An antenna radiates a primary signal of 500 watts output. If there is a 2nd harmonic output of 0.5 watt, what attenuation of the 2nd harmonic has occurred?

An antenna radiates a primary signal of 500 watts output. If there is a 2nd harmonic output of 0.5 watt, what attenuation of the 2nd harmonic has occurred?

• A. 10 dB
• B. 20 dB
• C. 30 dB
• D. 40 dB

Answer: (C)

To understand why the correct answer indicates a 30 dB attenuation of the second harmonic, we need to comprehend how to calculate the attenuation in decibels (dB) based on the power levels given. Attenuation in dB can be calculated with the formula: \[ \text{Attenuation (dB)} = 10 \cdot \log_{10} \left(\frac{P_{\text{input}}}{P_{\text{output}}}\right) \] In this case, the input power (the primary signal output) is 500 watts, and the output power (the second harmonic output) is 0.5 watts. Plugging these values into the formula gives: \[ \text{Attenuation (dB)} = 10 \cdot \log_{10} \left(\frac{500}{0.5}\right) \] First, calculate the power ratio: \[ \frac{500}{0.5} = 1000 \] Next, apply the logarithm: \[ \log_{10}(1000) = 3 \] Now, multiply by 10 to find the attenuation: \[ \text{Attenuation (dB)} = 10

To understand why the correct answer indicates a 30 dB attenuation of the second harmonic, we need to comprehend how to calculate the attenuation in decibels (dB) based on the power levels given.

Attenuation in dB can be calculated with the formula:

[ \text{Attenuation (dB)} = 10 \cdot \log_{10} \left(\frac{P_{\text{input}}}{P_{\text{output}}}\right) ]

In this case, the input power (the primary signal output) is 500 watts, and the output power (the second harmonic output) is 0.5 watts. Plugging these values into the formula gives:

[ \text{Attenuation (dB)} = 10 \cdot \log_{10} \left(\frac{500}{0.5}\right) ]

First, calculate the power ratio:

[ \frac{500}{0.5} = 1000 ]

Next, apply the logarithm:

[ \log_{10}(1000) = 3 ]

Now, multiply by 10 to find the attenuation:

[ \text{Attenuation (dB)} = 10