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 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:

2. 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

3. 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

4. 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?

5. 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?

6. –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?

7. The electromagnetic spectrum: F(Hz) 10^410^510^610^810^710^910^1110^1010^1210^1310^1510^14 Twisted pair coax AM FM TV satellite Fiber optics Conclusion?

8. 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

9. 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.

10. 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?

11. –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

12. 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.

13. 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.....