ANALOG TRANSMISSION
• Digital-to-AnalogConversion
– Amplitude Shift Keying (ASK)
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– Frequency Shift Keying (FSK)
– Phase Shift Keying (PSK)
– Quadrature Amplitude Modulation (QAM)
• AnalogtoAnalogConversion – Amplitude Modulation
– Frequency Modulation
– Phase Modulation
Digital-to-Analog Conversion
• Digital-to-analogconversionistheprocessofchanging one of the characteristics of an analog signal based on the information in digital data.
– Change Amplitude – Change Frequency – Change Phase
Digital-to-Analog Conversion
Types of Digital-to-Analog Conversion
Bit Rate vs. Baud Rate
• Bitrateisthenumberofbitspersecond.Baudrateis the number of signal elements per second.
• Intheanalogtransmissionofdigitaldata,thebaudrate is less than or equal to the bit rate.
• The ratio r between bit rate N and baud rate S is determined for each modulation scheme
• Ananalogsignalcarries4bitspersignalelement.If 1000 signal elements are sent per second, find the bit rate.
• Solution:Inthiscase,r=4,S=1000,andNis unknown. We can find the value of N from
• Ananalogsignalhasabitrateof8000bpsandabaud rate of 1000 baud. How many data elements are carried by each signal element? How many signal elements do we need?
• Solution: In this example, S = 1000, N = 8000, and r and L are unknown. We find first the value of r and then the value of L.
Amplitude Shift Keying (ASK)
• BinaryAmplitudeShiftKeying(BASK)–on-offkeying (OOK)
– Bandwidth = (1+d) S, where d is mostly 1
– ddependsonthemodulationandfilteringprocess.
– The value of d is between 0 and 1.
Implementation of Binary ASK
• Wehaveanavailablebandwidthof100kHzwhich spans from 200 to 300 kHz. What are the carrier frequency and the bit rate if we modulated our data by using ASK with d = 1?
• Solution:Thecarrierfrequencycanbeatthemiddleof the bandwidth is located at (200+300)/ 2 = 250 kHz.
• Wecanusetheformulaforbandwidthtofindthebit rate (with d = 1 and r = 1).
• Indatacommunications,wenormallyusefull-duplex links with communication in both directions. We need to divide the bandwidth into two with two carrier frequencies, as shown in Figure, the positions of two carrier frequencies and the bandwidths.
• Theavailablebandwidthforeachdirectionisnow50 kHz, which leaves us with a data rate of 25 kbps in each direction.
Frequency Shift Keying (FSK)
• Infrequencyshiftkeying,thefrequencyofthecarrier signal is varied to represent data
• Thefrequencyofthemodulatedsignalisconstantfor the duration of one signal element, but changes for the next signal element if the data element changes.
• Bothpeakamplitudeandphaseremainconstantforall signal elements
Binary FSK
• Wehaveanavailablebandwidthof100kHzwhich spans from 200 to 300 kHz. What should be the carrier frequency and the bit rate if we modulated our data by using FSK with d = 1?
• Solution: The midpoint of the band is at 250 kHz. We choose 2Δf to be 50 kHz; this means
Bandwidth of Binary FSK
• WecanthinkofFSKastwoASKsignals,eachwithits own carrier frequency (f1 or f2).
• Ifthedifferencebetweenthetwofrequenciesis2Δf, then the required bandwidth is:
– B=(1+d)xS+2Δf
FSK Implementation
• TherearetwoimplementationsofBFSK
• Non-coherent BFSK, there may be discontinuity in the phase when one signal element ends and the next begins
– canbeimplementedbytreatingBFSKastwoASK modulations and using two carrier frequencies
• coherentBFSK,thephasecontinuesthroughthe boundary of two signal elements
– can be implemented by using one voltage-controlled oscillator (VCO) that changes its frequency according to the input voltage
FSK Implementation
unipolar NRZ signal
Multilevel FSK
• Multilevelmodulation(MFSK)isnotuncommonwiththe FSK method. We can use more than two frequencies
– wecanusefourdifferentfrequenciesf1,f2,f3,and f4 to send 2 bits at a time
– wecanuseeightfrequenciestosend3bitsata time
– the frequencies need to be 2Δf apart
– For the proper operation of the modulator and demodulator, it can be shown that the minimum value of 2Δf needs to be S, the bandwidth is
–B=(1+d)xS +(L-1)x2Df→B=LxS
• Weneedtosenddata3bitsatatimeatabitrateof3 Mbps. The carrier frequency is 10 MHz. Calculate the number of levels (different frequencies), the baud rate, and the bandwidth.
• Solution:WecanhaveL=23 =8.
– ThebaudrateisS=3MHz/3=1Mbaud.
– This means that the carrier frequencies must be 1 MHz apart (2Δf = 1 MHz).
– ThebandwidthisB=8×1000=8000.
Phase Shift Keying (PSK)
• Inphaseshiftkeying,thephaseofthecarrierisvaried to represent two or more different signal elements.
• Bothpeakamplitudeandfrequencyremainconstantas the phase changes
• Today,PSKismorecommonthanASKorFSK.
• QAM,whichcombinesASKandPSK,isthedominant method of digital-to-analog modulation
Binary PSK
The bandwidth is the same as that for binary ASK, but less than that for BFSK
Implementation of Binary PSK
polar NRZ signal
Quadrature PSK (QPSK)
• ThesimplicityofBPSKenticeddesignerstouse2bitsat a time in each signal element, thereby decreasing the baud rate and eventually the required bandwidth.
• TheschemeiscalledquadraturePSKorQPSKbecause it uses two separate BPSK modulations
– oneisin-phase,
– the other quadrature (out-of-phase)
• Theincomingbitsarefirstpassedthroughaserial-to- parallel conversion that sends one bit to one modulator and the next bit to the other modulator
Quadrature PSK (QPSK)
sin(θ+45) = sinθ cos45 + cosθ sin45 = (sinθ+cosθ)/√2
• Findthebandwidthforasignaltransmittingat12Mbps for QPSK. The value of d = 0.
• Solution: For QPSK, 2 bits is carried by one signal element. This means that r = 2.
– Sothesignalrate(baudrate)isS=N×(1/r)=6 Mbaud.
– Withavalueofd=0,wehaveB=(1+d)S=6MHz.
Constellation Diagram
• Aconstellationdiagramcanhelpusdefinethe amplitude and phase of a signal element, particularly when we are using two carriers (one in-phase and one quadrature)
• usefulwhenwearedealingwithmultilevelASK,PSK,or QAM
• Thediagramhastwoaxes.ThehorizontalXaxisis related to the in-phase carrier; the vertical Y axis is related to the quadrature carrier
Constellation Diagram
• Foreachpointonthediagram,fourpiecesof information can be deduced:
– The projection of the point on the X axis defines the peak amplitude of the in-phase component
– the projection of the point on the Y axis defines the peak amplitude of the quadrature component
– The length of the line (vector) that connects the point to the origin is the peak amplitude of the signal element
– the angle the line makes with the X axis is the phase of the signal element
Constellation Diagram
• ShowtheconstellationdiagramsforanASK(OOK), BPSK, and QPSK signals.
Quadrature Amplitude Modulation
• Theideaofusingtwocarriers,onein-phaseandthe other quadrature, with different amplitude levels for each carrier is the concept behind quadrature amplitude modulation (QAM).
• Quadratureamplitudemodulationisacombinationof ASK and PSK
• ThepossiblevariationsofQAMarenumerous
Quadrature Amplitude Modulation
a. using a unipolar NRZ signal to modulate each carrier b. using a polar NRZ signal to modulate each carrier
c. using signal with two positive levels to modulate each of the two carriers
d. 16-QAM constellation of a signal with eight levels, four positive and four negative
Analog to Analog Conversion
• Analog-to-analogconversionistherepresentationof analog information by an analog signal.
• Onemayaskwhyweneedtomodulateananalog signal; it is already analog.
• Modulationisneededifthemediumisbandpassin nature or if only a bandpass channel is available to us.
Amplitude Modulation (AM)
• InAMtransmission,thecarriersignalismodulatedso that its amplitude varies with the changing amplitudes of the modulating signal.
• Thefrequencyandphaseofthecarrierremainthe same; only the amplitude changes to follow variations in the information
Amplitude Modulation (AM)
Amplitude Modulation Bandwidth
• Themodulationcreatesabandwidththatistwicethe bandwidth of the modulating signal and covers a range centered on the carrier frequency
• Thesignalcomponentsaboveandbelowthecarrier frequency carry exactly the same information. For this reason, some implementations discard one-half of the signals and cut the bandwidth in half
– The total bandwidth required for AM can be determined from the bandwidth of the audio signal:
AM radio stations frequency band
Frequency Modulation (FM)
• InFMtransmission,thefrequencyofthecarriersignal is modulated to follow the changing voltage level (amplitude) of the modulating signal.
• Thepeakamplitudeandphaseofthecarriersignal remain constant, but as the amplitude of the information signal changes, the frequency of the carrier changes correspondingly
• FMisnormallyimplementedbyusingavoltage- controlled oscillator as with FSK
Frequency Modulation (FM)
Frequency Modulation Bandwidth
The actual bandwidth is difficult to determine exactly, but it can be shown empirically that it is several times that of the analog signal or 2(1 + β)B where β is a factor that depends on modulation technique with a common value of 4.
ThetotalbandwidthrequiredforFMcanbedetermined from the bandwidth of the audio signal:
– BFM=2(1+β)B.
FM radio stations frequency band
Phase Modulation (PM)
• InPMtransmission,thephaseofthecarriersignalis modulated to follow the changing voltage level (amplitude) of the modulating signal. The peak amplitude and frequency
Phase Modulation (PM)
Phase Modulation Bandwidth
• Theactualbandwidthisdifficulttodetermineexactly, but it can be shown empirically that it is several times that of the analog signal.
• AlthoughtheformulashowsthesamebandwidthforFM and PM, the value of β is lower in the case of PM (around 1 for narrowband and 3 for wideband).
• FigureThetotalbandwidthrequiredforPMcanbe determined from the bandwidth and maximum amplitude of the modulating signal:
– BPM=2(1+β)B.
References
• DataCommunicationsandNetworking5thedition– 2013, Behrouz A. Forouzan; Chapter 5
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