Question

Suppose a radio device receives a signal with the power of 6.3 W, the noise power is 100 mW (milliwatt), and the channel has a bandwidth B=50MHz. According to Shannon formula, the channel capacity is Mbits/s.

Answer 1

Using the Shannon formula, the channel capacity of a radio device with a signal power of 6.3 W, a noise power of 100 mW, and a bandwidth of 50 MHz is calculated to be 300 megabits per second (Mbps).

Step-by-step explanation:

The question is asking for the channel capacity in megabits per second, which is a concept in communications technology that relates to the maximum rate at which information can be transmitted over a communication channel. According to the Shannon formula (or Shannon-Hartley theorem), the channel capacity C can be calculated using the formula C = B * log2(1 + S/N) where C is the channel capacity in bits per second, B is the bandwidth of the channel in Hz, S is the signal power in Watts, and N is the noise power in Watts.

Given the signal power of 6.3 W, noise power of 100 mW (0.1 W), and a channel bandwidth of 50 MHz (50 x 10⁶ Hz), we can calculate the channel capacity as follows:

C = 50 x 10⁶ * log2(1 + 6.3 / 0.1) bits per second

First, calculate the signal-to-noise ratio (S/N):

S/N = 6.3 W / 0.1 W = 63

Then incorporate it into the Shannon formula:

C = 50 x 10⁶ * log2(64)

The log base 2 of 64 is 6, so:

C = 50 x 10⁶ * 6 bits per second

C = 300 x 10⁶ bits per second or 300 megabits per second (Mbps).

Therefore, the channel capacity of the radio device is 300 Mbps.