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: Understanding the Concept:
The decibel (dB) is a logarithmic unit used to express the ratio of two values of a physical quantity, often power or intensity. It is used because it can represent very large or very small ratios in a more manageable way and because logarithmic scales often correspond better to human perception.
Step 2: Key Formula or Approach:
The formula to convert a power gain (which is a ratio, \( A_p = P_{out}/P_{in} \)) to decibels is:
\[ G_{dB} = 10 \log_{10}(A_p) \] For voltage or current gain (\( A_v \) or \( A_i \)), the formula is \( 20 \log_{10}(A_v) \) or \( 20 \log_{10}(A_i) \), assuming the impedance is constant.
Step 3: Detailed Explanation:
1. Identify the given value:
The power gain is given as a ratio, \( A_p = 100 \).
2. Substitute the value into the decibel formula for power:
\[ G_{dB} = 10 \log_{10}(100) \] 3. Evaluate the logarithm:
We know that \( \log_{10}(100) = \log_{10}(10^2) = 2 \).
4. Calculate the final value:
\[ G_{dB} = 10 \times 2 = 20 \, \text{dB} \] Step 4: Final Answer:
A power gain of 100 is equivalent to 20 dB.