Transcript Chapter 6: Data Transmission

University of Palestine
Faculty of Information Technology
ITGD4103 Data Communications and Networks
Lecture-9: Communication Techniques ,Spectrum and bandwidth
week 10- q-2/ 2008
Dr. Anwar Mousa
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Communication Techniques
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Major topics
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This chapter
 explains the differences between analog and digital
transmission
 discusses Transmission impairments and channel
capacity
 describes how digital data can be encoded by means
of a modem so that they can be transmitted over
analog telephone lines
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Analog and digital
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Analog data vs. digital data
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Analog signal vs. digital signal
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Analog: Continuous values on some interval (sound, light,
temperature, pressure)
Digital: Discrete values (text, integers, binary)
Analog: Continuously varying electromagnetic wave
Digital: Series of voltage pulses (square wave)
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Analog Data-->Signal Options
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Analog data to analog signal
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Inexpensive, easy conversion (e.g. telephone)
Data may be shifted to a different part of the available
spectrum (multiplexing)
Used in traditional analog telephony
Analog data to digital signal (PCM)
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Requires a codec (encoder/decoder)
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sender converts the voice data by a bit stream, receiver
reconstruct the bit stream to the analog data
Allows use of digital telephony
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Digital Data-->Signal Options
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Digital data to analog signal
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Digital data to digital signal (LINE CODING)
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Requires modem (modulator/demodulator)
Allows use of PSTN (public-switched Telephone
Network) to send data
Necessary when analog transmission is used
More reliable because no conversion is involved
Less expensive when large amounts of data are involved
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Advantages of
digital transmission (1/2)
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Cost
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Data integrity
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the LSI and VLSI technologies has caused a continuing
drop in the cost and size of digital circuitry
the maintenance costs for digital circuitry are a fraction of
those for analog circuitry
with the use of digital repeaters, the effects of noise and
other signal impairments are not cumulative
it is possible to transmit data longer distances and over
lesser-quality lines while maintaining the integrity of the
data
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Advantages of
digital transmission (2/2)
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Capacity utilization
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it has become economical to build transmission links
of very high bandwidth
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it is more easily and cheaply achieved with digital
transmission for a high degree of multiplexing to
effectively utilize the capacity
Security and privacy
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including satellite channels and optical fiber
encryption techniques can be readily applied to digital
data and to analog data that have been digitized
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Digital signals
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Digital signals
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Digital signals usually refers to the
transmission of electromagnetic pulses
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that represent two binary digits, 1 and 0
binary information is generated and then
converted into digital voltage pulses for
transmission
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EXAMPLES
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1. Given the frequencies listed below, calculate the
corresponding periods.
a. 24 Hz
b. 8 MHz
c.140 KHz
Solution
a.
F = 24 Hz
T= 1/f
T= 1/24 = 0.0416= 41.6 ms
b.
F = 8 MHz
T= 1/f
T= 1/(8 x 106)= 0.125µs
c.
F = 140kHz T= 1/f
T= 1/(140 x 1000)= 7.14 µs
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2. Given the following periods, calculate the corresponding
frequencies.
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a. 5 s
b. 12 μs
c. 220 ns
Solution
a.
f=1/T = 1/5= 0.2 Hz
b. f=1/T = 1/(12 x 10-6) = 0.083 x 106 Hz
c. f=1/T = 1/(220 x 10-9) = 4.5 MHz
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4. What
is the bandwidth of a signal that can be decomposed into
five sine waves with frequencies at 20,50,100, and 200 Hz?
All peak amplitudes are the same. Draw the bandwidth.
Solution
B = fh - fl = 200 - 20 = 180 Hz
20
50
100
200
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200
20
180
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5. What is the transmission time of a message sent by a
station if the length of the message one million bytes
(each byte consists of 8 bits) and the data transmission
rate is 200 kbps?
Solution
Transmission time = Message size / data transmission rate =
1 x 106 x 8/ 200 x 103
= 40s
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6. What is the total delay (latency) for a message of size 5 million
bits that is being sent on a link? The length of the link is
2000Km. The speed of electromagnetic wave inside the link is 2
× 108 m/s and the data transmission rate is 5Mbps?
Solution
Message size = 5 million bits = 5 x 106
Distance = 2000Km = 2 x 106 m
Speed = 2 × 108 m/s
data transmission rate = 5 × 106 b/s
Latency = Propagation time + Transmission time
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Propagation time = Distance / Propagation speed
= 2 x 106 /2 × 108 =1/100 = 0.01s
Transmission time = Message size / data transmission rate = 5 x
106 / 5 x 106 = 1 s
Latency = Propagation time + Transmission time
= 0.01
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= 1.01s
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8. Shown in the next figure, two signals with the same
amplitude and phase, but different frequencies. Compare
between the calculated periods for each signal.
Solution
Figure (a) :
T = 1/F
F =12
T = 1/12 = 0.083 s
= 83 ms
Figure (b) :
T = 1/F
F=6
T = 1/6 = 0.166 s
= 166 ms
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9. A nonperiodic composite signal has a bandwidth of 200
kHz, with a middle frequency of 140 kHz and peak
amplitude of 20 V. The two extreme frequencies have
amplitude of 0. Draw the frequency domain of the signal.
Solution
Bandwidth = 200 kHz = Fh - Fl
Where fh is the highest frequency,
and fl is the lowest frequency.
Then
Fh – Fl = 200
(Fh+Fl)/2 = 140
The lowest frequency must be at 40 kHz
and the highest at 240 kHz.
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20 V
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Unit
Seconds (s)
Equivalent
1s
Unit
hertz (Hz)
Equivalent
1 Hz
Milliseconds (ms)
10–3 s
kilohertz (KHz)
103 Hz
Microseconds (ms)
10–6 s
megahertz (MHz)
106 Hz
Nanoseconds (ns)
10–9 s
gigahertz (GHz)
109 Hz
Picoseconds (ps)
10–12 s
terahertz (THz)
1012 Hz
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