Basics of Data TransmissionOur Objective is to understand

Signals, bandwidth, data rate conceptsTransmission impairmentsChannel capacityData Transmission

SignalsA signal is generated by a transmitter and transmitted over a mediumfunction of time function of frequency, i.e., composed of components of different frequenciesAnalog signalvaries smoothly with timeE.g., speechDigital signalmaintains a constant level for some period of time, then changes to another level E.g., binary 1s and 0s

Periodic vs. Aperiodic Signals Periodic signalPattern repeated over times(t+T) = s(t)

Aperiodic signalPattern not repeated over time

Sine WaveThe fundamental periodic signalPeak Amplitude (A)maximum strength of signalvoltsFrequency (f)Rate of change of signalHertz (Hz) or cycles per secondPeriod = time for one repetition (T)T = 1/fPhase ()Relative position in time

Signals in Frequency DomainSignal is made up of many componentsComponents are sine waves with different frequenciesIn early 19th century, Fourier proved that Any periodic function can be constructed as the sum of a (possibly infinite) number of sines and cosines This decomposition is called Fourier seriesf is called the fundamental frequencyan, bn are amplitude of nth harmonicc is a constant

Frequency Domain (contd)Fourier Theorem enables us to represent signal in Frequency Domaini.e., to show constituent frequencies and amplitude of signal at these frequencies Example 1: sine wave:

s(t) = sin(2ft)

Time and Frequency Domains: Example 2Time domain s(t)Frequency domain S(f)

Frequency Domain (contd)So, we can use Fourier theorem to represent a signal as function of its constituent frequencies, and we know the amplitude of each constituent frequency. So what?

We know the spectrum of a signal, which is the range of frequencies it contains, and Absolute bandwidth = width of the spectrum

Q: What is the bandwidth of the signal in the previous example? [sin(2ft) + sin(23ft)] A: 2f Hz

Frequency Domain (contd)Q. What is the absolute bandwidth of square wave?Hint: Fourier tells you thatAbsolute BW = (ooops!!)But, most of the energy is contained within a narrow band (why?) we refer to this band as effective bandwidth, or just bandwidth

Approximation of Square WaveA. BW = 6*f HzUsing the first 3 harmonics, k=1, 3, 5Using the first 4 harmonics, k=1, 3, 5, 7

Q. What is BW in each case?A. BW = 4*f HzCool applet on Fourier Series

Signals and ChannelsSignalcan be decomposed to components (frequencies)spectrum: range of frequencies contained in signal(effective) bandwidth: band of frequencies containing most of the energy Communications channel (link)has finite bandwidth determined by the physical properties (e.g., thickness of the wire) truncates (or filters out) frequencies higher than its BWi.e., it may distort signals can carry signals with bandwidth channel bandwidth

Bandwidth and Data RateData rate: number of bits per second (bps)Bandwidth: signal rate of change, cycles per sec (Hz)Well, are they related?Ex.: Consider square wave with high = 1 and low = 0 We can send two bits every cycle (i.e., during T = 1/f sec)Assume f =1 MHz (fundamental frequency) T = 1 usecNow, if we use the first approximation (3 harmonics)BW of signal = (5 f 1 f) = 4 f = 4 MHzData rate = 2 / T = 2 Mbps

So we need a channel with bandwidth 4 MHz to send at date rate 2 Mbps

Bandwidth and Data Rate (contd)But, if we use the second approx. (4 harmonics)BW of signal = (7 f 1 f) = 6 f = 6 MHzData rate = 2 / T = 2 MbpsWhich one to choose? Can we use only 2 harmonics (BW = 2 MHz)? It depends on the ability of the receiver to discern the difference between 0 and 1Tradeoff: cost of medium vs. distortion of signal and complexity of receiver

Bandwidth and Data Rate (contd)Now, let us agree that the first appox. (3 harmonics) is good enoughData rate of 2 Mbps requires BW of 4 MHzTo achieve 4 Mbps, what is the required BW? data rate = 2 (bits) / T (period) = 4 Mbps T = 1 /2 usec f (fundamental freq) = 1 /T = 2 MHz BW = 4 f = 8 MHz

Bottom line: there is a direct relationship between data rate and bandwidthHigher data rates require more bandwidthMore bandwidth allows higher data rates to be sent

Bandwidth and Data Rate (contd)Nyquist Theorem: (Assume noise-free channel)If rate of signal transmission is 2B then signal with frequencies no greater than B is sufficient to carry signal rate, OR alternativelyGiven bandwidth B, highest signal rate is 2BFor binary signals Two levels we can send one bit (0 or 1) during each period data rate (C) = 1 x signal rate = 2 B That is, data rate supported by B Hz is 2B bpsFor M-level signalsM levels we can send log2M bits during each period C= 2B log2M

Bandwidth and Data Rate (contd)Shannon Capacity: Considers data rate, (thermal) noise and error rateFaster data rate shortens each bit so burst of noise affects more bitsAt given noise level, high data rate means higher error rateSNR Signal to noise ration SNR = signal power / noise powerUsually given in decibels (dB): SNRdB= 10 log10 (SNR)Shannon proved that: C = B log2(1 + SNR)This is theoretical capacity, in practice capacity is much lower (due to other types of noise)

Bandwidth and Data Rate (contd)Ex.: A channel has B = 1 MHz and SNRdB = 24 dB, what is the channel capacity limit?SNRdB = 10 log10 (SNR) SNR = 251C = B log2(1 + SNR) = 8 MbpsAssume we can achieve the theatrical C, how many signal levels are required?C = 2 B log2M M = 16 levels

Transmission ImpairmentsSignal received may differ from signal transmittedAnalog - degradation of signal qualityDigital - bit errorsCaused byAttenuation and attenuation distortionDelay distortionNoise

AttenuationSignal strength falls off with distanceDepends on mediumReceived signal strength:must be enough to be detectedmust be sufficiently higher than noise to be received without errorAttenuation is an increasing function of frequency attenuation distortion

Delay DistortionOnly in guided mediaPropagation velocity varies with frequencyCritical for digital dataA sequence of bits is being transmittedDelay distortion can cause some of signal components of one bit to spill over into other bit positions intersymbol interference, which is the major limitation to max bit rate

Noise (1)Additional signals inserted between transmitter and receiverThermalDue to thermal agitation of electronsUniformly distributed across frequencies White noiseIntermodulationSignals that are the sum and difference of original frequencies sharing a medium

Noise (2)CrosstalkA signal from one line is picked up by anotherImpulseIrregular pulses or spikes, e.g. external electromagnetic interferenceShort durationHigh amplitude

Data and SignalsData Entities that convey meaningAnalog: speechDigital: text (character strings)Signalselectromagnetic representations of dataAnalog: continuousDigital: discrete (pulses)TransmissionCommunication of data by propagation and processing of signals

Analog Signals Carrying Analog and Digital Data

Digital Signals Carrying Analog and Digital Data

Analog TransmissionAnalog signal transmitted without regard to contentMay be analog or digital dataAttenuated over distance Use amplifiers to boost signalBut, it also amplifies noise!

Digital TransmissionConcerned with contentIntegrity endangered by noise, attenuationRepeaters usedRepeater receives signalExtracts bit patternRetransmitsAttenuation is overcomeNoise is not amplified

Advantages of Digital TransmissionDigital technologyLow cost LSI/VLSI technologyData integrityLonger distances over lower quality linesCapacity utilizationHigh bandwidth links economicalHigh degree of multiplexing easier with digital techniquesSecurity & PrivacyEncryptionIntegrationCan treat analog and digital data similarly

SummarySignal: composed of components (Fourier Series)Spectrum, bandwidth, data rateShannon channel capacityTransmission impairmentsAttenuation, delay distortion, noiseData vs. signalsDigital vs. Analog Transmission