3b_Mod

8/14/2019 3b_Mod

http://slidepdf.com/reader/full/3bmod 1/42

1

Lecture 3Data Encoding and Signal Modulation

• Introduction

• Data Encoding and Signal Modulation

• Advantages of Signal Modulation

• Amplitude Modulation

• Amplitude Modulation of Digital Signals

• Reducing Power and Bandwidth of AM Signals

• Frequency Modulation• Power and Bandwidth of FM Signals

• Conclusion

8/14/2019 3b_Mod


2

Introduction

A network and transport layers basic services

comprise the end-to-end transport of the bit streamsover a set of routers (switches). They are producedusing six basic mechanisms:

a. Multiplexingb. Switchingc. Error control

d. flow controle. congestion controlf. and resource allocation.

8/14/2019 3b_Mod


3

Multiplexing

Combines data streams of many such users into

one large bandwidth stream for long duration.Users can share communication medium.

a. Switch b. Multiplexer/Demultiplexer

a. Fully connected network

b. network with shared links.

8/14/2019 3b_Mod


4

Data Encoding is• the process of preparing data for efficient and accurate transmission.

• Because analog data is altered by noise in transmission, the processof analog-to-digital conversion allows more accurate transmission odata. The conversion process will not change the analog data, inconversion the information contained in the analog data will not bedestroyed by the conversion process. Once the data is in digitalform, it can be encoded as a sequence of voltage levels.

Signal Modulation is• the process of encoding a baseband source signal S

m(t) onto a carrier

signal.• The carrier waveform is varied in a manner directly related to the

baseband signal. The carrier can be a sinusoidal signal or a pulsesignal. The result of modulating the carrier signal is called themodulated signal.

8/14/2019 3b_Mod


5

Sampling using PAM.

Original Signal

PulseAmplitude

Modulated

S(t)

S(t)

t

t

Sampling Interval

Ω s > 2 B

8/14/2019 3b_Mod


6

Signal Modulation

S c ( t ) = A · cos ( 2π f t+ φ )

A. Amplitude Modulation (AM), or Amplitude Shift

Keying (ASK): The amplitude A of the carrier signal

changes in direct proportion to the baseband signal.

B. Phase Modulation (PM), or Phase Shift Keying

(PSK): The phase φ of the carrier signal changes in

direct proportion to the baseband signal.C. Frequency Modulation (FM), or Frequency Shift

Keying (FSK): The frequency f of the carrier changes

in direct proportion to the baseband signal.

8/14/2019 3b_Mod


7

Amplitude, Phase, and Frequency Modulation of a digital baseband signal

AM

10 0 0 1 0 1 1 0 0

PM

10 0 0 1 0 1 1 0 0

FM

10 0 0 1 0 1 1 0 0

8/14/2019 3b_Mod


8

S m t

θ n

An

t

d n

T 2T

+<−<−

+>−−<−=

22

if

2or

2if 0

)(n

nn

nn

nn

nn

m d nT t

d A

d nT t

d nT t

t S

φ φ

φ φ

Amplitude, duration, and position modulationusing a pulse carrier.

8/14/2019 3b_Mod


9

Advantages of Signal Modulation

• FDM

• WDM• Radio transmission of the signal.

m103

s

11000

s

m103

;

s

11000 5

8

×=×

===

f

c f λ

Sc ( t ) = A · cos ( 2π fc ) range (fc + f) and (fc - f) .

m30

s

11010

s

m103

;s11010

6

8

6 =×

×==×=

f c f λ

5GHz 6 cm

8/14/2019 3b_Mod


10

FDM

Original Signals

M

S c1 (t )

S m1 (t )

M

S c2 (t )

S m2 (t )

M

S c3 (t )

S m3 (t )

DM

S c1 (t )

DM

S c2 (t )

DM

S c3 (t )

Received Signals

S m1 (t )

S m2 (t )

S m3 (t )

Transmission

Line

8/14/2019 3b_Mod


11

Amplitude Modulation

• AM is simply the multiplication of the baseband signal

with the carrier signal. )2(cos)( t f At S ccc π ⋅=

( )[ ] )2(cos)(

1

)( t f t S Ak t S cmc π ⋅⋅⋅=

( )[ ]

)2(cos)()2(cosoffsetDC

)2(cos)(offsetDC)(

11

1

t f t S At f A

t f t S At S

cmck

cmck

cmmck

π π

π

⋅⋅⋅+⋅⋅⋅=⋅+⋅⋅=

•Modulation coefficient k describes

how efficiently the modulator device multiplies the two signals.

Sm ( t ) X

8/14/2019 3b_Mod


12

DSBTC AM• DSBTC AM. DC = Ac .

• Modulation index i of a DSBTC AM signal

(without DC offset)

• The simplest case of DSBTC AM, where the

baseband signal is a sinusoidal signal with: DC = Ac

( )

c

m

A

t S i

)(max=

Sm ( t ) = Ac + Am · cos ( 2π fm t)

8/14/2019 3b_Mod


13

)2(cos)2(cos1)( 1 t f t f A

A A At S cm

c

mcck

π π ⋅

+⋅⋅⋅=

c

m

A

Ai

=[ ] )2(cos)2(cos1)(

2

t f t f ik

At S cm

cπ π ⋅⋅+⋅=

[ ])(cos)(cos2

1)(cos)(cos y x y x y x ++−=⋅

[ ]

[ ]

component frequencycarrier Noncomponent frequencyCarrier

t f f t f f ik

At f

k

A

t f t f ik At f

k A

t f t f ik

At S

mcmcc

cc

cmc

cc

cmc

−+=

++−⋅⋅+⋅=

⋅⋅⋅+⋅=

⋅⋅+⋅=

))(2cos())(2cos(2

1)2cos(

)2cos()2cos()2cos(

)2(cos)2(cos1)(

22

22

2

π π π

π π π

π π

8/14/2019 3b_Mod


14

Power spectrum of DSBTC AM signalwith sinusoidal baseband signal,

i = 100%, k = 1. V rms

f

frequency f c + f m f c - f m

2

f c

Ac2

2· 2

Ac2

Ac2

2· 2

S ( )

8/14/2019 3b_Mod


15

S m (t)

t

t

S c (t)

t

S(t)

Ac

S ( )

8/14/2019 3b_Mod


16

S 50% (t)

t

t

S 100% (t)

t

S 150% (t)

i = 50%

i = 100%

i = 150%

8/14/2019 3b_Mod


17

[ ] )2(cos)2(cos)( t f t f A At S cmcc π π ⋅⋅⋅=

[ ]))(2cos())(2cos(2

1

)2(cos)2(cos)(

2

2

t f f t f f A

t f t f At S

mcmcc

cmc

++−⋅=

⋅⋅=

π π

π π

V rms

f

frequency f c + f m f c - f m f c

Ac2

2· 2

Ac2

2· 2

Sm ( t ) = Ac · cos ( 2π fm t)

DSBSC AM

S (t)

8/14/2019 3b_Mod


18

S m (t)

t

t

S c (t)

t

S(t)

8/14/2019 3b_Mod


19

Amplitude Modulation of Digital Signals

Baseband Signal

Baseband Signal with DC shift

( ) ( ) ( )

−+−

⋅= t f t f t f t S mmmm π π π

π 10cos

5

16cos

3

12cos

4V5)(

( )t f t S cc π 2cosV5)( ⋅=

Carrier Signal

( ) ( ) ( )

−+−

⋅+= t f t f t f t S mmmm π π π

π

10cos5

16cos

3

12cos

4V5V5)(

8/14/2019 3b_Mod


20

Transmitted Signal

( )( ) ( )( ) ( )( )

( )( ) ( )( ) ( )( )

sideband upper

t f f t f f t f f k

V

sideband lower

t f f t f f t f f k

V

frequencycarrier

t f k

V

t S

mcmcmc

mcmcmc

c

−+++−++

−−+−−−+

=

...52cos5

132cos

3

12cos

92.15

...52cos5

132cos

3

12cos

92.15

)2cos(

25

)(

2

2

2

π π π

π π π

π

DSBTC AM signal with

8/14/2019 3b_Mod


21

DSBTC AM signal with square wave baseband signal, i = 100%,

k = 10.

V rms

f

frequency f c - f m

2

f c

2.50V

f c - 3 f m

2

1.59V

2

1.59V

f c + 3 f m f c + f m

2

0.53V

2

0.53V

20.32V

20.32V

f c - 5 f m f c + 5 f m

8/14/2019 3b_Mod


22

S m (t)

t

t

S c (t)

t

S(t)

5V

8/14/2019 3b_Mod


23

DSBSC AM( ) ( ) ( )

−+−

⋅= t f t f t f t S mmmm π π π

π 10cos

5

16cos

3

12cos

4V5)(

( )( ) ( )( ) ( )( )

( )( ) ( )( ) ( )( )

sideband upper

t f f t f f t f f k

V

sideband lower

t f f t f f t f f k

V t S

mcmcmc

mcmcmc

−+++−++

−−+−−−=

...52cos5

132cos

3

12cos

92.15

...52cos5

132cos

3

12cos

92.15)(

2

2

π π π

π π π

V rms

f frequency f c - f m f c f c - 3 f m

2

1.59V

2

1.59V

f c + 3 f m f c + f m

2

0.53V

2

0.53V

2

0.32V

2

0.32V

f c - 5 f m f c + 5 f m

S (t)

8/14/2019 3b_Mod


24

t

S m (t)

t

S c (t)

t

S(t)180 degree phase shifts

8/14/2019 3b_Mod


25

Binary Phase Shift Keying and

Quadrature Phase Shift Keying

( )

+=

−=

=

+⋅=

V5)(if 180

V5)(if 0

)(

)(2cosV25

)(

t S

t S

t

t t f k

t S

m

m

c

φ

φ π

Since the phase of the transmitted signal can take one of

two values (phase jumping), this type of modulation iscalled binary phase-shift keying (BPSK). It is a constant-

amplitude method of modulation.

If dd t th t BPSK i l th t ff t b 90

8/14/2019 3b_Mod


26

If we add together two BPSK signals that are offset by a 90

degree phase shift, there are four possible phase jumping during

every sampling period. Instead of one possible shift of 180º,

there are three possible transitions: +90º, 180º, and 90º (+270º).‑

The signal can be represented by the following formula:

( )

+=−=

+=+=

−=+=

−=−=

=

−⋅=

V5)(V,5)(if 315

V5)(V,5)(if 225

V5)(V,5)(if 135

V5)(V,5)(if 45

)(

)(2cosV25

)(

21

21

21

21

t S t S

t S t S

t S t S

t S t S

t

t t f

k

t S

mm

mm

mm

mm

c

φ

φ π

S1 (t)

8/14/2019 3b_Mod


27

t

S(t) phase shifts

t

S 1 (t)

t

S 2 (t)

0 1 0 1 0

0 1 1 0 0

00 11 01 10 00180º 90º 180º -90º

Time domainrepresentation of

QPSK signal,

created:

by combining BPSK

signals S1 and S2 .

R d i P d B d idth f AM Si l

8/14/2019 3b_Mod


28

Reducing Power and Bandwidth of AM Signals

• In DSBTC AM, the carrier frequency component carries noinformation.

• In all forms of amplitude modulation, the two sidebands containidentical information.

• Eliminating one of the two sidebands.

• There are four possibilities for transmitting an amplitudemodulated signal

• A. Dual sideband, transmitted carrier (DSBTC AM)• B. Vestigial sideband

• C. Dual sideband, suppressed carrier (DSBSC AM)

• D. Single sideband

• Selecting the type of transmission can depend on the following

factors:• Power limitations

• Bandwidth limitations

• Simplicity of Demodulation Equipment

• Th i d f t itti i id l

8/14/2019 3b_Mod


29

• The power required for transmitting a sinusoidal

baseband signal with DSBTC AM is:

P t : Transmitted power, Ac :Carrier Signal Amplitude

• i : Modulation index

• i=100%, the Pt =1.5 times the power squared of the

carrier frequency. Each sideband contributes 0.25times the power of the carrier frequency.

• Vestigial sideband uses only one of the two sidebands of

the signal.

+=

×××+

×=

21

222

12

22

1

2

2

222

i P

A A A P

c

cmc

t

F ibl t t it AM i l

8/14/2019 3b_Mod


30

Four possible ways to transmit an AM signal.

f c

V rms

(a) DSBTC AM

f f c

V rms

(c) DSBSC AM

f

f c

Gain

f

f c

V rms

f

Bandpass Filter

f c

V rms

(b) Vestigial sideband AM

f

f c

V rms

(d) Single sideband AM

f

Bandpass Filter

F M d l

8/14/2019 3b_Mod


31

Frequency Modulation

• Additive noise

• Frequency modulation is a special case of phasemodulation, where the phase of the transmitted

signal accumulates according to the amplitude

of the baseband signal:

k : Constructional coefficient of the modulator

⋅+⋅= ∫

∞−

t

mcc d S k t f At S ξ ξ π )(2cos)(

8/14/2019 3b_Mod


32

• To find the maximum possible frequency deviation of an FM

signal, we can assume the baseband signal is a constant voltage of

+ Am

, then the integral will reduce to a linear function of t :

( )

⋅

⋅+⋅⋅=⋅⋅+⋅= t Ak f At Ak t f At S m

ccmccπ

π π

22cos2cos)(

( )

π 2

maxm

c

Ak f

⋅=∆

Sm ( t ) = Am · sin (2π fm )

)2(cos2

)2(sin)()(

0

t f f

Ak d f Ak d S k t m

m

m

t

mm

t

m π π

ξ ξ π ξ ξ φ ⋅

=⋅−=⋅= ∫ ∫ ∞−

m

m f

f Ak k ⋅⋅=

π 2

[ ])2(cos2cos)( t f k t f At S m f cc

π π +⋅=

• E ti b d d i Bessel's trigonometric

8/14/2019 3b_Mod


33

• Equation can be expanded using Bessel's trigonometric

identities into a series of phase-shifted sinusoidal terms.

• The amplitude of each term is determined by the Bessel function

of the first kind J n ( k f ), where k f is the modulation index.

[ ] ( )∑∞

−∞=

++⋅=+

n

mc f nm f c

n f n f t k J t f k t f

22cos)()2(cos2cos

π π π π

( )∑∞

−∞=

++⋅⋅=

n

mc f nc

n f n f t k J At S

22cos)()(

π π

)()1()( x J x J n

n

n ⋅−=−

( )

( )

( )

+⋅+⋅+

+⋅−⋅⋅⋅+

⋅⋅=

∑∞

=

22cos

22cos)(

2cos)()(

1

0

π π

π π

π

n

t f n f

nt f n f k J A

t f k J At S

mc

n

mc f nc

c f c

8/14/2019 3b_Mod


34

k f

1.0

0.6

0.4

0.8

0.2

0

-0.2

-0.4

J 0 ( k f )

J 1 ( k f )

J 2 ( k f )

J 3 ( k f ) J 4 ( k f ) J 5 ( k f )

106 842

First five Bessel coefficients for varying values of kf .

8/14/2019 3b_Mod


35

Power and Bandwidth of FM Signals

• Calculating the power of an FM signal is much simpler in the

time domain.

⋅+⋅== ∑

∞

1

220

22

)(2)(2

1

2f n f c

c FM k jk j A

A P

12 +⋅⋅= f T k B B

In this equation B is effective bandwidth of baseband signal,

but BT is absolute bandwidth of FM signal

A cos ( 2π f + k cos ( 2π f ) ) at ario s al es of k

8/14/2019 3b_Mod


36

Ac · cos ( 2π fc + k f

cos ( 2π fm ) ) at various values of k f .

V rms

f

k f = 0.5

V rms

f

k f = 1

V rms

f

k f = 2

V rms

f

f c +6 f m

k f = 10

f c +9 f m f c -9 f m f c -6 f m

f c f c + 3 f m f c - 3 f m

f c f c + 3 f m f c - 3 f m

f c f c + 3 f m f c - 3 f m

f c f c +3 f m f c -3 f m

S(t)

8/14/2019 3b_Mod


37

Original

Signal

Quantizing of sampled

amplitudes

S(t)

S(t)

t

t

Recovery of

Original

Signal

S(t)

t

8/14/2019 3b_Mod


38

• For instance, we want to use only 8 bits to store each

sample, and the maximum possible value of the analog

signal is 10, then we can quantize the signal into multiples

of 10/256 (0.03906). Each number will be represented as

an 8-bit binary number between 0 and 255. To

approximately reconstruct the original analog signal, each

8-bit sample must be multiplied by 0.03906.

• When sending a digital signal over a transmission line, the

sampling rate of the signal determines the bandwidth

required for successful transmission.

• Level detection and regeneration of a digital signal allows

the signal to be transmitted with a significantly smaller

bandwidth than the absolute bandwidth of the signal.

8/14/2019 3b_Mod


39

PCM Signal

Input VoltageV max 0.25·V max 0.5·V max 0.75·V max

64

128

192

256

Conversion from signal voltage to PCM levelin DS1 telephone system.

8/14/2019 3b_Mod


40

• The Nyquist criterion determines the

sampling rate necessary to capture all the

information of an analog signal with adiscrete sequence (taken by sampling the

signal at a certain rate).

• “The signal must be sampled at a rate of twice the signal bandwidth

Ω s > 2 B

• (Ω s is the sampling frequency, B is

bandwidth of analog signal’s.):

Digital Data Encoding Schemes

8/14/2019 3b_Mod


41

Digital Data Encoding Schemes• When digital data is transmitted, it must be

mapped to a signal pattern. The signal pattern

should make transmission as reliable as possible.

• In the simplest encoding:

“1” is converted to signal voltage high, and

“0” is converted to signal voltage low.

This is called Non-Return to Zero-Level (NRZ-L)

encoding.

1 1 1 00

NRZ-L

0 1 10 1 0

Bit rate is twice the frequency

M h E di Bi h i L l

8/14/2019 3b_Mod


• Manchester Encoding, or Biphasic-Level:

0:Signal low (first half) / Signal high (second half)

1:Signal high (first half) / Signal low (second half)

1 1 1 00

Manchester

0 1 10 1 0

Bit rate = signal frequency

3b_Mod

Documents

Transcript of 3b_Mod