The Signal-to-Noise Ratio (S/N ratio) is a measure of the strength of the signal relative to the noise affecting it. It can be calculated using the formula:

$S/N = \frac{V_{\text{signal}}}{V_{\text{noise}}}$

Where:

• $V_{\text{signal}}$ is the RMS voltage of the signal (1.98 volts in this case).
• $V_{\text{noise}}$ is the RMS voltage of the noise (30 mV or 0.03 volts in this case).

Now, let's compute the S/N ratio:

$S/N = \frac{1.98}{0.03} = 66$

The Signal-to-Noise Ratio (S/N) is 66, meaning the signal is 66 times stronger than the noise.

If you'd like the S/N ratio in decibels (dB), it can be calculated using the formula:

$S/N_{\text{dB}} = 20 \cdot \log_{10} \left(\frac{V_{\text{signal}}}{V_{\text{noise}}}\right)$

Let's calculate that too:

$S/N_{\text{dB}} = 20 \cdot \log_{10} \left(\frac{1.98}{0.03}\right)$

$S/N_{\text{dB}} = 20 \cdot \log_{10}(66)$

I will compute the decibel value for you now.The Signal-to-Noise Ratio (S/N) in decibels (dB) is approximately 36.39 dB.

Would you like any further details or have any other questions?

Here are some related questions you might find interesting:

1. How does the S/N ratio affect the quality of transmitted signals?
2. What are the methods to reduce noise in a signal?
3. How does the S/N ratio differ between analog and digital signals?
4. What are some real-world applications where a high S/N ratio is crucial?
5. Can signal processing techniques improve the S/N ratio?

Tip: Increasing the signal power or reducing noise levels can help improve the Signal-to-Noise Ratio.