Chapter 4c Pcm[1]

8/3/2019 Chapter 4c Pcm[1]

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CHAPTER 4: PULSE CODE MODULATION (PCM)


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Chapter 4 / Pulse CodeModulation

2

TXN SYSTEM USING PCM


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3

PCM means modulating a signal by converting it into pulses andthen coding them.

PCM modifies the pulses created by PAM to create a completely

digital signal.To do so, PCM first quantizes the PAM pulses.

Quantization is a method of assigning integral values in a specificrange to sampled instances or process of assigning each actualsample height to its nearest numbered level.

Advantages: TDM possible, Less corruptible and easilyprocessed.

Disadvantages: much larger bandwidth and does not degradegracefully

INTRODUCTION


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PCM is different from PAM, PWM or PPM because PCM signal is sent to thetransmission line in binary code.

PCM signal consists of digital coded that represents the amplitude of modulatingsignal.

The original analogue baseband is first sampled using a sample-and-holdmethod.

Then the signal is applied to an analogue-to-digital conversion circuit whichconverts each sample height to the nearest sequential binary number using thetwo processes of quantization and encoding carried together.

The binary digits are then transform into a digital signal using one of the digital-

to-digital encoding technique.PCM is actually made up of four separate processes: Sampling (PAM)

Quantization

Binary encoding

Digital-to-digital encoding

PCM SIGNAL


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OBTAINING A PCM SIGNAL


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Quantization is a process of converting the sampled signal (PAM signal) to

a discrete form according to its level numbers.

Quantization level, L will be determined by the number of bits that

represented by each sample and it is given by:

The whole possible range of sample heights will be divided into a number

of voltage steps known as quantization interval.

Saying that a modulating signal is in the range of (-Vp, Vp) and it can bedivided into several step sizes, (Vgiven by:

QUANTIZATION

ebits/samplofnumberwhere2 !! nLn

n

p

n

pp VVV

L

VVV

2

2

2

)(minmax

!

!

!


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7

QUANTIZATION


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Chapter 4 / Pulse CodeModulation 8

QUANTIZATION


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The number of quantization levels are depending on their

applications. For example:

Digital telephone

Every samples represent by 8 bits where n=8 bits and L=256 levels

Compact disk (CD) system

Every samples represent by 16 bits where n=16 bits and L=65 536 levels

The higher the bits number, the better the recovered signal and the

lower the quantization noise would be.

QUANTIZATION


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Figure show a simple method of assigning sign and magnitude values to

quantized samples. Each value is translated into its seven bit binary equivalent.

The eighth bit indicates the sign.

EXAMPLE


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Quantization noise is the different between the original signal (input)

and recovered signal (output). It is given by:

The maximum quantization noise is half of the quantization interval

(step size). For example:

Therefore, the bigger the step size is the higher the noise that might

occurs.

QUANTIZATION NOISE

valueSampledvalueQuantized -QN !

VQVV

VQVV

N

N

5.0)1(211ii)

5.2)5(2

15i)

max

max

!!@!

!!@!


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V

4

0

t1 t2 t3 t4 t5 t6 t7 t

A signal having an amplitude in the range of 0-4V and thesampled signals are shown in figure below. By using the binarycode with n=2 bits, determine:

a) The quantization interval

b) Sampled value at each sampling time

c) Quantized value at each sampling time

d) Quantization noise at t2 and t5

e) The PCM signal in the bit string

EXAMPLE


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The signal-to-noise ratio for quantization (SNR)Q can be found as:

QUANTIZATION NOISE

R

V

R

V

R

V

N

VVVQ

R

VS

N

SSNR

n

n

n

Q

nnN

Q

Q

Q

Q

2

1x

22

22

2

22

2

2

1

2know,We

2Where,

powernoiseonquantizatiMax.

powersignalonquantizatiMax.)(

2

2

2

2

2

max

!

!

!@

!

!!

!

!!


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Hence, the (SNR)Q is:

QUANTIZATION NOISE

[dB]02.6

2log20

2log)2(10

2log10)(

dB,In

2

1

2x

2x

2

)(

2

)(

2

2

22

n

n

n

SNR

R

VR

V

N

SSNR

n

dBQ

nn

Q

Q

Q

!!

!

!

!!!


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Determine the number of levels, L and SNR for quantization both in linear anddecibel for n=8 bits and n=16 bits. Comment on the signal performance for bothconditions.

The higher the bit numbers, the better the performance of the signal yielding to ahigher SNR.

EXAMPLE

dB48.166.02(8)02.6)(

OR,dB16.48

65536log10)(iii)

65536222)(ii)

levels2562i)bits,8F

or

)(

)(

16)8(22

8

!!!

!

!

!!!!

!!

!

nSNR

SNR

SNR

Ln

dBQ

dBQ

n

Q

dB33.696.02(16)02.6)(

OR,dB33.96

10x295.4log10)(iii)

10x295.4222)(ii)

levels655362i)bits,16

For

)(

9

)(

932)16(22

8

!!!

!

!

!!!!

!!

!

nSNR

SNR

SNR

Ln

dBQ

dBQ

n

Q


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After sampling, the minimum sampling frequency is

Where fmax is the maximum baseband frequency.

After encoding, therefore the bit rate, is obtained.

Since the bandwidth of a PCM signal is much larger than that of

the original baseband and also larger than the corresponding PAM

signal, thus bandwidth for PCM is equal to the bit rate obtained.

It is given by:

PCM BANDWIDTH

max2ffs !

max2nfnff sb !!

max2nffBW bPCM !!


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For a digital telephone system, given that the baseband signal is 4

kHz and the number of bits used is 8 bits. Determine the bit rate

and the maximum bandwidth if PCM scheme is used.

Solution:

EXAMPLE

kHz64

)4)(8(2

2 max

!

!

!!

k

nffBW bPCM


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In PCM, the number of levels can be calculated ad below:

where: n = number of bits per sample.

Example:

In PCM, calculate the number of levels if the number of bit

Per sample is 32 as in compact disc video system.

Solution:

L = 2n = 232 = ? levels

No. of Levels, L = 2n


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maximum quantization noise,

QuantizationInterval

No. of bits for each sample, n = log2 L


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Example:

Amodulating signal Vm (t) is limited its frequency to 5 kHz

and sampled at a rate of 20% lower than a minimum of

Nyquist rate. The maximum quantization noise that can be

accepted is0.4%

from the peak amplitude,E

m of the signal.The quantized samples are then coded into binary digits.

Calculate:

a) The required sampling frequency, fs

b) The quantization levels, Lc) The number of bits for each sample, n

d) The maximum bandwidth , BWPCM and bit rate, fb of the

signal.

e) The signal to noise ratio for quantization, SNRQ in dB.


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One analogue signal with frequency of 4 kHz is limited to 0-8V of amplitude

voltage. If the signal is modulated by using Pulse Code Modulation (PCM) with

each sample is represented by 3 bits:

a. Calculate the quantization level, L and its quantization interval,(

V.b. Sketch and label completely the quantization levels of the signal.

c. If given the bit string of the PCM signal as

001010100101101100011010010011, sketch the analogue signal on the

quantization levels drawn in part (b).

d. Determine the signal-to-noise ratio of the quantization, (SNR)Q in dB.e. Find the maximum bandwidth and bit rate of the PCM signal

EXAMPLE

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Chapter 4c Pcm[1]