SME
Communication Systems · Problem type
Cascaded noise figure (Friis)
Use this for “A 3 dB noise figure amplifier with 20 dB of gain feeds a 10 dB mixer. Find the noise figure of the pair.”
Chart
Total noise figure: 3.0 + 0.2 = 3.2 dB
This is an example. Type your own values to start a new problem.
F₁
=
10
/
10
,
F₂
=
10
/
10
F
=
F₁
+
−
1
10
/
10
NF (dB)
=
10
log₁₀
F
ΔNF
Rise from stage 2
worked out (not typed) · dB
Formulas
NF₁ = 10 log₁₀(F₁)
3 = 10 log₁₀(1.9953)
NF₂ = 10 log₁₀(F₂)
10 = 10 log₁₀(10)
F
=
F₁
+
F₂ − 1
10
G₁ ÷ 10
2.0853
=
1.9953
+
10 − 1
10
20 ÷ 10
NF
=
10
log₁₀(
F
)
3.1916 = 10 log₁₀(2.0853)
ΔNF
=
10
log₁₀(
F
F₁
)
0.1916
=
10
log₁₀(
2.0853
1.9953
)
Assumptions
•
Two stages in a chain, each matched; Friis’s formula works with plain ratios, never decibels.
•
A noise figure NF in dB is 10 log₁₀ of the noise factor F, and a gain G₁ in dB is 10 log₁₀ of its power ratio.
•
The second stage’s added noise is divided by the first stage’s gain, because it meets an already amplified signal.
Step-by-step
WE KNOW
NF₁ = 3 dB, G₁ = +20 dB, NF₂ = 10 dB
FIND
stage 1 noise factor (F₁), stage 2 noise factor (F₂), total noise factor (F), total noise figure (NF), rise from stage 2 (ΔNF)
Step 1 · Find stage 1 noise factor (F₁)
Use NF₁ = 10 log₁₀(F₁)
Undo the decibels: 10 to the noise figure over 10.
F₁
=
10
NF₁ ÷ 10
F₁
=
10
3 ÷ 10
F₁ = 1.9953
Step 2 · Find stage 2 noise factor (F₂)
Use NF₂ = 10 log₁₀(F₂)
Undo the decibels: 10 to the noise figure over 10.
F₂
=
10
NF₂ ÷ 10
F₂
=
10
10 ÷ 10
F₂ = 10
Step 3 · Find total noise factor (F)
Use F = F₁ + (F₂ − 1) ÷ 10G₁ ÷ 10
Stage 2’s extra noise, F₂ − 1, counts at the input once it is divided by the gain in front of it.
F
=
1.9953
+
10 − 1
10
20 ÷ 10
F
=
1.9953
+
9
100
F
=
1.9953
+
0.09
F
=
2.0853
Step 4 · Find total noise figure (NF)
Use NF = 10 log₁₀(F)
A noise factor is a power ratio, so 10 times its common log.
NF = 10 log₁₀(2.0853)
NF = 3.1916 dB
Step 5 · Find rise from stage 2 (ΔNF)
Use ΔNF = 10 log₁₀(F ÷ F₁)
How many decibels stage 2 adds to stage 1’s noise figure.
ΔNF
=
10
log₁₀(
2.0853
1.9953
)
ΔNF = 0.1916 dB
Check
Put the numbers back in. Both sides should match.
3 = 10 log₁₀(1.9953) ✓
10 = 10 log₁₀(10) ✓
2.0853
=
1.9953
+
10 − 1
10
20 ÷ 10
✓
3.1916 = 10 log₁₀(2.0853) ✓
0.1916
=
10
log₁₀(
2.0853
1.9953
)
✓
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