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.
a) 12 MHz
b) 16 MHz
c) 10 kHz
d) 14 kHz
To find the bandwidth of the communication link, we can use the following equation:
Noise power (dBm) = -174 dBm/Hz + 10 * log10(Bandwidth)
Where -174 dBm/Hz is the noise power density at room temperature.
Given that the noise power at the receiver is -102 dBm, we can rewrite the equation as:
-102 dBm = -174 dBm/Hz + 10 * log10(Bandwidth)
Now we can solve for the bandwidth:
10 * log10(Bandwidth) = -102 dBm + 174 dBm/Hz
10 * log10(Bandwidth) = 72 dB
log10(Bandwidth) = 7.2
Bandwidth = 10^7.2
Bandwidth ≈ 15.8 MHz
The closest option to this value is:
b) 16 MHz