Transcript Computer Networks and Internets

Ch 6 Long-Distance
Communication
Carriers, Modulation, and Modems
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Sending Signals across Long
Distances
Important fact:
Current becomes weaker as it travels (signal loss)
 A continuous, oscillating signal travels farther than
direct current

For long-distance communication
Send a sine wave (called a carrier wave)
 Modifies (modulate) the carrier to encode date

modulated carrier
technique used for telephone,
radio, and television
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Illustration of a Carrier
Carrier
Usually a sine wave
 Oscillates continuously
 Frequency of carrier fixed (f = 1/T)

T
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Types of Modulation
Amplitude modulation (used in AM
radio)
Frequency modulation (used in FM radio)
Phase shift modulation (used for data)
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Illustration of AM
 Strength of
signal encodes 0 (1/3 full strength) or 1 (2/3 strength)
 Receiver monitors incoming carrier, detects modulation,
reconstruct the original data, and discard the carrier
 At least a half cycle of wave needed for each bit
 Data rate limited by carrier bandwidth
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Ex. The wave form diagram below shows the transmission of 8 bits
using amplitude modulation where a high amplitude represents a 1 bit
and a low amplitude represents a 0 bit. Each bit is transmitted in half
of a wavelength. Give the value of the 8 bits being transmitted.
The answer is 01101001
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Illustration of
Phase-Shift Modulation
 A change in phase (phase shift) encodes K>1
Data
bits
rate higher than carrier bandwidth
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Phase-Shift Example
 Section
of wave is omitted at phase shift
 Data bits determine size of omitted section

8 Patterns vs. 8 possibilities of 3-bit strings
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Ex. A communication system transmits information by sending
one of 8 different voltage levels (patterns) down a wire every
millisecond.
(1) What is the baud rate of the system?
(2) How many bits per second are being transmitted?
Answer:
(1) The baud rate is the signals per second or 1000.
(2) Each signal can be any one of 8 values. It takes 3 bits to
represent the 8 different values since log2(8) = 3.
There are 3000 bits per second being transmitted.
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Modem

Name abbreviates modulator / demodulator
transforms data into a modulated signal
 extracts data from modulated signal


Hardware device that contains separate
circuitry for
Modulation of outgoing signal
 Demodulation of incoming signal


Used for long-distance communication

One modem at each end
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Modem
 Modulator on
one modem connects to demodulator
on other (4-wire circuit)
 Telephone companies allow companies to lease a
circuit between any two locations
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Types of Modems
Conventional
Use four wires
 Transmit modulated electrical wave

Optical
Use glass fibers
 Transmit modulated light

Wireless
Use air / space
 Transmit modulated RF wave

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Dialup Modems
The modem can simulate lifting the
handset,
dialing, or hanging up the phone
Transmit modulated audio tone

Carrier is the tone heard if one lifts the handset
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Modem Terminology
Full-duplex modem
Provides
2-way communication
Allows simultaneous transmission
Uses four wires
Half-duplex modem
Does
provide 2-way communication
Transmits in one direction at any time
Uses only two wires
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Recall

Bandwidth


Maximum times per second signal can change
Throughput
The number of bits per second that can be
transmitted
 Related to underlying hardware bandwidth

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Relationship Between Digital
Throughput and Bandwidth
Given by Nyquist’s theorem:
D = 2 B log2 K
where
D
is maximum data rate
B is hardware bandwidth
K is number of values used to encode data
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Applications of Nyquist’s
Theorem
For RS-232
K
is 2 because RS-232 only uses two values,
+15 or –15 volts, to encode data bits
D is 2 B log2 2 = 2 B
For phase-shift encoding
Suppose
K is 8 (possible shifts)
D is 2 B log2 8 = 2 B x 3 = 6 B
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More Bad News
Physics tells us
that real systems emit and
absorb energy (e.g., thermal)
Engineers call unwanted energy noise
Nyquist’s theorem
Assumes
a noise-free system
Only works in theory
Shannon’s theorem corrects
for noise
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Shannon’s Theorem
 Gives
capacity in presence of noise:
C = B log2 (1 + S/N)
( B*(dB/3.01) ……Thompson formula )
 where
C
is the effective channel capacity in bits per second
 B is hardware bandwidth
B=log10(S/N) bels
 S is the average power (signal)
decibel (dB): 1dB=0.1 bel
 N is the noise
eg,
 S/N
is signal-to-noise ratio
30dB3=log10(S/N)S/N=1000
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Application of Shannon’s Theorem
Conventional telephone system
Engineered
for voice
Bandwidth is 3000 Hz
Signal-to-noise ratio is approximately 30DB (1000)
Effective capacity is
3000 log2 (1 + 1000) = ~30000 bps
Conclusion: dialup modems have
of exceeding 28.8 Kbps
little hope
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Ex. Consider a network connection with 200K Hz of bandwidth.
(1) If the channel is perfectly noiseless and you use a transmission
method that uses 4 different states, what is the maximum transmission
speed you can achieve?
(2) If the channel has 28dB of noise, what it the maximum
transmission speed you can achieve?
(3) How can it be that the noisy channel can send more data than the
ideal channel?
Answer:
(1) Speed = 2 * 200K * log24 = 800K bits/sec
(2) Speed = B * dB / 3.01 = 200K * 28 / 3.01 = 1.86 M bits/sec
(3) The ideal channel is slower because it is not using enough
different states. If it used more states, it would run faster.
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The Bottom Line
Nyquist’s theorem means finding a
way to
encode more bits per cycle improves the data
rate
Shannon’s theorem means that no amount of
clever engineering can overcome the
fundamental limits of a real transmission
system
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General Concept of Multiplexing
 Separate
pairs of communications travel across
shared channel
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Multiplexing Terminology
Multiplexor
Device that accepts data from multiple sources
 Sends data across shared channel

Demultiplexor
Device that extracts data from shared channel
 Sends to correct destination

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Time Division Multiplexing (TDM)
 Slotted
TDM
 Statistical TDM: if a given source does not have data to
send, the multiplexor skips the source (suitable for
burst traffic)
frames
A4 A3 A2 A1
B4 B3 B2 B1
C4 C3 C2 C1
D4 D3 D2 D1
M
M
U
X
A3B3C3D3
A2B2C2D2
A1B1C1D1
U
A4 A3 A2 A1
B4 B3 B2 B1
C4 C3 C2 C1
X
D4 D3 D2 D1
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Frequency Division Multiplexing (FDM)
Multiple items
transmitted simultaneously
Uses different carrier frequencies, or “channels”
 E.g., Radio, cable TV
M
guard
band
M
channel 3
U
channel 2
X
channel 1
U
X
guard
band
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Wave Division Multiplexing

When applied to light, FDM is called
wavelength division multiplexing (WDM)

Informally called color division multiplexing
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Scientific Principle Behind
Frequency Division Multiplexing
Two or more signals that use different carrier
frequencies can be transmitted over a single
medium simultaneously without interference
 Note:
this is the same principle that allows a cable
TV company to send multiple television signals
across a single cable
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