b5 radio digital modulation -comm theorem

8/14/2019 b5 Radio Digital Modulation -comm theorem

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SZE 3533SZE 3533

Topic V Radio Digital ModulationTopic V Radio Digital Modulation

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5.0 Digital Transmission5.0 Digital Transmission

Digital signal can also be transmitted through free space if theanalog carrier signal is used.

There are three techniques that can be used:

(ASK Amplitude-Shift Keying)

(FSK) Frequency-Shift Keying) (PSK Phase-Shift Keying)

Pemodulatan Digi

Pulse signals from the previous modulation or coding method mormallyare not transmitted in their original form (baseband signal). They will

modulate a carriers that are suitable with the channels used.

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5.1 Radio Digital Modulation

This modulation technique is similar to the analog modulation in which

the modulating signal will vary the ampltud, frequency or phase of the

carrier signal.

The only difference is that the modulating signal is digital signal :

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Analog ADCLine

coding

Digital

transmission

Sampling

QuantizationCoding

RZ, NRZ, AMI ASK, FSK,PSK

Block diagram for digital transmission system

Digital data is used to

modulate the carrier.The task of the carrier is to

shift the baseband signal

spectrum (digital data) to a

higher spectrum (around

the carrier signal).

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5.2 Amplitude Shift Keying (ASK)5.2 Amplitude Shift Keying (ASK)

Pemodulatan Digi

ASK

1 0 1 1 0

X

=

Unipolar

m m3 m5

Amplitud, V

w(rad/s)

c

Amplitud, V

w(rad/s)

mc 5

mc 5+

c

Amplitud, V

w(rad/s)

-Digital signal is

used to switch thecarrier ampiltude

(low and high).

-Also called on-off

keying (OOK) andinterrupted

continous wave

(ICW).

The string of pulses

from digital signal

will change the

amplitude of the

carrier signal. 5


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5.2.1 ASK signal generation5.2.1 ASK signal generation

Mathematical Analysis :

Vm(t)

Vc(t)

VASK (t)

ASK signal generation

+++= ...)5cos

5

13cos

3

1cos

2

2

1)( ttttv mmmm

mcccc wheretEtv >>= cos)(

+++= ....5cos5

13cos

3

1cos

2

2

1cos)( ttttEtv mmmccASK

Therefore :

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If only the 1st 5 harmonics are considered ;

BWASK

= 5 x fb

=> Bit rate,; fb

= 2 fm

Spectrum of ASK signal

Time domain

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

For unperiodic signal : BWASK

= fb

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vm(t)

vc(t)

vASK

(t)

ASK generation/trasmission :

Using multiplier

vm(t)

vc(t)

vASK

(t)

Using ON and OFF switch

ASK receiver :

vASK

(t)vm(t)

Using envelope detector Using coherent detector

vASK

(t)vm(t)

vc(t)

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Noise or error in digital communication system is measured using Bit Error

Rate (BER).

BER is measured based on the differences between the send and received

bits in period To .

Normally the BER will depends on other factors such the modulation

techniques and SNR (Eb/No) as shown in igure 5.19 (text book).

( )

( )sidedsingle:densityspectrumpowernoise

bitperenergy

oN

ESNR b=

o

be

N

EerfcP

42

1=

jalursatuhingarkuasaspektrumketumpatanadalah

bitsetiapbagitenagapurataadalah

oN

Eb

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5.3 Pemodulatan Anjakan Frekuensi (FSK)5.3 Pemodulatan Anjakan Frekuensi (FSK)

Pemodulatan Digi

FSK

1 0 1

X

=

0 1 0

X

=

+ASK

1

ASK2

Unipolar

m m3 m5

Amplitud, V

w(rad/s)

Amplitud, V

w(rad/s)

mc 5

2+2c

Amplitud, V

w(rad/s)

2c

1c

mc 5

1

1c

The generation of an

FSK waveform at the

Transmitter can be

achieved by generatingtwo ASK waveform and

adding them together

with a summing

amplifier.

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5.3.1 Penjanaan FSK5.3.1 Penjanaan FSK

)()()( 21 tvtvtvFSK +=

v1(t)

v2(t)

Mathematical Analysis :

( ) ( ) ( ) ( ) ( )1dan 12211 tvtvtvtvtv mmm ===Where :

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1jalurdasaruntuk....5cos5

13cos

3

1cos

2

2

1)(1

+++= ttttv mmmm

Taking Fourier series :

++= ....5cos5

13cos

3

1cos

2

2

1)(2 ttttv mmmm

( ) ( )1 12 tvtv mm =And ;

Therefore :

mcccc tEtv >>= ;cos)(And we have ;

Therefore :

+++= ...5cos

5

13cos

3

1cos

2

2

1cos)( 1 ttttEtv mmmccFSK

+++ ...5cos5

13cos

3

1cos

2

2

1cos 2 ttttE mmmcc

)()()(21

tvtvtvmmFSK

+=

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Spectrum for FSK signal

fff

fffBW

b

m

mmFSK

+==

++=

236

323

BW for FSK signal if up to 3rd harmonics are considered is

given by:

For unperiodic signal :

ffBW bFSK += 2

Where: fb

= 2 fm

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Generation of FSK signal :

Using switch

fc1

fc2

fm

vFSK

Using multiplier

The swichting actions will

produce 2 different

frequencies accordingly

LPF VCOfm vFSK

Using VCO

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Example 5.1 :

An unsynchronous FSK system is modulated by a digital signal and

operates at 10 MHz. The frequency deviation is 850 Hz and bit rate is 110

bit/s. The peak to peak carrier signal is 2 V and noise power spectrumdensity (2 bands) is 1 x 10-4 volts2/Hz. Calculate the BER for the system.

Solution :

voltsE

Epp

c 0.12

0.2

2===

sf

Tb

31009.9110

11 ===

( ) ( )

Joule

TEE cb

3

3

1055.4

2

1009.91

2

=

==

HzvoltsNo /1012

24=

HzvoltsNo /10224=

( ) 38.1110221055.4

2

1

2

1 43

===

eeBERPe

61074.5 =16


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Synchronous FSK receiver (coherent detector)

cos(c+ )t

cos( c- )t

vPSK

(t)LPF vm(t)-

+

The output fromthe amplifier is : ( ) ( )[ ]ttt

EEtv cc

ccD ++= 2cos2cos2cos

22

( ) ( )tEtv ccFSK += cosIf the FSK signalreceived is:

( ) ( )tEtv ccFSK = cos

( ) ( )[ ]tttEE

tv cccc

D ++

= 2cos2cos2cos22

LPF will allow eitherEc/2or

Ec/2which will represent the

output digital signal.

=o

be

NEerfcP 6.0

21

Probability of error (BER)

for synchronous detector :

Eb

, energy per bit

/ 2noise powerspectrum density :double

sided

erfc, error function

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Example

5.2 :FSK signal in Example 5.1 is fed to synchronous receiver. Calculate the new

BER and compare with that of BER which uses unsynchronous receiver.

Solution :

43 1021055.4 == ob NdanE

( ) ( )69.32

1

102

1055.46.0

2

14

3

erfcerfcBERPe =

==

From erfc table (3.69) ~ 1.9 x 10-7

87

105.92

109.1

=

=eP

(From Example 5.1)

6107.5 =eP

For unsyhcronous receiver :

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5.4 Phase Shift Keying (PSK)5.4 Phase Shift Keying (PSK)

Pemodulatan Digi

PSK

1 0 1 1 0

X

=

Bipolar

m

m3

m5

Amplitud, V

w(rad/s)c

Amplitud, V

w(rad/s)

mc 5

mc 5+

c

Amplitud, V

w(rad/s)

The phase of the carrier is

set to 0o

or 180o

depending on the digital

signal.

( ) ( )[ ]ttEtv ccPSK += cos

FSK signal can be

represented by :

(t) = 0 => 1

(t) = 180o => 0 , so ;

( ) 1cos


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BWPSK

= fb

= 2fm

PSK signal PSK spectrum

BWPSK

= BWBPSK

= fb

= 2fm The same as BWASK;

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vPSK

(t)

5.4.1 PSK generation5.4.1 PSK generation

Uses inverter to convert binary 1 to -1signal.

Both signals will be fed to one switch to

produce PSK signal.

vPSK

(t)

fc

vm(t)

-1

Using switch

LPF

vm(t)

fc

Using multiplier

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5.4.2 PSK receiver5.4.2 PSK receiverFKS demodulator must use coherent detector.

Advantages of PSK are :

Immune to noise

The same BW with ASK

Multilevel

Using multiplier

+1 atau -1

vPSK

(t) vm(t)

fc

The probability of error in the receiver

of PSK :

=

o

be

N

EerfcP

2

1

2

TEE cb =

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Example 5.3 :

Calculate the probability of error (BER) for Example 5.1 using PSK and

compare the BER with the synchronous and unsynchronous FSK.

Solution :

43 1021055.4 == ob NdanE From Example 5.1

( )77.421

1021055.4

21

4

3

erfcerfcPe =

=

From the table :

erfc (4.77) ~ 1.55 x 10-11

1211

1075.72

1055.1

=

=eP

8108 =eP6107.5 =eP

Comparison :

Synchronous

Unsynchronous

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5.4.3 Summary of Digital Modulation5.4.3 Summary of Digital Modulation

)(1

)(0

2

1

tkosA

tkosA

cc

cc

)(1

00

tkosA cc

Pemodulatan Digi

)(1

)(0

+

tkosA

tkosA

cc

cc

1 0 1 1 0

ASK

FSK

PSK

=

=

=

[

+

+=

...55

13

3

1

1

2

1)(

tkostkos

tkostm

mm

m

=

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Dibit Phase11 45

01 135

00 225

10 315

5.5 Quadrature Phase Shift Keying (QPSK)

One symbol represens 2 bits.

For example, for 4 different symbols they can be represented by

combination of 2 bits (00, 01, 10 & 11) and the phase will vary from 00 to

3600.

For binary PSK the phase changes from 00 dan 1800.

QPSK changes from ( /4) 450, (3 /4) 1350, (5 /4) 2250 to (7 /4) 3150. QPSK is better than PSK because of the efficeicy in the frequency

spectrum, = 2 bps/Hz. The code used in QPSK is Grey code .

Mapping of QPSK signal

using Grey code

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Mathematically they can be written as follow :

( ) ( )4/cos11 += tEtv cc

( ) ( )4/3cos01 += tEtv cc

( ) ( )4/5cos00 += tEtv cc

( ) ( )4/7cos10

+= tEtvcc

Mapping is done to separate the

input bits into 2 components, Iand Q.

I => Inphase

Q => Quadrature

QPSK signal is given

as :S

QPSK(t) = A cos (

ct + [2m 1]/4)

Where m = 1 , 2 , 3 , 4 dan 0 t Ts= 2T

b

The carrier

phase varies forevery 2T

b

11 01 00 10

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x(t)

t

ao a1 a2 a3 a4 a5

event

ao a2 a4 a1(t)

todd

a1 a3 a5a2(t)

I => Inphase

Q => Quadrature

2Tb

QPSK is actually 2 BPSK system

which has a 90o phase shift

between them.

i) Channel I is BPSKIwith

the phase of 0o and 180o

ii) Channel Q is BPSKQ withthe phase of 90o and

270o

180o 0o

I

R

90o

270o

R

I

+

Channel I Channel Q

(-1) (1)

(1)

(-1)

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135o 45o

315o225o

(-1,1) 10

(-1,-1) 00 01 (1,-1)

11 (1,1)CONSTELLATION

DIAGRAM

BPSKI

+ BPSKQ

= QPSK

QPSK spectrum is half of that of

BPSK with the same bit rate.

sin ct - /4

sin ct

sin ct + /4cos ct

28

-sin ct

-cos ct

sin ct +3 /4

sin ct -3 /4


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QPSK TransmitterQPSK Transmitter QPSK ReceiverQPSK Receiver

BPF is used to reduce the unwanted signals (noise, etc). The output from

BPF => I and Q signals. Both signals will be demodulated with oscillator

ofcos ct and sin

ct signals.

LPF will filter out the high frequency signals after demodulation process.

Output from the comparator is logic 1 if the sample value is positive and

logic 0 if negative.

Binary signal will be produced by the parallel to serial converter.

QPSK Receiver Operation :

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=

o

be

N

EerfcP

2

1

Probability of erroror BER for QPSK :

bo

b

f

BW

N

C

N

E=

BER is considered as the ratio of carrier to noise power (C/N) at the

receiver input

The relationship between / and C/N is given by the followingequation :

C/Nis the ration of carrier and

noise power

BWis the bandwidth for noise atthe receiver

fbis the bit rate

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Example 5.4 :

Compare the ratio of carrier to noise power (C/N) needed to send data at 120

Mb/s using BPSK and QPSK if BER required is 10-7 .

Solution :

BPSK :

=

o

be

N

EerfcP

2

1

=

o

b

N

Eerfc2

1107

5.13

o

b

N

EFrom the table:

spectrum BPSK : = 1bps/Hz

BW = fb

/ = 120 / 1 = 120 MHz

bo

b

f

BW

N

CE=

So :

=

o

bb E

BW

f

N

C

[ ] dBN

C3.115.135.13

120

120==

=

From :

For QPSK : 5.13o

b

N

E(= BPSK)

spectrum QPSK : = 2bps/HzBW = f

b/ = 120 / 2 = 60 MHz

[ ] dBN

C3.140.275.13

60

120==

=

QPSK system > 3 dB than BPSK to have

the same BER 31


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5.6 Phase Shift Keying M-ARY

Phase shift keying M-ary refers to the symbol used for modulation system.

M-ary system used includes 8 PSK, 16 PSK, 32 PSK, 64 PSK and so on.

In general every symbol can be represented by several bits :

M= 2n M = symbol or leveln = number of bits

For 16 levels system, every level or symbol can be represented by 4 bits

as follows :

0000, 0001, 0010 ..............1111

The larger the no. of level => more complex circuit & higherC/N.

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Table 5.2 is the summary of BER with 10-7 for M-ary system. In the M-ary

system bit rate is normally written as symbols/s orbaud rate.

Modulation

techniques

Spectrum

efficiency,C/N (dB) (P

e= 10-7)

BPSK 1 b/s/Hz 11.5

QPSK 2 b/s/Hz 14.5

8PSK 3 b/s/Hz 19.5

16PSK 4 b/s/Hz 25.5

32PSK 5 b/s/Hz 32.5

The differences for different modulation techniques, spectrum efficiency, &C/N

The increases in energy per bit, Ebwill reduce the BER and

increases the performance of the system.

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b5 radio digital modulation -comm theorem

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