SME
Engineering Electromagnetics · Problem type
Shorted stub
Use this for “A 50 Ω shorted stub is λ/8 long. Find its input reactance.”
Chart
X(45) = 50
This is an example. Type your own values to start a new problem.
βℓ
°
=
360°
×
ℓ (cm)
λ (cm)
X (Ω)
=
Z₀ (Ω)
tan(
βℓ
°)
Formulas
βℓ
=
360°
×
ℓ
λ
45
=
360°
×
2.5
20
X
=
Z₀
×
tan(βℓ)
+50 = 50 × tan(45°)
Assumptions
•
A lossless line shorted at its far end; its input is a pure reactance jX.
•
βℓ = 2πℓ ÷ λ is the electrical length, written here in degrees: λ/8 is 45°.
•
X > 0 is inductive while ℓ < λ/4; past λ/4 up to λ/2 it is capacitive.
•
At ℓ = λ/4 the short looks like an open circuit, so X is unbounded there.
Step-by-step
WE KNOW
Z₀ = 50 Ω, ℓ = 2.5 cm, λ = 20 cm
FIND
electrical length (βℓ), input reactance (X)
Step 1 · Find electrical length (βℓ)
Use βℓ = 360° × ℓ ÷ λ
One wavelength of line turns the phase through 360°.
βℓ
=
360°
×
2.5
20
βℓ = 900° ÷ 20
βℓ = 45°
Step 2 · Find input reactance (X)
Use X = Z₀ × tan(βℓ)
Back from the short, the voltage grows as sin βℓ and the current shrinks as cos βℓ.
X
=
50
×
tan(45°)
X
=
50
×
1
X
=
+50
Ω
Check
Put the numbers back in. Both sides should match.
45
=
360°
×
2.5
20
✓
+50 = 50 × tan(45°) ✓
Topic check
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