Data Communication Networks Exam #1

Name:

 

The following exam is worth 20% of your course grade. It consists of xxx questions, each question's point total is given below. Exam #1 covers material from chapters 1, 2, and 3, of the text, and any material discussed/given/assigned during the course thus far.

  1. True or false. (30 points total - two points each). Circle the best answer. If I cannot distinguish your answer, you will not receive the points!
  2. i.

    The Open Systems Interconnection (OSI) model consists of seven layers.

    True

    False

    ii.

    Four layers of the OSI model deal primarily with network operations

    True

    False

    iii.

    Datagrams are better than virtual circuits in dealing with congestion in networks.

    True

    False

    iv.

    A store-and-forward network uses message switching.

    True

    False

    v.

    Packets are units of a message.

    True

    False

    vi.

    Using odd or even parity increases the number of bits transmitted by one.

    True

    False

    vii.

    Microwave transmission is a type of satellite transmission.

    True

    False

    viii.

    Frequency is inversely related to the amplitude.

    True

    False

    ix.

    Analog signals vary continuously.

    True

    False

    x.

    Lasers and light emitting diodes (LEDs) are sources of light.

    True

    False

    xi.

    A periodic signal is a signal which may change over time.

    True

    False

    xii.

    Sputnik traveled in the geosynchronous orbit.

    True

    False

    xiii.

    Quadrature Amplitude Modulation (QAM) is a technique wherein a group of bits is assigned a signal defined by its amplitude and phase shift.

    True

    False

    xiv.

    In a Huffman code, the characters with the highest frequency have the longest codes.

    True

    False

    xv.

    It is impossible to have a collision without contention.

    True

    False

  3. Match the columns. Write the best match (item from right) in the box next to the term on left. (20 points total - two points each.)
  4.  

    i.

    Digital-to-Analog Conversion

    AM

    ii.

    Analog-to-Digital Conversion

    PM

    iii.

    Period

    Frequency

    iv.

    Transport Layer

    Buffering

    v.

    Multiplexing

    Time Division

    vi.

    Listen to medium protocol

    CSMA

    vii.

    Air medium protocol

    Aloha

    viii.

    Signal value: High = 0, Low = 1

    NRZ

    ix.

    7-bit code

    ASCII

    x.

    Data Link Layer

    Error Detection

     

     

     

    Aloha Protocol

    Amplitude Modulation (AM)

    ASCII

    Buffering

    Carrier Sense Multiple Access

    Differential Manchester

    EBCDIC

    Error Detection

    Frequency

    Manchester

    Non-Return-to-Zero (NRZ)

    Pulse Amplitude Modulation (PM)

    Routing

    Time Division

    Wavelength

  5. (25 points) What is one problem with world-wide mobile communication and how can it be solved?
  6. Communication frequencies are all different!

    Solve by: 1. Change all frequencies to the same, or 2. Handset changes in different geographic areas.

     

  7. (5 points) What is the difference between a quantum-bit (qubit) and a binary digit (bit)?
  8. A binary digit has two possibilities, 0 and 1, but not at the same time. A qubit has two states, up and down, or 0 and 1, at the same time.

  9. (10 points) Given the following function, draw the signal and show amplitude and period.
  10.  

    Amplitude = 4, period = 2p /2 - 4 = p - 4, 1/2 period = p /2 - 4.

    Not shown here!

     

  11. (10 point) Given a device operating at 125 baud. Give the bit rate if:

1.Binary symbols are used.

Answer:

When two symbols are used, the bit rate and baud rate are the same. That is because:

2 = 21, so n = 1 bit/symbol

bit rate = baud rate x n = 125 symbols/sec x 1 bit/symbol = 125 bits/sec = 125 bps

2.Multivalued symbols (16 symbols) are used.

Answer:

When more than two symbols are used, the bit rate and baud rate are the different. For 16 symbols:

16 = 24, so n = 4 bits/symbol

bit rate = baud rate x n = 125 symbols/sec x 4 bits/symbol = 500 bits/sec = 500 bps

Kathleen Kaplan, © 2001. Howard University.