Question

A power gain of 100 in decibel (db) is:

• A. 20 db
• B. 40 db
• C. 2 db
• D. 30 db

Hint:

Memorize some common logarithmic values for quick calculations:
\(\log_{10}(1) = 0\) \(\implies\) 0 dB gain (no change)
\(\log_{10}(2) \approx 0.3\) \(\implies\) 3 dB gain (power doubles)
\(\log_{10}(10) = 1\) \(\implies\) 10 dB gain (power increases by 10x)
\(\log_{10}(100) = 2\) \(\implies\) 20 dB gain (power increases by 100x)
For every factor of 10 increase in power gain, you add 10 dB.

Answer 1

The Correct Option is A

Step 1: Concept Definition:
The decibel (dB) is a logarithmic unit for expressing ratios of physical quantities, typically power or intensity. This logarithmic scale is employed for its ability to represent extreme ratios manageably and its closer alignment with human sensory perception.
Step 2: Formula/Methodology:
The decibel conversion for power gain (\( A_p = P_{out}/P_{in} \)) is given by:
\[ G_{dB} = 10 \log_{10}(A_p) \] For voltage or current gain (\( A_v \) or \( A_i \)), assuming constant impedance, the formula is \( 20 \log_{10}(A_v) \) or \( 20 \log_{10}(A_i) \).
Step 3: Calculation Breakdown:
1. Input Value Identification:
The power gain provided is \( A_p = 100 \).
2. Decibel Formula Application (Power):
Substitute the value into the decibel formula for power:
\[ G_{dB} = 10 \log_{10}(100) \] 3. Logarithm Evaluation:
Calculate the logarithm: \( \log_{10}(100) = \log_{10}(10^2) = 2 \).
4. Final Calculation:
Determine the final decibel value:
\[ G_{dB} = 10 \times 2 = 20 \, \text{dB} \] Step 4: Conclusion:
A power gain of 100 corresponds to 20 dB.