SME
Engineering Electromagnetics
Antennas
Use this for “A 1 W, 2.4 GHz link with 3 dBi antennas spans 1 km. Find the received power.”
Chart
Received power: 30.0 + 3.0 + 3.0 − 100.0 = −64.0 dBm
This is an example. Type your own values to start a new problem.
λ (m)
=
c
÷
f (GHz)
Lfs (dB)
=
20
log₁₀
4π
d (km)
λ (m)
Pᵣ (dBm)
=
Pₜ (dBm)
+
Gₜ (dBi)
+
Gᵣ (dBi)
−
Lfs (dB)
Formulas
λ
=
3 ×
10
8
f
0.125
=
3 ×
10
8
2,400,000,000
Lfs
=
20
log₁₀(4π
×
d
λ
)
100.046
=
20
log₁₀(4π
×
1,000
0.125
)
Pᵣ = Pₜ + Gₜ + Gᵣ − Lfs
−64.046 = 30 + 3 + 3 − 100.046
Formulas are worked in Hz, m. The step-by-step shows the conversions.
Assumptions
•
Free space and the far field, with the two antennas’ polarizations matched.
•
Powers are in dBm, decibels above 1 mW (1 W is 30 dBm); gains are in dBi, over an isotropic antenna.
•
The wave travels at c = 3.00 × 10⁸ m/s.
Step-by-step
WE KNOW
Pₜ = +30 dBm, Gₜ = 3 dBi, Gᵣ = 3 dBi, f = 2.4 GHz, d = 1 km
FIND
wavelength (λ), free-space path loss (Lfs), received power (Pᵣ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to Hz, m).
f = 2.4 GHz = 2,400,000,000 Hz (1 GHz = 10⁹ Hz)
d = 1 km = 1,000 m (1 km = 1000 m)
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
2,400,000,000
λ
=
0.125
m
Step 3 · Find free-space path loss (Lfs)
Use Lfs = 20 log₁₀(4π × d ÷ λ)
The
power
spreads
over
a
sphere
of
area
4πd²,
and
the
receiving
antenna
catches
a
share
set
by
λ
.
Lfs
=
20
log₁₀(4π
×
1,000
0.125
)
Lfs = 100.046 dB
Step 4 · Find received power (Pᵣ)
Use Pᵣ = Pₜ + Gₜ + Gᵣ − Lfs
In decibels the Friis product becomes a sum: add both gains to the power sent, take off the path loss.
Pᵣ = 30 + 3 + 3 − 100.046
Pᵣ = −64.046 dBm
Check
Put the numbers back in. Both sides should match.
0.125
=
3 ×
10
8
2,400,000,000
✓
100.046
=
20
log₁₀(4π
×
1,000
0.125
)
✓
−64.046 = 30 + 3 + 3 − 100.046 ✓
Topic check
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