To determine the minimum signal power in watts, given that the minimum signal-to-noise ratio (SNR) is 14 dB, we can use the following formula:

$\text{SNR (dB)} = 10 \cdot \log_{10} \left( \frac{P_{\text{signal}}}{P_{\text{noise}}} \right)$

where:

• $P_{\text{signal}}$ is the signal power,
• $P_{\text{noise}}$ is the noise power, given as 0.013 W,
• SNR is the signal-to-noise ratio in decibels, given as 14 dB.

Step-by-Step Solution

1. Convert SNR from dB to a ratio:

   Rearranging the formula to solve for the ratio $\frac{P_{\text{signal}}}{P_{\text{noise}}}$:

   $14 = 10 \cdot \log_{10} \left( \frac{P_{\text{signal}}}{0.013} \right)$

2. Isolate the logarithmic term:

   $\log_{10} \left( \frac{P_{\text{signal}}}{0.013} \right) = \frac{14}{10} = 1.4$

3. Solve for the power ratio:

   Rewrite in exponential form to get the ratio $\frac{P_{\text{signal}}}{0.013}$:

   $\frac{P_{\text{signal}}}{0.013} = 10^{1.4}$

4. Calculate $10^{1.4}$:

   $10^{1.4} \approx 25.12$

   So,

   $\frac{P_{\text{signal}}}{0.013} = 25.12$

5. Solve for $P_{\text{signal}}$:

   $P_{\text{signal}} = 25.12 \times 0.013$

   $P_{\text{signal}} \approx 0.32656 \, \text{W}$

Answer

The minimum signal power $P_{\text{signal}}$ should be approximately:

$P_{\text{signal}} \approx 0.327 \, \text{W}$

Would you like further details or have any questions?

Here are some related questions to deepen understanding:

1. How do you convert between dB and a power ratio?
2. Why is the signal-to-noise ratio critical in communication systems?
3. What impact would increasing the noise power have on the required signal power?
4. How does the SNR affect the quality of a received signal?
5. Could you apply similar calculations if the SNR was in a different frequency band?

Tip: When working with dB values, remember that each increase of 10 dB represents a tenfold increase in the power ratio.