1. 1 Ch 6 Long-Distance Communication Carriers, Modulation, and Modems

2. 2 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

3. 3 Illustration of a Carrier  Carrier Usually a sine wave Oscillates continuously Frequency of carrier fixed (f = 1/T) T

4. 4 Types of Modulation  Amplitude modulation (used in AM radio)  Frequency modulation (used in FM radio)  Phase shift modulation (used for data)

5. 5 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

6. 6 The answer is 01101001 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.

7. 7 Illustration of Phase-Shift Modulation  A change in phase (phase shift) encodes K>1 bits  Data rate higher than carrier bandwidth

8. 8 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

9. 9 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.

10. 10 Modem  Name abbreviates mo dulator / dem odulator 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

11. 11  Modulator on one modem connects to demodulator on other (4-wire circuit)  Telephone companies allow companies to lease a circuit between any two locations Modem

12. 12 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

13. 13 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

14. 14 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

15. 15 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

16. 16 Relationship Between Digital Throughput and Bandwidth  Given by Nyquist’s theorem: D = 2 B log 2 K where D is maximum data rate B is hardware bandwidth K is number of values used to encode data

17. 17 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 log 2 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

18. 18 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

19. 19 Shannon’s Theorem  Gives capacity in presence of noise: C = B log 2 (1 + S/N) (  B*(dB/3.01) ……Thompson formula )  where C is the effective channel capacity in bits per second B is hardware bandwidth S is the average power ( signal ) N is the noise  S/N is signal-to-noise ratio B=log 10 (S/N) bels decibel (dB): 1dB=0.1 bel eg, 30dB  3=log10(S/N)  S/N=1000

20. 20 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 little hope of exceeding 28.8 Kbps

21. 21 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 * log 2 4 = 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.

22. 22 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

23. 23 General Concept of Multiplexing  Separate pairs of communications travel across shared channel

24. 24 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

25. 25 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) MUXMUX MUXMUX A3B3C3D3A3B3C3D3 A2B2C2D2A2B2C2D2 A1B1C1D1A1B1C1D1 frames A 4 A 3 A 2 A 1 B 4 B 3 B 2 B 1 D 4 D 3 D 2 D 1 C 4 C 3 C 2 C 1 A 4 A 3 A 2 A 1 B 4 B 3 B 2 B 1 C 4 C 3 C 2 C 1 D 4 D 3 D 2 D 1

26. 26 Frequency Division Multiplexing (FDM)  Multiple items transmitted simultaneously  Uses different carrier frequencies, or “channels”  E.g., Radio, cable TV MUXMUX MUXMUX guard band channel 3 channel 2 channel 1

27. 27 Wave Division Multiplexing  When applied to light, FDM is called wavelength division multiplexing (WDM) Informally called color division multiplexing

28. 28 Scientific Principle Behind Frequency Division Multiplexing  Note: this is the same principle that allows a cable TV company to send multiple television signals across a single cable Two or more signals that use different carrier frequencies can be transmitted over a single medium simultaneously without interference