For an amplifier operating at 27 °C with.a bandwidth of 20KHz.determine a) thermal noise power in watts and dBm, b) Calculatenoise power in watts if the bandwidth is doubled, c) RMS noisevoltage for 50 ohm load resistance and 50 ohm internal resistance

a) To determine the thermal noise power in watts, we can use the formula:

P = k * T * B

where P is the power in watts, k is Boltzmann's constant (1.38 x 10^-23 J/K), T is the temperature in Kelvin, and B is the bandwidth in Hz.

First, we need to convert the temperature from Celsius to Kelvin:

T = 27 + 273 = 300 K

Next, we can calculate the thermal noise power:

P = (1.38 x 10^-23 J/K) * (300 K) * (20,000 Hz)

P ≈ 8.28 x 10^-17 W

To convert the power to dBm, we can use the formula:

P(dBm) = 10 * log10(P/1mW)

P(dBm) = 10 * log10(8.28 x 10^-17 W / 1 x 10^-3 W)

P(dBm) ≈ -166.6 dBm

b) If the bandwidth is doubled to 40 KHz, we can use the same formula to calculate the new thermal noise power:

P' = (1.38 x 10^-23 J/K) * (300 K) * (40,000 Hz)

P' ≈ 1.66 x 10^-16 W

c) To calculate the RMS noise voltage for a 50 ohm load resistance and 50 ohm internal resistance, we can use the formula:

V = sqrt(4 * k * T * R * B)

where V is the RMS noise voltage, k is Boltzmann's constant, T is the temperature in Kelvin, R is the resistance, and B is the bandwidth in Hz.

Using the given values:

V = sqrt(4 * (1.38 x 10^-23 J/K) * (300 K) * (50 ohm) * (20,000 Hz))

V ≈ 1.92 x 10^-7 V