Pulse Modulation Objectives To explain sampling theorem and analyze sampling process To study pulse...

Pulse ModulationPulse Modulation

Objectives• To explain sampling theorem and analyze sampling process• To study pulse amplitude modulation (PAM)• To illustrate time division multiplexing (TDM) method• To study pulse code modulation (PCM)• To describe quantization process• To determine quantization noise• To describe encoding process• To determine transmission bandwidth• To study differential pulse code modulation (DPCM)• To study delta modulation (DM)• To compare PCM and DPCM, PCM and DM systems


In amplitude modulation and angle modulation, some parameter of sinusoidal carrier wave is varied continuously in accordance with the message signal.

This is referred to as analog modulation.


If the carrier consists of (discrete) pulse trains, some parameter of the pulse train is varied in accordance with themessage signal, it is called pulse modulation.

After pulse modulation, the message signal becomes discrete.

One of the major concern:How to recover the message signal from the discrete pulse

trains?


Sampling TheoremA bandwidth limited signal having no frequency components higher than fm Hz may be completely recovered from its samples taken uniformly at a rate at least 2fm samples per second, i.e. the sampling frequency is fs = 2fm Hz.

How to prove the sampling theorem?

According to our knowledge, a pulse train is a periodic function. What is the spectrum of a pulse train?


Fourier transform of a periodic functionA periodic function f(t), of period Ts, or angular frequency s,can be expressed in a Fourier series

where Fn is the Fourier coefficient defined by

( ) sjn tn

n

f t F e

/ 2

/ 2

1( )

s

s

s

Tjn t

ns T

F f t e dtT


Since 1 2()from frequency shifting property, we have

then for a periodic function

its Fourier transform can be expressed as

2 ( )sjn tse n

( ) 2 ( )n sn

F F n

( ) sjn tn

n

f t F e


In particular, if the periodic function is a set of impulse trains,

its Fourier coefficient then becomes

thus

the Fourier transform of a set of impulse trains becomes

( ) ( ) ( )sT s

n

f t t t nT

/ 2 / 2

/ 2 / 2

1 1 1( ) ( )

s s

s s

s s

T Tjn t jn t

n sns s sT T

F t nT e dt t e dtT T T

1

( ) s

s

jn tT

ns

t eT

2( )s

ns

nT


For a message signal g(t) whose spectrum is band-limited to fm Hz (or m = 2fm radian/s), in the sampling process, g(t) is multiplied by a set of impulsetrain,

the product becomes

( ) ( )sT s

n

t t nT

( ) ( ) ( )

( ) ( ) ( ) ( )

ss T

s s sn n

g t g t t

g t t nT g nT t nT


So the sampled signal is a product of message signal and a set of impulse trains

According to the convolution theorem, the sampled signal Spectrum is

where we used the properties g(t) (t) = g(t) and g(t) (t – T) = g(t – T)

Proof

( ) ( ) ( )ss Tg t g t t

1 2 1( ) ( ) ( ) ( )

2s s sn ns s

G G n G nT T

( ) ( ) ( ) ( ) ( ) ( ) ( )g t t g t d g t t d g t




The signal spectrum G() can be recovered by the use of a low-pass filter if there is no overlap betweenthe successive cycles of Gs(). This requires

s 2m or fs 2fm or the sampling interval

Ts 1 / 2fm

Up to this point, we have proved the sampling theorem.

*The minimum sampling rate 2fm is also called the Nyquist rate.


What does sampling theorem tell us?• to convey the information contained in a band-

limited signal it is necessary to send only a finite number of discrete samples.

• a message signal that is band-limited to fm Hz is completely specified by its values at intervals spaced no greater than TS = 1/2fm seconds apart.

In pulse modulation, these discrete samples are used to vary a parameter of a pulse waveform.


Discrete samples play the role of message signal.

Notice that these discrete samples are discrete in time but continuous in amplitude which means our pulse modulation is an analog pulse modulation.

The pulse amplitude takes an analog value.


Analog pulse modulation

1. Pulse amplitude modulation (PAM)

2. Pulse width modulation (PWM)

3. Pulse position modulation (PPM).


Pulse amplitude modulation (PAM)

The amplitude of a train of constant-width pulses is varied in proportion to the sample values of the message signal.

In this scheme, the message is multiplied by a pulse train, which is similar to DSB-SC system.


Pulse amplitude modulation (PAM)The PAM signal spectrum is

A distortion is introduced because of the shape of the sampling pulse so that the spectral density G() has lost its original shape.

How to correct such a distortion?

1( ) ( ) ( ) ( )s s

n

G Q G n QT


A method of signal recovery is to use a filter that has a transfer function

Equalization

The technique of correcting the frequency response of a system for a known distortion is called equalization.

Equalization is often used in correcting distortions which are known but over which one has little control.

mw <w1/ ( )( )

0eq

QH

elsewhere



The message is recoverable by lowpass filtering if fS > 2m, where m is the highest frequency component contained in the message.

If fS < 2m, spectral overlap occurs and lowpass filtering can recover only a distorted form of the message.

PAM signals can be multiplexed in the time domain.


Time division multiplexing (TDM)

The transmission of several sampled signals at a time-sharing basis is called time division multiplexing (TDM).

To recover the individual message at the receiver, it is necessary to sample in a synchronous manner to that done at the transmitter.

Can analog signal be usedin TDM?


It is quite easy to see that the minimum bandwidth for TDM transmission is proportional to the product of the message signal bandwidth and the number of the multiplexed signals. (This assumes that all signals have the same bandwidth.)


Comparison between FDM and TDMIn TDM, multiple incoming signals are sliced into small

time intervals, whereas in FDM the incoming signals are placed on different frequency ranges.

Therefore, in the time domain, all the signals overlap in FDM, whereas signals may overlap in the frequency domain in TDM.

This implies that an analog modulation system cannot use TDM unless sampling is performed (to change the system into a pulse modulation system.)


Comparison between FDM and TDMIn TDM, all the channels require identical circuits, thus providing an advantage in simplicity to TDM. In FDM, different carriers are generated for different channels. Also, different bandpass filters are requiredbecause each channel occupies a different frequency band.

However, in TDM, sampling needs to be done at high speeds and synchronization of timing between the transmitter and the receiver must be achieved. This is a disadvantage.


PAM is still an analog pulse modulation. It is not completely digital because the amplitudes of the pulses takes analog value. In analog pulse modulation, information is transmitted in analog form, but the transmission takes place at discrete times.

If the message signal is represented in a form that is discrete in both time and amplitude, then we have digital pulse modulation. In digital pulse modulation, the signal transmission is in digital form, as a sequence of coded pulses.


In digital pulse modulation, PAM signals need to be further digitized and then encoded for transmission. This is achieved in a pulse code modulation (PCM) system.

Binary PCM (where the pulses have only two permissible values) is the most common.

We will confine our discussion to binary PCM system.

How to produce a PCM signal?


A PCM signal is produced by an analog-to-digitalconversion process.


Pulse code modulation (PCM) system

QuantizationQuantization is the process of transforming the sampled amplitude of a message signal into a discrete level taken from a finite set of possible amplitudes.


How to perform the quantization?1. The amplitudes of signal m(t) lie in the range (- mp, mp), which i

s partitioned into L intervals, each of magnitude = 2mp/L. 2. Each sample amplitude is approximated by the midpoint value o

f the interval in which the sample falls. Quantized samples of m(t) = m (t) (at the middle point of

q

m(t)

t

Quantization of sampled analog signal

- mp

mp

L = number of intervals

= 2m /Lp

m (t)q

m(t)

q(t)

0


The amplitude range:(- mp, mp),

[mp is not necessarily the peak amplitude of m(t)]

Interval: = 2mp/L

In quantization process, a sampling value is approximated by the midpoint of the interval. This introduces an error q(t), defined as the difference between the message signal m(t) and the corresponding quantized sample mq(t),

q(t) = m(t) – mq(t)This error is called the quantization noise.


Quantization noise

Since q(t) is uniformly distributed over the interval (-/2, /2), i.e., the error has equal probability to lie in the range (-/2, /2), the probability density is then 1/, hence the mean square value of q(t) is given by

/ 2 / 2 22 2 2

/ 2 / 2

1 1 ( )

12q q dq q dq


Quantization noise powerSince we can assume that the time average is equal to the statistical average, the quantization noise power is then

the output signal-to-noise ratio is

22 ( )( )

12qN q t

2 2 2

2 2

12 ( ) 3 ( )

( )Qp

m t L m tSNR

m


From above discussion, it is clear that quantizationresults in a loss of information.

(Information can also be lost in PAM due to noise)

Such an information lost due to quantization may be reduced by increasing the number of levels used, L.

e.g. 8 to 16 levels are sufficient for speech communication.

Example: The digital audio compact optical disc (CD) system uses 16 bit quantization and a sampling rate of 44.1 kHz per channel. Assuming the audio signal has a peak to mean power ratio

of 13 dB, occupies the frequency band 0 to 20 kHz and that the recovery filter has an effective bandwidth, allowing for the finite cut-off rate of a practical filter of 22 kHz, estimate the signal to quantization noise ratio attainable.

Solution:fs = 44.1 kHz, n = 16, thus we have L = 216 quantization levels.

So we have

2 2

2 2

10 log( / ) 13 ,

/ 20

p

p

m m dB thus

m m

2

2 ( )

pm

m t

2 2 328

2

3 ( ) 3 26.44 10 88.1

20Qp

L m tSNR dB

m


EncodingAfter sampling and quantization, the analog message signal becomes discrete in values, but it is still not in the form best suited for transmission.

In order that the signal is best suited for transmission, i.e. more robust to noise and interference, an encoding process is required to translate the discrete samples to a more appropriate form, such as the binary digits.


Encoding

m(t)

2m /Lp

000

001

010

011

100

101

110

111

-mp

mp

L = 8

Binary code

Signal quantization and binary code assignment


Encoding

T T0 t

T T0 t

76

5

{ { {7 6 5

1 1 1 1 1 0 1 0 1

Binary coding of samples

PAM signal

Coded samples


Transmission bandwidthDigital signals use much more bandwidth than analog

signals. This can be explained as follows: • The binary digits must be transmitted in the sampling

interval originally allotted to one sample, the binary pulse widths are correspondingly narrower and thereforeoccupy a larger bandwidth according to the inverse time–bandwidth relationship.

• The transmission bandwidth increases proportionately to the number of binary pulses needed.


Transmission bandwidth

For a binary PCM, a distinct group of binary digits (bits) is assigned to each of the L quantization levels.

As n binary digits can be arranged in 2n distinct patterns, L 2n or n log2L

Each quantized sample is thus encoded into n bits.


Transmission bandwidthAccording to sampling theorem, a signal m(t) band-lim

ited to B Hz requires a minimum of 2B samples per second, a total of 2nB bits per second (bps) is required, that is, 2nB piece of information per second. Because a unit bandwidth (1 Hz) can transmit a maximum of two pieces of information (1 or 0) per second, a minimum channel bandwidth is given by

BT = nB Hz

This is the theoretical minimum transmission bandwidth required to transmit the PCM signal.


Bandwidth – Signal-to-noise ratio trade-offAssuming that L = 2n, the output signal-to-noise ratio can be expre

ssed as

where

since n = BT/B, we have

It shows that the signal-to-noise ratio increases exponentially with the transmission bandwidth BT.

2 2 22 2

2 2

3 ( ) 3 ( )(2) no

o p p

S L m t m tL c

N m m

2

2

3 ( )

p

m tc

m

2 /(2) TB Bo

o

Sc

N


If n increases, then BT = nB also increases, this leads to that the signal-to-noise ratio increases. In other words, a larger channel bandwidth corresponds to a higher signal-to-noise ratio.

If you want to improve the signal-to-noise ratio, you have to use a larger channel bandwidth.

--- There is a trade-off between the SNR and the channel bandwidth.

2 /(2) TB Bo

o

Sc

N


Demodulation of PCM signalWhen the PCM signal is demodulated, the signal-to-noise ratio obtained at the receiver should be identical to that at the transmitter. Any noise which may be added during transmission can be eliminated because the binary signal can have only two known values, 1 and 0.

If the value of the pulse is different from the set values,we know that it is due to external noise and we can readjust it to its original value.


Key advantage of PCMIn analog system, a message signal suffers the channel noise and the signal distortion, which are cumulative. Amplification is of little help because it enhances the signal and thenoise in the same proportion. The analog signal can not cleaned periodically, and thus the transmission is not reliable.

In PCM system, the new, clean signals can be completely regenerated at repeater stations because all the information is contained in the code. The PCM signal can then be transmitted over long distance withgreat reliability.


Other advantages of PCM and digital communications

1. Allow us to use computer as a tool for communications. (Computers generate digital signals.)

2. Digital communications systems use a type of coding (error correction code) which can minimize noise and interference, thus producing high quality signals.

3. Digital systems can use both types of multiplexing (FDM and TDM) so that many different sources of information can be handled efficiently.


4. Transmission media (such as optical fibres) that have wide bandwidths are available so we can cope with the large bandwidth requirements of digital systems.

5. Digital signal processing has become well established. Digital electronic circuits are now easy to design and to implement in integrated circuit (IC) form.


Disdvantages of digital communications:

• Requires wider bandwidths than analog transmission

• Requires synchronization between receiver and transmitter.


Minimum Information Capacity (Bit Rate) of PCM SystemsThe information capacity is defined as the number of bits that

can be transmitted per second (bit rate). Since we are using the Nyquist rate for sampling, the minimum bit rate transmitted for a binary system is

That is, the minimum bit rate is equal to double the product of the signal bandwidth and the number of binary pulses.

min

. .

.

2 /

no of samples no of bitsC bit rate

second no of samples

nBbits second


Example:Plain-old-telephone system (POTS)

Voice bandwidth limited to: 3.4 kHzBabdwidth including guard band: 4 kHzSampling frequency: 8 kHzSampling rate: 8,000 samples/sCoding: binary

8 bits per sample(L = 28 = 256 quantization

levels)

Bit rate :./000,6488000 sbits

sample

bits

s

samples


Speech waveform

Speech waveform for part of the word "compute"

cslu.cse.ogi.edu/tutordemos/ SpectrogramReading/waveform.html


When audio or video signals are sampled, it is usually found that adjacent samples are close to the same value.

the difference signal is much less in amplitude than the actual sample less number of quantization levels are needed.

the number of bits per code is reduced resulting in a reduced bit-rate.

the bandwidth required in this case is less than the one required in PCM.

Differential pulse code modulation (DPCM)• to transmit PCM signals corresponding to the difference in

adjacent sample values.


If m[k] is the kth sample, we transmit the difference

d[k] = m[k] – m[k-1].

At the receiver,knowing d[k] and t

he previous sample

value m[k-1] gives m[k].

S ig n a l ra n g e

D iffe re n c es ig n a l ra n g e

m ( t)

S a m p lin gp u lse tra in

D e la y T

Q u a n tiz e r E n c o d e r D P C M o u tp u t+

-

D e la y e d s a m p le sm [k -1 ]

S a m p le d s ig n a lm [k ] d [k ] d [k ]q


Comparison between PCM and DPCM• PCM system has a relatively large signal range, i.e. mp(PCM) >

mp(DPCM)

•

if a constant SNR can be maintained, then from n = log2L and BT = nB

we have L(PCM) > L(DPCM) n(PCM) > n(DPCM) BT(PCM) > BT(DPCM)

To maintain the same S/N, DPCM system requires smaller channel bandwidth than PCM system.

The modulator and demodulator circuits for DPCM are more complicated than those in PCM.

2 2 2

22

12 ( ) 3 ( )

( )o

o p

S m t L m tSNR

N m


Delta modulation (DM)In DPCM, if the signal change is represented by just one bit, thatbit being used for the sign of the sample difference, then we have delta modulation (DM).

In DM, since only 1 bit/sample is employed, it transmits information to only indicate whether the analog signal it encodes is to “go up” or “go down”.

DM has high noise as 1 bit/sample is used, thus fs is much higher than Nyquist frequency 2fm.


Delta modulation system

D e la y T

Q u a n tiz e r

+-m [k -1 ]q

m [k ] d [k ]

m [k ]q

+

+ d [k ]q

D e la y T

-

m [k -1 ]q

m [k ]q+d [k ]q


Delta modulation (DM)

C lo c k p u lse s

d [k ]q

m (t) m [k ]q

I l lu s tra tio n o f d e lta m o d u la tio n


Delta modulation (DM)In DM, mq[k] = mq[k - 1] +dq[k]hence mq[k - 1] = mq[k - 2] +dq[k - 1]then mq[k] = mq[k - 2] +dq[k] + dq[k - 1] = …

= mq[0] +dq[k] + dq[k - 1] + … + dq[1]proceeding iteratively in this manner, we have

and assuming zero initial condition, i.e. mq[0] = 0, yields

The receiver is just an accumulator (adder)! Simple circuit!

0

[ ] [ ]k

q qm

m k d m

0

[ ] (0) [ ]k

q q qm

m k m d m


Delta modulation (DM)

DM system have an advantage in that the electronic circuitry required for modulation at the transmitter and demodulation at the receiver is substantially simpler than that required for other PCM systems.


Comparison between PCM and DMPCMRelatively complicated systemGood signal to quantization noise ratioDMSimple system (simple encoder/decoder and does not require

synchronisation)poor signal to quantization noise ratioHigh sampling rate (4 times of the Nyquist rate)Commonly used in the system with small capacity and low quality requirement and military system

DPCM uses n binary digit to represent the signal difference.DM uses only one binary digit to represent the signal difference.

Questions (Pulse Modulation)Questions (Pulse Modulation)

Questions1. How to sample a bandlimited analog signal to ensure distortion-free recon

struction?2. What is Nyquist rate and Nyquist interval?3. What is time division multiplexing (TDM)?4. How to perform quantization in a PCM system?5. What is quantization noise?6. What is the difference between PAM and PCM? What type of signal PAM

signal is? What type of signal PCM signal is?7. What are the advantages of PCM?8. What is non-uniform quantization?9. To maintain the same S/N, which system requires smaller channel bandwi

dth, PCM or DPCM?10.How to obtain delta modulation (DM) signal?

Exercise Problems (Pulse Exercise Problems (Pulse Modulation)Modulation)

1. For a given signal f(t) = cos1t + cos21t, (a) Draw the time waveform and the spectrum of the signal; (b) Determine the minimum sampling frequency.

2. If the signal f(t) = 10cos20t cos200t, the sampling frequency is 450Hz,

(a) Determine the spectrum of the sampling signal; (b) If an ideal low-pass filter is used to recover f(t) from the sam

pled signal, determine the bandwidth of the low-pass filter required;

(c) What is the Nyquist sampling rate for f(t)?


3. A TDM system consists of 24 transmission channels and a synchronization channel, a sampling rate of 8kHz is used. The bandwidth of the signal for each channel is below 3.3kHz. Determine the minimum channel bandwidth required to transmit TDM signal in the system.

4. Five signals are combined to be transmitted in a TDM system, the combined signals will pass through a low-pass filter. Three channels are used to transmit the signals of frequency range between 300 to 3300 Hz and the rest two channels transmit the signals of 50 Hz to 10 kHz range.

(a) What is the minimum sampling rate required?(b) What is the minimum bandwidth of the low-pass filter

required corresponding to the minimum sampling rate?


5. The information in an analog voltage waveform is to be transmitted over a PCM system with a 0.1% accuracy (full scale). The analog waveform has an absolute bandwidth of 100 Hz and an amplitude range of –10 to 10 V.

(a) Determine the minimum sampling rate needed;(b) Determine the number of bits needed in each PCM word;(c) Determine the minimum bit rate required in PCM signal;(d) Determine the minimum channel bandwidth required for

transmission of this PCM signal.


6. For a signal f(t) = 9 + Amcosmt, with Am 10, to be quantized into exactly 41 binary levels, with one level set at the smallest value of f(t).

(a) Determine the number of bits needed in each PCM word;(b) What are the values of extreme quantized levels Vmax and Vmin

if the quantized levels are centered to [f(t)max + f(t)min] / 2; (c) If Am = 10 V, find the signal to quantized noise ratio.


7. A speech signal with frequency range between 50 to 3300 Hz, the sampling rate used is 8 kHz, the sampled signal is transmitted through a PAM or PCM system.

(a) Determine the minimum bandwidth required by PAM system;

(b) If a binary PCM system is used and the number of quantization level is 8, determine the transmission channel bandwidth;

(c) If the number of quantization level is now 128, recalculate the transmission channel bandwidth.

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Pulse Modulation Objectives To explain sampling theorem and analyze sampling process To study pulse...

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