SME
Circuit Analysis II · Problem type
Gains in decibels
Use this for “A filter gives 1 V out for 10 V in. Find its gain in decibels, and the power out when 5 W goes in.”
Chart
f(AV) = 20 log₁₀(AV)
A log x-axis: each decade gets the same room.
This is an example. Type your own values to start a new problem.
AV
=
Vₒᵤₜ (V)
Vᵢₙ (V)
G (dB)
=
20
log₁₀(
AV
)
=
10
log₁₀(
AP
)
AP
=
Pₒᵤₜ (W)
Pᵢₙ (W)
Formulas
AV
=
Vₒᵤₜ
Vᵢₙ
0.1
=
1
10
G
=
20
×
log₁₀(AV)
−20 = 20 × log₁₀(0.1)
G
=
10
×
log₁₀(AP)
−20 = 10 × log₁₀(0.01)
AP
=
Pₒᵤₜ
Pᵢₙ
0.01
=
0.05
5
Assumptions
•
A voltage ratio in decibels is 20 log₁₀ of it and a power ratio 10 log₁₀, since power goes as voltage squared.
•
Input and output are across equal resistances, so both ratios give the same gain G.
•
A negative G is a loss: a power ratio of 2 is 3.01 dB, and −20 dB is a tenth of the voltage.
Step-by-step
WE KNOW
Vᵢₙ = 10 V, Vₒᵤₜ = 1 V, Pᵢₙ = 5 W
FIND
voltage ratio (AV), gain (G), power ratio (AP), output power (Pₒᵤₜ)
Step 1 · Find voltage ratio (AV)
Use AV = Vₒᵤₜ ÷ Vᵢₙ
The ratio says how many times the input the output is.
AV
=
1
10
AV = 0.1
Step 2 · Find gain (G)
Use G = 20 × log₁₀(AV)
A voltage ratio in decibels is 20 log₁₀ of it.
G
=
20
×
log₁₀(0.1)
G
=
20
×
(−1)
G
=
−20
dB
Step 3 · Find power ratio (AP)
Use G = 10 × log₁₀(AP)
Undo the decibels: raise 10 to the gain in dB over 10.
AP
=
10
G
÷ 10
AP
=
10
−20 ÷ 10
AP = 0.01
Step 4 · Find output power (Pₒᵤₜ)
Use AP = Pₒᵤₜ ÷ Pᵢₙ
Multiply both sides by Pᵢₙ.
Pₒᵤₜ = AP × Pᵢₙ
Pₒᵤₜ = 0.01 × 5
Pₒᵤₜ = 0.05 W
Check
Put the numbers back in. Both sides should match.
0.1
=
1
10
✓
−20 = 20 × log₁₀(0.1) ✓
−20 = 10 × log₁₀(0.01) ✓
0.01
=
0.05
5
✓
Topic check
See what you know: 5 questions on this topic
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