Question

#22. If a system has an input power level of 8000, and an output power level of 5000, what is the difference when expressed in dB? #23. If a system has an average power level of 350, an average noise level of 50, and a bandwidth of 100 MHz, what is the effective limit on channel capacity?

Answer 1

22) Input power level (IN) = 8000

Output power level (OUT) = 5000

Difference expressed in dB = 10 log₁₀ [OUT/IN]

or, dB = 10 log₁₀ [5000/8000]

or, dB = 2.05 (Approx.)

The difference is 2.05 dB.

23) Average power level (S) = 350

Average noise level (N) = 50

Bandwidth (B) = 100 MHz

Channel capacity C = B log₂ (1 + S/N)

or, C = 100 log₂ (1 + 350/50)

or, C = 100 log₂ (1 + 7)

or, C = 100 log₂ 8

or, C = 100 log₂ 2³

or, C = 100*3 log₂ 2

or, C = 100 * 3 * 1 [as log₂ 2 = 1]

or, C = 300 bps

So, the channel capacity is 300 bps.

Hope this helps.