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Pulse Amplitude Modulation dan Pulse Code Modulation oleh Risanuri Hidayat

Pulse Amplitude Modulation dan Pulse Code Modulation

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Pulse Amplitude Modulation dan Pulse Code Modulation. oleh Risanuri Hidayat. t. 0. T. t. 0. t. 0. T. 0. t. t. 0. Pulse Amplitude Modulation. Low Pass. Band Pass. 0. 0. *make sure that Sinc function is big and flat by reduce time. By using very narrow. Demodulation of PAM. - PowerPoint PPT Presentation

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Page 1: Pulse Amplitude Modulation  dan Pulse Code Modulation

Pulse Amplitude Modulationdan

Pulse Code Modulation

oleh Risanuri Hidayat

Page 2: Pulse Amplitude Modulation  dan Pulse Code Modulation

190/4 /2 3

-PAM PCM 2

Pulse Amplitude Modulation tf

t

t

t

t

t

n

imp nTttf

tftftf imps

n

nTtnTf

T

T

trecttr

2T 2T

trtft sPAM

n

nTtrnTf

2

sincT

TR

nimp T

n

TF

22

imps FFF 2

1

n T

nF

T

21

0

0

F

B2 B2

0

B2

0

B2B2

T

2

T

2

T

2

T

2

T

0

RFF sPAM

n

RT

nF

T 21

Page 3: Pulse Amplitude Modulation  dan Pulse Code Modulation

190/4 /2 3

-PAM PCM 3

Demodulation of PAM

0

Low Pass Band Pass

ns T

nF

TF

21

LowpassnFT

0;1

0

T

T

RFF sPAM

n

RT

nF

T 21

RFT

n 1

;0

By using very narrow 1.0

T

t

T

trect*make sure that Sinc function

is big and flat by reduce time

Page 4: Pulse Amplitude Modulation  dan Pulse Code Modulation

190/4 /2 3

-PAM PCM 4

-Multiplexing PAM TDM

LPF

LPF

LPFClock

Pulsegenerator

Pulsegenerator

Sampler

Sampler

switch

tf1

tf2

PAM1

PAM2

PAM-TDM before filtering

PAM-TDM to thetransmission line

Switch : determining the synchronization and sequence of the channelsClock : determine the timing of the overall systemPulse generator : produces narrow rectangular pulses to drive the sampler

Page 5: Pulse Amplitude Modulation  dan Pulse Code Modulation

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-PAM PCM 5

Nyquist Sampling

T

2

TTx Two signal are sampled equally

Tx=the time interval between adjacent channels or samples

For n channel

n

TTx

Nyquist interval for one signal

Page 6: Pulse Amplitude Modulation  dan Pulse Code Modulation

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-PAM PCM 6

Bandwidth Requirement

S1 S2 S3 S4

tTx

BF-BF0

Tx

With condition of equal BW and sampling equallytherefore the Total BW Requirement is n x BW

If BW of each channel is not equalTherefore the Total BW Requirement is n x largest BW

Fx B

T2

1

xF T

B2

1

Page 7: Pulse Amplitude Modulation  dan Pulse Code Modulation

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-PAM PCM 7

Transmitting Analog sign al in Digital format

• Advantages Immunity to noise : with some amount of noise digital can withstand

while analog failed to provide virtually error free transmission.

Reduce signal to noise ratio

Error control coding ; parity check, Hamming code make more reliable

Signal can be completely regenerated at intermediate regenerator for long haul system.

More compatible with computer system for signal processing and digital memories for data storage.

More elaborate code can be used.

Ideal for integrated services digital network (ISDN)

Page 8: Pulse Amplitude Modulation  dan Pulse Code Modulation

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-PAM PCM 8

Pulse Code Modulation• m(t) is sampled, each sample value is rounded off to

the nearest allowable value. This value is digitally encoded as a sequence of binary digits

• There are three process of Digitization 1. Sampling

2. ( )Quantization devide into level of voltage oo ooooooooo ooooo oo ooooo oooo() oooo ooo oo o oooooooo ooooooooo oooooo

3. Encoding Each quantization level is encoded into N binary digits

NM 2 MN 2logWhere N is the number of binary digit per code word

M is the number of quantization level

Page 9: Pulse Amplitude Modulation  dan Pulse Code Modulation

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-PAM PCM 9

Quantization

0

-V

V

t

v M Steps

Sampling Signal

v

VM

2 Where M = no. of steps = quantization stepv

Page 10: Pulse Amplitude Modulation  dan Pulse Code Modulation

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-PAM PCM 10

Encoding

0

-V

V

t

000

001

010

011

100

101

110

111

NM 2

1 1 1 1 1 0 1 0 0 0 1 1 0 1 0 1 0 0 1 0 1 1 1

Page 11: Pulse Amplitude Modulation  dan Pulse Code Modulation

190/4 /2 3

-PAM PCM 11

Quantization Error

0

2

v

2

v 2

v

Quantization Noise

2

v2

v

vv1

Uniform distribution

tftfte Q 22

vte

v

Page 12: Pulse Amplitude Modulation  dan Pulse Code Modulation

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-PAM PCM 12

Signal to Noise Ratio

lPowerErrorSigna

rSignalPoweSNR

Q

dttfT

PT

21lim

2

2

22 1T

T

dtteT

te

dvvpv2

2

v2

v

vv1

The average power

Time average

Continuous RV

time

Page 13: Pulse Amplitude Modulation  dan Pulse Code Modulation

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-PAM PCM 13

Signal to Noise Ratio[1]

2

2

2

2

322

3

11v

v

v

v

v

vdv

vvte

883

1 33 vv

v

12

2v

M

Vvtf

vte

tfSNR

Q

2;

12 222

2

tfMV

2224

12

2

2

22 33 M

V

tfM AvgPower

PeakPower

tf

V

2

2

where

Page 14: Pulse Amplitude Modulation  dan Pulse Code Modulation

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-PAM PCM 14

Signal to Noise Ratio[2]

23MSNR

Q

In dB )(log10log203log10 101010 dBMSNRQ

)(log10log2077.4 1010 dBM

Encoding : each quantization level is encoded into N binary digit

NM 2

MN 2log No.of level

No.of binary digit per code word

b

aab

10

10

log

loglog

Page 15: Pulse Amplitude Modulation  dan Pulse Code Modulation

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-PAM PCM 15

- TDM PCM (E1 standard)

300 3.4k

3.1kvoice time

0 500 800300 3.4k

MAN WOMAN

kHzFs 8 (2x3.4k=6.8k)Nyquist

Each sample is quantized and encoded into 8 bits

sec1258

18

kHzTkHzFs

sFBit rate = x 8 = 64kbps ; with 32 channels sec25.4886432

1n

kT

1channel = 8 bits therefore 8 x 488.25nsec = 3.9 sec

Total bit length = 8 bit x 32 channels = 256 bits

MbpsDataRate 048.2sec125

256

nTimePropagatio

LengthBit