Transcript Lecture 7: Physical Layer 1

• Review: The application layer.
– Network Applications see the network as the
abstract provided by the transport layer:
• Logical full mesh among network end-points
• Reliable and unreliable communication among endpoints
– Application layer protocol issues closely related
to the particular application.
– Some example applications and their protocols
• Email system: SMTP, POP3
• Web: HTTP
• DNS:
• The physical layer.
– Physical layer issues: how to transfer bits correctly.
• How to physically connect computers (what kind of connectors
should we use?)
• How to represent 0’s and 1’s? Timing?
– Components:
• transmitter
• transmission medium
• receiver
• Example: Telephone network.
– transmitter: converts sound waves into vibrating currents
==> electromagnetic waves down to the wire.
– receiver: convert vibrating currents to voice.
– Telephone network: analog transmission of analog signal
• Transmission medium
• Important parameters for the transmission medium
are the capacity and the distance.
• Capacity depends on distance.
Capacity
Distance
----------------------------------------------------------------------------unshield twisted pair
4000Hz
< 10 km
(10Mbps)
(20m??)
coaxial cable
baseband(50-ohm ThinNet) 10-100Mbps
broadband (75-ohm ThickNet) 10-100Mbps
200m (10Base2)
500m
optical fiber
multi-mode
100 Mbps
30km
single-mode
10Gbps
30km
• Bandwidth and Capacity
– Bandwidth: width of the frequency range of
signal or transmission (Hz) e.g. human voice:
100 ~ 3300 Hz, bandwidth 3200, twisted pair:
4kHz.
– Capacity: rate in bits per second
• Baud rate = how many symbols per second
• Bit rate = number of bits / symbol * Baud rate
• How to determine the number of bits per symbol?
– Number of bits/symbol = log_2(number of symbols)
• E.g: eight voltage outputs, how many bits per
symbol?
• Bandwidth and Capacity
– Nyquist's theorem: maximum baud rate for
noiseless channel.
max baud rate = 2 * Bandwidth
– Implication:
(1) max bit rate = 2 * Bandwidth * # of bits /symbol
(2) also applies to the noisy channels.
– Example: A 10kHz bandwidth channel is used
to send binary signals, what is the maximum bit
rate?
– Shannon's theorem: maximum bit rate for noisy
channel.
C = Bandwidth * log_2 (1 + S/N)
(S: strength of signal, N: strength of noise)
S/N is given in the units of decibel(dB), 10log_10(S/N)
signal_to_noise ratio = 20 dB, S/N = ?
– The typical local loop telephone line:
Bandwidth = 4000 Hz, C = ?
S/N=1000,
– Based on Nyquist and Shannon theorems, what is the
key for the transmission media to achieve high data
rate?
The electromagnetic spectrum:
F(Hz)
10^4
10^5
10^6 10^7 10^8 10^9 10^10 10^11 10^12 10^13
satellite
Twisted pair
coax
AM
FM
TV
Conclusion?
10^14
10^15
Fiber
optics
• Analog vs. Digital
– analog: continuous, digital: discrete
– different contexts:
Analog
Digital
---------------------------------------------------------------------------Data: something that has a meaning
voice
text
---------------------------------------------------------------------------Signal: encoded data
continuously sequence
varying
of pulses
electromagnetic (1's, 0's)
wave
-----------------------------------------------------------------------------Transmission: how data transmitted propagate
propagate
wave’s
pulses
-----------------------------------------------------------------------------Computer networks: transmit digital data
Telephone networks: transmit voice
• Data Encoding: map data into signals
– Digital data to digital signal
– NRZ: high 1, low 0
– Manchester: 1: low-high transition, 0: high-low transition
– Digital data to analog signals (example modem)
• Square wave (digital signal) suffers from strong
attenuation and delay distortion.
• modulation: -- make analog signals.
– Amplitude modulation: use two different voltage levels to
represent 0 and 1.
– Frequency modulation: use two different tones to represent
0 and 1.
– Phase modulation: carrier wave is shifted at different
intervals to represent 0 and 1.
• high speed modem:
– Bandwidth in the local loop: 3000HZ,
– maximum baud rate ???,
– how to achieve higher speed (56Kbps modem)?
» many bits per baud,
» a combination of modulation techniques.
» more amplitude levels and more phase intervals -QAM (quadrature Amplitue Modulation)
» Using 2400 baud rate, how many symbols are needed
to achieve 56kbps?
– Analog data to digital signals (example digital
voice)
• 300 - 3400 HZ human voice
• PCM: 3000 HZ with protection: 4000 HZ Samples:
Nyquist Theorem: 8000 samples per second
• Digitization: 8 or 7 bits per sample (logarithmically
spaced) 64 kbps or 56 kbps
• methods to reduce the number of bits per sample
– differential pulse code
– delta modulation
– predictive encoding
– Analog data to analog signals
• radio, TV, telephone
• Simplex and duplex communication
• simplex communication: data travel in one direction
• half-duplex communication: data travel in either
direction, but not simultaneously
• full-duplex communication: data travel in both
direction simultaneously.
• Multiplexing
– combines slow channels into faster channels.
• two schemes:
– Time Division Multiplexing: time domain is divided into
slots, put channels in different time domain.
– Frequency division Multiplexing: frequency spectrum is
divided into logical channels.
– Code division multiplexing.
• In the telephone system:
– basic voice channel: 64kbps
– T1 line: 24 basic channels + 1 bits per 24*8 bits
» 193 bits per 125 us 193*8000 = 1.544Mbps
– T2 line: 4 T1 line (96 basic channels) + extra bits for
framing 6.176 Mbps, actually 6.312Mbps
– T3 line: 7 T2 line (7*96 basic channels) + extra bits for
framing 44.184Mbps, actually 44.736Mbps
– T4 line 6 T3 line .......
– OC-1: 51.84Mbps
– OC-3: 3 OC-1 155.52
– OC-9 .....