Question

A receiver has a noise power bandwidth of 12 Khz. A transistor which matches the receiver input impedance is connected across the antenna terminal of the receiver. Calculate the noise power contributed by the transistor within the receiver bandwidth assuming an operation temperature of 30୭ C. Express the obtained value in units of dBm.

Answer 1

The noise power contributed by the transistor in the receiver within the bandwidth can be calculated using the formula Noise Power = kTB, where k is Boltzmann's constant, T is the temperature in Kelvin, and B is the noise power bandwidth. Plugging in the given values, the noise power is determined to be -174.6 dBm.

Step-by-step explanation:

The noise power contributed by the transistor can be calculated using the formula:

Noise Power = kTB

Where k is Boltzmann's constant (1.38 x 10^-23 J/K), T is the temperature in Kelvin, and B is the noise power bandwidth. In this case, the temperature is given as 30୭ C, which is equivalent to 303.15 K. The noise power bandwidth is given as 12 kHz, which is equivalent to 12 x 10^3 Hz.

Plugging in the values:

Noise Power = (1.38 x 10^-23 J/K) x (303.15 K) x (12 x 10^3 Hz)

Noise Power = 5.06 x 10^-18 W

To convert the noise power to dBm:

Noise Power (dBm) = 10log10(Noise Power / 10^-3)

Noise Power (dBm) = 10log10(5.06 x 10^-18 W / 10^-3)

Noise Power (dBm) = -174.6 dBm