CS 536 ParkInformation Transmission under NoiseUncertainty introduced by noise:−→ encoding/decoding: a 7→ wa7→ w 7→ [?]−→ wagets corrupted, i.e., becomes w−→ if w = wb, incorrectly conclude b as symbol−→ detect w is corrupted: error detection−→ correct w to wa: error correctionCS 536 ParkWould like: If received code word w = wcfor some sym-bol c ∈ Σ, then probability that actual symbol sent isindeed c is high.−→ Pr{symbol sent = c | w = wc}≈1−→ noiseless channel: special case (prob = 1)In practice, w may not match any legal code word:−→ for all c ∈ Σ, w 6= wc−→ then what?CS 536 ParkFundamental limitation to reliable data transmission:Channel capacity C: maximum achievable reliable datatransmission rate (bps) over a noisy channel (dB) withbandwidth W (Hz).Channel Coding Theorem (Shannon): Given band-width W , signal power PS, noise power PN, channel sub-ject to white noise,C = W logµ1+PSPN¶bps.PS/PN: signal-to-noise ratio (SNR)−→ upper bound achieved by using longer codes−→ detailed set-up/conditions omittedCS 536 ParkIncreasingly important for modern day networking:• Power control (e.g., pocket PCs)→ trade-off w.r.t. battery power→ trade-off w.r.t. multi-user interference→ signal-to-interference ratio (SIR)• Recent trend: software radio→ hardware-to-software migration→ configurableCS 536 ParkSignal-to-noise ratio (SNR) is expressed asdB = 10 log10(PS/PN).Example: Assuming a decibel level of 30, what is thechannel capacity of a telephone line?Answer: First, W = 3000 Hz, PS/PN= 1000. UsingChannel Coding Theorem,C = 3000 log 1001 ≈ 30 kbps.−→ compare against 28.8 kbps modems−→ what about 56 kbps modems?−→ DSL lines?CS 536 ParkDigital vs. Analog TransmissionTwo forms of transmission:• digital transmission: data transmission using squarewaves• analog transmission: data transmission using all otherwavesFour possibilities to consider:• analog data via analog transmission→ “as is” (e.g., radio)• analog data via digital transmission→ sampling (e.g., voice, audio, video)• digital data via analog transmission→ broadband & wireless (“high-speed networks”)• digital data via digital transmission→ baseband (e.g., Ethernet)CS 536 ParkWhy consider digital transmission?Common to both: problem of attenuation.• decrease in signal strength as a function of distance• increase in attenuation as a function of frequencyRejuvenation of signal via amplifiers (analog) and re-peaters (digital).CS 536 ParkDelay distortion: different frequency components travelat different speeds.Most problematic: effect of noise−→ thermal, interference, ...• Analog: Amplification also amplifies noise—filteringout just noise, in general, is a complex problem.• Digital: Repeater just generates a new square wave;more resilient against ambiguitity.CS 536 ParkAnalog Transmission of Digital DataThree pieces of information to manipulate: amplitude,frequency, phase.• Amplitude modulation (AM): encode bits using am-plitude levels.• Frequency modulation (FM): encode bits using fre-quency differences.• Phase modulation (PM): encode bits using phase shifts.00011 1CS 536 ParkFM radio uses ... FM!AM radio uses ... AM!iPod & radio experiment uses ... ?Why is FM radio clearer (“high fidelity”) than AM radio?Broadband uses ... ?CS 536 ParkBaud Rate vs. Bit RateBaud rate: Unit of time within which carrier wave canbe altered for AM, FM, or PM.−→ signalling rate−→ e.g., clockNot synonymous with bit rate: e.g., AM with 8 levels,PM with 8 phases−→ bit rate (bps) = 3 × baud rate... less than one bit per baud?CS 536 ParkBroadband vs. BasebandPresence or absence of carrier wave: allows many channelsto co-exist at the same time−→ frequency division multiplexing (FDM)MUXChannel F1Channel F2Channel F3Channel F4DEMUXEx.: AM radio (535 kHz–1705 kHz)−→ tuning to specific frequency: Fourier transform−→ coefficient (magnitude) carries bit informationCS 536 ParkEx.: FM radio−→ 88 MHz–108 MHz−→ 200 kHz slices−→ how does it work?−→ better or worse than AM?Ex.: Digital radio−→ digital audio radio service−→ GEO satellites (a.k.a. satellite radio)−→ uses 2.3 GHz spectrum (a.k.a. S-band)−→ e.g., XM, SiriusCS 536 ParkIn the absence of carrier wave, can still use multiplexing:−→ time-division multiplexing (TDM)M12341 1223344UXXUEDM• digital transmission of analog data→ first digitize→ PCM (e.g., PC sound cards), modem• digital transmission of digital data→ e.g., telephony backbone networkCS 536 ParkExample: T1 carrier (1.544 Mbps). . .Channel 1 Channel 2 Channel 2412345678 1234567812345678frame bit control bitOne Frame (193 bits)• 24 simultaneous users• 7 bit quantizationAssuming 4 kHz telephone channel bandwidth, SamplingTheorem dictates 8000 samples per second.−→ 125 µsec inter-sample intervalBandwidth = 8000 × 193 = 1.544 MbpsCS 536 ParkDigital transmission of digital dataDirect encoding of square waves using voltage differen-tials; e.g., -15V–+15V for RS-232-C.• NRZ-L (non-return to zero, level)• NRZI (NRZ invert on ones)• Manchester (biphase or self-clocking codes)00011 1−→ baud rate vs. bit rate of Manchester?CS 536 ParkTrade-offs:• NRZ codes—long sequences of 0’s (or 1’s) causes syn-chronization problem; need extra control line (clock)or sensitive signalling equipment.• Manchester codes—synchronization achieved throughself-clocking; however, achieves only 50% efficiencyvis-`a-vis NRZ codes.4B/5B codeEncode 4 bits of data using 5 bit code where the codeword has at most one leading 0 and two trailing 0’s.0000 ↔ 11110, 0001 ↔ 01001, etc.−→ at most three consecutive 0’s−→ efficiency: 80%CS 536 ParkMultiplexing techniques:• TDM• FDM• mixture (FDM + TDM); e.g., TDMA• CDMA (code division multiple access) or spread spec-trum→ wireless communication→ competing scheme with