- 2072 Asoj · 2+8 marks
Admittance chart and single stub for 80+j100 Ω
Short answer
An admittance chart is the Smith chart rotated 180° reading normalized admittance g+jb. For ZL=80+j100 Ω on 50 Ω line, single shunt stub: short stub at d=0.2295λ, l=0.0867λ; open stub at d=0.3697λ, l=0.1633λ.
What is admittance chart? A load impedance of ZL = 80+j100 is connected to a microstrip transmission line. Find the size and placement of the matching stub. Use single stub shunt tuning short and open stubs.
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 toward the generator from the short-circuit point (y = ∞; with y plotted on an ordinary Smith chart, this is the right end, 0.25λ WTG).
- For an open-circuited stub: y = j tan βl, so l = (1/2π)·tan⁻¹(b). Measure toward the generator from the open-circuit point (y = 0; the left end, 0λ WTG).
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
| Solution | Stub position d | Short stub l | Open stub l |
|---|---|---|---|
| 1 | 0.2295λ | 0.0867λ | 0.3367λ |
| 2 | 0.3697λ | 0.4133λ | 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).
✓ Answer checked by Claude Opus 5.5 (high effort)
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