Question

Modulation and Encoding

1. Telephone channels have a bandwidth of about 3.1 kHz. Answer the following questions. (If you like, you can use Excel to do the calculations and then cut/paste your analysis into your homework).

a) If a telephone channel’s signal-to-noise ratio (SNR) is 2,000 (the signal strength is 2,000 times as much as the noise strength), use the Shannon Equation to calculate the maximum data rate of a telephone channel?

b) How fast could a telephone channel carry data if the SNR were increased massively, from 2,000 to 20,000? (Note: This would not be realistic in practice.)

c) With an SNR of 2,000, how fast could a telephone channel carry data if the bandwidth were increased to 4 kHz? Show your work or no credit.

d) What do you think is a better way to increase data rate, and why?

Answer 1

a) Shannon's Equation to calculate the maximum data rate or capacity of a noisy channel.

It is expressed as:

Capacity = Bandwidth × log₂( 1+SNR )

Here, Capacity is the maximum data rate of the channel in bps

Bandwith = 3.1 KHz = 3100 Hz

SNR = 2000

Capacity = 3100 × log₂( 1+2000 )

= 3100 x 10.966505451905741

= 33996.16690089 bps = 33.996 Kbps

b. Bandwith = 3.1 KHz = 3100 Hz

SNR = 20000

Capacity = 3100 × log₂( 1+20000 )

= 3100 x 14.287784512498186

= 44292.13198999 bps = 44.292 Kbps

The Capacity of the Channel will be increased , if the SNR is increased from 2000 to 20000

c. Bandwith = 4 KHz = 4000 Hz

SNR = 2000

Capacity = 4000 × log₂( 1+2000 )

= 4000 x 10.966505451905741

= 43866.0218204 bps = 43.866 Kbps

d. The Better way to increase the Data rate is to increase the Bandwidth rather than increasing the SNR

Because, small variation in the bandwidth yields higher variation in the data rate. Moreover the clarity will be good if we increase the Bandwidth