Pulse Modulation 1. Introduction In Continuous Modulation C.M. a parameter in the sinusoidal signal...

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Transcript of Pulse Modulation 1. Introduction In Continuous Modulation C.M. a parameter in the sinusoidal signal...

Pulse Modulation

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Introduction

• In Continuous Modulation C.M. a parameter in the sinusoidal signal is proportional to m(t)

• In Pulse Modulation P.M. a parameter in the pulse train is proportional to m(t)

• In analog P.M. the parameter (amplitude, position, duration) is varied in a continuous manner

• In digital P.M. the values are discrete values• P.M. is a transition between analog modulation

and digital modulation2

Kinds of Pulse Modulation

• PAM Pulse Amplitude Modulation• PDMPulse Duration Modulation (or) PWM Pulse Width Modulation• PPMPulse Position Modulation• PCM Pulse Code Modulation

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Natural Sampling

• The sampled signal consists of a sequence of pulses of varying amplitude whose tops are not flat but follow the shape of the waveform of the signal m(t).

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• The sampled signal consists of a sequence of pulses of flat tops amplitude.

• It will make distortion for recovered signal, but the distortion will not be noticeable when the number of samples are large,

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Flat-Top Sampling

Pulse Amplitude Modulation(PAM)

• The amplitudes of regularly spaced pulses are varied in proportion to the corresponding sampling values of a continuous message

• Similar to natural sampling: the message signal is multiplied by a train of rectangular pulses. The top of each modulated rectangle is maintained not flat

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• Two operations evaluated in the generation of PAM:

- Instantaneous sampling every Ts

- Lengthening the duration of the sample to some constant value T.

• To reconstruct the signal, we need an equalizer (since the sample and hold filter used at the transmitter alter the shape of the signal) plus the reconstruction filter

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Reconstruction filter

EqualizerPAM signal

• The reconstruction filter is an ideal L.P.F having a cut off frequency equals to the signal bandwidth.

• The equalizer frequency response:

Heq(f) = 1/ | H0(f)| = 1 / [ T sinc (fT )

= π f / [ sin (π f T) ] 10

Quantization Process• Quantization is to approximate each sample

to the nearest level.• Amplitude quantization: The process of

transforming the sample amplitude m(nTs) of a message signal m(t) at time t = nTs into a discrete amplitude v(nTs) taken from a finite set of possible amplitudes

Quantizerg(.)

Continuous sample m

Discrete sample v

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Define:• L Total number of amplitude levels used in

the quantizer• mk discrete amplitudes k = 1,2 …L

(decision levels, thresholds )• vk representation (or reconstruction) levels

• Ik quantizer interval

mk-1 mk mk+1mk+2

vk-2 vk-1 vk vk+1

Ik13

• Step size (quantum): spacing between two representation levels

• The quantizer O/P v = vk if the input signal m belongs to the interval Ik

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Uniform (linear) Quantization

The representation levels are uniformly spaced Mid-tread:

Origin in the middle of staircase Mid-rise:

Origin in the middle of rising part of staircase

Mid-tread Mid-rise15

Quantization Noise for Uniform Quantization

• The use of quantization introduces an error between the I/P signal m(t) and the signal at quantizer output mq(t).

• This error is known as: quantization error or (e), where:

e = m(t) – mq(t)

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• If m(t) is in the range ( - mmax, mmax )Then the Step size of the quantizer :

S = 2 mmax / L • Assume that the quantization error is

uniformly distributed within each quantization range, then its p.d.f f(m) will be: f1(m)+f2(m)+ f3(m)+….+ fL(m) = 1 / S

f(m)

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Quantization Noise for Uniform Quantization

f2(m)f1(m) f4m(

f3(m)f1(m)

fL(m)

ms

• σQ2 =Quantization noise

= mean square value of the error

= mmax2 / [ 3 L2 ] [S=2mmax/L]

• If ‘n’ is the number of bits per sample

L = 2n

S = 2mmax / 2n

σQ2 = mmax

2 2-2n / 318

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232/

2/

2

2/2

2/2

2

22

2/1

2/1

2

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ss

sdxx

s

dmvmmfdmvmmf

s

s

sv

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sv

sv

• Let P = average power of the message signal m(t)

• Assuming m(t) is uniformly distributed from –mmax to mmax

• Output signal to noise ratio =

SNRo/p = P / σQ2 = 3 P 22n / mmax

2

= 3 K (2)2n

• Where:

K = P / mmax2 = 1/3

• [SNRo/p]= 10 log [(2)2n] = 6n (dB)19

max

max

max2

max

2

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1)(

m

m

mdm

mtmP

Non-Uniform quantization(Companding)

• Why?• Low amplitudes happens more frequently

than larger ones• Since the mean square error is proportional

to the step size, we need to decrease the step size for lower values than larger values.

• This is accomplished by using a compressor at the transmitter and an expander at the receiver

• The combination of a compressor and an expander is called compander

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• Two types of compressors may be used:• A law and μ law refer to the parameter which

appear in the equation of compression and expansion.

• -Law

A-Law21

Non-Uniform quantization(Companding)

)1log(

)1log(

xy

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log1

log1

1

log1

xA

forA

xAy

Axfor

A

xAy

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• Dynamic Range: is the difference in dB between max. signal level, and min. signal level having accepted output SNR .

• To improve dynamic range Companding is used to keep SNR high for low signal level as well as for high signal level

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Non-Uniform quantization(Companding)

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Non-Uniform quantization(Companding)

Output SNR (dB) Companded

Un-Companded

Si (dB)normalized input signal

power in dB

30

-18-48 0

10 dB48

• Example:• For acceptable voice transmission the received

signal have a ratio SNR > 30 dB.• The companded system has dynamic range of

input signal=48 dB.• The un-companded system has dynamic range

of input signal=18 dB for the same condition of SNR>30 dB

• The penalty paid is at max. amplitude of input signal, SNR is less 10 dB after companding.

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Non-Uniform quantization(Companding)