This question asks about the effect of a 3 dB increase in power level on the actual power.

Understanding Decibels (dB) and Power

Decibels (dB) are a logarithmic unit used to express the ratio between two values of a physical quantity, such as power or intensity. In electronics and communications, decibels are commonly used to represent gains or losses in signal power.

Calculating Power Change with dB Increase

The formula relating a power ratio to decibels (dB) is:

$ \text{dB} = 10 \log_{10} \left( \frac{P_{\text{out}}}{P_{\text{in}}} \right) $

Where:

• $ \text{dB} $ is the difference in decibels.
• $ P_{\text{out}} $ is the output power.
• $ P_{\text{in}} $ is the input power.
• $ \log_{10} $ denotes the base-10 logarithm.

We are given an increase of 3 dB. We want to find the ratio $ \frac{P_{\text{out}}}{P_{\text{in}}} $ when $ \text{dB} = 3 $.

Step-by-Step Calculation

1. Set the decibel value in the formula:

   $ 3 = 10 \log_{10} \left( \frac{P_{\text{out}}}{P_{\text{in}}} \right) $

2. Isolate the logarithm term by dividing both sides by 10:

   $ \frac{3}{10} = \log_{10} \left( \frac{P_{\text{out}}}{P_{\text{in}}} \right) $

   $ 0.3 = \log_{10} \left( \frac{P_{\text{out}}}{P_{\text{in}}} \right) $

3. Convert the logarithmic equation to an exponential equation. Remember that $ \log_{10}(x) = y $ is equivalent to $ 10^y = x $:

   $ \frac{P_{\text{out}}}{P_{\text{in}}} = 10^{0.3} $

4. Calculate the value of $ 10^{0.3} $:

   $ 10^{0.3} \approx 1.995 $

   This value is very close to 2.

5. Interpret the result:

   $ \frac{P_{\text{out}}}{P_{\text{in}}} \approx 2 $

   This means that $ P_{\text{out}} \approx 2 \times P_{\text{in}} $. The output power is approximately double the input power.

Conclusion

Therefore, with every increase of 3 dB in the power level, the power approximately doubles.