• 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 end-points

– 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 200m (10Base2) broadband (75-ohm ThickNet) 10-100Mbps 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: S/N=1000, Bandwidth = 4000 Hz, C = ?

– 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 10^810^7 10^9 10^1110^10 10^12 10^13 10^1510^14

Twisted pair

coax

AM FM

TV

satellite Fiberoptics

Conclusion?

• 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 .....