SME
Communication Systems · Problem type
Noise temperature
Use this for “A receiver has a noise factor of 2. Find its noise temperature.”
Chart
This is an example. Type your own values to start a new problem.
Tₑ (K)
=
(
F
−
1)
×
290
K
Tₛᵧₛ (K)
=
TA (K)
+
Tₑ (K)
NF
Noise figure
calculated · dB
Formulas
NF
=
10
log₁₀(
F
)
3.0103 = 10 log₁₀(2)
Tₑ
=
(
F
−
1)
×
290
290 = (2 − 1) × 290
Tₛᵧₛ = TA + Tₑ
340 = 50 + 290
Assumptions
•
The receiver’s added noise is pictured as thermal noise from a resistor at Tₑ, its noise temperature, at the input.
•
The noise factor is measured from the reference T₀ = 290 K, so F = 1 + Tₑ ÷ T₀.
•
The antenna’s noise temperature TA adds to it: the system works at Tₛᵧₛ = TA + Tₑ.
•
A noise figure NF in dB is 10 log₁₀ F.
Step-by-step
WE KNOW
F = 2, TA = 50 K
FIND
noise figure (NF), noise temperature (Tₑ), system noise temperature (Tₛᵧₛ)
Step 1 · Find noise figure (NF)
Use NF = 10 log₁₀(F)
A noise factor is a power ratio, so 10 times its common log.
NF = 10 log₁₀(2)
NF = 3.0103 dB
Step 2 · Find noise temperature (Tₑ)
Use Tₑ = (F − 1) × 290
F
−
1
is
the
receiver’s
noise
over
the
reference
noise,
so
times
T₀
=
290
K.
Tₑ = (2 − 1) × 290
Tₑ = 290 K
Step 3 · Find system noise temperature (Tₛᵧₛ)
Use Tₛᵧₛ = TA + Tₑ
Noise powers add, and each is k times its temperature times B, so the temperatures add.
Tₛᵧₛ = 50 + 290
Tₛᵧₛ = 340 K
Check
Put the numbers back in. Both sides should match.
3.0103 = 10 log₁₀(2) ✓
290 = (2 − 1) × 290 ✓
340 = 50 + 290 ✓
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