transmissions method

8/9/2019 transmissions method

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Exponential Carrier Wave Modulation

S-72.1140 Transmission Methods inTelecommunication Systems (5 cr)


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2 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen

Exponential modulation:

Frequency (FM) and phase (PM) modulation

s FM and PM waveforms

s Instantaneous frequency and phase

s Spectral properties

narrow band

arbitrary modulating waveform

tone modulation - phasor diagram

wideband tone modulation

Transmission BW

s Generating FM - signals de-tuned tank circuit

narrow band mixer modulator

indirect modulators


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Contents (cont.)

s Detecting FM/PM

FM-AM conversion followed by envelope detector

Phase-shift discriminator

Zero-crossing detection (tutorials)

PLL-detector (tutorials)

s Effect of additive interference on FM and PM

analytical expressions and phasor diagrams

implications for demodulator designs FM preemphases and deemphases filters


4/41


Linear and exponential modulation

s In linear CW (carrier wave) modulation:

transmitted spectra resembles modulating spectra

spectral width does not exceed twice the modulating

spectral width destination SNR can not be better than the baseband

transmission SNR (lecture: Noise in CW systems)s In exponential CW modulation:

baseband and transmitted spectra does not carry a simplerelationship

usually transmission BW>>baseband BW

bandwidth-power trade-off (channel adaptation-suppressionof in-band noise)


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Assignment


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Phase modulation (PM)s Carrier Wave (CW) signal:

s In exponential modulation the modulation is in the exponent or

in the angle

s Note that in exponential modulation superposition does not apply:

s In phase modulation (PM) carrier phase is linearly proportional to

the modulationx(t):

s Angular phasor has the

instantaneous

frequency (phasor rate)

x t A t t C C C

tC

( ) cos( ( ))

( )

= +

x t A t t PM C C

x t

tC

( ) cos( ( ) )

( ),

( )

= +

3

( ) cos( ( )) Re[e ( )xp( )]C C C CC

x t A t j tA = =

2 ( )C Cf t =

Ct

[ ]{ }[ ]

1 2

1 2

( ) cos ( ) ( )cos cos ( ) ( )

C C f

C f

x t A t k a t a t A t A k a t a t

= + + + +


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Instantaneous frequency

s Angular frequency (rate) is the derivative of the phase (thesame way as the velocity v(t) is the derivative of distance s(t))

s For continuously changing frequency instantaneous frequency

is defined by differential changes:

( )t

/t s

2

Angle modulated carrier

Constant frequency carrier:

0 02 rad/s, 1Hzf = =

1st=

( )( )

d tt

dt

= ( ) ( )

t

t d =

2 1

2 1

( ) ( ) ( )( )

ds t s t s t v t

dt t t

=

Compare to

linear motion:

( )i

v t

( )q

v t


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Assignment

( )( )

d tt

dt

= ( ) ( )

t

t d =


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Frequency modulation (FM)

s In frequency modulation carriers instantaneous frequency is

linearly proportional to modulation frequency:

s Hence the FM waveform can be written as

s

Therefore for FM

and for PM

( ) 2 [ ] ( ) /

( ) 2 [

( )

]

( )

( )

C C C

C C

t f d t dt

t

t f x t

f x f dt t

= + = + =

= +

x t A t f x d t t C C C t

t

C t

( ) cos( ( ) ),

( )

= + z

20 0

( ) ( )Cf t f f x t = +

integrate

( ) ( )t

t d =

( ) ( )t x t =


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AM, FM and PM waveforms

x t A t f x d FM C C

t

( ) cos( ( ) )= + z 2

x t A t x t PM C C

( ) cos( ( ))= +

constant frequency followsderivative of the modulation waveform

Instantaneous frequency directly

proportional to modulation

waveform


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Assignment

s Briefly summarize what is the main difference between FM and

PM ?

s How would you generate FM by using a PM modulator?


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Narrowband FM and PM (small modulation

index, arbitrary modulation waveform)s

The CW presentation:s The quadrature CW presentation:

s Narrow band (small angle) condition:

s Hence the Fourier transform ofXC(t) is

x t A t t C C C( ) cos[ ( )]

= +

x t x t t x t t C ci C cq C ( ) ( )cos( ) ( )sin( )=

x t A t A t ci C C ( ) cos ( ) [ ( / !) ( ) ...]= = + 1 1 2 2

x t A t A t t cq C C ( ) sin ( ) [ ( ) ( / !) ( ) ...]= = + 1 33

( ) 1t

1

2 20

cos( ) cos( ) cos( )

sin( )sin( )

+ =

[ ] [ ]cos( ) ( )sin( )( ) C C CC CA tx A t tt F F

[ ]

[ ]

0

0 0

cos(2 ) ( )

1 ( )exp( ) ( )exp( )2

f t x t

X f f j jX f f j

+

= + +

F[ ]

[ ]

0

0 0

cos(2 )

1 ( ) ( )2

f t

f f f f

= + +

F


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Narrow band FM and PM spectra

s Remember the instantaneous phase in CW presentation:

s The small angle assumption produces compact spectral

presentation both for FM and AM:

( ) [ ( )]

( ),PM

( ) / ,FM

f t

X f

jf X f f

=

=

F

X f A f f j

A f f f C C C C C ( ) ( ) ( ), + >

1

2 20

PM

FM t

t

t x t

t f x d t t

( ) ( )

( ) ( ) ,

=

= z

2 0 0

x t A t t C C C( ) cos[ ( )]= +

0

(0) ()

) )(

(t

t

GGg d

j

+

What does it mean to set this

component to zero?


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s Assume: 1( ) sinc2 ( )

2 2

fx t Wt X f

W W

= = 1

( ) ( ) ( ), 02 2

C C C C C

jX f A f f A f f f + >

( ) [ ( )] ( ) PM PM f F t X f = =

X fPM

( )

( ) [ ( )] ( ) / FM FM f F t jf X f f = =

1( ) ( ) , 02 4 2

C PM C C C

j f f X f A f f A f W W

+ >

1( ) ( ) , 0

2 4 2

C FM C C C

C

f f f X f A f f A f

f f W W

+ >

X fFM

( )

Examp

le


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Assignment


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Solution


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Tone modulation with PM and FM:

modulation index s

Remember the FM and PM waveforms:

s Assume tone modulation

s Then


t

t

( ) cos[ ( ) ]

( )

= + z

2

x t A t x t PM C C

t

( ) cos[ ( )]

( )

= +

x tA t

A t

m m

m m

( )sin( ),

cos( ),= RST

PM

FM

{( ) sin( ),PM

( )2 ( ) ( / )sin( ),FM

m m

m m mt

x t A t

tf x d A f f t

=

= =

142 43


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Tone modulation in frequency domain:

Phasors and spectra for narrowband cases Remember the quadrature presentation:

s For narrowband assume

x t x t t x t t C ci C cq C ( ) ( )cos( ) ( )sin( )=

x t A t A t ci C C ( ) cos ( ) [ ( / !) ( ) ...]= = + 1 1 2 2

x t A t A t t cq C C

( ) sin ( ) [ ( ) ( / !) ( ) ...]= = + 1 3 3

x t A t t C C C( ) cos[ ( )]= +


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Narrow band tone modulation:

spectra and phasors

s Phasors and spectra resemble AM:

x t A t A

t

At

C C C

C

C m

C

C m

( ) cos( ) cos( )

cos( )

=

+ +

2

2


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Assignment

( ) cos( )cos( ) cos( )

cos( ) cos( )2 2

cos( )

C C m m C C C

C m C mC m C m

C C

x t A A t t A t

A A A At t

A t

= +

= + +

+

x t A t A

t

A

t

C C C

C

C m

C

C m

( ) cos( ) cos( )

cos( )

=

+ +

2

2

(i) Discuss the phasor diagrams and explain phasor positions based on

analytical expressions (ii) Comment amplitude and phase modulation

in both cases

AM

NB-FM


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FM and PM with tone modulation and

arbitrary modulation index

s Time domain expression for FM and PM:

s Remember:

s Therefore:

x t A t t C C C m( ) cos[ sin( )]= +

cos( ) cos( )cos( )

sin( )sin( )

+ =

x t A t t

A t t

C C m C

C m C

( ) cos( sin( ))cos( )

sin( sin( ))sin( )

=

cos( sin( )) ( ) ( ) cos( )

sin( sin( )) ( )sin( )

m O mn

m mn

t J J n t

t J n t

= +

=

2

2

neven

nodd

Jn is the first kind,ordern Bessel function

/

PM m

FM m m

A

A f f

=

=


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Wideband FM and PM spectra

s After simplifications we can write:

x t A J n t C C n C mn( ) ( )cos( )= +=

note: ( ) ( 1) ( )n

n nJ J =

,PM

/ ,FM

m

m m

A

A f f

=


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Assignment

x t A J n t C C n C mn( ) ( )cos( )= +=

,PM

/ ,FM

m

m m

A

A f f

=

Assuming other parameters fixed, what happens to spectral

width and line spacing when

(i) Frequency deviation increases in FM

(ii) Modulation frequency increases in FM

(iii) Phase deviation decreases in PM


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Determination of transmission bandwidth

s The goal is to determine the number of significant sidebands

s Thus consider again how Bessel functions behave as the

function of , e.g. we considers Significant sidebands:

s Minimum bandwidth includes 2 sidebands (why?):

s Generally:

Jn( ) >

B fT m,min

= 2

B M f M T m= 2 1( ) , ( ) M( ) + 2

A f W m m

1,

( ) 0.01MJ >

( ) 0.1MJ > ,PM

/ ,FM

m

m m

A

A f f

=


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Assignment


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Transmission bandwidth and deviationD

s Tone modulation is extrapolated into arbitrary modulating signal

by defining deviation by

s Therefore transmission BW is also a function of deviation

s For very large D and small D with

s that can be combined into

1,/ /

m mm m A f W

A f f f W D = =

= =

B M D W T

= 2 ( )

1,2(

2 1

2)

,

mT m D f W

B D f

DW D

>> = +

>>

2 ( )

2 , (a single pair of sideband1 s)

TB M

D

D W

W

=

>=


27/41


Example: Commercial FM bandwidths

Following commercial FM specifications

s High-quality FM radios RF bandwidth is about

s Note that

under estimates the bandwidth slightly

f W

D f W

B D W T

=

= == +

75

5

2 2 210

kHz, 15 kHz

kHz,(D > 2)

/

( )

BT

200 kHz

B D W DT

= 2 1 180 kHz, >> 1


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A practical FM modulator circuit

s A tuned circuit oscillator

biased varactor diode capacitance

directly proportional tox(t)

other parts:

input transformer RF-choke

DC-block


29/41


Generating FM/PMs Capacitance of a resonant circuit can be made to be a function

of modulation voltage.

s That can be simplified by the series expansion

( - )1 12

3

81

1 2

2 2

kxkx k x

kx = + +


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Assignment

Ph d l t


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s Integrating the input signal to a phase modulator produces frequency modulation! Thus PM modulator applicablefor FM

s Also, PM can be produced by an FM modulator bydifferentiating its input

s Narrow band mixer modulator:

Phase modulators:

narrow band mixer modulator


t

( ) cos( ( ) )= + z 2

x t A t x t PM C C

( ) cos( ( ))= +

1( ) ( ) ( ), 02 2

( ) ( ) PM

( ) / 2 ( ) 2 [ ( )] FM

C C C C C

C

jX f A f f A X f f f

t x t

d t dt f t f f x t

+ >

=

= = + Narrow bandFM/PM spectra


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Frequency detection

s Methods of frequency detection

FM-AM conversion followed by envelope detector

Phase-shift discriminator

Zero-crossing detection (tutorials)

PLL-detector (tutorials)

s FM-AM conversion is produced by a transfer function havingmagnitude distortion, as the time derivative (other possibilities?):

s As for example

x t A t t

dx t

dt

A t t d t dt

C C

C

C C C

( ) cos( ( ))

( )sin[ ( )]( ( ) / )

= +

= + +

0

d t dt f t f f x t C

( ) / ( ) [ ( )]= = +2 2

FM


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35 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen25


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Assignment


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Indirect FM transmitters FM/PM modulator with high linearity and modulation index difficult to realize

s One can first generate a small modulation index signal that is then applied into a

nonlinear circuit

s Therefore applying FM/PM wave into non-linearity increases modulation index

should be filtered away, how?

1 4 4 4 4 4 2 4 4 4 4 4 3

Math

ematica

-exp

ressions


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In[14]:= TrigReduce Cos w0t +AmSin wmt2

Out[14]=1

21 + Cos 2 Sin t wm Am + 2 t w0

Indirect FM transmitter:circuit realizations The frequency multiplier produces n-fold multiplication of

instantaneous frequency

s Frequency multiplication of tone modulation increases

modulation index but the line spacing remains the same

2 1

1

( ) ( )

c

f t nf t

nf n

== + FM: 2 ( )

t

n f x d

=

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