 + 8.30 + 2.21 = 16.71 dB

Unilateral maximum gain

GS,max = 1/(1 − 0.72²) = 2.076 → 3.17 dB
G0     = 6.76                → 8.30 dB
GL,max = 1/(1 − 0.73²) = 2.141 → 3.31 dB
GTU,max = 30.05 → 14.78 dB

Comparison

Unilateral figure of merit U = 0.72×0.03×2.6×0.73/(0.4816×0.4671) = 0.182, so the error band is −1.45 dB to +1.75 dB. The actual gain (16.71 dB) is 1.93 dB higher than the unilateral estimate (this is just beyond the band because U is large). The unilateral model is therefore not accurate for this HEMT; the bilateral design should be used.

Answer: unconditionally stable (K = 1.195, |Δ| = 0.488); GT,max = 16.71 dB (bilateral) vs GTU,max = 14.78 dB (unilateral).

In the insertion loss method, the filter response is defined by the power loss ratio:

PLR = P_available / P_delivered = 1 / (1 − |Γ(ω)|²)
IL (dB) = 10 log PLR

A physically realisable PLR must be a ratio of even polynomials in ω, so the designer chooses a polynomial that gives the wanted response.

Procedure for a Butterworth LPF prototype

1. Specify the filter: cut-off frequency fc, source/load impedance R0, and the minimum attenuation A (dB) at some stopband frequency ω.
2. Select the maximally flat response:

PLR = 1 + k²(ω/ωc)^(2N)

k² = 10^(Lr/10) − 1 sets the loss at ωc; for the usual 3 dB cut-off k = 1. The response is flat at ω = 0 and rolls off at 20N dB/decade.

3. Find the order N:

N ≥ log10(10^(A/10) − 1) / (2·log10(ω/ωc))
Example: A = 20 dB at ω = 2ωc
N ≥ log10(99)/(2×0.301) = 3.31  → N = 4

4. Get the normalised element values (R0 = 1 Ω, ωc = 1 rad/s) from the table, or

g0 = 1,  gk = 2 sin[(2k − 1)π/(2N)],  g(N+1) = 1

5. Draw the ladder: gk alternate between shunt C and series L. Two equivalent forms exist: starting with a shunt C (π type) or a series L (T type).

   Rs   L2       L4        (π-type, N=4)
 o-/\/-+--UUU--+--UUU--+--o
       |       |       |
      C1      C3       RL
       |       |       |
 o-----+-------+-------+--o

6. Impedance and frequency scaling to R0 and ωc:

L = gk·R0/ωc,   C = gk/(R0·ωc),   RL = g(N+1)·R0

Example (N = 3, R0 = 50 Ω, fc = 2 GHz): C1 = C3 = 1.59 pF, L2 = 7.96 nH.

7. Transform (if needed) to high-pass, band-pass or band-stop using the standard frequency transformations.
8. Implement at microwave frequency: Richards' transformation (L → short-circuited stub, C → open-circuited stub, λ/8 at fc) with Kuroda identities, or stepped-impedance lines. Simulate and tune.

Butterworth is chosen when a flat passband is needed; Chebyshev gives a sharper cut-off for the same N at the cost of ripple.

The transducer gain is split into the gain of the input network GS, the transistor gain G0 and the output network gain GL. Here the source and load (25 Ω and 40 Ω) are connected to the device, so GS and GL show how much these terminations add or lose.

Reflection coefficients

Γs = (Zs − Z0)/(Zs + Z0) = (25 − 50)/(25 + 50) = −0.3333
ΓL = (ZL − Z0)/(ZL + Z0) = (40 − 50)/(40 + 50) = −0.1111

Stability (check first)

Δ = S11S22 − S12S21 = 0.2555∠−60.56°
K = (1 − 0.1444 − 0.16 + 0.0653)/(2×0.385) = 0.988

K < 1, so the device is potentially unstable. With these terminations:

Γin  = S11 + S12S21ΓL/(1 − S22ΓL) = 0.3653∠−152.04°
Γout = S22 + S12S21Γs/(1 − S11Γs) = 0.5452∠−42.98°

|Γin| < 1 and |Γout| < 1, so the amplifier is stable with Zs = 25 Ω and ZL = 40 Ω.

Gains of the input and output networks (bilateral)

GS = (1 − |Γs|²)/|1 − ΓinΓs|² = 0.8889/0.7997
   = 1.1115 → 0.46 dB
G0 = |S21|² = 3.5² = 12.25    → 10.88 dB
GL = (1 − |ΓL|²)/|1 − S22ΓL|² = 0.9877/1.0670
   = 0.9257 → −0.34 dB
GT = 1.1115 × 12.25 × 0.9257 = 12.60 → 11.01 dB

Other gains for the same terminations: power gain G = 13.09 (11.17 dB), available gain GA = 19.84 (12.97 dB).

With the unilateral approximation (S12 = 0): GS = (1−|Γs|²)/|1−S11Γs|² = 1.138 (0.56 dB), so GTU = 12.90 (11.11 dB).

If conjugate matching networks are used

The largest gains the matching networks can give (unilateral) are:

GS,max = 1/(1 − 0.38²) = 1.169 → 0.68 dB
GL,max = 1/(1 − 0.40²) = 1.190 → 0.76 dB
GTU,max = 1.169 × 12.25 × 1.190 = 17.04 → 12.32 dB

Since K < 1, the conjugate points must be checked against the stability circles before use.

Answer: GS = 1.11 (0.46 dB), GL = 0.93 (−0.34 dB), G0 = 10.88 dB, GT = 11.01 dB with the given 25 Ω/40 Ω terminations.

Design flowchart

 +------------------------------+
 | Specs: f, gain, Z0, BW, NF   |
 +------------------------------+
               |
 +------------------------------+
 | Choose transistor, bias;     |
 | get [S] at design frequency  |
 +------------------------------+
               |
 +------------------------------+
 | Compute Δ, K (and μ)         |
 +------------------------------+
               |
      K>1 and |Δ|<1 ?
     yes |          | no
         v          v
 Uncond. stable   Draw stability circles;
         |        pick Γs, ΓL in stable
         |        region (or stabilise)
         +----+-----+
              |
   Unilateral? (U small, S12≈0)
   yes: Γs=S11*, ΓL=S22*
   no : Γs=ΓMS, ΓL=ΓML (bilateral)
              |
 +------------------------------+
 | Compute GS, G0, GL -> GT     |
 +------------------------------+
              |
 +------------------------------+
 | Design input/output matching |
 | (stubs or L-C) on Smith chart|
 +------------------------------+
              |
 | Bias network, layout, simulate
              |
   Specs met? -- no --> revise
              | yes
            Build & test

In the insertion loss method, the filter response is defined by the power loss ratio:

PLR = P_available / P_delivered = 1 / (1 − |Γ(ω)|²)
IL (dB) = 10 log PLR

A physically realisable PLR must be a ratio of even polynomials in ω, so the designer chooses a polynomial that gives the wanted response.

Prototype design steps (both responses)

1. Specify fc, R0, ripple Lr (Chebyshev) and stopband attenuation.
2. Choose PLR and find the order N.
3. Get prototype values g1 … gN+1 (R0 = 1 Ω, ωc = 1 rad/s).
4. Scale: L = g·R0/ωc, C = g/(R0·ωc).
5. Realise with microstrip lines and simulate.

Butterworth (maximally flat)

PLR = 1 + k²(ω/ωc)^(2N)
gk  = 2 sin[(2k − 1)π/(2N)]

Flat passband, monotonic roll-off of 20N dB/decade.

Chebyshev (equal ripple)

PLR = 1 + k²·TN²(ω/ωc),  TN = Chebyshev polynomial
k² = 10^(Lr/10) − 1   (ripple Lr dB)
β = ln[coth(Lr/17.37)],  γ = sinh[β/(2N)]
g1 = 2a1/γ,  gk = 4a(k−1)ak / [b(k−1)g(k−1)]
ak = sin[(2k−1)π/2N],  bk = γ² + sin²(kπ/N)

Ripple in the passband, but a sharper cut-off for the same N. For odd N, g(N+1) = 1.

Double π-section (N = 5) values

A double π-section is C–L–C–L–C (two π sections sharing the middle C). Assume fc = 2 GHz, R0 = 50 Ω, ripple 0.5 dB.

Microstrip implementation (stepped impedance, Chebyshev)

Shunt C → low-Z line (20 Ω), βl = g·Zℓ/R0; series L → high-Z line (120 Ω), βl = g·R0/Zh. Assumed εr = 4.2, h = 1.58 mm.

(For the Butterworth version: C1 = C5 → 14.16°, L2 = L4 → 38.63°, C3 → 45.84°.)

 50Ω  ____        ______        ____  50Ω
 ====|    |------|      |------|    |====
     | C1 |  L2  |  C3  |  L4  | C5 |
     |____| thin |______| thin |____|

The middle capacitor line is long (βl > 45°), so in practice it is split or tuned in simulation, and step discontinuities are included.

Stability on the Smith chart is studied with stability circles: the locus of Γs (or ΓL) for which |Γout| = 1 (or |Γin| = 1). One side of each circle is stable, the other unstable.

Since the figure is missing, assume a transistor with S11 = 0.894∠−60.6°, S12 = 0.020∠62.4°, S21 = 3.122∠123.6°, S22 = 0.781∠−27.6°.

Steps

1. Compute Δ and K: Δ = 0.6964∠−83.07°, K = 0.607 < 1 → potentially (conditionally) unstable.
2. Compute circle centres and radii:

CS = (S11 − ΔS22*)* / (|S11|² − |Δ|²)
RS = |S12S21| / ||S11|² − |Δ|²|
CL = (S22 − ΔS11*)* / (|S22|² − |Δ|²)
RL = |S12S21| / ||S22|² − |Δ|²|

CS = 1.132∠68.46°,  RS = 0.199
CL = 1.363∠46.69°,  RL = 0.500

3. Plot the circles on the chart: each is centred outside the unit circle and cuts the chart edge near the upper-right.
4. Find the stable side by testing the chart centre (Γ = 0). For ΓL = 0, |Γin| = |S11| = 0.894 < 1, so the centre is stable. The centre lies outside the load circle (|CL| − RL = 0.863 > 0), so the stable region is outside the load circle. The same test with |S22| = 0.781 < 1 shows the stable source region is outside the source circle.

          unit Smith chart
        .-----------.   ( CL circle,
      /        (####)    shaded = unstable)
     |    o    (####)
     | centre   (##)  ( CS circle )
      \  stable      /
        '-----------'

5. Synthesis: choose Γs and ΓL in the unshaded region and away from the circle edge for margin. For example S11* = 0.894∠60.6° is 0.275 from CS (> RS) and S22* is 0.675 from CL (> RL), so conjugate unilateral matching is allowed here.
6. Verify: compute Γin and Γout with the chosen terminations; both must have magnitude < 1 (here |Γin| = 0.912, |Γout| = 0.848).

Design flowchart

 +------------------------------+
 | Specs: f, gain, Z0, BW, NF   |
 +------------------------------+
               |
 +------------------------------+
 | Choose transistor, bias;     |
 | get [S] at design frequency  |
 +------------------------------+
               |
 +------------------------------+
 | Compute Δ, K (and μ)         |
 +------------------------------+
               |
      K>1 and |Δ|<1 ?
     yes |          | no
         v          v
 Uncond. stable   Draw stability circles;
         |        pick Γs, ΓL in stable
         |        region (or stabilise)
         +----+-----+
              |
   Unilateral? (U small, S12≈0)
   yes: Γs=S11*, ΓL=S22*
   no : Γs=ΓMS, ΓL=ΓML (bilateral)
              |
 +------------------------------+
 | Compute GS, G0, GL -> GT     |
 +------------------------------+
              |
 +------------------------------+
 | Design input/output matching |
 | (stubs or L-C) on Smith chart|
 +------------------------------+
              |
 | Bias network, layout, simulate
              |
   Specs met? -- no --> revise
              | yes
            Build & test

Stability

Δ = S11S22 − S12S21
  = (0.0981 + j0.0557) − (−0.0039 + j0.2806)
  = 0.1020 − j0.2248 = 0.2469∠−65.59°
K = (1 − 0.4303 − 0.0296 + 0.0610)/(2×0.2806)
  = 0.6010/0.5612 = 1.071

K = 1.071 > 1, |Δ| = 0.247 < 1 → unconditionally stable.

Maximum gain: bilateral model

GT,max = (2.30/0.122)(1.071 − √(1.071² − 1))
       = 18.85 × 0.6876 = 12.96 → 11.13 dB
B1 = 1.3398, C1 = S11 − ΔS22* = 0.6612∠150.37°
ΓMS = 0.8504∠−150.37°
B2 = 0.5383, C2 = S22 − ΔS11* = 0.2467∠−76.27°
ΓML = 0.6548∠76.27°
GS = 3.613 (5.58 dB), G0 = 5.29 (7.23 dB),
GL = 0.678 (−1.69 dB)  → GT = 11.13 dB

Maximum gain: unilateral model

GS,max = 1/(1 − 0.656²) = 1.755 → 2.44 dB
G0     = 2.30² = 5.29         → 7.23 dB
GL,max = 1/(1 − 0.172²) = 1.030 → 0.13 dB
GTU,max = 9.57 → 9.81 dB
U = 0.057 → error −0.48 dB to +0.51 dB

The bilateral maximum (11.13 dB) is 1.32 dB higher, because |S12| = 0.122 is not small.

Γin, Γout and Zin, Zout

With the bilateral match, Γin = ΓMS* and Γout = ΓML*:

Γin  = 0.8504∠150.37°
zin  = (1+Γin)/(1−Γin) = 0.0864 + j0.2627
Zin  = 50·zin = 4.32 + j13.13 Ω
Γout = 0.6548∠−76.27°
zout = 0.511 − j1.138
Zout = 25.55 − j56.90 Ω

(Unilateral model: Γin = S11 → Zin = 11.27 + j14.25 Ω; Γout = S22 → Zout = 40.90 − j12.91 Ω.)

Tracing on the Smith chart

1. Draw a circle of radius 0.850 (of the chart radius) about the centre and mark the angle 150.37° on the outer angle scale. The point lies near r = 0.086, x = +0.263: upper half, inductive, very close to the short-circuit end. Read Zin = 50(0.086 + j0.263).
2. Draw a circle of radius 0.655 and mark −76.27°. The point lies near r = 0.51, x = −1.14 in the lower (capacitive) half. Read Zout = 50(0.51 − j1.14).
3. The source must present Zs = Zin* = 4.32 − j13.13 Ω and the load ZL = Zout* = 25.55 + j56.90 Ω; matching networks are designed from these points to the chart centre.

Answer: GT,max = 12.96 (11.13 dB) bilateral; GTU,max = 9.57 (9.81 dB) unilateral; Zin = 4.32 + j13.13 Ω, Zout = 25.55 − j56.90 Ω.

The input matching network sets Γs. Its stability is judged from the source (input) stability circle: the set of Γs that make |Γout| = 1. Since the sketched chart is not available, the steps are shown with an assumed transistor: S11 = 0.894∠−60.6°, S12 = 0.020∠62.4°, S21 = 3.122∠123.6°, S22 = 0.781∠−27.6° (Z0 = 50 Ω).

Step 1: Stability parameters

Δ = S11S22 − S12S21 = 0.6964∠−83.07°,  |Δ| < 1
K = (1 − |S11|² − |S22|² + |Δ|²)/(2|S12S21|) = 0.607

K < 1 → conditionally stable; the input stability circle must be drawn.

Step 2: Input (source) stability circle

CS = (S11 − ΔS22*)* / (|S11|² − |Δ|²)
RS = |S12S21| / ||S11|² − |Δ|²|
CL = (S22 − ΔS11*)* / (|S22|² − |Δ|²)
RL = |S12S21| / ||S22|² − |Δ|²|

CS = 1.132∠68.46°,  RS = 0.199

Step 3: Locate the stable region

• |CS| − RS = 0.933 > 0, so the chart centre (Γs = 0) is outside the circle.
• With Γs = 0, |Γout| = |S22| = 0.781 < 1, so the centre is stable.
• Hence the stable region is outside the circle; the small shaded zone inside the circle, near the top-right edge of the chart (inductive, high-|Γ| region), must be avoided.

           .----------.
         /       (##)  \   ## = unstable Γs
        |   o           |  o  = centre (stable)
        |  stable       |
         \             /
           '----------'

Step 4: Synthesis of the input match

1. Target Γs = S11* = 0.894∠60.6° for maximum unilateral gain. Distance from CS = 0.275 > RS, so it is in the stable region, but close (margin 0.076). A practical design may back off to a constant-gain circle (smaller |Γs|) for a safer margin.
2. Zs = 50(1+Γs)/(1−Γs) = 10.89 + j84.52 Ω.
3. Network: from the 50 Ω centre, add a shunt stub to reach the |Γ| = 0.894 circle, then a series line to rotate to 60.6°; check that the whole path stays at the required end point.
4. Verify: with the chosen Γs and ΓL, |Γout| = 0.848 < 1 and |Γin| = 0.912 < 1, so the amplifier is stable.

GS,max = 1/(1 − 0.894²) = 4.98 (6.97 dB) is the gain contributed by this input network.

Microwave filters are designed mainly by the insertion loss method: a low-pass prototype is synthesised for the required response, then scaled, transformed and built with transmission-line elements.

Design steps

1. Specifications: type (LP/HP/BP/BS), cut-off fc, ripple, stopband attenuation, R0, size/substrate.
2. Choose the response: Butterworth PLR = 1 + (ω/ωc)^(2N) (flat), Chebyshev PLR = 1 + k²TN²(ω/ωc) (sharper, rippled), or linear phase.
3. Find the order N from the stopband attenuation.
4. Low-pass prototype values g1…gN+1 from tables (R0 = 1 Ω, ωc = 1).
5. Scaling and transformation:

LPF: L = gR0/ωc,  C = g/(R0ωc)
HPF: series C = 1/(ωc·R0·g),  shunt L = R0/(ωc·g)
BPF/BSF: use ω0 and fractional bandwidth Δ

6. Implementation: Richards' transformation (L → short stub, C → open stub), Kuroda identities (convert series stubs to shunt stubs), stepped-impedance lines, or coupled lines for BPF.
7. Simulate, optimise, fabricate and measure.

Example: passive HPF using microstrip

Specs assumed: fc = 2 GHz, R0 = 50 Ω, Butterworth N = 3 (g1 = 1, g2 = 2, g3 = 1). The LP prototype series L – shunt C – series L becomes series C – shunt L – series C:

C1 = C3 = 1/(2π×2×10⁹ × 50 × 1) = 1.592 pF
L2      = 50/(2π×2×10⁹ × 2)     = 1.989 nH

Microstrip realisation (εr = 4.2, h = 1.58 mm assumed):

• Series capacitors C1, C3: gap-coupled or interdigital capacitors in the 50 Ω line (width 3.13 mm), sized to give 1.59 pF.
• Shunt inductor L2: short-circuited stub (via hole to ground). For a 50 Ω stub, Z·tan(βl) = ωcL2 = 25 Ω → βl = 26.57°, length ≈ 6.19 mm at 2 GHz.

  50Ω   C1  ||     C3  ||   50Ω
 =======|| |===+===|| |=======
          (gap) |  (gap)
                | L2: shorted
                | stub 6.19 mm
               _|_
              (via) ground

The response gives 3 dB loss at 2 GHz and about 18 dB at 1 GHz. Because a stub is resonant, a spurious stopband appears near the frequency where the stub is λ/2; this is checked in simulation.

Δ   = S11·S22 − S12·S21
K   = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
Unconditionally stable if K > 1 and |Δ| < 1
GTU,max = 1/(1−|S11|²) · |S21|² · 1/(1−|S22|²)
GT,max  = (|S21|/|S12|)·(K − √(K² − 1))
B1 = 1 + |S11|² − |S22|² − |Δ|²,  C1 = S11 − Δ·S22*
B2 = 1 + |S22|² − |S11|² − |Δ|²,  C2 = S22 − Δ·S11*
ΓMS = [B1 − √(B1² − 4|C1|²)] / (2C1)
ΓML = [B2 − √(B2² − 4|C2|²)] / (2C2)

Stability

S11·S22 = 0.78×0.81∠(−113°−54°) = 0.6318∠−167°
        = −0.6156 − j0.1421
S12·S21 = 0.028×2.60∠(247°+76°) = 0.0728∠323°
        = 0.0581 − j0.0438
Δ = −0.6737 − j0.0983 = 0.6809∠−171.70°
K = (1 − 0.6084 − 0.6561 + 0.4636)/(2×0.0728)
  = 0.1991/0.1456 = 1.367

K = 1.367 > 1 and |Δ| = 0.681 < 1 → unconditionally stable (μ = 1.097 > 1).

Bilateral maximum gain (simultaneous conjugate match)

GT,max = (|S21|/|S12|)(K − √(K² − 1))
       = (2.60/0.028)(1.367 − √(1.367² − 1))
       = 92.86 × 0.4348 = 40.37 → 16.06 dB
B1 = 0.4887, C1 = 0.2347∠−101.90° → ΓMS = 0.7517∠101.90°
B2 = 0.5841, C2 = 0.2840∠−45.19°  → ΓML = 0.7891∠45.19°
GS = 2.299 (3.62 dB)
G0 = 6.76  (8.30 dB)
GL = 2.597 (4.15 dB)   → GT = 16.06 dB

Unilateral maximum gain

GS,max = 1/(1 − 0.78²) = 2.554 → 4.07 dB
G0     = 2.60² = 6.76         → 8.30 dB
GL,max = 1/(1 − 0.81²) = 2.908 → 4.64 dB
GTU,max = 50.20 → 17.01 dB

Comparison

U = 0.78×0.028×2.6×0.81 / (0.3916×0.3439) = 0.342
Error band: −2.55 dB to +3.63 dB

The unilateral formula gives 0.95 dB more than is really possible. Because |S11| and |S22| are large, U = 0.34 and the unilateral model is unreliable; the bilateral value 16.06 dB is the true maximum gain and the design should use ΓMS and ΓML.

Answer: unconditionally stable (K = 1.367, |Δ| = 0.681); GT,max = 16.06 dB (bilateral), GTU,max = 17.01 dB (unilateral estimate).

Flow diagram for FET amplifier design

 +------------------------------+
 | Specs: f, gain, Z0, BW, NF   |
 +------------------------------+
               |
 +------------------------------+
 | Choose transistor, bias;     |
 | get [S] at design frequency  |
 +------------------------------+
               |
 +------------------------------+
 | Compute Δ, K (and μ)         |
 +------------------------------+
               |
      K>1 and |Δ|<1 ?
     yes |          | no
         v          v
 Uncond. stable   Draw stability circles;
         |        pick Γs, ΓL in stable
         |        region (or stabilise)
         +----+-----+
              |
   Unilateral? (U small, S12≈0)
   yes: Γs=S11*, ΓL=S22*
   no : Γs=ΓMS, ΓL=ΓML (bilateral)
              |
 +------------------------------+
 | Compute GS, G0, GL -> GT     |
 +------------------------------+
              |
 +------------------------------+
 | Design input/output matching |
 | (stubs or L-C) on Smith chart|
 +------------------------------+
              |
 | Bias network, layout, simulate
              |
   Specs met? -- no --> revise
              | yes
            Build & test

Notes: the FET is biased (e.g. VDS, IDS from the datasheet) before reading [S]; matching networks transform 50 Ω to the required Γs and ΓL; RF chokes and DC blocks isolate bias from RF.

Conditional stability

An amplifier is conditionally (potentially) stable if |Γin| > 1 or |Γout| > 1 for some passive source or load. It happens when K < 1 (or |Δ| > 1). Then only certain Γs and ΓL are allowed, and these are found with stability circles.

Self-defined example

Assume a GaAs FET with S11 = 0.894∠−60.6°, S12 = 0.020∠62.4°, S21 = 3.122∠123.6°, S22 = 0.781∠−27.6°.

Δ = 0.6964∠−83.07°,   K = 0.607 < 1
→ conditionally stable
Output (load) circle:  CL = 1.363∠46.69°,  RL = 0.500
Input (source) circle: CS = 1.132∠68.46°,  RS = 0.199

Using the Smith chart

1. Plot CL with radius RL on the ΓL plane and CS with radius RS on the Γs plane.
2. Test the centre: for ΓL = 0, |Γin| = |S11| = 0.894 < 1 → centre stable. The centre is outside both circles, so the stable region is outside the circles.
3. The parts of the circles that fall inside the unit chart (near the top-right, inductive high-|Γ| region) are the unstable regions; Γs and ΓL must not be placed there.

             .-----------.
           /          (###)    ### = unstable
          |   o        (##)      region (inside
          | centre             circle and chart)
          | stable           |
           \               /
             '-----------'

4. Select Γs, ΓL in the stable region with some margin and check |Γin| < 1 and |Γout| < 1. For Γs = S11*, ΓL = S22*: |Γin| = 0.912, |Γout| = 0.848, so this choice is stable and gives GTU = 20.95 dB.
5. If the desired point is unstable, add resistive loading (series/shunt resistor at input or output) or feedback to make the FET stable.

In the insertion loss method, the filter is synthesised directly from a chosen power loss ratio PLR(ω), so the complete frequency response is controlled from the start.

Justification

Hence it gives optimum, repeatable designs, and it is the standard method for microwave filters.

Description of the method

1. Specify fc (or band edges), R0, passband ripple and stopband attenuation.
2. Choose the response:

PLR = 1/(1 − |Γ(ω)|²),  IL = 10 log PLR
Butterworth:   PLR = 1 + k²(ω/ωc)^(2N)
Chebyshev:     PLR = 1 + k²·TN²(ω/ωc)
Linear phase:  flat group delay

3. Find the order N from the stopband specification. Example: Butterworth, 15 dB at 2fc: N ≥ log(10^1.5 − 1)/(2 log 2) = 2.47 → N = 3.
4. Low-pass prototype values gk from tables (R0 = 1, ωc = 1): for N = 3 Butterworth, 1, 2, 1.
5. Scaling and transformation: L = gR0/ωc, C = g/(R0ωc); LP → HP, BP, BS by frequency substitution. For fc = 2 GHz, R0 = 50 Ω: C = 1.59 pF, L = 7.96 nH.
6. Implementation: lumped elements become transmission-line sections using Richards' transformation, Kuroda identities, stepped-impedance lines or coupled lines.
7. Simulate and tune, including junction and step discontinuities, then fabricate.

Δ   = S11·S22 − S12·S21
K   = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
Unconditionally stable if K > 1 and |Δ| < 1
GTU,max = 1/(1−|S11|²) · |S21|² · 1/(1−|S22|²)
GT,max  = (|S21|/|S12|)·(K − √(K² − 1))
B1 = 1 + |S11|² − |S22|² − |Δ|²,  C1 = S11 − Δ·S22*
B2 = 1 + |S22|² − |S11|² − |Δ|²,  C2 = S22 − Δ·S11*
ΓMS = [B1 − √(B1² − 4|C1|²)] / (2C1)
ΓML = [B2 − √(B2² − 4|C2|²)] / (2C2)

Stability

S11·S22 = 0.64×0.22∠(−169°−82°) = 0.1408∠109°
        = −0.0458 + j0.1331
S12·S21 = 0.03×10.11∠(50°+91°) = 0.3033∠141°
        = −0.2357 + j0.1909
Δ = 0.1899 − j0.0577 = 0.1985∠−16.92°
K = (1 − 0.4096 − 0.0484 + 0.0394)/(2×0.3033)
  = 0.5814/0.6066 = 0.958
μ = 0.960

K = 0.958 < 1 (|Δ| = 0.198 < 1) → potentially unstable (conditionally stable).

Stability circles:

CS = 1.800∠165.96°,  RS = 0.819
CL = 34.60∠62.75°,   RL = 33.64

Both circles lie mostly outside the chart; since |S11|, |S22| < 1 the chart centre is stable and lies outside both circles, so stable regions are outside the circles.

Bilateral model

Since K < 1, simultaneous conjugate matching is not possible (ΓMS, ΓML would lie outside the chart), so GT,max does not exist. The upper limit is the maximum stable gain:

MSG = |S21|/|S12| = 10.11/0.03 = 337 → 25.28 dB

To use the bilateral formula, the device must be modified: e.g. add a small series/shunt resistor (resistive loading) or feedback so that the modified S-parameters give K > 1.

Unilateral model (S12 modified to 0)

GS,max = 1/(1 − 0.64²) = 1.694 → 2.29 dB
G0     = 10.11² = 102.21       → 20.10 dB
GL,max = 1/(1 − 0.22²) = 1.051 → 0.22 dB
GTU,max = 181.93 → 22.60 dB
U = 0.076 → error −0.64 dB to +0.69 dB

Check that the unilateral terminations are stable:

• Γs = S11* = 0.64∠169°: distance from CS = 1.162 > RS = 0.819 → stable side.
• ΓL = S22* = 0.22∠82°: distance from CL = 34.39 > RL = 33.64 → stable side.
• With both: |Γin| = 0.700, |Γout| = 0.528, both < 1.

So the unilateral design is usable.

Answer: K = 0.958 → potentially unstable; bilateral: MSG = 25.28 dB (GT,max undefined); unilateral: GTU,max = 181.9 = 22.60 dB.

In the insertion loss method, the filter response is defined by the power loss ratio:

PLR = P_available / P_delivered = 1 / (1 − |Γ(ω)|²)
IL (dB) = 10 log PLR

A physically realisable PLR must be a ratio of even polynomials in ω, so the designer chooses a polynomial that gives the wanted response.

Steps of the method

1. Specify: fc, R0, ripple and minimum stopband attenuation.
2. Choose the response: Butterworth PLR = 1 + k²(ω/ωc)^(2N) for a flat passband; Chebyshev PLR = 1 + k²TN²(ω/ωc) for a sharper cut-off with ripple; linear-phase for flat group delay.
3. Find N from the stopband requirement.
4. Prototype values g0 … gN+1 from tables (R0 = 1 Ω, ωc = 1 rad/s).
5. Scale impedance and frequency: L = gR0/ωc, C = g/(R0ωc); transform to HP/BP/BS if needed.
6. Realise in transmission lines (Richards + Kuroda, stepped impedance), then simulate and tune.

Example: passive LPF using microstrip

Specs: fc = 2 GHz, R0 = 50 Ω, Chebyshev 0.5 dB ripple, N = 3 (π type: shunt C – series L – shunt C).

g1 = g3 = 1.5963,  g2 = 1.0967,  g4 = 1.0
C1 = C3 = 1.5963/(50×2π×2×10⁹) = 2.541 pF
L2      = 1.0967×50/(2π×2×10⁹) = 4.364 nH

Stepped-impedance microstrip (εr = 4.2, h = 1.58 mm, Zℓ = 20 Ω, Zh = 120 Ω):

C: βl = g·Zℓ/R0     L: βl = g·R0/Zh

 50Ω    ______        ______    50Ω
 ======|      |------|      |======
       |  C1  |  L2  |  C3  |
       |______| thin |______|
        wide          wide

All lines are shorter than λ/8 (45°), so the lumped approximation holds well. The 50 Ω feed lines are 3.13 mm wide.

Δ   = S11·S22 − S12·S21
K   = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
Unconditionally stable if K > 1 and |Δ| < 1
GTU,max = 1/(1−|S11|²) · |S21|² · 1/(1−|S22|²)
GT,max  = (|S21|/|S12|)·(K − √(K² − 1))
B1 = 1 + |S11|² − |S22|² − |Δ|²,  C1 = S11 − Δ·S22*
B2 = 1 + |S22|² − |S11|² − |Δ|²,  C2 = S22 − Δ·S11*
ΓMS = [B1 − √(B1² − 4|C1|²)] / (2C1)
ΓML = [B2 − √(B2² − 4|C2|²)] / (2C2)

Justification of conditional stability

S11·S22 = 0.894×0.781∠(−60.6°−27.6°) = 0.6982∠−88.2°
        = 0.0219 − j0.6979
S12·S21 = 0.020×3.122∠(62.4°+123.6°) = 0.0624∠186°
        = −0.0621 − j0.0065
Δ = 0.0840 − j0.6913 = 0.6964∠−83.07°
K = (1 − 0.7992 − 0.6100 + 0.4850)/(2×0.0624)
  = 0.0758/0.1249 = 0.607
μ = 0.863

• |Δ| = 0.696 < 1 but K = 0.607 < 1 (and μ < 1), so the device is not unconditionally stable.
• Since |S11| < 1 and |S22| < 1, it is stable for a 50 Ω source and load. It is therefore conditionally stable: stable for some Γs, ΓL and unstable for others.

Stability circles show the unstable zones:

CL = 1.363∠46.69°,  RL = 0.500
CS = 1.132∠68.46°,  RS = 0.199

Both circles cut the unit Smith chart, so parts of the chart (near the top-right edge) are unstable; the stable region is outside the circles.

Maximum unilateral gain

GS,max = 1/(1 − 0.894²) = 4.981 → 6.97 dB
G0     = 3.122² = 9.747          → 9.89 dB
GL,max = 1/(1 − 0.781²) = 2.564 → 4.09 dB
GTU,max = 124.47 → 20.95 dB

Check that Γs = S11* and ΓL = S22* are in the stable region: distance of S11* from CS = 0.275 > 0.199 and of S22* from CL = 0.675 > 0.500, so both are stable (|Γin| = 0.912, |Γout| = 0.848).

U = 0.557 (error band −3.84 dB to +7.07 dB), so the unilateral value is only an estimate.

Answer: K = 0.607 < 1, |Δ| = 0.696 → conditionally stable; GTU,max = 124.5 ≈ 20.95 dB.

The question is read as a short note on the insertion loss method of filter design, as used for the cavity and other filters around a low-noise amplifier (LNA) front end.

Insertion loss method: the filter is synthesised from a chosen power loss ratio:

PLR = 1/(1 − |Γ(ω)|²),   IL = 10 log PLR (dB)
Butterworth: PLR = 1 + k²(ω/ωc)^(2N)
Chebyshev:   PLR = 1 + k²·TN²(ω/ωc)

Steps:

1. Specify cut-off/band, ripple, stopband rejection and R0.
2. Choose the response and order N.
3. Take low-pass prototype values gk from tables.
4. Scale and transform (LP → BP for a receiver pre-select filter).
5. Realise with transmission lines, coupled lines, or cavity resonators (high Q), then tune.

Why it matters for an LNA: a band-pass filter is placed before or after the LNA to reject out-of-band interferers and image signals. Filter loss before the LNA adds directly to the system noise figure (NF_total ≈ IL_filter + NF_LNA in dB), so a low-loss, high-Q cavity filter is preferred at the input. The insertion loss method lets the designer choose the lowest order that meets the rejection need, keeping passband loss small.

Microwave filter parameters describe how well a filter passes wanted and rejects unwanted frequencies:

LPF prototyping (insertion loss method):

1. Choose the response: Butterworth PLR = 1 + (ω/ωc)^(2N), or Chebyshev PLR = 1 + k²TN²(ω/ωc).
2. Find the order N from the stopband attenuation.
3. Read normalised values g0 … gN+1 (R0 = 1 Ω, ωc = 1 rad/s), e.g. Butterworth N = 3: 1, 2, 1.
4. Draw the ladder (shunt C first, π type; or series L first, T type).
5. Scale: L = gR0/ωc, C = g/(R0ωc). For R0 = 50 Ω, fc = 2 GHz: C = 1.59 pF, L = 7.96 nH.
6. Realise with stepped-impedance microstrip or Richards' stubs.

  o--+---UUU---+--o
     |   L2    |
    C1        C3     N = 3, π-type
     |         |
  o--+---------+--o

An amplifier and an oscillator use the same active device, but they need opposite stability conditions: an amplifier must never oscillate, while an oscillator is deliberately made unstable at one frequency.

Stability is measured by the Rollett factor and Δ:

Δ = S11S22 − S12S21
K = (1 − |S11|² − |S22|² + |Δ|²)/(2|S12S21|)

Amplifier: if K < 1, stability circles are drawn and Γs, ΓL are chosen in the stable region; resistive loading may be added. The aim is GT with no oscillation at any frequency.

Oscillator: a transistor with K < 1 is chosen, or made so by adding series/shunt feedback (e.g. inductor in the common terminal). The terminating network ΓT is chosen inside the unstable region so that Rin < 0 and the oscillation condition holds:

Rin + RL = 0 and Xin + XL = 0
(start-up: |Rin| > RL, about 3×)

So the stability factor is a "design for stability" target in an amplifier and a "design for instability" target in an oscillator.

Δ   = S11·S22 − S12·S21
K   = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
Unconditionally stable if K > 1 and |Δ| < 1
GTU,max = 1/(1−|S11|²) · |S21|² · 1/(1−|S22|²)
GT,max  = (|S21|/|S12|)·(K − √(K² − 1))
B1 = 1 + |S11|² − |S22|² − |Δ|²,  C1 = S11 − Δ·S22*
B2 = 1 + |S22|² − |S11|² − |Δ|²,  C2 = S22 − Δ·S11*
ΓMS = [B1 − √(B1² − 4|C1|²)] / (2C1)
ΓML = [B2 − √(B2² − 4|C2|²)] / (2C2)

Assumptions: Z0 = 50 Ω; the stated S-parameters apply at the design frequency; lossless matching networks.

Stability

S11·S22 = 0.55×0.45∠(150°−30°) = 0.2475∠120°
        = −0.1238 + j0.2143
S12·S21 = 0.04×2.82∠(20°+180°) = 0.1128∠200°
        = −0.1060 − j0.0386
Δ = −0.0178 + j0.2529 = 0.2535∠94.02°
K = (1 − 0.3025 − 0.2025 + 0.0643)/(2×0.1128)
  = 0.5593/0.2256 = 2.479

K = 2.479 > 1 and |Δ| = 0.254 < 1 → unconditionally stable.

Bilateral maximum gain

GT,max = (2.82/0.04)(2.479 − √(2.479² − 1))
       = 70.5 × 0.2106 = 14.85 → 11.72 dB
B1 = 1.0357, C1 = 0.4502∠156.37° → ΓMS = 0.5819∠−156.37°
B2 = 0.8357, C2 = 0.3303∠−19.34° → ΓML = 0.4903∠19.34°
GS = 1.512 (1.80 dB), G0 = 7.952 (9.00 dB),
GL = 1.235 (0.92 dB)

Matching impedances: Zs = 13.75 − j9.70 Ω, ZL = 120.52 + j51.53 Ω.

Unilateral maximum gain

GS,max = 1/(1 − 0.55²) = 1.434 → 1.56 dB
G0     = 2.82² = 7.952          → 9.00 dB
GL,max = 1/(1 − 0.45²) = 1.254 → 0.98 dB
GTU,max = 14.30 → 11.55 dB
U = 0.050 → error −0.43 dB to +0.45 dB

The two differ by only 0.17 dB because |S12| is small (U = 0.05), so the unilateral model is acceptable here.

Answer: GT,max = 14.85 (11.72 dB) bilateral; GTU,max = 14.30 (11.55 dB) unilateral.

Flow diagram of microwave amplifier design

 +------------------------------+
 | Specs: f, gain, Z0, BW, NF   |
 +------------------------------+
               |
 +------------------------------+
 | Choose transistor, bias;     |
 | get [S] at design frequency  |
 +------------------------------+
               |
 +------------------------------+
 | Compute Δ, K (and μ)         |
 +------------------------------+
               |
      K>1 and |Δ|<1 ?
     yes |          | no
         v          v
 Uncond. stable   Draw stability circles;
         |        pick Γs, ΓL in stable
         |        region (or stabilise)
         +----+-----+
              |
   Unilateral? (U small, S12≈0)
   yes: Γs=S11*, ΓL=S22*
   no : Γs=ΓMS, ΓL=ΓML (bilateral)
              |
 +------------------------------+
 | Compute GS, G0, GL -> GT     |
 +------------------------------+
              |
 +------------------------------+
 | Design input/output matching |
 | (stubs or L-C) on Smith chart|
 +------------------------------+
              |
 | Bias network, layout, simulate
              |
   Specs met? -- no --> revise
              | yes
            Build & test

Stability from the given circles

Source (input) circle: CS = 1.15∠10°, RS = 0.85. Load (output) circle: CL = 1.10∠80°, RL = 1.10.

Source circle:

|CS| − RS = 1.15 − 0.85 = 0.30  (< 1)
|CS| + RS = 1.15 + 0.85 = 2.00  (> 1)

• The circle cuts the unit Smith chart, so part of the Γs plane is unstable → the device is conditionally (potentially) stable.
• |CS| > RS, so the chart centre (Γs = 0) is outside the circle. Assuming |S22| < 1 (normal for a transistor), Γs = 0 is stable, so the stable source region is outside the circle. The unstable part is a lens near the right edge, around angle 10°, covering Γs from about 0.30∠10° outward.

Load circle:

|CL| = RL = 1.10  →  the circle passes through Γ = 0

• The load circle passes exactly through the chart centre. On a stability circle |Γin| = 1, and at ΓL = 0, Γin = S11; so |S11| = 1. A 50 Ω load is therefore on the edge of instability.
• The circle also cuts the chart, covering a large region in the upper half around 80°. Loads inside it are unstable; loads outside it (e.g. the lower half of the chart, such as ΓL = −0.5, which is 1.285 from CL > RL) are stable, taking the outside as the stable side.

          ΓL plane (unit chart)
         .-------------.
       /   (#########)   \   # = inside load
      |   (###########)   |      circle (unstable)
      |  (#####o######)   |   o = centre, on the
      |                   |      circle boundary
       \   stable here   /
         '-------------'

Conclusion: both circles intersect the Smith chart, so the transistor is conditionally stable, not unconditionally stable. Γs must be chosen outside the source circle and ΓL outside the load circle, with a safety margin; the 50 Ω load must be avoided, or the device stabilised with resistive loading.

Microwave filters are normally designed by the insertion loss method, which gives exact control of the passband and stopband.

Procedure:

1. Specifications: filter type (LP, HP, BP, BS), cut-off or centre frequency, bandwidth, ripple, stopband attenuation, R0.
2. Choose the response from the power loss ratio PLR = 1/(1 − |Γ|²):
   • Butterworth: PLR = 1 + k²(ω/ωc)^(2N), flat passband.
   • Chebyshev: PLR = 1 + k²TN²(ω/ωc), equal ripple, sharp cut-off.
   • Linear phase: flat group delay.
3. Order N from the stopband attenuation (graphs or formula).
4. Low-pass prototype: g-values from tables for R0 = 1 Ω, ωc = 1 rad/s.
5. Scaling and transformation: L = gR0/ωc, C = g/(R0ωc); LP → HP (ω → −ωc/ω), BP and BS using ω0 and fractional bandwidth.
6. Implementation: Richards' transformation (L → short stub, C → open stub), Kuroda identities, stepped-impedance lines, coupled-line or resonator filters.
7. Simulation, tuning and measurement (S21, S11 on a network analyzer).

Specs → Response → N → g-values → Scale/
Transform → Line realisation → Simulate → Build

Example: Butterworth N = 3 LPF at 2 GHz, 50 Ω → C = 1.59 pF, L = 7.96 nH, C = 1.59 pF.

Δ   = S11·S22 − S12·S21
K   = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
Unconditionally stable if K > 1 and |Δ| < 1
GTU,max = 1/(1−|S11|²) · |S21|² · 1/(1−|S22|²)
GT,max  = (|S21|/|S12|)·(K − √(K² − 1))
B1 = 1 + |S11|² − |S22|² − |Δ|²,  C1 = S11 − Δ·S22*
B2 = 1 + |S22|² − |S11|² − |Δ|²,  C2 = S22 − Δ·S11*
ΓMS = [B1 − √(B1² − 4|C1|²)] / (2C1)
ΓML = [B2 − √(B2² − 4|C2|²)] / (2C2)

Step 1: Δ and K

S11·S22 = 0.656×0.172∠(146.7°−117.1°) = 0.1128∠29.6°
        = 0.0981 + j0.0557
S12·S21 = 0.122×2.3∠(46.1°+44.7°) = 0.2806∠90.8°
        = −0.0039 + j0.2806
Δ = 0.1020 − j0.2248 = 0.2469∠−65.59°
K = (1 − 0.4303 − 0.0296 + 0.0610)/(2×0.2806)
  = 0.6010/0.5612 = 1.071

K = 1.071 > 1 and |Δ| = 0.247 < 1 → unconditionally stable, so simultaneous conjugate matching is possible.

Step 2: Maximum transducer gain (bilateral)

GT,max = (|S21|/|S12|)(K − √(K² − 1))
       = (2.3/0.122)(1.071 − √(1.071² − 1))
       = 18.852 × 0.6876 = 12.96  → 11.13 dB

Step 3: Matching reflection coefficients

B1 = 1 + 0.4303 − 0.0296 − 0.0610 = 1.3398
C1 = S11 − ΔS22* = 0.6612∠150.37°
ΓMS = 0.8504∠−150.37°   (Zs = 4.32 − j13.13 Ω)
B2 = 1 + 0.0296 − 0.4303 − 0.0610 = 0.5383
C2 = S22 − ΔS11* = 0.2467∠−76.27°
ΓML = 0.6548∠76.27°     (ZL = 25.55 + j56.90 Ω)

(The other roots, 1.176 and 1.527 in magnitude, are outside the chart and are rejected.)

Step 4: Gain blocks (check)

GS = 1/(1 − |ΓMS|²)             = 3.613 → 5.58 dB
G0 = |S21|² = 5.29                     → 7.23 dB
GL = (1 − |ΓML|²)/|1 − S22ΓML|² = 0.678 → −1.69 dB
GT = 5.58 + 7.23 − 1.69 = 11.13 dB  ✓

Unilateral comparison

GTU,max = 1/(1−0.4303) × 5.29 × 1/(1−0.0296)
        = 1.755 × 5.29 × 1.030 = 9.57 → 9.81 dB

The unilateral formula underestimates by 1.32 dB because |S12| = 0.122 is large, so the bilateral value is the true maximum.

Answer: Gmax = GT,max = 12.96 ≈ 11.13 dB (with ΓMS = 0.850∠−150.4°, ΓML = 0.655∠76.3°).

For maximum gain the transistor must be conjugate matched at both ports at once (bilateral design): Γs = ΓMS and ΓL = ΓML, with lossless matching networks to the 50 Ω source and load.

Δ   = S11·S22 − S12·S21
K   = (1 − |S11|² − |S22|² + |Δ|²) / (2|S12·S21|)
Unconditionally stable if K > 1 and |Δ| < 1
GTU,max = 1/(1−|S11|²) · |S21|² · 1/(1−|S22|²)
GT,max  = (|S21|/|S12|)·(K − √(K² − 1))
B1 = 1 + |S11|² − |S22|² − |Δ|²,  C1 = S11 − Δ·S22*
B2 = 1 + |S22|² − |S11|² − |Δ|²,  C2 = S22 − Δ·S11*
ΓMS = [B1 − √(B1² − 4|C1|²)] / (2C1)
ΓML = [B2 − √(B2² − 4|C2|²)] / (2C2)

Step 1: Stability

S11·S22 = 0.72×0.73∠(−116°−54°) = 0.5256∠−170°
        = −0.5176 − j0.0913
S12·S21 = 0.03×2.60∠(57°+76°) = 0.0780∠133°
        = −0.0532 + j0.0570
Δ = −0.4644 − j0.1483 = 0.4875∠−162.29°
K = (1 − 0.5184 − 0.5329 + 0.2377)/(2×0.0780)
  = 0.1864/0.1560 = 1.195

K = 1.195 > 1 and |Δ| = 0.488 < 1 → unconditionally stable (μ = 1.040). Stability circles (CS = 1.320∠123.41°, RS = 0.278; CL = 1.304∠61.03°, RL = 0.264) lie completely outside the Smith chart.

Step 2: Source and load reflection coefficients

B1 = 1 + |S11|² − |S22|² − |Δ|² = 0.7478
C1 = S11 − ΔS22* = 0.3704∠−123.41°
ΓMS = [B1 − √(B1² − 4|C1|²)]/(2C1) = 0.872∠123.41°
B2 = 1 + |S22|² − |S11|² − |Δ|² = 0.7768
C2 = S22 − ΔS11* = 0.3850∠−61.03°
ΓML = [B2 − √(B2² − 4|C2|²)]/(2C2) = 0.876∠61.03°

Corresponding impedances: Zs = 4.41 + j26.76 Ω, ZL = 12.63 + j83.43 Ω. Check: Γin with ΓML = 0.872∠−123.41° = ΓMS*, so the input is conjugate matched.

Step 3: Gains

GS = 1/(1 − |ΓMS|²)             = 4.166 → 6.20 dB
G0 = |S21|² = 2.60²             = 6.76  → 8.30 dB
GL = (1 − |ΓML|²)/|1 − S22ΓML|² = 1.665 → 2.21 dB
GT,max = 6.20 + 8.30 + 2.21 = 16.71 dB
Check: (2.60/0.03)(1.195 − √(1.195²−1)) = 46.88
     → 16.71 dB  ✓

Step 4: Matching networks (shunt open stub + series 50 Ω line)

Each network: from the 50 Ω termination, a shunt open-circuited stub moves the point to the circle |Γ| = 0.872 (or 0.876), then a series 50 Ω line rotates it to the required angle.

For y = 1 + jb to lie on |Γ| = m: b = √(4m²/(1 − m²)).

Input network (Γs = 0.872∠123.4°):

b = +3.559 → stub: tan βl = 3.559 → l = 0.206λ (open)
Γ after stub = 0.872∠−150.67°
Series line rotates (toward generator = transistor):
2βd = −150.67° − 123.41° + 360° = 85.92°
d = 85.92°/720° = 0.119λ

Output network (ΓL = 0.876∠61.0°):

b = +3.638 → stub: l = 0.207λ (open)
Γ after stub = 0.876∠−151.20°
2βd = −151.20° − 61.03° + 360° = 147.77°
d = 147.77°/720° = 0.205λ

50Ω src  0.119λ line     0.205λ line  50Ω load
 o---+---=========--[FET]--=========---+---o
     |                                 |
  open stub                        open stub
   0.206λ                           0.207λ

Step 5: Bias and realisation

• DC bias (e.g. VDS ≈ 3–5 V for a GaAs FET) through λ/4 high-impedance lines or RF chokes; DC blocking capacitors at the ports.
• On a microstrip substrate, λ is replaced by the guided wavelength λg at 4 GHz to get physical lengths.
• Simulate over the band: the gain is maximum (16.7 dB) at 4.0 GHz and falls away from it because the stubs are frequency dependent.

Unilateral comparison

GTU,max = 2.076 × 6.76 × 2.141 = 30.05 (14.78 dB). U = 0.182, so the unilateral model is inaccurate; the bilateral design above is required for true maximum gain.

Answer: K = 1.195 (stable), ΓMS = 0.872∠123°, ΓML = 0.876∠61°, GT,max = 46.9 ≈ 16.7 dB; input: stub 0.206λ + line 0.119λ; output: stub 0.207λ + line 0.205λ.