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EE4512 Analog and Digital Communications Chapter 8

Chapter 8Chapter 8

AnalogAnalog--toto--Digital andDigital andDigital to AnalogDigital to AnalogConversionConversion


Chapter 8Chapter 8

AnalogAnalog--toto--Digital andDigital andDigitalDigital--toto--AnalogAnalogConversion Conversion •• Sampling and QuantizationSampling and Quantization

•• Pages 390Pages 390--391391


•• Traditional Traditional analog transmissionanalog transmission (AM, FM and PM) are less (AM, FM and PM) are less complex than digital data transmission have been the basis complex than digital data transmission have been the basis of broadcasting and communication for 100 years.of broadcasting and communication for 100 years.

Analog television signal Analog television spectrum

S&M Figure 8S&M Figure 8--1a1a


•• Digital data transmissionDigital data transmission (PAM, ASK, PSK, FSK and QAM) (PAM, ASK, PSK, FSK and QAM) is more complex but (perhaps) offers higher performance is more complex but (perhaps) offers higher performance with control of accuracy and easier storage, simpler signal with control of accuracy and easier storage, simpler signal processing for noise reduction, error detection and processing for noise reduction, error detection and correction and encryption.correction and encryption.

S&M Figure 8S&M Figure 8--1b1b


•• Digital data transmission requires analogDigital data transmission requires analog--toto--digital (ADC) digital (ADC) and digitaland digital--toto--analog (DAC) converters. The ADC process analog (DAC) converters. The ADC process utilizes sampling and quantization of the continuous analog utilizes sampling and quantization of the continuous analog signal.signal.

ADCADC

DACDAC



•• ADC sampling occurs at a uniform rate (the ADC sampling occurs at a uniform rate (the sampling ratesampling rate) ) and has a continuous amplitude.and has a continuous amplitude.

Continuousamplitude

Uniformsampling rate

Analog signal

S&M Figure 8S&M Figure 8--2a,b2a,b


•• The continuous amplitude sample is then quantized to The continuous amplitude sample is then quantized to n n bits or resolution for the bits or resolution for the full scalefull scale input or input or 22nn levelslevels..

Continuousamplitude

Uniformsampling rate

Quantized

S&M Figure 8S&M Figure 8--2b,c2b,c

Quantizedamplitude


•• Here n =Here n = 4 and there are 24 and there are 244 = 16 = 16 levels for a full scale input levels for a full scale input of 2 V (of 2 V (±± 1 V). The 1 V). The step size step size = 2 V / 16 = 0.125 V and the = 2 V / 16 = 0.125 V and the quantized valuequantized valueis theis the midpointmidpointof the of the voltagevoltagerangerange. .

S&M Table 8.1S&M Table 8.1


Chapter 8Chapter 8

AnalogAnalog--toto--Digital andDigital andDigitalDigital--toto--AnalogAnalogConversion Conversion •• Sampling BasebandSampling BasebandAnalog SignalsAnalog Signals

•• Pages 392Pages 392--399399


•• The analog signal The analog signal x(tx(t) which is continuously, uniformly ) which is continuously, uniformly sampled is represented by:sampled is represented by:

Multiplication in the temporal domain is convolution in the Multiplication in the temporal domain is convolution in the frequency domain and the frequency domain representation frequency domain and the frequency domain representation is: is:

∞

−∞

= −∑sk =

x (t) x(t) δ(t k )ST S&M Eq. 8.1S&M Eq. 8.1

∞

−∞

∞

−∞

∞

−∞

= ∗ −

= ∗ −

= −

∑

∑

∑

sk =

sk =

sk =

X (f) X(f) δ(t k )

X (f) X(f) δ(f k )

X (f) X(f k )

S

S S

S S

T

f f

f f

F

S&M Eq. 8.2S&M Eq. 8.2


•• Temporal and spectral representation of the continuous Temporal and spectral representation of the continuous sampling process for a sum of three sinusoids.sampling process for a sum of three sinusoids.

∞

−∞

−∑k =

δ(t k )ST

∞

−∞

= −∑sk =

x (t) x(t) δ(t k )ST

∞

−∞

= −∑sk =

X (f) X(f k )S Sf f

∞

−∞

−∑k =

δ(f k )Sf

S&M Figure 8S&M Figure 8--33


•• 2 V, 202 V, 20°° initialinitialphase,phase, 500 Hz500 Hzsinusoidsinusoidsampled atsampled at5 k samples/sec5 k samples/sec



•• Aliased samplesAliased samplescan becan bereconstructedreconstructedfor a 4500 Hzfor a 4500 Hzand a 5500 Hzand a 5500 Hzsinusoid thatsinusoid thatappears to beappears to bea 500 Hza 500 Hzsinusoidsinusoid

S&MS&MFigure 8Figure 8--4a,c,d4a,c,d


•• The The aliasingaliasing ofofthe signal canthe signal canbe predicted bybe predicted bythe magnitudethe magnitudespectrum of thespectrum of theoriginal 500 Hzoriginal 500 Hzsampled signal.sampled signal.

If the 4500 HzIf the 4500 Hzand 5500 Hzand 5500 Hzsignals are thensignals are thensampled at sampled at 5 k samples/sec5 k samples/secaliasing at occurs at | 4500 aliasing at occurs at | 4500 –– 5000 | and (5500 5000 | and (5500 –– 5000) Hz5000) Hz



•• The sum of The sum of three sinusoidsthree sinusoidsdoes not havedoes not haveany aliasedany aliasedfrequenciesfrequenciessince thesince thesamplingsamplingfrequency frequency ffSSis greater thanis greater thantwice thetwice thehighesthighestfrequencyfrequency ffmaxmax

ffSS > 2 > 2 ffmaxmax

S&M Figure 8S&M Figure 8--4a,c4a,c



•• The frequencyThe frequency2 2 ffmax max is calledis calledthe the NyquistNyquistfrequencyfrequency..

Harry Nyquist,Harry Nyquist,who contributedwho contributedto the understanding of thermal noiseto the understanding of thermal noisewhile at Bell Labs, is also rememberedwhile at Bell Labs, is also rememberedin electrotechnology for his analysis of in electrotechnology for his analysis of sampled data signals. sampled data signals.

S&M Figure 8S&M Figure 8--4a4a

Harry Nyquist Harry Nyquist 18891889--19761976


•• The analog signal is The analog signal is reconstructedreconstructed from the quantized from the quantized samples by a DAC and a low pass filter (LFP).samples by a DAC and a low pass filter (LFP).



•• For For practical signals practical signals ffSS > 2 > 2 ffmax max using a using a guard band guard band for for LPFsLPFs


ffSS = 2 = 2 ffmaxmax

ffSS > 2 > 2 ffmaxmax

guard bandguard band


•• With With outout--ofof--band noise band noise and sample signals, aliases of the and sample signals, aliases of the noise now appear noise now appear inin--band band and should be filtered before the and should be filtered before the sampling process.sampling process.



Chapter 8Chapter 8

AnalogAnalog--toto--Digital andDigital andDigitalDigital--toto--AnalogAnalogConversionConversion•• Sampling BasebandSampling BasebandAnalog SignalsAnalog Signals

•• Pages 149Pages 149--182182


•• The periodic basebandThe periodic basebandsignal consisting ofsignal consisting ofthree sinusoids isthree sinusoids isimpulse sampledimpulse sampled,,sampledsampled--andand--heldheld,,processed by an processed by an 88--bit ADCbit ADC--DACDACand a and a quantizerquantizerin in SimulinkSimulink..

MS Figure 4.1MS Figure 4.1


•• The periodic baseband signal is the sum of a 1 V 500 Hz, a The periodic baseband signal is the sum of a 1 V 500 Hz, a 0.5 V 1.5 kHz and a 0.2 V 2.5 kHz sinusoid.0.5 V 1.5 kHz and a 0.2 V 2.5 kHz sinusoid.



•• The power spectral density (PSD) of the periodic baseband The power spectral density (PSD) of the periodic baseband signal has the expected peaks at 0.5, 1.5 and 2.5 kHz.signal has the expected peaks at 0.5, 1.5 and 2.5 kHz.

MS Figure 4.3MS Figure 4.3Three sinusoidsThree sinusoids


•• The periodic baseband signal is overlaid with the continuous The periodic baseband signal is overlaid with the continuous amplitude sampleamplitude sample--andand--hold signal with hold signal with ffSS = 8 kHz.= 8 kHz.


0.125 msec0.125 msec


•• The analog signal The analog signal x(tx(t) here is sampled and held rather than ) here is sampled and held rather than impulse sampled:impulse sampled:

The power spectral density (PSDThe power spectral density (PSDss--hh) of the sample and hold ) of the sample and hold operation is: operation is:

= − = ≤ ≤

=

∑s-hn

y (t) x(n ) h(t n ) h(t) 1 0 t

h(t) 0

S S ST T T

otherwise MS Eq. 4.3MS Eq. 4.3

( )

( )

∞

−∞

∞

−∞

= −

= −

∑

∑

2 2 2 2s-h S

k =

2 2s-h

k =

PSD | X(f k ) | T sinc 2π f

PSD | X(f k ) | sinc 2π f

S S S

S S

f f T

f TMS Eq. 4.4MS Eq. 4.4


•• However, if the analog signal However, if the analog signal x(tx(t) is impulse sampled:) is impulse sampled:

Then the power spectral density (PSD) does not have a Then the power spectral density (PSD) does not have a sincsinc22 term:term:

The PSDThe PSDss--h h does have the sincdoes have the sinc2 2 term: term:

MS Eq. 4.1MS Eq. 4.1

( )

∞

−∞

∞

−∞

= −

= −

∑

∑

2 2

k =

2 2s-h

k =

PSD | X(f k ) |


S S

S S

f f

f T


= −∑n

x(n ) x(t) δ(t n )S ST T



•• The PSD of the impulse sampled sum of three sinusoid The PSD of the impulse sampled sum of three sinusoid signal with signal with ffSS = 8 kHz is:= 8 kHz is:

MS Figure 4.5MS Figure 4.58 kHz8 kHz 16 kHz16 kHz

No sincNo sinc22 termterm

∞

−∞

= −∑2 2

k = PSD | X(f k ) |S Sf f MS Eq. 4.2MS Eq. 4.2


•• The PSD of the continuous amplitude sample and hold The PSD of the continuous amplitude sample and hold sum of three sinusoid signal with sum of three sinusoid signal with ffSS = 8 kHz is:= 8 kHz is:

MS Figure 4.6MS Figure 4.68 kHz8 kHz 16 kHz16 kHz

sincsinc22 termterm

( )

( )

∞

−∞

∞

−∞

= −

= −

∑

∑

2 2 2 2s-h

k =

2 2s-h

k =



S S S S

S S

f f T T

f TMS Eq. 4.4MS Eq. 4.4


Chapter 8Chapter 8

AnalogAnalog--toto--Digital andDigital andDigitalDigital--toto--AnalogAnalogConversion Conversion •• Sampling BandpassSampling BandpassAnalog SignalsAnalog Signals

•• Pages 399Pages 399--400400


•• A bandpass signal does not need to be sampled at 2 A bandpass signal does not need to be sampled at 2 ff22. . NyquistNyquist’’s bandpass sampling theory states that the s bandpass sampling theory states that the sampling rate sampling rate ffSS > 2(> 2(ff22 −− ff11) which is substantially less than) which is substantially less than2 2 ff22


ff11 ff22ffSS = 20 ksamples/sec= 20 ksamples/sec

ffSS = 7 ksamples/sec= 7 ksamples/sec

LPF 10 kHzLPF 10 kHz

BPF 8BPF 8--10 kHz10 kHz

8 10 kHz8 10 kHz


Chapter 8Chapter 8

AnalogAnalog--toto--Digital andDigital andDigitalDigital--toto--AnalogAnalogConversionConversion•• Sampling BandpassSampling BandpassAnalog SignalsAnalog Signals

•• Pages 180Pages 180--181181


•• The The SimulinkSimulink simulation uses the DSB AM modulation simulation uses the DSB AM modulation block and the sum of three sinusoids source.block and the sum of three sinusoids source.

MS Figure 4MS Figure 4--3232



•• The The SimulinkSimulink simulation initially uses a sampling rate of simulation initially uses a sampling rate of 5 MHz and results in 4 194 304 = 25 MHz and results in 4 194 304 = 222 22 sampling points. The sampling points. The PSD shows the DSBPSD shows the DSB--LC AM signal with the LSB and USB.LC AM signal with the LSB and USB.


LSB USBLSB USB

ffCC

Scaled PSD Scaled PSD ffmax max = 50 kHz = 50 kHz


•• The bandwidth of the bandpass signal is The bandwidth of the bandpass signal is ff22 −− ff11 = 22.5 = 22.5 −−17.5 = 5 kHz and the 17.5 = 5 kHz and the SimulinkSimulink sampling rate is set tosampling rate is set to50 kHz50 kHz and results in only 32 768 = 2and results in only 32 768 = 21515 sampling points.sampling points.


LSB USBLSB USB

ffCC

Scaled PSD Scaled PSD ffmaxmax = 50 kHz= 50 kHzAliased frequency range > 25 kHzAliased frequency range > 25 kHz


Chapter 8Chapter 8

AnalogAnalog--toto--Digital andDigital andDigitalDigital--toto--AnalogAnalogConversion Conversion •• Quantizing Process:Quantizing Process:Uniform QuantizationUniform Quantization

•• Pages 400Pages 400--404404


•• The The quantizing process quantizing process divides thedivides the range range ((±± full scale) into full scale) into 22nn (n = 4 here) regions which are assigned an n(n = 4 here) regions which are assigned an n--bit binary bit binary code.code.



•• The The errorerror associated with the quantizing process is associated with the quantizing process is assumed to have a uniform probability density function. assumed to have a uniform probability density function. The maximum error for uniform quantization is:The maximum error for uniform quantization is:

The quantizer range is The quantizer range is ±± VVmaxmaxand the uniform quantizerand the uniform quantizervoltage step size is:voltage step size is:

The mean square quantizing EThe mean square quantizing Eqq is the normalized is the normalized quantizing noise power: quantizing noise power:


= =

max maxn n

2 V Vq ± 0.5 ±2 2

∆ = =max maxn n-1

2 V V2 2

( ) ( )/ 2

/ 2

∆

−∆

= = = =∫2 22

2 max maxq 2 2nn

V V1 ∆E q dq∆ 12 3 23 2




•• The The signal to quantizing noise power signal to quantizing noise power (SNR(SNRqq) is:) is:

PPSS is the normalized power of the signal that is quantized.is the normalized power of the signal that is quantized.

For the ADC here For the ADC here ∆∆ = 10 mV and n = 8. The sum of three = 10 mV and n = 8. The sum of three sinusoids as the input signal has a peak amplitude of 1.1 Vsinusoids as the input signal has a peak amplitude of 1.1 Vand the quantizing noise has a peak amplitude of 10 mV.and the quantizing noise has a peak amplitude of 10 mV.


MS Eq. 4.8MS Eq. 4.8( )= = 2nS Sq 2 2

max

12 P PSNR 3 2∆ V

10 mV10 mV


Chapter 8Chapter 8

AnalogAnalog--toto--Digital andDigital andDigitalDigital--toto--AnalogAnalogConversion Conversion •• Quantizing Process:Quantizing Process:Nonuniform QuantizationNonuniform Quantization

•• Pages 400Pages 400--404404


•• Nonuniform quantization divides the dynamic range of an Nonuniform quantization divides the dynamic range of an analog signal into nonuniform quantization regions. Lower analog signal into nonuniform quantization regions. Lower magnitudes have smaller quantization regions than high magnitudes have smaller quantization regions than high magnitudes.magnitudes.

The quantization of speech benefits from nonuniform The quantization of speech benefits from nonuniform quantization since the perception of hearing is quantization since the perception of hearing is logarithmicallogarithmicalrather than linear.rather than linear.


•• UniformUniformquantizationquantization(top) results in(top) results ina large amounta large amountof error forof error forsmall samplesmall sampleamplitude.amplitude.

NonNon--uniformuniformquantizationquantization(bottom)(bottom)reduces thereduces theerror for smallerror for smallsamplesampleamplitudes.amplitudes. S&M Figure 8S&M Figure 8--1313

Uniform quantizationUniform quantization

Nonuniform quantizationNonuniform quantization


•• Uniform quantization is simpler to implement so a Uniform quantization is simpler to implement so a compressorcompressor (a non(a non--linear transfer function) is used before linear transfer function) is used before the quantizer.the quantizer.The The µµ--LawLawcompressorcompressoris used inis used intelephonytelephonywith MS Eq. 4.9with MS Eq. 4.9µµ = 255. At the receiver= 255. At the receiveran expander has thean expander has theinverse noninverse non--linearlineartransfer function andtransfer function andresults in results in compandingcompanding((COMpressingCOMpressing andandexPANDINGexPANDING).).

( )

≤ ≤

in

max inout max

max

Vln 1+µV VV = V 0 1

ln 1+µ V


Chapter 8Chapter 8

AnalogAnalog--toto--Digital andDigital andDigitalDigital--toto--AnalogAnalogConversionConversion•• CompandingCompanding

•• Pages 157Pages 157--159159


•• The The µµ--Law compander concept can be simulated in Law compander concept can be simulated in Simulink Simulink with the with the µµ--Law Compressor and Law Compressor and µµ--Law Expander Law Expander blocks. The Ablocks. The A--Law Compressor and ALaw Compressor and A--Law Expander Law Expander blocks are included for comparison.blocks are included for comparison.



•• The The µµ--Law compressor voltage transfer function is Law compressor voltage transfer function is sigmoidalsigmoidal (S(S--shaped).shaped).



Chapter 8Chapter 8

AnalogAnalog--toto--Digital andDigital andDigitalDigital--toto--AnalogAnalogConversionConversion•• Pulse Code ModulationPulse Code Modulation

•• Pages 171Pages 171--175175


•• The The pulse code modulator (PCM) transmitter utilizes a pulse code modulator (PCM) transmitter utilizes a SimulinkSimulink µµ--Law compressor block, an 8Law compressor block, an 8--bit ADC subsystem, bit ADC subsystem, an 8an 8--bit DAC subsystem and a bit DAC subsystem and a µµ--Law expander block.Law expander block.



•• The The Simulink Simulink 88--bit ADC subsystem has a samplebit ADC subsystem has a sample--andand--hold hold block controlled by a block controlled by a samplingsampling pulse generator, an 8pulse generator, an 8--bit bit encoder block, an integerencoder block, an integer--toto--bit converter block which bit converter block which provides an 8provides an 8--bit bit vectorvector to a demultiplexer block and a to a demultiplexer block and a multiport switch.multiport switch. An 8An 8--level level staircase staircase subsystem subsystem sequences the multiport switch to select 1 of the 8 inputs forsequences the multiport switch to select 1 of the 8 inputs forbit serial output. bit serial output.



•• The 8The 8--level level staircase Simulink staircase Simulink subsystem sequences the subsystem sequences the multiport switch with a 3multiport switch with a 3--bit counter and a 3bit counter and a 3--bit DAC for the bit DAC for the output. output.


33--bit DACbit DAC

33--bit counterbit counter


•• The 8The 8--bit DAC bit DAC Simulink Simulink subsystem for the PCM system subsystem for the PCM system uses a 8uses a 8--bit bit shift registershift register and an 8and an 8--bit DAC.bit DAC.

MS Figure 4.24MS Figure 4.2488--bit DACbit DAC

88--bit shift bit shift registerregister


•• Analog input signal to the PCM systemAnalog input signal to the PCM system




•• 88--bit DAC output after 8bit DAC output after 8--bit ADC and bit ADC and µµ--Law CompressorLaw Compressor




•• µµ--Law Expander block Law Expander block output of the PCM systemoutput of the PCM system




•• LPF LPF output of the PCM systemoutput of the PCM system




•• Analog input signal to the PCM systemAnalog input signal to the PCM system

MS Figure 4.25MS Figure 4.25•• LPF output of the PCM systemLPF output of the PCM system

startupstartup


Chapter 8Chapter 8

AnalogAnalog--toto--Digital andDigital andDigitalDigital--toto--AnalogAnalogConversion Conversion •• Differential Pulse CodeDifferential Pulse CodeModulationModulation

•• Pages 407Pages 407--411411


•• Sampled speech data are Sampled speech data are highly correlated highly correlated and and differential differential pulse code modulationpulse code modulation (DPCM) exploits this to lower the(DPCM) exploits this to lower theoverall dataoverall datarate.rate.DPCMDPCMuses auses apredictorpredictorto subtractto subtracta predicteda predictedvalue fromvalue fromthe input.the input.The The errorerrordifferencedifferenceis sent.is sent.



•• The predictor is a recursive equation, for example:The predictor is a recursive equation, for example:SS(n(n) = 0.75 s(n) = 0.75 s(n−−1) + 0.2 s(n1) + 0.2 s(n−−2) +0.05 s(n2) +0.05 s(n−−3)3)

where where SS(n(n))is theis thepredictedpredictedvalue of thevalue of then n thth samplesampleand and s(ns(n--i)i)is the is the nn--i i ththsample.sample.The errorThe errorsignal issignal iss(ns(n) ) −− SS(n(n))



•• A typical continuous analog signal is sampled and results in A typical continuous analog signal is sampled and results in a discrete signal a discrete signal s(ns(n), The discrete predicted signal ), The discrete predicted signal SS(n(n) is ) is recursively computed. The discrete error signal is recursively computed. The discrete error signal is transmitted and has less quantizing bits than the actual transmitted and has less quantizing bits than the actual discrete signal. discrete signal.



•• A DPCM example of actual discrete values, predicted A DPCM example of actual discrete values, predicted values and the error terms: values and the error terms:

S&M Table 8S&M Table 8--33


Chapter 8Chapter 8

AnalogAnalog--toto--Digital andDigital andDigitalDigital--toto--AnalogAnalogConversionConversion•• Differential Pulse CodeDifferential Pulse CodeModulationModulation

•• Pages 175Pages 175--180180


•• A 4A 4--bit first order bit first order differential pulse code modulatordifferential pulse code modulator (DPCM) (DPCM) can be simulated in can be simulated in Simulink. Simulink.



•• The first order linear predictor MetaSystem determines the The first order linear predictor MetaSystem determines the error signal: error signal: e(ne(n) = s(n+1) ) = s(n+1) −− 2 2 s(ns(n) + s(n) + s(n--1)1)


error signalerror signal

inputinput

ADC conversion commandADC conversion command


•• The The Simulink Simulink 44--bit ADC subsystem of the DPCM system is bit ADC subsystem of the DPCM system is similar to the 8similar to the 8--bit ADC of the PCM system and illustrates bit ADC of the PCM system and illustrates design reuse.design reuse.



•• The The Simulink Simulink 44--bit DAC subsystem of the DPCM system is bit DAC subsystem of the DPCM system is also similar to the 8also similar to the 8--bit DAC of the PCM system and bit DAC of the PCM system and again illustrates again illustrates design reuse.design reuse.

MS Figure 4.29MS Figure 4.29 44--bit DACbit DAC

44--bit shift bit shift registerregister


•• The first order linear predictor The first order linear predictor SimulinkSimulink subsystem subsystem reconstructs an estimate of the signal reconstructs an estimate of the signal ssee(n(n) from the error ) from the error signal signal e(ne(n) received and past estimates:) received and past estimates:

ssee(n+1) = e(n+1) + 2 (n+1) = e(n+1) + 2 ssee(n(n) ) −− ssee(n(n−−1)1)


reconstructed reconstructed signalsignal

inputinput


•• Analog inAnalog input signal of the DPCM systemput signal of the DPCM system




•• Output of the 4Output of the 4--bit first order bit first order DPCM systemDPCM system




•• Analog input signal to the DPCM systemAnalog input signal to the DPCM system

MS Figure 4.31MS Figure 4.31•• Output of the 4Output of the 4--bit first order DPCM systembit first order DPCM system

startupstartup


•• Output of the 8Output of the 8--bit PCM systembit PCM system

MS Figure 4.31MS Figure 4.31•• Output of the 4Output of the 4--bit first order DPCM systembit first order DPCM system

startupstartup

startupstartup



Chapter 8Chapter 8

AnalogAnalog--toto--Digital andDigital andDigitalDigital--toto--AnalogAnalogConversion Conversion •• Delta ModulationDelta Modulation

•• Pages 411Pages 411--415415


•• Delta modulationDelta modulation is an extreme example of DPCMis an extreme example of DPCM using using 11--bit data representing bit data representing ±± ∆∆::

SS(n(n) = ) = SS(n(n−−1) + 1) + ∆∆ bbii = 1 if = 1 if SS(n(n−−1) 1) ≤≤ s(ns(n−−1)1)SS(n(n) = ) = SS(n(n−−1) 1) −− ∆∆ bbii = 0 if = 0 if SS(n(n−−1) >1) > s(ns(n−−1)1)

S&M Eq. 8.10S&M Eq. 8.10


DM transmitterDM transmitter

DM receiverDM receiver


•• The reconstructed signal increments The reconstructed signal increments ±± ∆∆ on each on each transmitted bit.transmitted bit.

bbii = 1 = 1 SS(n(n) = ) = SS(n(n−−1) + 1) + ∆∆ bbii = 0 = 0 SS(n(n) = ) = SS(n(n−−1) 1) −− ∆∆

S&M Figure 8S&M Figure 8--19194 1s 4 0s4 1s 4 0s


Chapter 8Chapter 8

AnalogAnalog--toto--Digital andDigital andDigitalDigital--toto--AnalogAnalogConversionConversion•• Delta ModulationDelta Modulation

•• Pages 72Pages 72--7575


•• Delta modulation (DM) can be simulated in Delta modulation (DM) can be simulated in SimulinkSimulink. The . The DM receiver utilizes a sample and hold token as an DM receiver utilizes a sample and hold token as an accumulator and the step size accumulator and the step size ∆∆ = 20 mV.= 20 mV.


DM transmitterDM transmitter DM receiverDM receiver

f = 2 Hzf = 2 HzA = 1 VA = 1 V

ffSS = 2 kHz= 2 kHzTTSS = 0.5 msec= 0.5 msec


•• DM can be subject to DM can be subject to slope overload slope overload which occurs when:which occurs when:∆∆ / / TTSS < max | d < max | d m(tm(t) / ) / dtdt | SVU Eq. 2.61 modified| SVU Eq. 2.61 modified

Here the sinusoid has A = 1 but f = 10 Hz and:Here the sinusoid has A = 1 but f = 10 Hz and:∆∆ / / TTSS = 20 mV / 0.5 msec = 40 < max | d = 20 mV / 0.5 msec = 40 < max | d m(tm(t) / ) / dtdt | = 80| = 80ππ

and slope overload occurs.and slope overload occurs.

MS Figure 2.63MS Figure 2.63sinusoidalsinusoidalinput signalinput signal


•• Granular noise Granular noise occursoccurs inin DM because if the input DM because if the input m(tm(t) is ) is constant the received signal constant the received signal oscillatesoscillates by by ±± ∆∆ because there because there is no is no 00 possible. possible. ClockingClocking occurs at the DM symbol interval occurs at the DM symbol interval TTSS = 0.5 msec.= 0.5 msec.

MS Figure 2.64MS Figure 2.64∆∆ = 20 mV= 20 mV

±± ∆∆ = = ±± 20 mV20 mV

TTSS = 0.5 msec= 0.5 msec


•• The tradeoff between slope overload and granular noise is The tradeoff between slope overload and granular noise is that a large value of that a large value of ∆∆ (to avoid slope overload) would (to avoid slope overload) would increase granular noise. A decrease in increase granular noise. A decrease in TTss (again to avoid (again to avoid slope overload) would increase the data rate slope overload) would increase the data rate rrSS..

•• The step size The step size ∆∆ = 20 mV and = 20 mV and TTSS = 0.5 msec (= 0.5 msec (rrSS = 2 kb/sec) = 2 kb/sec) here.here.


•• For a 10 Hz sinusoidal input signal to a DM: For a 10 Hz sinusoidal input signal to a DM: m(tm(t) = sin (2) = sin (2ππ 10t)10t)max | d max | d m(tm(t) / ) / dtdt || = 20= 20ππ

•• If step size If step size ∆∆ = 20 mV and = 20 mV and TTSS = 0.5 msec then= 0.5 msec then∆∆ / T/ TSS = 40 < max | d = 40 < max | d m(tm(t) / ) / dtdt | = 20| = 20ππ so so slope overload is slope overload is predicted to occurpredicted to occur..


•• For the 10 Hz sinusoidal input signal to a DMFor the 10 Hz sinusoidal input signal to a DMm(tm(t) = sin (2) = sin (2ππ 10t)10t)max | d max | d m(tm(t) / ) / dtdt || = 20= 20ππ

and the step size remains and the step size remains ∆∆ = 20 mV but = 20 mV but TTSS = 0.25 m= 0.25 msec sec then then ∆∆ / T/ TSS = 80 > max | d = 80 > max | d m(tm(t) / ) / dtdt | and slope overload is | and slope overload is mitigated but mitigated but rrSS = 4 kb/sec.= 4 kb/sec.


•• In comparison, an 8In comparison, an 8--bit PCM system sampling a 10 Hz bit PCM system sampling a 10 Hz sinusoid at a reasonable sampling rate of 500 Hz (50 sinusoid at a reasonable sampling rate of 500 Hz (50 sampling points/period) has sampling points/period) has rrbb = 8(500) = 4 kb/sec or = 8(500) = 4 kb/sec or rrb b = r= rSSbut PCM is more complicated than DM.but PCM is more complicated than DM.


End of Chapter 8End of Chapter 8

AnalogAnalog--toto--Digital andDigital andDigitalDigital--toto--AnalogAnalogConversionConversion

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