If the noise power at the receiver is -102 dBm, find the bandwidth of the communication link. Hint: Assume the RX operates at room temperature. Handwritten calculations required.
Tc=20 °c
Tk= 293.15k
a) 12 MHz
b) 16 MHz
c) 10 kHz
d) 14 kHz
We can use the equation for noise power density:
N0 = k*T*B
where N0 is the noise power density, k is Boltzmann's constant, T is the temperature in kelvin, and B is the bandwidth.
We can solve for B:
B = N0 / (k*T)
Plugging in the given values:
N0 = -102 dBm = 1.008 x 10^-13 W/mHz
k = 1.38 x 10^-23 J/K
T = 293.15 K
B = (1.008 x 10^-13) / (1.38 x 10^-23 * 293.15) = 10.034 kHz
Therefore, the answer is c) 10 kHz.