SME
Engineering Electromagnetics · Problem type
Dish gain, beamwidth and far field
Use this for “A 1 m dish at 12 GHz has an aperture efficiency of 0.6. Find its gain, beamwidth and far-field distance.”
Diagram
λ
=
c
f
=
0.025
m;
G
=
e
(
πD
λ
)
2
=
0.6
×
(π ×
1
0.025
)
2
=
9,475
=
39.77
dBi
θ
≈
70λ
D
=
70
×
0.025
1
=
1.75°,
drawn
wider
to
show
it;
r
=
2D²
λ
=
80
m
(not
to
scale)
The dish turns the feed’s spreading wave into a plane wave across D: the wider the dish in wavelengths, the narrower the beam.
This is an example. Type your own values to start a new problem.
λ (m)
=
c
÷
f (GHz)
G
=
e
π
D (m)
λ (m)
2
GdBi
Gain in dBi
calculated · dBi
θ
Beamwidth
calculated · °
r
Far-field distance
calculated
Gideal
Ideal gain
worked out (not typed) · dBi
Lₑ
Efficiency in dB
worked out (not typed) · dB
Formulas
λ
=
3 ×
10
8
f
0.025
=
3 ×
10
8
12,000,000,000
G
=
e
×
(π ×
D
λ
)
2
9,474.8202
=
0.6
×
(π ×
1
0.025
)
2
GdBi
=
10
log₁₀(
G
)
39.7657 = 10 log₁₀(9,474.8202)
θ
=
70
×
λ
D
1.75
=
70
×
0.025
1
r
=
2
×
D
2
λ
80
=
2
×
1
2
0.025
Gideal
=
20
log₁₀(π
×
D
λ
)
41.9842
=
20
log₁₀(π
×
1
0.025
)
Lₑ
=
10
log₁₀(
e
)
−2.2185 = 10 log₁₀(0.6)
Formulas are worked in Hz. The step-by-step shows the conversions.
Assumptions
•
A parabolic dish many wavelengths across; the aperture efficiency e is the share of the ideal gain it reaches.
•
The half-power beamwidth θ ≈ 70°λ ÷ D is a rule of thumb for a dish.
•
The far field starts near r = 2D² ÷ λ; the gain and the beam hold past it.
•
c = 3.00 × 10⁸ m/s, and a gain in dBi is 10 log₁₀ G.
Step-by-step
WE KNOW
D = 1 m, f = 12 GHz, e = 0.6
FIND
wavelength (λ), gain (G), gain in dBi (GdBi), beamwidth (θ), far-field distance (r), ideal gain (Gideal), efficiency in dB (Lₑ)
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 = 12 GHz = 12,000,000,000 Hz (1 GHz = 10⁹ 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
12,000,000,000
λ
=
0.025
m
Step 3 · Find gain (G)
Use G = e × (π × D ÷ λ)²
An
ideal
aperture’s
gain
is
4π
times
its
area
over
λ
2
,
which
is
(
πD
λ
)
2
;
the
efficiency
scales
it.
G
=
0.6
×
(π ×
1
0.025
)
2
G
=
0.6
×
125.66
2
G
=
0.6
×
15,790.4356
G
=
9,474.82
Step 4 · Find gain in dBi (GdBi)
Use GdBi = 10 log₁₀(G)
A gain is a power ratio, so 10 times its common log.
GdBi = 10 log₁₀(9,474.82)
GdBi = 39.7657 dBi
Step 5 · Find beamwidth (θ)
Use θ = 70 × λ ÷ D
A wider dish, in wavelengths, makes a narrower beam.
θ
=
70
×
0.025
1
θ
=
1.75°
Step 6 · Find far-field distance (r)
Use r = 2 × D² ÷ λ
Past
this
distance
the
paths
from
the
dish’s
middle
and
rim
differ
by
under
λ
16
.
r
=
2
×
1
2
0.025
r
=
2
0.025
r
=
80
m
Step 7 · Find ideal gain (Gideal)
Use Gideal = 20 log₁₀(π × D ÷ λ)
The
ideal
gain
(
πD
λ
)
2
in
dBi:
a
squared
ratio
is
20
times
its
log.
Gideal
=
20
log₁₀(π
×
1
0.025
)
Gideal = 41.9842 dBi
Step 8 · Find efficiency in dB (Lₑ)
Use Lₑ = 10 log₁₀(e)
The efficiency in dB: what the dish gives up from the ideal gain.
Lₑ = 10 log₁₀(0.6)
Lₑ = −2.2185 dB
Check
Put the numbers back in. Both sides should match.
0.025
=
3 ×
10
8
12,000,000,000
✓
9,474.82
=
0.6
×
(π ×
1
0.025
)
2
✓
39.7657 = 10 log₁₀(9,474.82) ✓
1.75
=
70
×
0.025
1
✓
80
=
2
×
1
2
0.025
✓
41.9842
=
20
log₁₀(π
×
1
0.025
)
✓
−2.2185 = 10 log₁₀(0.6) ✓
Topic check
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