SME
Engineering Electromagnetics · Problem type
Half-wave dipole
Use this for “How long should a half-wave dipole for 100 MHz be cut?”
Diagram
λ
=
c
f
=
3 ×
10
8
1
×
10
8
=
3
m;
L
=
λ
2
=
1.5
m
ℓ = kL = 0.95 × 1.5 = 1.425 m: cut short, the tips’ fields make it act longer
It radiates most broadside and nothing along the wire; half power 39° either side of broadside.
This is an example. Type your own values to start a new problem.
λ (m)
=
c
÷
f (MHz)
L (m)
=
λ (m)
2
ℓ (m)
=
k
×
L (m)
Formulas
λ
=
3 ×
10
8
f
3
=
3 ×
10
8
100,000,000
L
=
λ
2
1.5
=
3
2
ℓ
=
k
×
L
1.425 = 0.95 × 1.5
Formulas are worked in Hz. The step-by-step shows the conversions.
Assumptions
•
A thin, center-fed dipole: its current is 0 at both tips and largest at the feed, half a wavelength in all.
•
A half-wave dipole has a radiation resistance near 73 Ω and a gain of 2.15 dBi.
•
The wire is cut about 5% short in practice (k ≈ 0.95), because the fields at its tips make it act longer.
•
The wave travels at c = 3.00 × 10⁸ m/s.
Step-by-step
WE KNOW
f = 100 MHz, k = 0.95
FIND
wavelength (λ), half-wave length (L), cut length (ℓ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to Hz).
f = 100 MHz = 100,000,000 Hz (1 MHz = 10⁶ Hz)
Step 2 · Find wavelength (λ)
Use λ = 3 × 10⁸ ÷ f
A wave moves one wavelength each cycle, so the speed of light over the frequency.
λ
=
3 ×
10
8
100,000,000
λ
=
3
m
Step 3 · Find half-wave length (L)
Use L = λ ÷ 2
The current fits half a wavelength: 0 at each tip, a peak at the feed.
L
=
3
2
L
=
1.5
m
Step 4 · Find cut length (ℓ)
Use ℓ = k × L
Trim
the
ideal
length
by
the
shortening
factor,
so
the
wire
resonates
at
f
.
ℓ
=
0.95
×
1.5
ℓ
=
1.425
m
Check
Put the numbers back in. Both sides should match.
3
=
3 ×
10
8
100,000,000
✓
1.5
=
3
2
✓
1.425 = 0.95 × 1.5 ✓
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