Question

1. How many bits per second can be sent over a communications system that has a baud rate of 4800 with 32 signal levels?

2. (a) (10 points) What is the maximum data rate for a 64-level system with an analog bandwidth of 400 MHz?

(b) (10 points) What is the maximum data rate if that same system is subject to noise, with an average power level of 600 and an average noise level of 200?

Include units in your final answer.

Answer 1

Solution:

Q1 Bit rate = baud rate x log₂S

= 4800 x log₂32

= 4800 x 5

= 24000 bits/s

Q2 (a) According to Nyquist Bit Rate formula,

Bit Rate = 2 x bandwidth x log₂Levels

  = 2 x 4 x 10⁶ Hz x log₂(64)

= 8 x 10⁶ x 6

= 48 x 10⁶ bits/s = 48 Mbps

(b) According to Shannon Capacity,

Capacity(Bit Rate) = bandwidth x log₂(1 + SNR) {SNR -> Singnal to Noise Ratio}

  = 4 x 10⁶ Hz x log₂(1 + (600/200))

= 4 x 10⁶ x log₂(4)  

= 4 x 10⁶ x 2

= 8 x 10⁶ = 8 Mbps