chap11--sampling & pulse modulation

8/3/2019 Chap11--Sampling & Pulse Modulation

http://slidepdf.com/reader/full/chap11-sampling-pulse-modulation 1/59

Prof. J.F. Huang, Fiber-Optic Communication Lab. National Cheng Kung University, Taiwan

1

Chapter 11.

Sampling and Pulse Modulation

• The Sampling Theorem

• PAM -- Natural and Flat-Top SamplingTime-Division Multiplexing (TDM ) Intersymbol Interference ( ISI )

• Pulse Width and Pulse Position Modulation Demodulation

• Digital Modulation Pulse Code Modulation ( PCM )

Delta Modulation ( DM )• Qualitative Comparisons Of Pulse and Digital

Modulation Systems




2

The Sampling Theorem

Figure 11-1. Impulse sampling of an analog voltage.




3

• A sampler is a mixer with a train of very narrowpulses as the local oscillator input.

• If the analog input is sampled instantaneously at

regular intervals at a rate that is at least twice the

highest analog frequency

f s > 2 f a(max)

• then the samples contain all of the information of the original signal.





4

• The analog signal v( t) has a signal spectrum

represented by the Fourier transform V ( f ),

• and the sampling signal

consists of instantaneous impulses every nT s sec,

where n = 0, +1, +2, …

•

The Fourier transform of s( t) is

n

snT t t s

s

ns

nf f T

f S

1





5

• The time-domain product performed by the

sampler produces a sampled output

spectrum given by

• where this spectrum consists of replicas of

the analog signal spectrum V ( f ), translated

in frequency by each of the sampling

frequency harmonics.

s

ns

s nf f V T

f V

1





6

• The sampler is a wideband (harmonic) mixerproducing upper and lower sidebands at eachharmonic of the sampling frequency.

• Figure 11-2a illustrates the correct way to sample: if sampling is done at f s > 2 f A(max) the upper andlower sidebands do not overlap each other,

• and the original information can be recovered bypassing the signal through a low-pass filter (seeFigure 11-2c and d).





7

Figure 11-2. Sample spectra and their outputs. (a) f s > 2 f A(max)Nyquist criteria met. (b) f s < 2 f A(max) Frequency foldover of “aliasing” distortion occurs. (c) f s > 2 f A(max) and recovery of original information with low-pass filter. (d) The original analogsignal spectrum following recovery as in (c).





8

• However, if the sampling rate is less than the

Nyquist rate, f s < 2 f A(max) the sidebands overlap,

as shown in Figure 11-2b.

• The result is frequency-folding or aliasing distortion,

which makes it impossible to recover the original

signal without distortion.





9

Pulse Amplitude Modulation –

Natural and Flat-Top Sampling

• The circuit of Figure 11-3 is used to illustrate pulse

amplitude modulation (PAM). The FET is the

switch used as a sampling gate.

• When the FET is on, the analog voltage is shorted to

ground; when off, the FET is essentially open, so

that the analog signal sample appears at the output.

• Op-amp 1 is a noninverting amplifier that isolates

the analog input channel from the switching

function.




10

Figure 11-3. Pulse amplitude modulator,natural sampling.






11

• Op-amp 2 is a high input-impedance voltage

follower capable of driving low-impedance loads

(high “fanout”).

• The resistor R is used to limit the output current of

op-amp 1 when the FET is “on” and provides a

voltage division with rd of the FET. ( rd, the drain-to-

source resistance, is low but not zero)






12

• The most common technique for sampling voice in

PCM systems is to a sample-and-hold circuit.

• As seen in Figure 11-4, the instantaneous amplitude

of the analog (voice) signal is held as a constant

charge on a capacitor for the duration of the

sampling period T s.

• This technique is useful for holding the sample

constant while other processing is taking place, but

it alters the frequency spectrum and introduces an

error, called aperture error, resulting in an inability

to recover exactly the original analog signal.






13

• The amount of error depends on how mach the

analog changes during the holding time, called

aperture time.

• To estimate the maximum voltage error possible,

determine the maximum slope of the analog signal

and multiply it by the aperture time DT in Figure

11-4.






14

Figure 11-4. Sample-and-hold circuit andflat-top sampling.






15

• In the three-channel multiplexed PAM system of

Figure 11-6, each channel is filtered and sampled

once per revolution (cycle) of the commutator.

•

Notice that the commutator is performing both thesampling and the multiplexing.

• The commutator must operate at a rate that satisfies

the sampling theorem for each channel.

• Consequently, the channel of highest cutoff

frequency determines the commutation rate for the

system of Figure 11-6.

Time-Division Multiplexing



Prof. J.F. Huang, Fiber-Optic Communication Lab.

National Cheng Kung University, Taiwan

16

Figure 11-6. Time-division multiplex of three

information sources.






17

• As an example, suppose the maximum signal

frequency for the three input channels are

f A1(max) = 4 kHz, f A2(max) = 20 kHz,

and f A3(max) = 4 kHz.

• For the TDM system of Figure 11-6,

the multiplexing must proceed at

f > 2 f A(max) = 40 kHz

to satisfy the worst-case condition.






18

• We can calculate the transmission line pulse rate as

follows:

The commutator completes one cycle, called a frame,

every 1/40 kHz = 25 ms.

• Each time around, the commutator picks up a pulse

from each of the three channels. Hence, there are

3 pulses/frame x 40k frames/s = 120k pulses/s.






19

• The 4 kHz channel is being sampled at five times

the rate required by the sampling theorem. But if

we slow down the commutator, the 20-kHz channel

will be inadequately sampled.

• One the thought might be to multiplex at 8 k-

frames/sec and sample the 20-kHz channel 5 times

per frame.

•

If you sketch this, as is done in Figure 11-7,you discover that there are

7 pulses/frame x 8k frames/s = 56k pulses/s,

which looks good.






20

Figure 11-7. Possible TDM solution.






21


• The two missing samples stolen from the 20-kHzchannel results in inadequate sampling and periodicaliasing distortion.

• For no errors, the commutation rate must be 17.14 kHz,producing 120k samples/s on the transmission line.

• A better scheme is shown in Figure 11-8 with insertionof channel 1 and 3 between two samples of channel 2.

• With 12.5 ms/pulse and 7 pulses/frame, the multiplexingrate can be

(2 pulses/25ms)/(7 pulses/frame) = 11.428k frames/sand

(11.428k frames/s) x (7 pulses/frame) = 80k pulses/swith no errors.





22


Figure 11-8. TDM solution for minimumtransmission line pulse rate.





23

• In pulse width modulation (PWM), thewidth of each pulse is made directlyproportional to the amplitude of theinformation signal.

• In pulse position modulation, constant-widthpulses are used, and the position or time of occurrence of each pulse from somereference time is made directly proportional

to the amplitude of the information signal.• PWM and PPM are compared and

contrasted to PAM in Figure 11-11.

Pulse Width and Pulse PositionModulation





24

Figure 11-11. Analog/pulse modulation signals.






25

• Figure 11-12 shows a PWM modulator. This circuitis simply a high-gain comparator that is switchedon and off by the sawtooth waveform derived froma very stable-frequency oscillator.

• Notice that the output will go to +V cc

the instantthe analog signal exceeds the sawtooth voltage.

• The output will go to -V cc the instant the analogsignal is less than the sawtooth voltage. With thiscircuit the average value of both inputs should be

nearly the same.• This is easily achieved with equal value resistors to

ground. Also the +V and – V values should notexceed V cc.






26

Figure 11-12. Pulse width modulator.






27


• A 710-type IC comparator can be used for positive-only output pulses that are also TTL compatible.PWM can also be produced by modulation of various voltage-controllable multivibrators.

• One example is the popular 555 timer IC. Other(pulse output) VCOs, like the 566 and that of the565 phase-locked loop IC, will produce PWM.

• This points out the similarity of PWM to continuous

analog FM. Indeed, PWM has the advantages of FM---constant amplitude and good noise immunity---and also its disadvantage---large bandwidth.





28

• Since the width of each pulse in the PWM signalshown in Figure 11-13 is directly proportional to the

amplitude of the modulating voltage.

• The signal can be differentiated as shown in Figure

11-13 (to PPM in part a), then the positive pulses are

used to start a ramp, and the negative clock pulses

stop and reset the ramp.

• This produces frequency-to-amplitude conversion (or

equivalently, pulse width-to-amplitude conversion).

• The variable-amplitude ramp pulses are then time-

averaged (integrated) to recover the analog signal.

Demodulation





29

Figure 11-13. Pulse position modulator.






30

Demodulation

• As illustrated in Figure 11-14, a narrow clock pulsesets an RS flip-flop output high, and the next PPMpulses resets the output to zero.

• The resulting signal, PWM, has an average voltageproportional to the time difference between thePPM pulses and the reference clock pulses.

• Time-averaging (integration) of the outputproduces the analog variations.

• PPM has the same disadvantage as continuous

analog phase modulation: a coherent clockreference signal is necessary for demodulation.

• The reference pulses can be transmitted along withthe PPM signal.





31

• This is achieved by full-wave rectifying the PPMpulses of Figure 11-13a, which has the effect of reversing the polarity of the negative (clock-rate)pulses.

• Then an edge-triggered flipflop (J-K or D-type) canbe used to accomplish the same function as the RSflip-flop of Figure 11-14, using the clock input.

• The penalty is: more pulses/second will requiregreater bandwidth, and the pulse width limit thepulse deviations for a given pulse period.

Demodulation





32

Figure 11-14. PPM demodulator.

Demodulation





33

Pulse Code Modulation (PCM)

• Pulse code modulation (PCM) is produced by

analog-to-digital conversion process.

• As in the case of other pulse modulation techniques,

the rate at which samples are taken and encodedmust conform to the Nyquist sampling rate.

• The sampling rate must be greater than, or equal to,

twice the highest frequency in the analog signal,

f s > 2 f A(max)





34

• A simple example to illustrate the pulse code

modulation of an analog signal is shown in Figure

11-15.

•Here, an analog input sample becomes three binarydigits (bits) in a sequence which represents the

amplitude of the analog sample.

• At time t = 1, the analog signal is 3 V. This voltage

is applied to the encoder for a time long enough thatthe three-bit digital "word", 011, is produced.






35

• The second sample at t = 2 has an amplitude of 6 V,

which is encoded as 110.

• This particular example system is conveniently set

up so that the analog value (decimal) is encodedwith its binary equivalent.






36

Figure 11-15. A 3-bit PCM system showing

A/D conversion.






37

Delta Modulation (DM)

• Like PCM, a delta modulation system consists of an

encoder and a decoder;

• unlike PCM, however, a delta modulator generates

single-bit words that represent the difference (delta)

between the actual input signal and a quantized

approximation of the preceding input signal sample.

• This is represented in Figure 11-19 with a sample-

and-hold, comparator, up-down counter staircasegenerator, and a D-type flip-flop (D-FF) to derive

the digital pulse stream.





38

• The continuous analog signal is band-limited in the

low-pass filter (LPF) to prevent aliasing distortion,

as in any sampling system.

• The analog signal V A is then compared to its

discrete approximation V B .






39

Figure 11-19. Possible delta modulation encoder.






40

• If the amplitude of V A is greater than V B , the

comparator goes high, calling for positive going

steps from the staircase generator.

•

if, however, V B exceeds V A, the comparator goes low,calling for negative-going from the staircase

generator.

• The comparator also sets the D flip-flop (D-FF) and

the output will be properly clocked because theedge-triggered D-FF can change state only at rising

edges of the input clock.






41

• Decoding of the delta modulation (DM) signalcan be accomplished with an up-down staircasegenerator and a smoothing filter

• or simply by integrating the DM pulses as shownin Figure 11-20. The resulting demodulated signal

is illustrated as curve B.

• A practical implementation of a delta modulator isshown in Figure 11-21, where the up-down counterand digital-to-analog converter (DAC) comprise the

staircase generator of Figure 11-19.• The delta modulator of Figure 11-21 is usually

referred to as a tracking or servo analog-to-digitalconverter.






42Figure 11-20. DM demodulator.






43

Fig. 11-21. Up-down staircase generatorfor delta modulator.






44

• As seen in Figure 11-22, the critical parametersdetermining the quality of a system using a constantstep size are the designer’s choice of step size and

sampling period length

•With too small a step size, the analog signal changescannot be followed closely enough; this is calledslope overload (Figure 11-22a).

• With too large a step size, two problems arise: poor

signal approximation (resolution) and largequantization noise (Figure 11-22b). This condition iscalled granular noise.






45

Too long a period has the same problem as too smalla step size and poor resolution (Figure 11-22c).

When the period is too short, too much transmissionbandwidth is required.

Fig. 11-22. Critical design parameters in constantstep-size linear delta modulation.






46

Example: A 5-V pk, 4-kHz sinusoid is to be converted to

a digital signal by delta modulation.

The step size must be 10 mV. Determine the minimum clock rate that will

allow the DM system to follow exactly the

fastest input analog signal change,that is, to avoid slope overload.






47

Solution: The fastest rate of change of a sinewave, v( t) =

V sinw t is the slope at the zero crossover points

( t = 0 in Figure 11-23).

slope = dv( t)/ dt= ( d / dt).(V sinw t) = wV cosw t.

At t = 0, the slope is wV .cos0 = wV or Dv / D t =

2p fV , where f is the frequency of the analogsinusoid.

cyclesV Hz

mV

fV

vt / 079.0

5)4000(2

10

2m

p p

DD






48

D t is half the clock period because steps occur only atpositive transitions of the clock in a practical system.• Thus, T clock = 2D t = 1/ f clock, so that f clock = 1/2Dt =

1/(2x0.079x10-6 s/cycle) = 6.3 MHz.

Figure 11-23. Example problem.






49

Practical DM

• A circuit configuration in present use for telecom-munication applications involving filters, speechscramblers, instrumentation, and remote motorcontrol is shown in Figure 11-24.

• To demodulate the digital signal simple integratethe pulses. In fact, the integrator has the same RCtime constant as the modulator above.

• The integrated demodulator output is the same as

the B curve of Fig. 11-24. The integrator output isthen put through a sharp cutoff LPF to smooth outthe final gain amplifier (VGA) and decision logic asindicated in Fig. 11-25 for adaptive DM.





50

Fig. 11-24. Integrating linear delta modulatorblock diagram and signals.

Practical DM





51

Adaptive DM (ADM)

• A solution to the tradeoffs and compromisesof the simple delta modulation system aboveis to have a variable step size system.

•

This could be accomplished as in Fig. 11-25with a variable-gain amplifier at the outputof the D-type flip-flop.

• A decision circuit that counts the number of

+ and – steps taken over a given period of time and decides whether the step sizeshould be increased and by how much.





52Figure 11-25. Adaptive delta modulator.

Adaptive DM (ADM)





53

• As an example of how this circuit might operate,suppose that the adaptive algorithm (decisioncriteria) will be as follows: The preceding four bitsof ADM output are counted.

• If an equal number of 1s and 0s occur in this

interval (the last four bits), then the VGA gain willbe (the FET switch will be a short).

• If more 1s than 0s or more 0s than 1s are counted,the step size is doubled, the input to the integratorwill be doubled.

• The results are constructed for and analog inputand compared to the results for the linear DM inFigs. 11-26 and 11-27.

Adaptive DM (ADM)





54

Figure 11-26. ADM with step or double step sizes.Arrows are shown along the horizontal axis to indicatewhere the step size changes. These changes are basedon the number of 1s and 0s of the previous 4 bits.

Adaptive DM (ADM)

A i (A )





55

Fig. 11-27. Different DM results depending onstep size used.

Adaptive DM (ADM)

Qualitative Comparisons of





56

• Like PAM, PCM can be time-division multiplexedbecause the modulated samples maintain a fixedposition (slot) and duration in time.

• However, PCM is less noise-sensitive than PAM,

and PCM can use digital constant-amplitudecircuitry, unlike PAM, which requires linear

circuits.

• A disadvantage of PCM is its greater bandwidth

requirement. For example, in a simple 3-bit PCMsystem, three pulses must be transmitted, whereasonly one is transmitted for the PAM sample (seeFigure 11-30).

Qualitative Comparisons of Pulse Modulation Systems

Ad i DM (ADM)





57

Figure 11-30. Pulse and digital modulation waveforms.

Adaptive DM (ADM)



Qualitative Comparisons of




59

Qualitative Comparisons of Pulse Modulation Systems

• An additional disadvantage for PPM over all the

other techniques is that PPM, like continuous-phase

modulation, requires coherent demodulation.

• This usually means that a phase-locked loop and its

acquisition circuitry are required.

• In addition to these advantages for PCM over other

pulse modulation techniques, the use of digital

terminal equipment makes PCM more desirable in

today’s communications marketplace.

chap11--sampling & pulse modulation

Documents