What is the input-to-output power ratio of a coaxial cable that has 1 dB of attenuation?

What is the input-to-output power ratio of a coaxial cable that has 1 dB of attenuation?

• A. 1:1
• B. 1.5:1
• C. 1.2589:1
• D. 2:1

Answer: (C)

The input-to-output power ratio of a coaxial cable with 1 dB of attenuation can be determined using the formula that relates decibels (dB) to power ratios. The formula is: \[ \text{attenuation (in dB)} = 10 \log_{10} \left( \frac{P_{\text{in}}}{P_{\text{out}}} \right) \] where \( P_{\text{in}} \) is the input power and \( P_{\text{out}} \) is the output power. When the attenuation is given as 1 dB, we can rearrange the formula to find the power ratio: \[ 1 = 10 \log_{10} \left( \frac{P_{\text{in}}}{P_{\text{out}}} \right) \] To isolate the power ratio, we divide both sides by 10: \[ 0.1 = \log_{10} \left( \frac{P_{\text{in}}}{P_{\text{out}}} \right) \] By converting the logarithmic expression back to its exponential form, we find: \[ \frac{P_{\text{in}}}{P

The input-to-output power ratio of a coaxial cable with 1 dB of attenuation can be determined using the formula that relates decibels (dB) to power ratios. The formula is:

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

where ( P_{\text{in}} ) is the input power and ( P_{\text{out}} ) is the output power. When the attenuation is given as 1 dB, we can rearrange the formula to find the power ratio:

[ 1 = 10 \log_{10} \left( \frac{P_{\text{in}}}{P_{\text{out}}} \right) ]

To isolate the power ratio, we divide both sides by 10:

[ 0.1 = \log_{10} \left( \frac{P_{\text{in}}}{P_{\text{out}}} \right) ]

By converting the logarithmic expression back to its exponential form, we find:

[ \frac{P_{\text{in}}}{P