Infsci 1004 - Homework Assignment

Homework Assignment #2 Solutions

Questions

  1. Explain the significance of the Shannon-Hartley theorem.
  • Given a S/N ratio over a communication channel and the channel bandwidth for a particular distance, there is a maximum rate or capacity at which one can transmit data over that channel.
  • The capacity is given by

C = W log2 (!+S/N)

where, C = channel capacity in bits per second.

W = channel bandwidth in Hertz.

S/N = signal-to-noise power ratio.

  • One can transmit at rates higher than C if the extra errors are tolerable.
  1. Compare and contrast the time domain with the frequency domain.
  • The time-domain characteristics give the amplitude, frequency and phase of a signal with respect to time.
  • The frequency-domain characteristics give the frequency components that make up any arbitrary signal in the time-domain. It gives us the bandwidth that a particular type of signal will need.

Problems

  1. A telephone channel has a bandwidth of 3100 Hz.
  2. If the minimum frequency is 300Hz, what is the maximum frequency?

Maximum Frequency = Minimum Frequency + Bandwidth = 3400 Hz.

  1. What is the maximum capacity of the channel if the signal to noise ratio is 20dB? 30dB?

C = W log (1+S/N)

2

S/N = 10 ** (SNR/10)

SNR(dB) / S/N / Capacity (bits per sec)
20 / 100 / 20,600 (approx)
30 / 1000 / 30,800 (approx)
  1. Most stereo systems reproduce sound from about 20Hz to about 20KHz, based on the hearing ability of the human ear, which is much higher than the bandwidth of the telephone channel. Speculate why the telephone channel's bandwidth is so much lower than the human ear's audible bandwidth.

Human voice generates frequencies in the range of 300 to 3400 Hz. Hence, if we provide about 4 kHz of BW (including a guard band)per telephone channel, it would be sufficient for voice quality. Hence the main reasons are:

  • Acceptable Voice Quality when reproduced at the receiving end.
  • Low Cost.
  • Savings in Bandwidth.
  1. There are many tradeoffs inherent in the design of codes for information transmission. Explore these tradeoffs by answering the following:
  2. Compute the number of bits that would be necessary to transmit the uppercase alphabet, digits from 0 to 9, and the punctuation marks (.,'!)?.

There are 44 symbols in all. So it would require 6 bits per character since

2 ** 6 =64 > 44.

  1. Compare the bit efficiency of this code with the Baudot code and with the ASCII code for transmitting the strings DARTH VADER and R2D2.C3P0 (for Baudot, assume that the "." character is a "letter")?

Bit Efficiency / DARTH VADER /

R2D2.C3PO

Information Bits / Transmitted bits / Information Bits / Transmitted bits
6-bit code / 10 / 10 / 9 / 9
ASCII / 10 / 10 / 9 / 9
Baudot / 10 / 11 / 9 / 16
Bit Efficiency / DARTH VADER /

R2D2.C3PO

6-bit code / 100 / 100
ASCII / 100 / 100
Baudot / 91 / 56
  1. If the bit rate of a channel is 150bps, what is the difference in transmission times for the three codes? At 1200bps?

Transmission time / DARTH VADER /

R2D2.C3PO

150 bps / 1200 bps / 150 bps / 1200 bps.
6-bit code / 60/150=0.4 sec / 60/1200=0.05 sec / 54/150=0.36 sec / 54/1200=0.045 sec
ASCII / 70/150=0.46 sec / 70/1200=0.06 sec / 63/150=0.42 sec / 63/1200=0.052 sec
Baudot / 55/150= 0.36 sec / 55/1200= 0.045 sec / 80/150= 0.53 sec / 80/1200= 0.06 sec
  1. What do you conclude about the different strategies used in each of the codes from the previous calculations?
  • The lesser the number of bits, the smaller the transmission time.
  • The lesser the overhead the higher the efficiency.