Data Communication Networks Exam #1
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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.
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i. |
The Open Systems Interconnection (OSI) model consists of seven layers. |
True |
False |
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ii. |
Four layers of the OSI model deal primarily with network operations |
True |
False |
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iii. |
Datagrams are better than virtual circuits in dealing with congestion in networks. |
True |
False |
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iv. |
A store-and-forward network uses message switching. |
True |
False |
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v. |
Packets are units of a message. |
True |
False |
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vi. |
Using odd or even parity increases the number of bits transmitted by one. |
True |
False |
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vii. |
Microwave transmission is a type of satellite transmission. |
True |
False |
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viii. |
Frequency is inversely related to the amplitude. |
True |
False |
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ix. |
Analog signals vary continuously. |
True |
False |
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x. |
Lasers and light emitting diodes (LEDs) are sources of light. |
True |
False |
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xi. |
A periodic signal is a signal which may change over time. |
True |
False |
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xii. |
Sputnik traveled in the geosynchronous orbit. |
True |
False |
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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 |
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xiv. |
In a Huffman code, the characters with the highest frequency have the longest codes. |
True |
False |
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xv. |
It is impossible to have a collision without contention. |
True |
False |
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i. |
Digital-to-Analog Conversion |
AM |
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ii. |
Analog-to-Digital Conversion |
PM |
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iii. |
Period |
Frequency |
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iv. |
Transport Layer |
Buffering |
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v. |
Multiplexing |
Time Division |
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vi. |
Listen to medium protocol |
CSMA |
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vii. |
Air medium protocol |
Aloha |
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viii. |
Signal value: High = 0, Low = 1 |
NRZ |
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ix. |
7-bit code |
ASCII |
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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
Communication frequencies are all different!
Solve by: 1. Change all frequencies to the same, or 2. Handset changes in different geographic areas.
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.
Amplitude = 4, period = 2p /2 - 4 = p - 4, 1/2 period = p /2 - 4.
Not shown here!
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.