RF and Microwave Transmission Lines

An immittance chart (also called a ZY Smith chart) is a Smith chart with the impedance (Z) chart and the admittance (Y) chart drawn on the same Γ-plane. The admittance chart is the impedance chart rotated by 180°. One point can therefore be read as a normalised impedance z = r + jx or as a normalised admittance y = g + jb, without rotating the point by λ/4.

             +jx (inductive)
           .-~~~~~~~-.
        .'  Z-circles  '.
  SC   /  (r = const)   \  OC
 Γ=-1 o------- 1 --------o Γ=+1
       \ Y-circles      /
        '. (g = const) .'
           '-~~~~~~~-'
             -jx (capacitive)
r-circles touch at OC (right)
g-circles touch at SC (left)

Main features

• Constant-r circles pass through the open-circuit point (right end). Constant-g circles are their mirror images and pass through the short-circuit point (left end).
• Upper half: inductive (+jx and −jb). Lower half: capacitive (−jx and +jb).
• The centre is the matched point (z = y = 1).
• Usually the Z-chart is printed in one colour (red) and the Y-chart in another (blue).

Uses

• Lumped L-section matching: a series element moves the point along a constant-r circle, and a shunt element moves it along a constant-g circle. Both can be read directly on one chart.
• Design of amplifier matching networks and stub tuners where series and shunt elements are mixed.

Given: ZL = 110 + j110 Ω, Z0 = 100 Ω, stub spacing d = 3λ/8. Short-circuited shunt stubs are assumed. Assumption: the distance from the load to the first stub is not given, so the first stub is placed at the load (d1 = 0).

Step 1: Normalise and convert to admittance

zL = ZL/Z0 = 1.1 + j1.1
yL = 1/zL = (1.1 − j1.1)/(1.1² + 1.1²)
   = 0.4545 − j0.4545

On the Smith chart, zL is at 0.1706λ WTG (wavelengths toward generator). yL is diametrically opposite, at 0.4206λ WTG.

Step 2: Check the forbidden region

For spacing d the first-stub conductance must satisfy g ≤ 1/sin²(βd).

βd = 2π(3/8) = 135°,  1/sin²135° = 2
g = 0.4545 < 2  → matching is possible

Step 3: Rotated g = 1 circle

Rotate the g = 1 circle by 3λ/8 toward the load (270° anticlockwise). The first-stub point must lie on this rotated circle. Moving along the constant g = 0.4545 circle from yL, it meets the rotated circle at two points. Analytically (t = tan βd = −1):

b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.9091 − 0.2066 = 0.7025
√0.7025 = 0.8381
Solution A: b1 = 0.4545 − 0.1619 = +0.2927
Solution B: b1 = 0.4545 − 1.8381 = −1.3836

Step 4: Admittance after stub 1 and after moving 3λ/8

A: y1' = 0.4545 − j0.1619  (0.468λ WTG)
   move 3λ/8 → y2 = 1 − j0.8439 (0.343λ WTG)
B: y1' = 0.4545 − j1.8381  (0.326λ WTG)
   move 3λ/8 → y2 = 1 + j2.8439 (0.201λ WTG)

Both points lie on the g = 1 circle, as required.

Step 5: Second stub cancels the susceptance

A: b2 = +0.8439      B: b2 = −2.8439

Step 6: Stub lengths (short-circuited stub: y = −j cot βl)

Measure from the short-circuit point (y = ∞, 0λ on the WTG scale at the right end) toward the generator up to the required jb:

l = (1/2π)·cot⁻¹(−b), taken in 0 to λ/2

Solution A: b1 = +0.2927 → l1 = 0.2953λ
            b2 = +0.8439 → l2 = 0.3616λ
Solution B: b1 = −1.3836 → l1 = 0.0996λ
            b2 = −2.8439 → l2 = 0.0538λ

Result

Solution B uses shorter stubs, which gives less loss and wider bandwidth, so it is preferred.

 Z0=100Ω          3λ/8          ZL
 o----------+-------------+------[110+j110]
            |             |
         stub 2        stub 1
        0.0538λ        0.0996λ
          SC             SC

Check: with stub 1 = 0.0996λ, spacing 0.375λ and stub 2 = 0.0538λ, the input admittance at stub 2 is 1.000 + j0.000, so the load is perfectly matched.

Answer: first stub at the load, spacing 3λ/8 = 0.375λ, l1 = 0.0996λ, l2 = 0.0538λ (shortest set). The alternative is l1 = 0.2953λ, l2 = 0.3616λ.

Given: Z0 = 50 Ω, ZL = 75 + j100 Ω. The values below were computed exactly; Smith chart readings agree to about ±0.005λ.

zL = ZL/Z0 = (75 + j100)/50 = 1.5 + j2.0

Plot zL at the intersection of the r = 1.5 circle and the x = 2.0 arc. It reads 0.198λ on the WTG scale.

(a) Reflection coefficient ΓL

ΓL = (zL − 1)/(zL + 1) = (0.5 + j2)/(2.5 + j2)
   = 0.644 ∠ 37.3°

On the chart: |Γ| = OP/OR (distance from the centre to the point, divided by the chart radius) = 0.644. The angle is read on the "angle of reflection coefficient" scale: 37.3°.

(b) VSWR

VSWR = (1 + |Γ|)/(1 − |Γ|) = 1.644/0.356 = 4.62

On the chart: draw the constant-|Γ| circle through zL. It cuts the right-hand real axis at r = 4.62.

(c) Zin at 0.375λ from the load

Move 0.375λ toward the generator (clockwise) on the |Γ| circle:

0.198λ + 0.375λ = 0.573λ → 0.073λ WTG
zin = (zL + j tan βl)/(1 + j zL tan βl),  βl = 270°
    = 0.267 + j0.467
Zin = 50 × zin = 13.33 + j23.33 Ω

(d) Shortest length for a purely resistive impedance

Moving clockwise from 0.198λ, the |Γ| circle first meets the real axis at the voltage maximum (right side, 0.25λ):

dmax = 0.25λ − 0.198λ = 0.0518λ
(= θΓ/(4π) = 37.3°/720° × λ)

The next resistive point is the voltage minimum at dmin = 0.0518λ + 0.25λ = 0.3018λ.

(e) Value of this resistance

At dmax: R = Z0 × VSWR = 50 × 4.617 = 230.8 Ω
(at dmin: R = Z0/VSWR = 10.83 Ω)

Answer: ΓL = 0.644∠37.3°, VSWR = 4.62, Zin(0.375λ) = 13.33 + j23.33 Ω, shortest length = 0.0518λ, R = 230.8 Ω.

Self-defined data: Z0 = 50 Ω, ZL = 100 + j80 Ω, single shunt short-circuited stub with the same Z0 = 50 Ω. Aim: find the stub position d (from the load) and length l so that the line sees Z0 (Γ = 0).

Principle

A shunt stub adds susceptance, so the work is done with admittances. Move from the load toward the generator until the real part of the line admittance is y = 1 + jb. At that point, connect a stub of susceptance −jb. The total is then y = 1.

Step 1: Normalise and plot

zL = (100 + j80)/50 = 2 + j1.6
|ΓL| = 0.555, VSWR = 3.49

Step 2: Convert to admittance

Move zL by 180° (λ/4) on the constant-VSWR circle:

yL = 1/zL = 0.305 − j0.244  (0.4584λ WTG)

Step 3: Move toward the generator to the g = 1 circle

The VSWR circle cuts the g = 1 circle at two points:

y1 = 1 + j1.334  at 0.1718λ WTG
y2 = 1 − j1.334  at 0.3282λ WTG
d1 = 0.5 − 0.4584 + 0.1718 = 0.2134λ
d2 = 0.5 − 0.4584 + 0.3282 = 0.3697λ

Step 4: Required stub susceptance

Solution 1: bstub = −1.334
Solution 2: bstub = +1.334

Step 5: Stub length (short circuit at y = ∞, 0λ WTG)

For a short-circuited stub, y = −j cot βl. Start at the short-circuit point and move toward the generator to the required susceptance:

Sol. 1: −cot βl = −1.334 → l = 0.1024λ
Sol. 2: −cot βl = +1.334 → l = 0.3976λ

Result

Solution 1 is preferred because it uses the shorter line and the shorter stub, giving lower loss and wider bandwidth.

 Z0=50Ω           d = 0.2134λ
 o------------+----------------[ZL=100+j80]
              |
             stub l = 0.1024λ
              |
            short

Physical length example: at 2 GHz with an air line (λ = 15 cm), d = 3.20 cm and l = 1.54 cm.

Check: y at d1 = 1 + j1.334. The stub adds −j1.334, so y = 1 and the line is matched.

Impedance matching makes the load look like Z0 to the line, so that Γ = 0. This gives maximum power transfer, no standing waves, a better signal-to-noise ratio, and protection for the source.

Impedance matching techniques

1. Lumped L-section (L-network): one series and one shunt reactance (L or C). It is simple, but works only up to a few GHz because real L and C have parasitics.
2. Quarter-wave transformer: a λ/4 line with Z1 = √(Z0·RL) matches a real load. It is narrowband. Multisection (binomial or Chebyshev) versions give wider bandwidth.
3. Single-stub tuning: one shunt (or series) short or open stub at a distance d from the load. Both d and l must be adjustable.
4. Double-stub tuning: two shunt stubs at fixed positions (spacing usually λ/8, 3λ/8 or λ/4). Only the stub lengths are adjusted, which suits adjustable coaxial tuners. Some loads cannot be matched (forbidden region g > 1/sin²βd).
5. Triple-stub tuner: removes the forbidden region.
6. Tapered lines (exponential, triangular, Klopfenstein): very wideband.
7. Slide-screw tuner, E-H tuner: used in waveguides.

L-section       λ/4 transformer      single stub
-[jX]-+-        -[ Z1 , λ/4 ]-       --+--d--[ZL]
      |                                |
     jB                                l (stub)

S-matrix of a perfectly matched double-stub network

For a two-port with both ports referenced to Z0, the S-parameters follow from the normalised ABCD matrix:

S11 = (A + B − C − D)/Δ     S12 = 2(AD − BC)/Δ
S21 = 2/Δ                  S22 = (−A + B − C + D)/Δ
Δ = A + B + C + D

A double-stub tuner is a cascade of shunt stub 2, a line of length d, and shunt stub 1:

[ABCD] = |1    0| |cos βd   j sin βd| |1    0|
         |jb2  1| |j sin βd  cos βd | |jb1  1|

Building blocks:

Shunt jb:  S = 1/(2+jb) · | −jb   2  |
                          |  2   −jb |
Line βd:   S = |   0      e^(−jβd) |
               | e^(−jβd)    0     |

Condition for a perfect match: with the load ΓL on port 2,

Γin = S11 + S12 S21 ΓL/(1 − S22 ΓL) = 0

For a lossless, reciprocal tuner this holds when S22 = ΓL* (conjugate match at the load side). Unitarity then gives |S21|² = 1 − |ΓL|², and S12 = S21.

If port 2 is referenced to the load itself (the reference impedance is ZL), the matched network becomes an ideal matched, lossless, reciprocal two-port:

S = |    0      e^(−jφ) |
    | e^(−jφ)      0    |

Here φ is the total phase delay. S11 = S22 = 0 (no reflection) and |S21| = 1 (all power reaches the load).

Numerical example (ZL = 110 + j110 Ω on a 100 Ω line, first stub at the load, d = 3λ/8, b1 = −1.384, b2 = −2.844, stub lengths 0.0996λ and 0.0538λ):

S11 = 0.466∠145.5°   S12 = 0.885∠−45.8°
S21 = 0.885∠−45.8°   S22 = 0.466∠−57.2°
ΓL  = 0.466∠+57.2°  → S22 = ΓL*  ✓
|S11|² + |S21|² = 1 (lossless) ✓
Γin = 0  (VSWR = 1) ✓

Before matching, the line sees S11 = ΓL = 0.466∠57.2° (VSWR = 2.75). After matching, S11 at the input = 0.

Sketch: double-stub matching network in microstrip

Two shunt stubs are etched in the same microstrip as the main line. They are spaced d = 3λ/8 apart (λ is the guided wavelength in the microstrip). A shorted stub ends in a via hole to the ground plane, and an open stub simply ends.

Top view (microstrip, ground plane below)

   stub 2 (l2)        stub 1 (l1)
      ||                 ||
      ||                 ||
      ●via (short)       ●via (short)
      ||                 ||
=====##======== d =======##=====[ ZL ]
port 1 (Z0)    3λ/8      d1 to load

Electrical equivalent
 Z0 o---+-------- d --------+-- d1 --[ZL]
        |                   |
      jb2 (stub 2)        jb1 (stub 1)

Preparing its S-matrix

For a two-port with both ports referenced to Z0, the S-parameters follow from the normalised ABCD matrix:

S11 = (A + B − C − D)/Δ     S12 = 2(AD − BC)/Δ
S21 = 2/Δ                  S22 = (−A + B − C + D)/Δ
Δ = A + B + C + D

A double-stub tuner is a cascade of shunt stub 2, a line of length d, and shunt stub 1:

[ABCD] = |1    0| |cos βd   j sin βd| |1    0|
         |jb2  1| |j sin βd  cos βd | |jb1  1|

Building blocks:

Shunt jb:  S = 1/(2+jb) · | −jb   2  |
                          |  2   −jb |
Line βd:   S = |   0      e^(−jβd) |
               | e^(−jβd)    0     |

Condition for a perfect match: with the load ΓL on port 2,

Γin = S11 + S12 S21 ΓL/(1 − S22 ΓL) = 0

For a lossless, reciprocal tuner this holds when S22 = ΓL* (conjugate match at the load side). Unitarity then gives |S21|² = 1 − |ΓL|², and S12 = S21.

If port 2 is referenced to the load itself (the reference impedance is ZL), the matched network becomes an ideal matched, lossless, reciprocal two-port:

S = |    0      e^(−jφ) |
    | e^(−jφ)      0    |

Here φ is the total phase delay. S11 = S22 = 0 (no reflection) and |S21| = 1 (all power reaches the load).

Numerical example (ZL = 110 + j110 Ω on a 100 Ω line, first stub at the load, d = 3λ/8, b1 = −1.384, b2 = −2.844, stub lengths 0.0996λ and 0.0538λ):

S11 = 0.466∠145.5°   S12 = 0.885∠−45.8°
S21 = 0.885∠−45.8°   S22 = 0.466∠−57.2°
ΓL  = 0.466∠+57.2°  → S22 = ΓL*  ✓
|S11|² + |S21|² = 1 (lossless) ✓
Γin = 0  (VSWR = 1) ✓

Before matching, the line sees S11 = ΓL = 0.466∠57.2° (VSWR = 2.75). After matching, S11 at the input = 0.

Assumptions: the antenna is a half-wave dipole with ZL = 73 + j42.5 Ω. It is fed by a lossless Z0 = 50 Ω line at f = 1 GHz (air line, λ = 30 cm). A single shunt stub is used. Both short and open stubs are given.

A single-stub tuner places a shunt stub at a distance d from the load, where the line admittance is y = 1 + jb. The stub supplies −jb, so the total admittance becomes 1 (matched).

Step 1: Normalise and plot the load

zL = ZL/Z0 = 1.460 + j0.850
ΓL = (zL − 1)/(zL + 1) = 0.371∠42.5°
VSWR = (1 + |Γ|)/(1 − |Γ|) = 2.181

Draw the constant-VSWR circle through zL (zL is at 0.1909λ WTG).

Step 2: Convert to admittance (shunt stub)

Move zL by 180° on the VSWR circle, which is the same as moving λ/4:

yL = 1/zL = 0.512 − j0.298   (0.4409λ WTG)

Step 3: Move toward the generator to the g = 1 circle

The VSWR circle cuts the g = 1 circle at two points:

y1 = 1.000 + j0.800  at 0.1553λ WTG
y2 = 1.000 − j0.800  at 0.3447λ WTG
d1 = 0.2143λ from the load
d2 = 0.4038λ from the load

Step 4: Stub susceptance and length

The stub must cancel the susceptance: bstub = −b.

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.
• For an open-circuited stub: y = j tan βl, so l = (1/2π)·tan⁻¹(b); measure from the open-circuit point (y = 0, left end of the chart) toward the generator.

Sol 1: bstub = −0.7999
   short stub l = 0.1426λ
   open stub  l = 0.3926λ
Sol 2: bstub = +0.7999
   short stub l = 0.3574λ
   open stub  l = 0.1074λ

Physical lengths (λ = 30.000 cm):

Sol 1: d = 6.430 cm, short l = 4.279 cm, open l = 11.779 cm
Sol 2: d = 12.113 cm, short l = 10.721 cm, open l = 3.221 cm

Answer (choice): solution 1 with a short-circuited stub (d = 0.2143λ, l = 0.1426λ) uses the shortest lengths. It has the lowest loss and the widest bandwidth.

 Z0=50Ω        d = 0.2143λ (6.43 cm)
 o----------+------------------[Dipole 73+j42.5Ω]
            |
          stub l = 0.1426λ (4.28 cm)
            |
          short

Check: y(d) = 1 + j0.800. The stub adds −j0.800, so y = 1 and Γ = 0.

S-matrix

Treat the stub and the line section d as a two-port: port 1 faces the main line, and port 2 faces the load. Both ports are referenced to Z0. Use the normalised ABCD matrix [1 0; jb 1]·[cos βd j sin βd; j sin βd cos βd] and convert with S11 = (A + B − C − D)/Δ, S21 = 2/Δ, S22 = (−A + B − C + D)/Δ, Δ = A + B + C + D.

Before matching (line sees the load directly):
  S11 = ΓL = 0.371∠42.5°,  VSWR = 2.18
Matching network (solution 1):
  S11 = 0.371∠111.8°   S12 = 0.928∠-55.4°
  S21 = 0.928∠-55.4°   S22 = 0.371∠-42.5°
  S22 = ΓL* (conjugate match), |S11|²+|S21|² = 1
After matching (network + load):
  Γin = S11 + S12S21ΓL/(1 − S22ΓL) = 0, VSWR = 1

With port 2 referenced to the load, the matched lossless network has the ideal form S = [0 e^(−jφ); e^(−jφ) 0]: no reflection at either port, and all power is transferred.

Given: Z0 = 75 Ω coaxial line, normalised load zL = 0.4 + j0.85 (ZL = 30 + j63.75 Ω). The stubs are short-circuited 75 Ω coaxial lines. Assumptions (not given): the first stub is at the load, and the stub spacing is d = 3λ/8.

Step 1: Normalise and convert to admittance

zL = 0.400 + j0.850
yL = 1/zL = 0.453 − j0.963
ΓL = 0.635∠94.0°,  VSWR = 4.483

On the chart, zL is at 0.1195λ WTG and yL is diametrically opposite, at 0.3695λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.453 − j0.963.

Step 3: Forbidden-region check and rotated g = 1 circle

βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.4533  → matchable

Rotate the g = 1 circle by 0.375λ (270° on the chart) toward the load. The point just after stub 1 must lie on this rotated circle, so move from y1 along the constant g = 0.453 circle until it meets the rotated circle.

Step 4: Susceptance of stub 1

Analytically (Pozar), with t = tan βd:

t = tan 135° = −1.0000
b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.7011, √ = 0.8373
Sol A: b1 = −0.8741
Sol B: b1 = +0.8005

Step 5: Move 0.375λ to stub 2 and find b2

Sol A: y after stub 1 = 0.453 − j1.837
   (0.3263λ) → after 0.375λ: y2 = 1.000 + j2.847
   b2 = −2.8473
Sol B: y after stub 1 = 0.453 − j0.163
   (0.4681λ) → after 0.375λ: y2 = 1.000 − j0.847
   b2 = +0.8473

Both y2 values lie on the g = 1 circle, as required.

Step 6: Stub lengths

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.

Sol A short: l1 = 0.1357λ, l2 = 0.0538λ
Sol B short: l1 = 0.3574λ, l2 = 0.3619λ

Answer (choice): solution A (l1 = 0.1357λ, l2 = 0.0538λ) has the shorter stubs, so it is preferred.

Check: the input admittance at stub 2 is 1.000 + j0.000, so Γ = 0 and VSWR = 1.

Sketch using microstrip

In microstrip, each "short-circuited" stub is a strip that ends in a via hole to the ground plane. The coax line is replaced by a 75 Ω microstrip trace.

Top view (ground plane underneath)

     stub 2           stub 1
   l2=0.0538λ       l1=0.1357λ
       ||               ||
       ● via            ● via
       ||               ||
=======##====== 3λ/8 ===##[ZL]
in (75 Ω microstrip)       load

An immittance chart (ZY Smith chart) is a combined chart that shows both immittance forms, impedance and admittance, on the same reflection-coefficient (Γ) plane. It overlays the normal impedance Smith chart (constant r and x circles) and the admittance Smith chart (constant g and b circles). The admittance chart is the impedance chart rotated by 180°. Any point can therefore be read as z = r + jx or as y = g + jb directly, without the λ/4 rotation.

Sketch

               inductive (+jx, −jb)
              .----~~~~~~~~----.
           .'   r-circles  ↘    '.
          /   (touch at OC) ↘     \
   SC  o(----------( 1 )----------)o  OC
 y = ∞    \  g-circles ↗          /  z = ∞
 z = 0      '. (touch at SC)   .'    y = 0
              '----~~~~~~~~----'
               capacitive (−jx, +jb)

Properties

• Constant-r circles all touch at the open-circuit point (right). Constant-g circles all touch at the short-circuit point (left).
• Constant-x arcs leave the right point, and constant-b arcs leave the left point.
• The centre (z = y = 1) is the matched point.
• Upper half: inductive in both readings (+jx, −jb). Lower half: capacitive.
• A series element moves the point along a constant-r circle. A shunt element moves it along a constant-g circle.
• It is used for lumped L, T and π matching networks and for amplifier input and output matching.

Duality parameters used in microwave network design

In each pair, swapping one quantity for the other turns a valid circuit relation into another valid one.

Why duality matters: a design for a shunt (admittance) network can be reused for a series (impedance) network by swapping the dual quantities. On the immittance chart this swap is only a change of scale (Z or Y), not a recalculation.

Given: ΓL = 0.5∠51° (the "Ohm" in the question is a slip; Γ has no unit), Z0 = 50 Ω. Design a shunt single-stub match with (a) a short-circuited and (b) an open-circuited stub.

Step 0: Find the load from Γ

On the Smith chart, plot the point at radius 0.5 and angle 51°. It reads:

zL = (1 + ΓL)/(1 − ΓL) = 1.208 + j1.252
ZL = 50 × zL = 60.42 + j62.60 Ω

Step 1: Normalise and plot the load

zL = ZL/Z0 = 1.208 + j1.252
ΓL = (zL − 1)/(zL + 1) = 0.500∠51.0°
VSWR = (1 + |Γ|)/(1 − |Γ|) = 3.000

Draw the constant-VSWR circle through zL (zL is at 0.1792λ WTG).

Step 2: Convert to admittance (shunt stub)

Move zL by 180° on the VSWR circle, which is the same as moving λ/4:

yL = 1/zL = 0.399 − j0.414   (0.4292λ WTG)

Step 3: Move toward the generator to the g = 1 circle

The VSWR circle cuts the g = 1 circle at two points:

y1 = 1.000 + j1.155  at 0.1667λ WTG
y2 = 1.000 − j1.155  at 0.3333λ WTG
d1 = 0.2375λ from the load
d2 = 0.4042λ from the load

Step 4: Stub susceptance and length

The stub must cancel the susceptance: bstub = −b.

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.
• For an open-circuited stub: y = j tan βl, so l = (1/2π)·tan⁻¹(b); measure from the open-circuit point (y = 0, left end of the chart) toward the generator.

Sol 1: bstub = −1.1547
   short stub l = 0.1136λ
   open stub  l = 0.3636λ
Sol 2: bstub = +1.1547
   short stub l = 0.3864λ
   open stub  l = 0.1364λ

Answer (choice): solution 1 with the short stub (d = 0.2375λ, l = 0.1136λ). With an open stub, solution 2 (d = 0.4042λ, l = 0.1364λ) is the shorter choice.

(a) Short stub            (b) Open stub
 Z0  d=0.2375λ             Z0  d=0.4042λ
 o---+--------[ZL]         o---+--------[ZL]
     |                         |
    l=0.1136λ                 l=0.1364λ
     |                         |
   short                      open

Check: y(d) = 1 ± j1.155. The stub supplies ∓j1.155, so y = 1, Γ = 0 and VSWR = 1 (it was 3.0 before matching).

Advantages of double-stub over single-stub matching

1. The stub positions are fixed. Only the two stub lengths are varied, while a single stub needs both its position d and its length l to change.
2. It suits adjustable tuners (sliding shorts in coax or waveguide) for loads that vary. A single stub would need a movable stub along a slotted line, which is impractical.
3. It is easier to build in fixed microstrip or coax with a fixed junction. (Disadvantage: some loads cannot be matched, the forbidden region g > 1/sin²βd. A triple stub removes this.)

Steps for double-stub matching

1. Normalise ZL, plot zL, and convert it to yL (rotate 180°).
2. If the first stub is at d1 from the load, rotate yL toward the generator by d1 to get y1.
3. Draw the g = 1 circle rotated toward the load by the stub spacing d (chart angle 720°·d/λ).
4. From y1, move along its constant-g circle to the rotated g = 1 circle. This gives y1′ = g1 + jb′ and b1 = b′ − b(y1).
5. Rotate y1′ toward the generator by d. It lands on the g = 1 circle: y2 = 1 + jb2′.
6. Stub 2 must supply b2 = −b2′.
7. Convert b1 and b2 into stub lengths (from the short-circuit point for shorted stubs, or the open-circuit point for open stubs).
8. Check that y = 1 at stub 2.

Example

ZL = 60 − j80 Ω, Z0 = 50 Ω, short-circuited stubs, first stub at the load, spacing d = λ/8.

Step 1: Normalise and convert to admittance

zL = 1.200 − j1.600
yL = 1/zL = 0.300 + j0.400
ΓL = 0.593∠-46.8°,  VSWR = 3.911

On the chart, zL is at 0.3151λ WTG and yL is diametrically opposite, at 0.0651λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.300 + j0.400.

Step 3: Forbidden-region check and rotated g = 1 circle

βd = 360° × 0.125 = 45°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.3000  → matchable

Rotate the g = 1 circle by 0.125λ (90° on the chart) toward the load. The point just after stub 1 must lie on this rotated circle, so move from y1 along the constant g = 0.300 circle until it meets the rotated circle.

Step 4: Susceptance of stub 1

Analytically (Pozar), with t = tan βd:

t = tan 45° = +1.0000
b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.5100, √ = 0.7141
Sol A: b1 = +1.3141
Sol B: b1 = −0.1141

Step 5: Move 0.125λ to stub 2 and find b2

Sol A: y after stub 1 = 0.300 + j1.714
   (0.1675λ) → after 0.125λ: y2 = 1.000 − j3.380
   b2 = +3.3805
Sol B: y after stub 1 = 0.300 + j0.286
   (0.0481λ) → after 0.125λ: y2 = 1.000 + j1.380
   b2 = −1.3805

Both y2 values lie on the g = 1 circle, as required.

Step 6: Stub lengths

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.

Sol A short: l1 = 0.3965λ, l2 = 0.4542λ
Sol B short: l1 = 0.2319λ, l2 = 0.0998λ

Answer: solution B (l1 = 0.2319λ, l2 = 0.0998λ) gives the shorter total stub length.

 Z0=50Ω        λ/8
 o------+-------------+-------[60−j80 Ω]
        |             |
     stub 2        stub 1
     0.0998λ       0.2319λ
       SC            SC

Assumptions: an inductive antenna load ZL = 117 + j78 Ω on a Z0 = 78 Ω line at f = 1 GHz (air-filled coax, λ = 30 cm). Short-circuited 78 Ω stubs, the first stub at the load, and spacing d = 3λ/8.

Step 1: Normalise and convert to admittance

zL = 1.500 + j1.000
yL = 1/zL = 0.462 − j0.308
ΓL = 0.415∠41.6°,  VSWR = 2.420

On the chart, zL is at 0.1922λ WTG and yL is diametrically opposite, at 0.4422λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.462 − j0.308.

Step 3: Forbidden-region check and rotated g = 1 circle

βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.4615  → matchable

Rotate the g = 1 circle by 0.375λ (270° on the chart) toward the load. The point just after stub 1 must lie on this rotated circle, so move from y1 along the constant g = 0.462 circle until it meets the rotated circle.

Step 4: Susceptance of stub 1

Analytically (Pozar), with t = tan βd:

t = tan 135° = −1.0000
b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.7101, √ = 0.8427
Sol A: b1 = −1.5350
Sol B: b1 = +0.1503

Step 5: Move 0.375λ to stub 2 and find b2

Sol A: y after stub 1 = 0.462 − j1.843
   (0.3260λ) → after 0.375λ: y2 = 1.000 + j2.826
   b2 = −2.8257
Sol B: y after stub 1 = 0.462 − j0.157
   (0.4688λ) → after 0.375λ: y2 = 1.000 − j0.826
   b2 = +0.8257

Both y2 values lie on the g = 1 circle, as required.

Step 6: Stub lengths

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.

Sol A short: l1 = 0.0919λ, l2 = 0.0541λ
Sol B short: l1 = 0.2737λ, l2 = 0.3599λ

Physical lengths (λ = 30.000 cm):

d1 = 0.000 cm, spacing = 11.250 cm
Sol A short: l1 = 2.757 cm, l2 = 1.624 cm
Sol B short: l1 = 8.212 cm, l2 = 10.796 cm

Answer (choice): solution A (l1 = 0.0919λ = 2.76 cm, l2 = 0.0541λ = 1.62 cm). These are the shortest stubs, so the design has the lowest loss and the widest bandwidth.

Check: y at stub 2 = 1.000 + j0.000, so the line is matched.

Electrical connection

 Z0=78Ω          3λ/8 = 11.25 cm          
 o------+---------------+--------[117+j78Ω]
        |               |
     stub 2          stub 1
     l2=1.62 cm       l1=2.76 cm
       SC              SC

Physical connection

Two coaxial T-junctions are placed 11.25 cm apart on the main 78 Ω coax. The first is at the antenna terminals. Each side arm is a 78 Ω coaxial stub closed by an adjustable sliding short, set to l1 = 2.76 cm and l2 = 1.62 cm.

Physical connection (coaxial tuner)

        sliding short   sliding short
             ┃               ┃
  stub 2  ═══╋═══   stub 1 ══╋══
             ║               ║
 gen ════════╩═══════════════╩══════ antenna
      main coax     d            (load)

Assumptions: inductive load ZL = 50 + j100 Ω on a Z0 = 50 Ω line. Double shunt stubs, the first at the load, spacing d = λ/8. Both short- and open-circuited stubs are given.

Step 1: Normalise and convert to admittance

zL = 1.000 + j2.000
yL = 1/zL = 0.200 − j0.400
ΓL = 0.707∠45.0°,  VSWR = 5.828

On the chart, zL is at 0.1875λ WTG and yL is diametrically opposite, at 0.4375λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.200 − j0.400.

Step 3: Forbidden-region check and rotated g = 1 circle

βd = 360° × 0.125 = 45°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.2000  → matchable

Rotate the g = 1 circle by 0.125λ (90° on the chart) toward the load. The point just after stub 1 must lie on this rotated circle, so move from y1 along the constant g = 0.200 circle until it meets the rotated circle.

Step 4: Susceptance of stub 1

Analytically (Pozar), with t = tan βd:

t = tan 45° = +1.0000
b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.3600, √ = 0.6000
Sol A: b1 = +2.0000
Sol B: b1 = +0.8000

Step 5: Move 0.125λ to stub 2 and find b2

Sol A: y after stub 1 = 0.200 + j1.600
   (0.1619λ) → after 0.125λ: y2 = 1.000 − j4.000
   b2 = +4.0000
Sol B: y after stub 1 = 0.200 + j0.400
   (0.0625λ) → after 0.125λ: y2 = 1.000 + j2.000
   b2 = −2.0000

Both y2 values lie on the g = 1 circle, as required.

Step 6: Stub lengths

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.
• For an open-circuited stub: y = j tan βl, so l = (1/2π)·tan⁻¹(b); measure from the open-circuit point (y = 0, left end of the chart) toward the generator.

Sol A short: l1 = 0.4262λ, l2 = 0.4610λ
Sol A open : l1 = 0.1762λ, l2 = 0.2110λ
Sol B short: l1 = 0.3574λ, l2 = 0.0738λ
Sol B open : l1 = 0.1074λ, l2 = 0.3238λ

Answer (choice): with short stubs, solution B (l1 = 0.3574λ, l2 = 0.0738λ). With open stubs, solution B (l1 = 0.1074λ, l2 = 0.3238λ) or solution A (l1 = 0.1762λ, l2 = 0.2110λ), which are similar in total length.

Check: y at stub 2 = 1 + j0, so all standing waves on the main line are removed (VSWR goes from 5.83 to 1).

Electrical diagram

 Z0=50Ω          λ/8          
 o------+---------------+--------[50+j100Ω]
        |               |
     stub 2          stub 1
     l2=0.0738λ       l1=0.3574λ
       SC              SC

Physical connection (microstrip)

   stub 2 (l2)       stub 1 (l1)
      ||                ||
      ● via (short)     ○ open end
      ||                ||
======##===== λ/8 ======##====[ZL]
 in (50 Ω)                  load

A short stub ends in a via to ground. An open stub just ends (a small end-correction length is subtracted for fringing).

Admittance chart

An admittance chart is the Smith chart read in normalised admittance y = g + jb instead of z. It is the impedance chart rotated by 180°. Constant-g circles touch at the short-circuit point, and the upper half then means inductive susceptance (−jb). It is used for shunt elements (stubs, shunt L or C), because shunt admittances simply add. A point on the Z-chart becomes its admittance when moved λ/4 (180°) around the VSWR circle.

Single-stub design

Assumption: the microstrip line has Z0 = 50 Ω (not stated). ZL = 80 + j100 Ω. Shunt stub, both short and open designs.

Step 1: Normalise and plot the load

zL = ZL/Z0 = 1.600 + j2.000
ΓL = (zL − 1)/(zL + 1) = 0.637∠35.7°
VSWR = (1 + |Γ|)/(1 − |Γ|) = 4.503

Draw the constant-VSWR circle through zL (zL is at 0.2004λ WTG).

Step 2: Convert to admittance (shunt stub)

Move zL by 180° on the VSWR circle, which is the same as moving λ/4:

yL = 1/zL = 0.244 − j0.305   (0.4504λ WTG)

Step 3: Move toward the generator to the g = 1 circle

The VSWR circle cuts the g = 1 circle at two points:

y1 = 1.000 + j1.651  at 0.1799λ WTG
y2 = 1.000 − j1.651  at 0.3201λ WTG
d1 = 0.2295λ from the load
d2 = 0.3697λ from the load

Step 4: Stub susceptance and length

The stub must cancel the susceptance: bstub = −b.

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.
• For an open-circuited stub: y = j tan βl, so l = (1/2π)·tan⁻¹(b); measure from the open-circuit point (y = 0, left end of the chart) toward the generator.

Sol 1: bstub = −1.6508
   short stub l = 0.0867λ
   open stub  l = 0.3367λ
Sol 2: bstub = +1.6508
   short stub l = 0.4133λ
   open stub  l = 0.1633λ

Answer (choice): short stub, solution 1 (d = 0.2295λ, l = 0.0867λ). Open stub, solution 2 (d = 0.3697λ, l = 0.1633λ).

 Microstrip:
          stub (l)
            ||
            ● via (short) / open end
            ||
 ===========##==== d ====[80+j100 Ω]

Check: y(d) = 1 ± j1.651. The stub cancels ±j1.651, so the line is matched (VSWR goes from 4.50 to 1).

An immittance (ZY) chart overlays the impedance Smith chart and the admittance Smith chart (the Z-chart rotated by 180°) on one Γ-plane. It is usually printed in two colours.

            +jx / −jb (inductive)
          .-~~~~~~~~~~~~-.
        /  Z: r-circles   \
 SC  ( g-circles  1   →    ) OC
 z=0   \ ←touch SC        /  z=∞
 y=∞    '-~~~~~~~~~~~~-'    y=0
            −jx / +jb (capacitive)

Comparison of scales

The outer wavelength scales (WTG/WTL) and the radial scales (|Γ|, VSWR, return loss) are common to both.

Given: Z0 = 75 Ω, ZL = 78.27 + j60.93 Ω. Design (a) a shunt short-circuited stub and (b) a shunt open-circuited stub.

Step 1: Normalise and plot the load

zL = ZL/Z0 = 1.044 + j0.812
ΓL = (zL − 1)/(zL + 1) = 0.370∠65.2°
VSWR = (1 + |Γ|)/(1 − |Γ|) = 2.174

Draw the constant-VSWR circle through zL (zL is at 0.1594λ WTG).

Step 2: Convert to admittance (shunt stub)

Move zL by 180° on the VSWR circle, which is the same as moving λ/4:

yL = 1/zL = 0.597 − j0.464   (0.4094λ WTG)

Step 3: Move toward the generator to the g = 1 circle

The VSWR circle cuts the g = 1 circle at two points:

y1 = 1.000 + j0.796  at 0.1552λ WTG
y2 = 1.000 − j0.796  at 0.3448λ WTG
d1 = 0.2458λ from the load
d2 = 0.4355λ from the load

Step 4: Stub susceptance and length

The stub must cancel the susceptance: bstub = −b.

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.
• For an open-circuited stub: y = j tan βl, so l = (1/2π)·tan⁻¹(b); measure from the open-circuit point (y = 0, left end of the chart) toward the generator.

Sol 1: bstub = −0.7964
   short stub l = 0.1430λ
   open stub  l = 0.3930λ
Sol 2: bstub = +0.7964
   short stub l = 0.3570λ
   open stub  l = 0.1070λ

(a) Shunt short-circuited stub design

Solution 1 is preferred: stub at d = 0.2458λ from the load, with a short-circuited stub of length l = 0.1430λ.

(b) Open-circuited stub design

Solution 2 is preferred: stub at d = 0.4355λ, with an open stub of length l = 0.1070λ. (Solution 1 with an open stub would need l = 0.3930λ.)

Check: y(d) = 1 ± j0.796. The stub supplies ∓j0.796, so y = 1 and VSWR = 1 (it was 2.17 before matching).

Physical diagram (microstrip)

(a) short stub              (b) open stub
     ||  l=0.1430λ               ||  l=0.1070λ
     ● via to ground             ○ open end
     ||                          ||
=====##== d=0.2458λ ==[ZL]  =====##== d=0.4355λ ==[ZL]
 75 Ω microstrip             75 Ω microstrip

The stub is a 75 Ω strip of the same width as the main line, joined at a T-junction.

Assumptions: load ZL = 100 + j100 Ω on a Z0 = 50 Ω line. Short-circuited shunt stubs, the first stub at the load, and stub spacing d = λ/8.

Step 1: Normalise and convert to admittance

zL = 2.000 + j2.000
yL = 1/zL = 0.250 − j0.250
ΓL = 0.620∠29.7°,  VSWR = 4.266

On the chart, zL is at 0.2087λ WTG and yL is diametrically opposite, at 0.4587λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.250 − j0.250.

Step 3: Forbidden-region check and rotated g = 1 circle

βd = 360° × 0.125 = 45°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.2500  → matchable

Rotate the g = 1 circle by 0.125λ (90° on the chart) toward the load. The point just after stub 1 must lie on this rotated circle, so move from y1 along the constant g = 0.250 circle until it meets the rotated circle.

Step 4: Susceptance of stub 1

Analytically (Pozar), with t = tan βd:

t = tan 45° = +1.0000
b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.4375, √ = 0.6614
Sol A: b1 = +1.9114
Sol B: b1 = +0.5886

Step 5: Move 0.125λ to stub 2 and find b2

Sol A: y after stub 1 = 0.250 + j1.661
   (0.1649λ) → after 0.125λ: y2 = 1.000 − j3.646
   b2 = +3.6458
Sol B: y after stub 1 = 0.250 + j0.339
   (0.0548λ) → after 0.125λ: y2 = 1.000 + j1.646
   b2 = −1.6458

Both y2 values lie on the g = 1 circle, as required.

Step 6: Stub lengths

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.

Sol A short: l1 = 0.4233λ, l2 = 0.4574λ
Sol B short: l1 = 0.3347λ, l2 = 0.0869λ

Answer: first stub at the load, spacing λ/8 = 0.125λ. Solution B (l1 = 0.3347λ, l2 = 0.0869λ) has the shorter total stub length.

 Z0=50Ω          λ/8          
 o------+---------------+--------[100+j100Ω]
        |               |
     stub 2          stub 1
     l2=0.0869λ       l1=0.3347λ
       SC              SC

Check: with these lengths the admittance at stub 2 is 1.000 + j0.000, so the VSWR on the main line drops from 4.27 to 1.

Given: antenna load ZL = 300 + j300 Ω, Z0 = 300 Ω, stub spacing 3λ/8. Assumptions: short-circuited 300 Ω stubs, with the first stub at the antenna terminals.

Step 1: Normalise and convert to admittance

zL = 1.000 + j1.000
yL = 1/zL = 0.500 − j0.500
ΓL = 0.447∠63.4°,  VSWR = 2.618

On the chart, zL is at 0.1619λ WTG and yL is diametrically opposite, at 0.4119λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.500 − j0.500.

Step 3: Forbidden-region check and rotated g = 1 circle

βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.5000  → matchable

Rotate the g = 1 circle by 0.375λ (270° on the chart) toward the load. The point just after stub 1 must lie on this rotated circle, so move from y1 along the constant g = 0.500 circle until it meets the rotated circle.

Step 4: Susceptance of stub 1

Analytically (Pozar), with t = tan βd:

t = tan 135° = −1.0000
b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.7500, √ = 0.8660
Sol A: b1 = −1.3660
Sol B: b1 = +0.3660

Step 5: Move 0.375λ to stub 2 and find b2

Sol A: y after stub 1 = 0.500 − j1.866
   (0.3247λ) → after 0.375λ: y2 = 1.000 + j2.732
   b2 = −2.7321
Sol B: y after stub 1 = 0.500 − j0.134
   (0.4721λ) → after 0.375λ: y2 = 1.000 − j0.732
   b2 = +0.7321

Both y2 values lie on the g = 1 circle, as required.

Step 6: Stub lengths

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.

Sol A short: l1 = 0.1006λ, l2 = 0.0558λ
Sol B short: l1 = 0.3058λ, l2 = 0.3506λ

Answer (design): solution A, with the first stub at the load (l1 = 0.1006λ), the second stub 0.375λ away (l2 = 0.0558λ).

 Z0=300Ω          3λ/8          
 o------+---------------+--------[300+j300Ω]
        |               |
     stub 2          stub 1
     l2=0.0558λ       l1=0.1006λ
       SC              SC

Justification

• Matchable: the forbidden region for 3λ/8 spacing is g > 2. The load conductance g = 0.5 lies outside it, so the design is possible.
• Perfect match: the admittance at stub 2 becomes 1 + j0. VSWR drops from 2.62 to 1, and no power is reflected (Γ goes from 0.447 to 0).
• Choice of solution A: both stubs are short (about 0.1λ and 0.06λ). This gives the least stub loss, the widest bandwidth and the smallest size.
• Short-circuited stubs are preferred: they do not radiate from their ends and are easy to tune with sliding shorts.
• 3λ/8 spacing gives the same forbidden region as λ/8 but more physical room between the junctions.
• Solution B (l1 = 0.3058λ, l2 = 0.3506λ) is also valid but longer.

Given: ZL = 80 + j180 Ω, Z0 = 100 Ω, f = 3 GHz, spacing 3λ/8. Assumptions: short-circuited stubs, the first stub at the load, and an air line (λ = c/f = 3×10⁸/3×10⁹ = 10 cm).

Step 1: Normalise and convert to admittance

zL = 0.800 + j1.800
yL = 1/zL = 0.206 − j0.464
ΓL = 0.711∠51.3°,  VSWR = 5.931

On the chart, zL is at 0.1787λ WTG and yL is diametrically opposite, at 0.4287λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.206 − j0.464.

Step 3: Forbidden-region check and rotated g = 1 circle

βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.2062  → matchable

Rotate the g = 1 circle by 0.375λ (270° on the chart) toward the load. The point just after stub 1 must lie on this rotated circle, so move from y1 along the constant g = 0.206 circle until it meets the rotated circle.

Step 4: Susceptance of stub 1

Analytically (Pozar), with t = tan βd:

t = tan 135° = −1.0000
b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.3699, √ = 0.6082
Sol A: b1 = −1.1442
Sol B: b1 = +0.0721

Step 5: Move 0.375λ to stub 2 and find b2

Sol A: y after stub 1 = 0.206 − j1.608
   (0.3377λ) → after 0.375λ: y2 = 1.000 + j3.950
   b2 = −3.9496
Sol B: y after stub 1 = 0.206 − j0.392
   (0.4385λ) → after 0.375λ: y2 = 1.000 − j1.950
   b2 = +1.9496

Both y2 values lie on the g = 1 circle, as required.

Step 6: Stub lengths

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.

Sol A short: l1 = 0.1143λ, l2 = 0.0395λ
Sol B short: l1 = 0.2615λ, l2 = 0.4246λ

Physical lengths (λ = 10.000 cm):

d1 = 0.000 cm, spacing = 3.750 cm
Sol A short: l1 = 1.143 cm, l2 = 0.395 cm
Sol B short: l1 = 2.615 cm, l2 = 4.246 cm

Answer (design): solution A, with the stubs 3.75 cm apart, l1 = 0.1143λ = 1.143 cm, l2 = 0.0395λ = 0.395 cm.

Check: y at stub 2 = 1 + j0, so VSWR goes from 5.93 to 1.

Physical connection

 Z0=100Ω          3.75 cm          
 o------+---------------+--------[80+j180Ω]
        |               |
     stub 2          stub 1
     l2=0.395 cm      l1=1.143 cm
       SC              SC

Microstrip layout (dielectric: scale all lengths by 1/√εeff)
     l2            l1
     ||            ||
     ● via         ● via
     ||            ||
=====##== 3.75 ====##[ZL]

Double-stub tuner

A double-stub tuner is an impedance-matching network with two shunt stubs (usually short-circuited) at fixed positions on the line, a fixed distance apart (λ/8, 3λ/8 or λ/4). Only the stub lengths are adjusted. Stub 1 moves the admittance onto the rotated g = 1 circle. After travelling the spacing d this point lands on the g = 1 circle, and stub 2 cancels the remaining susceptance. It suits adjustable coaxial or waveguide tuners because nothing slides along the main line. Its limit is the forbidden region g > 1/sin²βd, which cannot be matched.

Design

Given: ZL = 75 + j75 Ω, Z0 = 50 Ω, spacing 3λ/8. Assumptions: short-circuited stubs, with the first stub at the load.

Step 1: Normalise and convert to admittance

zL = 1.500 + j1.500
yL = 1/zL = 0.333 − j0.333
ΓL = 0.542∠40.6°,  VSWR = 3.370

On the chart, zL is at 0.1936λ WTG and yL is diametrically opposite, at 0.4436λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.333 − j0.333.

Step 3: Forbidden-region check and rotated g = 1 circle

βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.3333  → matchable

Rotate the g = 1 circle by 0.375λ (270° on the chart) toward the load. The point just after stub 1 must lie on this rotated circle, so move from y1 along the constant g = 0.333 circle until it meets the rotated circle.

Step 4: Susceptance of stub 1

Analytically (Pozar), with t = tan βd:

t = tan 135° = −1.0000
b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.5556, √ = 0.7454
Sol A: b1 = −1.4120
Sol B: b1 = +0.0787

Step 5: Move 0.375λ to stub 2 and find b2

Sol A: y after stub 1 = 0.333 − j1.745
   (0.3309λ) → after 0.375λ: y2 = 1.000 + j3.236
   b2 = −3.2361
Sol B: y after stub 1 = 0.333 − j0.255
   (0.4559λ) → after 0.375λ: y2 = 1.000 − j1.236
   b2 = +1.2361

Both y2 values lie on the g = 1 circle, as required.

Step 6: Stub lengths

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.

Sol A short: l1 = 0.0981λ, l2 = 0.0477λ
Sol B short: l1 = 0.2625λ, l2 = 0.3917λ

Answer: stub 1 at the load, stub 2 at 3λ/8 = 0.375λ from it. Solution A: l1 = 0.0981λ, l2 = 0.0477λ (shortest). Solution B: l1 = 0.2625λ, l2 = 0.3917λ.

 Z0=50Ω          3λ/8          
 o------+---------------+--------[75+j75Ω]
        |               |
     stub 2          stub 1
     l2=0.0477λ       l1=0.0981λ
       SC              SC

Check: y at stub 2 = 1.000 + j0.000, so VSWR drops from 3.37 to 1.

Given: ΓL = 0.33∠66°, Z0 = 75 Ω (microstrip). A single shunt stub is used, and both short (via) and open designs are given.

Load impedance

zL = (1 + ΓL)/(1 − ΓL) = 1.060 + j0.717
ZL = 75 × zL = 79.52 + j53.80 Ω

Step 1: Normalise and plot the load

zL = ZL/Z0 = 1.060 + j0.717
ΓL = (zL − 1)/(zL + 1) = 0.330∠66.0°
VSWR = (1 + |Γ|)/(1 − |Γ|) = 1.985

Draw the constant-VSWR circle through zL (zL is at 0.1583λ WTG).

Step 2: Convert to admittance (shunt stub)

Move zL by 180° on the VSWR circle, which is the same as moving λ/4:

yL = 1/zL = 0.647 − j0.438   (0.4083λ WTG)

Step 3: Move toward the generator to the g = 1 circle

The VSWR circle cuts the g = 1 circle at two points:

y1 = 1.000 + j0.699  at 0.1518λ WTG
y2 = 1.000 − j0.699  at 0.3482λ WTG
d1 = 0.2434λ from the load
d2 = 0.4399λ from the load

Step 4: Stub susceptance and length

The stub must cancel the susceptance: bstub = −b.

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.
• For an open-circuited stub: y = j tan βl, so l = (1/2π)·tan⁻¹(b); measure from the open-circuit point (y = 0, left end of the chart) toward the generator.

Sol 1: bstub = −0.6992
   short stub l = 0.1529λ
   open stub  l = 0.4029λ
Sol 2: bstub = +0.6992
   short stub l = 0.3471λ
   open stub  l = 0.0971λ

Answer (design): for a microstrip, an open stub avoids a via hole. Solution 2: d = 0.4399λg, l = 0.0971λg. With a shorted stub (via to ground), solution 1: d = 0.2434λg, l = 0.1529λg. Here λg is the guided wavelength on the microstrip.

       open stub l = 0.0971λg
            ||
            ○
            ||
 ===========##====== d = 0.4399λg ====[patch]
  75 Ω feed

Scattering matrices

Treat the stub and the line section d as a two-port: port 1 faces the main line, and port 2 faces the load. Both ports are referenced to Z0. Use the normalised ABCD matrix [1 0; jb 1]·[cos βd j sin βd; j sin βd cos βd] and convert with S11 = (A + B − C − D)/Δ, S21 = 2/Δ, S22 = (−A + B − C + D)/Δ, Δ = A + B + C + D.

Before matching (line sees the load directly):
  S11 = ΓL = 0.330∠66.0°,  VSWR = 1.99
Matching network (solution 2):
  S11 = 0.330∠-109.3°   S12 = 0.944∠-177.6°
  S21 = 0.944∠-177.6°   S22 = 0.330∠-66.0°
  S22 = ΓL* (conjugate match), |S11|²+|S21|² = 1
After matching (network + load):
  Γin = S11 + S12S21ΓL/(1 − S22ΓL) = 0, VSWR = 1

With port 2 referenced to the load, the matched lossless network has the ideal form S = [0 e^(−jφ); e^(−jφ) 0]: no reflection at either port, and all power is transferred.

Given: f = 10 GHz, ΓL = 0.45∠60°, Z0 = 100 Ω. Assumptions: short-circuited stubs, the first stub at the antenna, spacing 3λ/8. Microstrip with effective permittivity εeff ≈ 1.9 (for example a 100 Ω line on RT/Duroid 5880), so λg = λ0/√εeff = 3 cm/√1.9 = 2.176 cm.

Load impedance

zL = (1 + Γ)/(1 − Γ) = 1.060 + j1.036
ZL = 105.98 + j103.58 Ω

Step 1: Normalise and convert to admittance

zL = 1.060 + j1.036
yL = 1/zL = 0.483 − j0.472
ΓL = 0.450∠60.0°,  VSWR = 2.636

On the chart, zL is at 0.1667λ WTG and yL is diametrically opposite, at 0.4167λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.483 − j0.472.

Step 3: Forbidden-region check and rotated g = 1 circle

βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.4826  → matchable

Rotate the g = 1 circle by 0.375λ (270° on the chart) toward the load. The point just after stub 1 must lie on this rotated circle, so move from y1 along the constant g = 0.483 circle until it meets the rotated circle.

Step 4: Susceptance of stub 1

Analytically (Pozar), with t = tan βd:

t = tan 135° = −1.0000
b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.7323, √ = 0.8557
Sol A: b1 = −1.3841
Sol B: b1 = +0.3274

Step 5: Move 0.375λ to stub 2 and find b2

Sol A: y after stub 1 = 0.483 − j1.856
   (0.3253λ) → after 0.375λ: y2 = 1.000 + j2.773
   b2 = −2.7732
Sol B: y after stub 1 = 0.483 − j0.144
   (0.4706λ) → after 0.375λ: y2 = 1.000 − j0.773
   b2 = +0.7732

Both y2 values lie on the g = 1 circle, as required.

Step 6: Stub lengths

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.

Sol A short: l1 = 0.0996λ, l2 = 0.0551λ
Sol B short: l1 = 0.3004λ, l2 = 0.3548λ

Physical lengths (λ = 2.176 cm):

d1 = 0.000 cm, spacing = 0.816 cm
Sol A short: l1 = 0.217 cm, l2 = 0.120 cm
Sol B short: l1 = 0.654 cm, l2 = 0.772 cm

Answer (design): solution A, with l1 = 0.0996λg = 0.217 cm, l2 = 0.0551λg = 0.120 cm, and spacing 0.375λg = 0.816 cm.

Physical diagram (microstrip)

   stub 2              stub 1
  l2=1.20 mm          l1=2.17 mm
     ||                  ||
     ● via               ● via
     ||                  ||
=====##=== 8.16 mm ======##[antenna]
 100 Ω microstrip feed

S-matrices before and after the design

Treat the tuner (stub 2, line 0.375λ, stub 1) as a two-port with both ports referenced to Z0. Port 1 faces the generator, and port 2 faces the load. The S-parameters come from the cascaded normalised ABCD matrix [1 0; jb2 1]·[cos βd j sin βd; j sin βd cos βd]·[1 0; jb1 1], with S11 = (A + B − C − D)/Δ, S21 = 2/Δ, S22 = (−A + B − C + D)/Δ, Δ = A + B + C + D.

Before matching:
  S11 = ΓL = 0.450∠60.0°, VSWR = 2.64
Tuner, solution A:
  S11 = 0.450∠145.9°   S12 = 0.893∠-47.1°
  S21 = 0.893∠-47.1°   S22 = 0.450∠-60.0°
  S12 = S21 (reciprocal), |S11|²+|S21|² = 1 (lossless)
  S22 = ΓL* (conjugate match)
After matching (tuner + load):
  Γin = S11 + S12S21ΓL/(1 − S22ΓL) = 0, VSWR = 1

Referred to the load (port 2 normalised to ZL), the perfectly matched network is S = [0 e^(−jφ); e^(−jφ) 0]. S11 = S22 = 0, and |S21| = 1, so all power reaches the load.

Given: Z0 = 100 Ω, ZL = 120 − j160 Ω (capacitive). A single shunt stub is used.

Reasoning: a shunt stub can only add susceptance. So, first move along the line to a point where the real part of the admittance equals Y0 (g = 1). Then cancel the imaginary part with the stub.

Step 1: Normalise and plot the load

zL = ZL/Z0 = 1.200 − j1.600
ΓL = (zL − 1)/(zL + 1) = 0.593∠-46.8°
VSWR = (1 + |Γ|)/(1 − |Γ|) = 3.911

Draw the constant-VSWR circle through zL (zL is at 0.3151λ WTG).

Step 2: Convert to admittance (shunt stub)

Move zL by 180° on the VSWR circle, which is the same as moving λ/4:

yL = 1/zL = 0.300 + j0.400   (0.0651λ WTG)

Step 3: Move toward the generator to the g = 1 circle

The VSWR circle cuts the g = 1 circle at two points:

y1 = 1.000 + j1.472  at 0.1755λ WTG
y2 = 1.000 − j1.472  at 0.3245λ WTG
d1 = 0.1104λ from the load
d2 = 0.2594λ from the load

Step 4: Stub susceptance and length

The stub must cancel the susceptance: bstub = −b.

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.
• For an open-circuited stub: y = j tan βl, so l = (1/2π)·tan⁻¹(b); measure from the open-circuit point (y = 0, left end of the chart) toward the generator.

Sol 1: bstub = −1.4720
   short stub l = 0.0950λ
   open stub  l = 0.3450λ
Sol 2: bstub = +1.4720
   short stub l = 0.4050λ
   open stub  l = 0.1550λ

Answer (design): solution 1, with the stub at d = 0.1104λ from the load and a short-circuited stub of l = 0.0950λ. These are the shortest lengths, giving the widest bandwidth. (Open-stub alternative: solution 2, d = 0.2594λ, l = 0.1550λ.)

 Z0=100Ω     d = 0.1104λ
 o---------+--------------[120 − j160 Ω]
           |
         l = 0.0950λ
           |
         short

S-matrix before and after the design

Treat the stub and the line section d as a two-port: port 1 faces the main line, and port 2 faces the load. Both ports are referenced to Z0. Use the normalised ABCD matrix [1 0; jb 1]·[cos βd j sin βd; j sin βd cos βd] and convert with S11 = (A + B − C − D)/Δ, S21 = 2/Δ, S22 = (−A + B − C + D)/Δ, Δ = A + B + C + D.

Before matching (line sees the load directly):
  S11 = ΓL = 0.593∠-46.8°,  VSWR = 3.91
Matching network (solution 1):
  S11 = 0.593∠126.4°   S12 = 0.805∠-3.4°
  S21 = 0.805∠-3.4°   S22 = 0.593∠46.8°
  S22 = ΓL* (conjugate match), |S11|²+|S21|² = 1
After matching (network + load):
  Γin = S11 + S12S21ΓL/(1 − S22ΓL) = 0, VSWR = 1

With port 2 referenced to the load, the matched lossless network has the ideal form S = [0 e^(−jφ); e^(−jφ) 0]: no reflection at either port, and all power is transferred.

Given: ΓL = 0.64∠58°, Z0 = 100 Ω (taking the unclear value as 100 Ω). Assumptions: short-circuited stubs, the first stub at the load, spacing 3λ/8.

Load impedance

zL = (1 + Γ)/(1 − Γ) = 0.807 + j1.484
ZL = 80.73 + j148.43 Ω

Step 1: Normalise and convert to admittance

zL = 0.807 + j1.484
yL = 1/zL = 0.283 − j0.520
ΓL = 0.640∠58.0°,  VSWR = 4.556

On the chart, zL is at 0.1694λ WTG and yL is diametrically opposite, at 0.4194λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.283 − j0.520.

Step 3: Forbidden-region check and rotated g = 1 circle

βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.2828  → matchable

Rotate the g = 1 circle by 0.375λ (270° on the chart) toward the load. The point just after stub 1 must lie on this rotated circle, so move from y1 along the constant g = 0.283 circle until it meets the rotated circle.

Step 4: Susceptance of stub 1

Analytically (Pozar), with t = tan βd:

t = tan 135° = −1.0000
b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.4856, √ = 0.6968
Sol A: b1 = −1.1769
Sol B: b1 = +0.2167

Step 5: Move 0.375λ to stub 2 and find b2

Sol A: y after stub 1 = 0.283 − j1.697
   (0.3333λ) → after 0.375λ: y2 = 1.000 + j3.464
   b2 = −3.4643
Sol B: y after stub 1 = 0.283 − j0.303
   (0.4497λ) → after 0.375λ: y2 = 1.000 − j1.464
   b2 = +1.4643

Both y2 values lie on the g = 1 circle, as required.

Step 6: Stub lengths

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.

Sol A short: l1 = 0.1121λ, l2 = 0.0447λ
Sol B short: l1 = 0.2840λ, l2 = 0.4046λ

Answer (design): solution A, with l1 = 0.1121λ, l2 = 0.0447λ, and spacing 0.375λ.

 Z0=100Ω          3λ/8          
 o------+---------------+--------[ZL]
        |               |
     stub 2          stub 1
     l2=0.0447λ       l1=0.1121λ
       SC              SC

S-matrix of the matched network

Treat the tuner (stub 2, line 0.375λ, stub 1) as a two-port with both ports referenced to Z0. Port 1 faces the generator, and port 2 faces the load. The S-parameters come from the cascaded normalised ABCD matrix [1 0; jb2 1]·[cos βd j sin βd; j sin βd cos βd]·[1 0; jb1 1], with S11 = (A + B − C − D)/Δ, S21 = 2/Δ, S22 = (−A + B − C + D)/Δ, Δ = A + B + C + D.

Before matching:
  S11 = ΓL = 0.640∠58.0°, VSWR = 4.56
Tuner, solution A:
  S11 = 0.640∠149.7°   S12 = 0.768∠-44.1°
  S21 = 0.768∠-44.1°   S22 = 0.640∠-58.0°
  S12 = S21 (reciprocal), |S11|²+|S21|² = 1 (lossless)
  S22 = ΓL* (conjugate match)
After matching (tuner + load):
  Γin = S11 + S12S21ΓL/(1 − S22ΓL) = 0, VSWR = 1

Referred to the load (port 2 normalised to ZL), the perfectly matched network is S = [0 e^(−jφ); e^(−jφ) 0]. S11 = S22 = 0, and |S21| = 1, so all power reaches the load.

Given: Z0 = 75 Ω coax, ZL = 109 + j120 Ω, short-circuited 75 Ω coaxial stubs. Assumptions: the first stub is at the load, and the stub spacing is 3λ/8.

Step 1: Normalise and convert to admittance

zL = 1.453 + j1.600
yL = 1/zL = 0.311 − j0.342
ΓL = 0.568∠41.1°,  VSWR = 3.627

On the chart, zL is at 0.1930λ WTG and yL is diametrically opposite, at 0.4430λ WTG.

Step 2: First stub at the load

The first stub is placed at the load, so y1 = yL = 0.311 − j0.342.

Step 3: Forbidden-region check and rotated g = 1 circle

βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.3111  → matchable

Rotate the g = 1 circle by 0.375λ (270° on the chart) toward the load. The point just after stub 1 must lie on this rotated circle, so move from y1 along the constant g = 0.311 circle until it meets the rotated circle.

Step 4: Susceptance of stub 1

Analytically (Pozar), with t = tan βd:

t = tan 135° = −1.0000
b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.5254, √ = 0.7248
Sol A: b1 = −1.3824
Sol B: b1 = +0.0673

Step 5: Move 0.375λ to stub 2 and find b2

Sol A: y after stub 1 = 0.311 − j1.725
   (0.3320λ) → after 0.375λ: y2 = 1.000 + j3.330
   b2 = −3.3302
Sol B: y after stub 1 = 0.311 − j0.275
   (0.4533λ) → after 0.375λ: y2 = 1.000 − j1.330
   b2 = +1.3302

Both y2 values lie on the g = 1 circle, as required.

Step 6: Stub lengths

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.

Sol A short: l1 = 0.0997λ, l2 = 0.0464λ
Sol B short: l1 = 0.2607λ, l2 = 0.3974λ

Answer (design): solution A, with l1 = 0.0997λ, l2 = 0.0464λ, and spacing 0.375λ.

 Z0=75Ω          3λ/8          
 o------+---------------+--------[109+j120Ω]
        |               |
     stub 2          stub 1
     l2=0.0464λ       l1=0.0997λ
       SC              SC

S-matrix

Treat the tuner (stub 2, line 0.375λ, stub 1) as a two-port with both ports referenced to Z0. Port 1 faces the generator, and port 2 faces the load. The S-parameters come from the cascaded normalised ABCD matrix [1 0; jb2 1]·[cos βd j sin βd; j sin βd cos βd]·[1 0; jb1 1], with S11 = (A + B − C − D)/Δ, S21 = 2/Δ, S22 = (−A + B − C + D)/Δ, Δ = A + B + C + D.

Before matching:
  S11 = ΓL = 0.568∠41.1°, VSWR = 3.63
Tuner, solution A:
  S11 = 0.568∠145.3°   S12 = 0.823∠-37.9°
  S21 = 0.823∠-37.9°   S22 = 0.568∠-41.1°
  S12 = S21 (reciprocal), |S11|²+|S21|² = 1 (lossless)
  S22 = ΓL* (conjugate match)
After matching (tuner + load):
  Γin = S11 + S12S21ΓL/(1 − S22ΓL) = 0, VSWR = 1

Referred to the load (port 2 normalised to ZL), the perfectly matched network is S = [0 e^(−jφ); e^(−jφ) 0]. S11 = S22 = 0, and |S21| = 1, so all power reaches the load.

Given: ZL = 60 − j80 Ω, Z0 = 50 Ω, open-circuited shunt stubs, spacing d = 3λ/8, first stub 0.4λ from the load.

Step 1: Normalise and convert to admittance

zL = 1.200 − j1.600
yL = 1/zL = 0.300 + j0.400
ΓL = 0.593∠-46.8°,  VSWR = 3.911

On the chart, zL is at 0.3151λ WTG and yL is diametrically opposite, at 0.0651λ WTG.

Step 2: Move to the first stub (0.4λ from the load)

Rotate yL clockwise (toward the generator) on its VSWR circle:

y1 = 0.268 − j0.208  (0.4651λ WTG)

Step 3: Forbidden-region check and rotated g = 1 circle

βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.2675  → matchable

Rotate the g = 1 circle by 0.375λ (270° on the chart) toward the load. The point just after stub 1 must lie on this rotated circle, so move from y1 along the constant g = 0.268 circle until it meets the rotated circle.

Step 4: Susceptance of stub 1

Analytically (Pozar), with t = tan βd:

t = tan 135° = −1.0000
b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.4635, √ = 0.6808
Sol A: b1 = −1.4730
Sol B: b1 = −0.1114

Step 5: Move 0.375λ to stub 2 and find b2

Sol A: y after stub 1 = 0.268 − j1.681
   (0.3341λ) → after 0.375λ: y2 = 1.000 + j3.545
   b2 = −3.5447
Sol B: y after stub 1 = 0.268 − j0.319
   (0.4477λ) → after 0.375λ: y2 = 1.000 − j1.545
   b2 = +1.5447

Both y2 values lie on the g = 1 circle, as required.

Step 6: Stub lengths

• For an open-circuited stub: y = j tan βl, so l = (1/2π)·tan⁻¹(b); measure from the open-circuit point (y = 0, left end of the chart) toward the generator.

Sol A open : l1 = 0.3449λ, l2 = 0.2938λ
Sol B open : l1 = 0.4824λ, l2 = 0.1586λ

Answer (design): solution A, with stub 1 at 0.4λ from the load (open, l1 = 0.3449λ) and stub 2 at 0.775λ from the load (open, l2 = 0.2938λ). Solution B (l1 = 0.4824λ, l2 = 0.1586λ) is equally valid, with almost the same total length.

 Z0=50Ω          3λ/8          0.4λ
 o------+---------------+--------[60−j80Ω]
        |               |
     stub 2          stub 1
     l2=0.2938λ       l1=0.3449λ
       OC              OC

S-matrix of the design

Treat the tuner (stub 2, line 0.375λ, stub 1, line 0.4λ) as a two-port with both ports referenced to Z0. Port 1 faces the generator, and port 2 faces the load. The S-parameters come from the cascaded normalised ABCD matrix [1 0; jb2 1]·[cos βd j sin βd; j sin βd cos βd]·[1 0; jb1 1] ·[line d1], with S11 = (A + B − C − D)/Δ, S21 = 2/Δ, S22 = (−A + B − C + D)/Δ, Δ = A + B + C + D.

Before matching:
  S11 = ΓL = 0.593∠-46.8°, VSWR = 3.91
Tuner, solution A:
  S11 = 0.593∠143.6°   S12 = 0.805∠-174.8°
  S21 = 0.805∠-174.8°   S22 = 0.593∠46.8°
  S12 = S21 (reciprocal), |S11|²+|S21|² = 1 (lossless)
  S22 = ΓL* (conjugate match)
After matching (tuner + load):
  Γin = S11 + S12S21ΓL/(1 − S22ΓL) = 0, VSWR = 1

Referred to the load (port 2 normalised to ZL), the perfectly matched network is S = [0 e^(−jφ); e^(−jφ) 0]. S11 = S22 = 0, and |S21| = 1, so all power reaches the load.

Given: Z0 = 50 Ω (Y0 = 0.02 ℧), YL = 0.00813 + j0.0065 ℧, spacing 3λ/8, first stub 0.01λ from the load. Short-circuited stubs are assumed (the usual choice; open-stub lengths are λ/4 different).

Step 1: Normalise and convert to admittance

yL = YL·Z0 = (0.00813 + j0.0065)×50 = 0.4065 + j0.325
zL = 1.501 − j1.200
yL = 1/zL = 0.406 + j0.325
ΓL = 0.469∠-41.7°,  VSWR = 2.765

On the chart, zL is at 0.3079λ WTG and yL is diametrically opposite, at 0.0579λ WTG.

Step 2: Move to the first stub (0.01λ from the load)

Rotate yL clockwise (toward the generator) on its VSWR circle:

y1 = 0.425 + j0.385  (0.0679λ WTG)

Step 3: Forbidden-region check and rotated g = 1 circle

βd = 360° × 0.375 = 135°
g must be ≤ 1/sin²βd = 2.000
g1 = 0.4250  → matchable

Rotate the g = 1 circle by 0.375λ (270° on the chart) toward the load. The point just after stub 1 must lie on this rotated circle, so move from y1 along the constant g = 0.425 circle until it meets the rotated circle.

Step 4: Susceptance of stub 1

Analytically (Pozar), with t = tan βd:

t = tan 135° = −1.0000
b1 = −B + [1 ± √((1+t²)g − g²t²)]/t
(1+t²)g − g²t² = 0.6694, √ = 0.8182
Sol A: b1 = −2.2031
Sol B: b1 = −0.5667

Step 5: Move 0.375λ to stub 2 and find b2

Sol A: y after stub 1 = 0.425 − j1.818
   (0.3273λ) → after 0.375λ: y2 = 1.000 + j2.925
   b2 = −2.9250
Sol B: y after stub 1 = 0.425 − j0.182
   (0.4655λ) → after 0.375λ: y2 = 1.000 − j0.925
   b2 = +0.9250

Both y2 values lie on the g = 1 circle, as required.

Step 6: Stub lengths

• For a short-circuited stub: y = −j cot βl, so l = (1/2π)·cot⁻¹(−b); measure from the short-circuit point (y = ∞, right end of the chart) toward the generator.

Sol A short: l1 = 0.0678λ, l2 = 0.0524λ
Sol B short: l1 = 0.1679λ, l2 = 0.3688λ

Answer (Smith chart values agree within reading accuracy): solution A, with l1 = 0.0678λ and l2 = 0.0524λ (shortest). Solution B: l1 = 0.1679λ, l2 = 0.3688λ.

 Z0=50Ω          3λ/8          0.01λ
 o------+---------------+--------[YL]
        |               |
     stub 2          stub 1
     l2=0.0524λ       l1=0.0678λ
       SC              SC

S-parameters of the matched network

Treat the tuner (stub 2, line 0.375λ, stub 1, line 0.01λ) as a two-port with both ports referenced to Z0. Port 1 faces the generator, and port 2 faces the load. The S-parameters come from the cascaded normalised ABCD matrix [1 0; jb2 1]·[cos βd j sin βd; j sin βd cos βd]·[1 0; jb1 1] ·[line d1], with S11 = (A + B − C − D)/Δ, S21 = 2/Δ, S22 = (−A + B − C + D)/Δ, Δ = A + B + C + D.

Before matching:
  S11 = ΓL = 0.469∠-41.7°, VSWR = 2.76
Tuner, solution A:
  S11 = 0.469∠106.4°   S12 = 0.883∠-15.9°
  S21 = 0.883∠-15.9°   S22 = 0.469∠41.7°
  S12 = S21 (reciprocal), |S11|²+|S21|² = 1 (lossless)
  S22 = ΓL* (conjugate match)
After matching (tuner + load):
  Γin = S11 + S12S21ΓL/(1 − S22ΓL) = 0, VSWR = 1

Referred to the load (port 2 normalised to ZL), the perfectly matched network is S = [0 e^(−jφ); e^(−jφ) 0]. S11 = S22 = 0, and |S21| = 1, so all power reaches the load.

Stripline and microstrip are planar transmission lines made by etching copper strips on dielectric substrates. They are used in microwave integrated circuits.

   Stripline (triplate)        Microstrip
 ==================== ground
 |   dielectric εr   |       ____ strip (w)
 |      ▬▬▬ strip    |      |  dielectric εr | h
 |                   |      ================= ground
 ==================== ground

Choice: microstrip is the most popular because it is cheap and components are easy to mount. Stripline is chosen where isolation, low radiation and no dispersion are needed, for example in high-performance couplers and filters.

A microstrip is a planar transmission line made of a thin metal strip of width W on one side of a dielectric substrate (thickness h, relative permittivity εr) with a full ground plane on the other side. It is the most common line for microwave integrated circuits (MICs) and printed RF boards.

        W
     <----->
     #######          <- conducting strip
 ---------------------
 |  dielectric εr    |  h
 ---------------------
 ##################### <- ground plane

Mode of propagation: The field lies partly in the dielectric and partly in the air above it, so a pure TEM mode cannot exist. The wave is quasi-TEM, and it travels as if in a uniform medium of effective permittivity εeff, where 1 < εeff < εr:

• εeff ≈ (εr + 1)/2 + ((εr − 1)/2) · 1/√(1 + 12h/W)
• Phase velocity vp = c/√εeff, guide wavelength λg = λ0/√εeff
• Characteristic impedance depends on W/h: a wide strip (large W/h) gives low Z0, a narrow strip gives high Z0. Typical values are 20–120 Ω; 50 Ω on FR-4 (εr ≈ 4.4) needs W ≈ 1.9h.

Losses: conductor loss (skin effect in strip and ground), dielectric loss (tan δ of substrate) and radiation loss (at bends, open ends and discontinuities, which increases with frequency and thick, low-εr substrates).

Advantages:

• Small, light, cheap; made by photolithography.
• Easy to mount active devices (transistors, diodes) and chip components on the surface.
• Good for integration of filters, couplers, matching networks and patch antennas.

Disadvantages: higher loss and lower power handling than waveguide or coax, radiation and dispersion at high frequency, and coupling between nearby lines.

Applications: MMICs/MICs, LNAs and power amplifiers, branch-line couplers, Wilkinson dividers, stub filters and patch antenna feeds.