Voltage ratio in decibels: For a filter with 12 V input and 6 V output at a certain frequency, compute the gain (Vout/Vin) in dB.

Introduction / Context:
Decibels (dB) are widely used to express voltage gains and losses in filters and amplifiers. For voltage ratios measured using the same impedance reference, the correct formula involves 20 times the base-10 logarithm of the ratio Vout/Vin.

Given Data / Assumptions:

• Vin = 12 V, Vout = 6 V.
• Same impedance conditions for input and output.
• We need V(dB) = 20 * log10(Vout / Vin).

Concept / Approach:
Since Vout < Vin, the result should be negative, indicating attenuation. A 2:1 voltage reduction typically corresponds to approximately -6.02 dB.

Step-by-Step Solution:
Ratio r = Vout/Vin = 6 / 12 = 0.5Gain(dB) = 20 * log10(0.5)log10(0.5) ≈ -0.3010Gain(dB) ≈ 20 * (-0.3010) ≈ -6.02 dB

Verification / Alternative check:
This is a classic benchmark: 0.5 voltage ratio → -6.02 dB. Similarly, a 2× increase would be +6.02 dB in voltage terms.

Why Other Options Are Wrong:

• +6.02 dB or +12.04 dB: These indicate amplification, not attenuation.
• -12.04 dB: Would correspond to a 0.25 voltage ratio, not 0.5.

Common Pitfalls:
Using 10 * log10 for voltage (reserved for power ratios), or forgetting that a value less than 1 yields negative dB. Always use 20 * log10 for voltage ratios with equal impedances.

Final Answer:
–6.02 dB