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Determine the decibel values of thermal noise power (n) and thermal noise power density (no) given the following: t=290 degrees kelvin, b = 48mhz?

User EnGassa
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Final answer:

The decibel value of thermal noise power (n) at 290 degrees Kelvin is approximately -174 dBm/Hz, and the decibel value of thermal noise power density (n₀) for a bandwidth of 48 MHz is approximately -134 dBm.

Step-by-step explanation:

The thermal noise power (n) can be calculated using the formula n = k * T * B, where k is the Boltzmann constant (1.38 ×
10^(-23) J/K), T is the temperature in Kelvin, and B is the bandwidth. Substituting the given values (T = 290 K and B = 48 MHz), we get n = 1.38 ×
10^(-23) * 290 * 48 × 10^6, which simplifies to approximately 1.92 ×
10^{(-14) W. Converting this to decibels using the formula 10 * log₁₀(n) gives us the decibel value of thermal noise power (n) as approximately -174 dBm/Hz.

The thermal noise power density (n₀) is calculated by dividing the thermal noise power (n) by the bandwidth (B), and in decibels, it is expressed as 10 * log₁₀(n/B). Substituting the given values, we get 10 * log₁₀(1.92 × 10^(-14)/48 ×
10^6), which simplifies to approximately -134 dBm. Therefore, the decibel value of thermal noise power density (n₀) for a bandwidth of 48 MHz is approximately -134 dBm.

Understanding the decibel values of thermal noise power and thermal noise power density is crucial in communication systems, especially when assessing signal-to-noise ratios and optimizing the performance of electronic devices in the presence of thermal noise.

User Alineat
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