tm 1 version liu v3 - uni-due.de

Transmission and Modulation of Signals

Slide 1 I. Willms

Lecture Slides


Lecture by Prof. Dr.-Ing. I. WillmsProf. Dr.-Ing. T. Kaiser


Slide 2 I. WillmsSome relevant facts

Lecture: 11:00-12:30hold by

Prof. Dr.-Ing. Ingolf Willms(BA 232,Tel. 0203 379 3276,email: willms @ nts.uni-duisburg-essen.de)

Exercise: 13:00-13:45hold by

Dipl.-Ing. Yao(BA 239, Tel. 0203 379 2547, email: Yao @ uni-due.de)


Slide 3 I. WillmsSome relevant facts

Lecture Script/Video/Exercises etc:@ Copy Shop Steeger, Bismarckstrasseor Website

Examination: 90 minutes, 2 exercises

Examination collection:available for download from the website


Slide 4 I. WillmsModel of a communication system

Transmitter

Receiver

Discrete

Source

Source

encoder

Channel

encoderModulator

Channel

DemodulatorChannel

decoder

Source

decoder

Discrete

sink


Slide 5 I. Willms

What is Modulation ?

Modulation is a process to adapt a given signal to a givenchannel. Most often it means „shifting in frequency“.

Why it is used?

• Besides, transmission of signals at lower frequencies is in general moredifficult.

• Perfect match to allowed bands (for wireless comm. e.g.) is achieved.

• For many baseband signals, the wavelengths are too large forreasonable antenna dimension (e.g. speech signals with 15 km wavelength).

Definition of Modulation


Slide 6 I. WillmsIdeal modulator

The carrier signal might be:

• The information is hidden in the carrier amplitude aAM (Linear Modulation).

• The information is hidden in the carrier phasePM (Non-linear Modulation).

• The information is hidden in the carrier frequencyFM (Non-linear Modulation).

( ) ( ) ( )x t m t s t=

( ) cos( )cm t a tω φ= +


Slide 7 I. WillmsModulation Methods

SinusoidalCarrier

Pulsed Carrier

Carrier

Analog

digital uncoded

digital coded

Source Signal

Delta-Sigma Modulation

Delta ModulationPulse Code ModulationPulse Phase Modulation

Pulse Frequency ModulationPulse Duration Modulation

Pulse Amplitude ModulationPhase Shift Keying

Frequency Shift KeyingAmplitude Shift KeyingFrequency Modulation

Phase ModulationVestigal Sideband AMSingle Side Band AM

Double Sideband AM sCDouble Sideband AM wC

Modulation Method

MM

PCMPPM

PFMPDMPAMPSKFSKASKFMPM

VSBAMSSBAM

DSBAMsCDSBAMwC

Abrev.

( )m t =cos( )ca tω φ+

( )m t =

( )m t T+


Slide 8 I. Willms

AM-Modulation

• Old technique, first used last third of 19th century• Helps to understand more sophisticated techniques• Application in low quality AM radio and analog television

(LW, MW, SW, USW and VHF+ UHF bands)• Usable range differs very much (SW waves travel

around the globe, USW (UKW radio), UHF only usable for regional transmission)

• AM radio is effected by different wave propagation effects


Slide 9 I. WillmsAmplitude Modulation (AM)

( ) cos( ),s s cs t tω ω ω= <<

1A Mm <

( ) (1 cos )cos( )AM AM s cx t A m t tω ω φ= + +

• Source signal

• AM signal

Amplitude modulated sinusoidal signal with modulation degree

oscillation with oscillation with

(1 cos )AM sA m tω+

sωcω (1 cos )AM sA m tω− +

(1 )AMA m+

(1 )AMA m−


Slide 10 I. WillmsAmplitude Modulation (AM)

Amplitude modulated sinusoidal signal with degree of modulation > 1

Phase reversalenvelope

envelope

( ) cos( ),s s cs t tω ω ω= <<( ) (1 cos )cos( )AM AM s cx t A m t tω ω φ= + +

• Source signal

• AM signal

(1 )AMA m+

(1 )AMA m−


Slide 11 I. Willms

Note:• Demodulation can be performed by pure envelope

demodulator• This demodulation fails (ambiguity) with a too large

degree of modulation (without detection of phase reversals)

( ) (1 cos )cos( )cos( ) cos( ) cos( )

AM AM s c

c AM s c

x t A m t tA t Am t t

ω ω φω φ ω ω φ

= + +

= + + +

Consideration in frequency domain of:


Slide 12 I. Willms

Fourier Transform of AM

( ) c o s ( ) (c o s (( ) ) )2

A MA M c s c

A mx t A t tω φ ω ω φ= + + − −

1Using cos cos (cos( ) cos( )) follows:2

α β α β α β= − + +

(c o s (( ) ))2

A Ms c

A m tω ω φ+ + +

0cos( )tω φ+ 0 0( ( ) ( ))j je eφ θπ δ ω ω δ ω ω− + + −

( ) ( ( ) ( ))j jAM c cX A e eφ φω π δ ω ω δ ω ω−= + + −

( ( ) ( ))2

j jAMs c s c

A m e eφ φπ δ ω ω ω δ ω ω ω−+ + − + − +

( ( ) ( ))2

j jAMs c s c

A m e eφ φπ δ ω ω ω δ ω ω ω−+ + + + − −


Slide 13 I. Willms

Power of AM with sinusoidalinput

( )2AMAm π ( )

2AMAm π

/ 22

/ 2

1lim ( )T

AM AMTT

P x t dtT→∞

−

= ∫

Fourier-transform of a modulated sinusoidal signal

( )Aπ ( )Aπ

( )2AMAm π ( )

2AMAm π

cω− cωc sω ω− − c sω ω− + c sω ω− c sω ω+

2 2 2 2 2 22 2

2 8 8 2 4AM AM AMA m A m A mA A

= + + = +

( )AMX ω


Slide 14 I. Willms

Note:• Sidebands have equal and full information about source

signal• xAM(t) is quasiperiodic• Carrier contains most part of power due to typical low

value of degree of modulation• Thus transmission with suppresed carrier is useful• Moreover: arbitrary source signals are considered


Slide 15 I. WillmsExample of Spectrum

0

0

11 ( )2sf

a ωβ ωω−

= − +0

1 aαω−

=

0 0

( ) ( )( ) ( )( )sf sfrS rect rect

ω ω ω ωω β αω β αω

ω ω− +

= + + −

Example for arbitrary source signal with representative spectrum:

where

sfω−sfω0

2sfωω− −

0

2sfωω− + 0

2sfωω − 0

2sfωω +

( )rS ω


Slide 16 I. Willms

Representative signal

Representative Signal

{ }( ) ( )r rs t S ω= -1F0 00 0/ 2 / 2

(( ) ) (( ) )2 2

sf sfsf sfsi t si t

ω ω ω ωω ωω ωπ π+ −

= + − −

0(1 )( ) ( )

2sf

sf

asi t si t

ω ωωπ−

−

( )rs t

0 ( 1)2aωπ+


Slide 17 I. Willms

• By varying the parameter either higher frequenciesor lower frequencies can be emphasized.

• The shift frequency enables either a low pass signalor a band pass signal. This is especially useful in orderto explain Single Sideband Amplitude Modulation (SSBAM).

Properties of representative signal/spectrum :

• Band limitation with cut-off frequency .

Representative Signal

0 / 2co sfω ω ω= +

sfω

• The representative spectrum as well as the signal are still easy enough.

( 1)a <( 1)a >


Slide 18 I. Willms

The general transmitted AM-signal is given by

Bandlimited Signal

min ( ) 1 or min ( ) 1,s t s t≥ − − ≤

( ) (1 ( ))cos( ).AM cx t A s t tω φ= + +

(1 ( )) 0,A s t+ ≥For envelope demodulation we require

Thus it follows

Comparison with

gives the degree of modulation:

1,AMm ≤

min ( ).AMm s t= −


Slide 19 I. Willms

( )AMX ω

{ } { }( ) ( ) (1 ( )) cos( )AM AM cX x t A s t tω ω φ= = + +F F

{ } { }cos( ) ( ) cos( )c cA t As t tω φ ω φ= + + +F F

The Fourier-transform of is obtained as

{ } { } { }1cos( ) ( ) * cos( )2c cA t A s t tω φ ω φπ

= + + +F F F

( )AMx t

( ( ) ( ))j jc cA e eφ φπ δ ω ω δ ω ω−= + + −

( ) * ( ( ) ( ))2

j jc c

A S e eφ φω δ ω ω δ ω ω−+ + + −

( ( ) ( ))j jc cA e eφ φπ δ ω ω δ ω ω−= + + −

( ( ) ( ) ).2

j jc c

A S e S eφ φω ω ω ω−+ + + −

Double-Side-Band AMwC for arbitrary spectrum


Slide 20 I. WillmsRepresentative spectrum

cω

, ( )AM rX ω

2A

envelopes

lower sideband

upper sideband

0

2sf cωω ω− − − 0

2sf cωω ω+ +

(1 (0))rA s+

, ( )AM rx t0 2 4Hzω π=

2 4sf Hzω π=

2 12c Hzω π=

0.75a =0φ =

cω−


Slide 21 I. Willms

Power of AM is given by (see exercises)

Power of AM signal

/ 22

/ 2

1lim ( )T

AM AMTT

P x t dtT→∞

−

= ∫/ 22

2

/ 2

1(1 lim 2 ( ) ( ) )2

T

TT

A s t s t d tT→ ∞

−

= + +∫

2

,2A

with a carrier power of

and a „mean“ power of

and a power needed for transmission of

/ 22

/ 2

1lim 2 ( ) ,2

T

TT

A s t dtT→∞

−∫

/ 222

/ 2

1lim ( ) .2

T

TT

A s t dtT→∞

−∫


Slide 22 I. WillmsBand spreading factor

( )AMX ω

A practical example (MW, 110 channels)

30Hz 4.5kHz 510kHz 520kHz 1600kHz

Band spreading factor

bandwidth of the modulated signalsum of bandwidth of all source signals

( )x t( )ns t


Slide 23 I. Willms

Demodulation of DSBAMwC signals can be achieved in two principaldifferent ways:

1- Using a time-variant system

An example is the multiplication of with a sinusoidalfunction

2- Using a non-linear systemAn example for a non-linear system is given by or

Demodulation of DSBAMwC

( )AMy t0cos( ).ctω φ+

2 ( )AMy t

( ) .AMy t

Main principles of these two versions to extract information of the envelope


Slide 24 I. WillmsDemodulation withtime-variant system

( ) ( ) cos( ) with DAM for Demodulation of AM and ( ) ( )

DAM AM c

AM AM

z t y t ty t x t

ω φ= +=

( ) cos( )AM cx t tω φ= +

(1 ( )) cos( ) cos( )c cA s t t tω φ ω φ= + + +

Conc. Type 1: Demodulation with a multiplier

1(1 ( )) (1 cos(2( )))2 cA s t tω φ= + + +

(1 ( ))(1 cos(2( ))),2 cA s t tω φ= + + +

1( ) (2 ( ) ( )) 2 2D AMAZ Sω πδ ω ω

π= +

2 2*(2 ( ) ( 2 ) ( 2 ) )j jc ce eφ φπδ ω π δ ω ω δ ω ω −⎡ ⎤+ − + +⎣ ⎦

2 22 ( ) ( 2 ) ( 2 )4

j jc c

A S S e S eφ φω ω ω ω ω −⎡= + − + +⎣2 24 ( ) 2 ( 2 ) 2 ( 2 ) .j j

c ce eφ φπδ ω πδ ω ω πδ ω ω − ⎤+ + − + + ⎦


Slide 25 I. Willms

This results from two times multiplication of source signals including DC signal by carrier signal:

Demodulation-Figure

2 2( ) 2 ( ) ( 2 ) ( 2 )4

j jDAM c c

AZ S S e S eφ φω ω ω ω ω ω −⎡= + − + +⎣

2 24 ( ) 2 ( 2 ) 2 ( 2 ) .j jc ce eφ φπδ ω πδ ω ω πδ ω ω − ⎤+ + − + + ⎦

( )DAMZ ω

( )LPH ω

( )2

Aπ ( )2

Aπ

2 cω− 2 cω2 c sω ω− + 2 c sω ω−sω− sω

Aπ

2A


Slide 26 I. Willms

Demodulated output signal is:

Conclusions

( ) ( ) ( ) (2 ( ) ( ))2

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

DAM DAM LP

DAM DAM LP

AG Z H S

Ag t z t h t s t

ω ω ω πδ ω ω= ∗ = +

= ∗ = +

• Why is the division of A by two?

• cannot be suppressed by an additional highpass filteras long as has non-zero mean, why?

• A channel with a possible delay and/or a scale factor , so

For DAM the receiver needs to know .

Is it possible?

( )δ ω( )s t

0 0 0( ) ( ) (1 ( )) cos( ( ) )AM AM cy t x t t A s t t t tα α ω φ= − = + − − +

0ctφ ω−


Slide 27 I. Willms

Now the influence of an unknown phase is considered.


Slide 28 I. Willms

Demodulation is considered using an ideal modulator with modulating function

Unknown Phase

cos( )ctω φ′+ cos( )ctω φ+instead of

FOURIER-transform of it gives:

( ) ( ) cos( )DAM AM cz t x t tω φ′ ′= +

(1 ( )) cos( ) cos( )c cA s t t tω φ ω φ′= + + +

[ ](1 ( )) cos( ) cos(2 )2 cA s t tφ φ ω φ φ′ ′= + − + + +

1( ) (2 ( ) ( )) (2 ( ) cos( ))2 2DAMAZ Sω πδ ω ω πδ ω φ φ

π′ ′= + ∗ −

( ) ( )( 2 ) ( 2 )j jc ce eφ φ φ φπ δ ω ω δ ω ω′ ′+ − +⎡ ⎤+ − + +⎣ ⎦

( ) ( )2 ( ) cos( ) ( 2 ) ( 2 )4

j jc c

A S S e S eφ φ φ φω φ φ ω ω ω ω′ ′+ − +′⎡= − + − + +⎣( ) ( )4 ( ) cos( ) 2 ( 2 ) 2 ( 2 ) .j j

c ce eφ φ φ φπδ ω φ φ πδ ω ω πδ ω ω′ ′+ − +′ ⎤+ − + − + + ⎦From this higher frequency bands are filtered out!


Slide 29 I. Willms

This gives: [2 ( )cos( ) 4cos( )]4

[ ( ) ( )]cos( )2

A S

A S

ω φ φ φ φ

ω πδ ω φ φ

′ ′− + −

′= + −

Is it possible to exactly reconstruct from this signal (or from this spektrum) s(t) without knowing the carrier phase?

Yes, by tricky applying 2 2cos ( ) sin ( ) 1x x+ =

For this we need the signal [ ( ) ( )]sin( ).2A S ω πδ ω φ φ′+ −


Slide 30 I. WillmsUnknown Phase

( ) (1 ( )) cos( )sin( )DAM c ct A s t t tω ω φ ω φ′ ′ ′= + + +

[ ](1 ( )) s in ( ) co s(2 ) .2 cA s t tφ φ ω φ φ′ ′= + − + + +

( ) ( ) ( )DAM DAM LPu t t h tω′ ′= ∗ (1 ( )) s in ( ).2A s t φ φ ′+ −

2 2( ) ( ( )) ( ( ))QADM DAM DAMg t g t u t′ ′= +

2 22 2 2 2(1 ( )) s in ( ) (1 ( )) cos( )

4 4A As t s tφ φ φ φ′ ′= + − + + −

2 21 ( ) sin( ) cos( ) 1 ( )2 2A As t s tφ φ φ φ′ ′= + − + − = +

This signal is applied according to:

The next figure shows the blocks of a corresponding demodulator.


Slide 31 I. WillmsQADM

( ) ( ) cos( )DAM AM cz t y t tω φ′ ′= + ( ) ( ) sin( )DAM AM ct y t tω ω φ′ ′= +

( ) ( ) ( )DAM DAM LPg t z t h t′ ′= ∗ ( ) ( ) ( )DAM DAM LPu t t h tω′ ′= ∗

2 2( ) ( ) ( )QADM DAM DAMg t g t u t′ ′= +

( )DAMz t′ ( )DAMg t′

( )AMy t

( )LPH ω

cos( )ctω φ ′+

sin( )ctω φ ′+

( )DAM tω′ ( )DAMu t′

2( ( ))DAMg t′

2( ( ))DAMu t′

( )QADMg t


Slide 32 I. Willms

• Advantages: Demodulation in presence of unknownphases

• Disadvantages: Suitable square root device needed• Moreover: Sign of 1 + s(t) is lost

Conclusions on QADM


Slide 33 I. Willms

QAM transmission• Due to orthogonal carrier signals when applying a sine

and cosine it is possible to transmit 2 signals s1(t) and s2(t) at one time!

• The following calculation shows the demodulation technique


Slide 34 I. WillmsQAM-transmission of 2 signals

1 1 2 2( ) (1 ( )) cos( ) (1 ( )) sin( )QAM c cx t A s t t A s t tω φ ω φ= + + + + +

1 1( ) (1 ( )) cos( ) cos( )Q AM c cz t A s t t tω φ ω φ ′= + + +

2 2(1 ( )) sin( ) cos( )c cA s t t tω φ ω φ ′+ + + +

Band spreading factor of QAM:

1sin cos (sin ( ) cos( ))2

α β α β α β= − + +

11(1 ( ))(cos( ) cos(2 ))

2 cA s t tφ φ ω φ φ′ ′= + − + + +

22(1 ( ))(sin( ) sin(2 )),

2 cA s t tφ φ ω φ φ′ ′+ + − + + +

( ) ( ) ( )QAM QAM LPg t z t h t= ∗

1 21 2(1 ( )) cos( ) (1 ( )) sin( ).

2 2A As t s tφ φ φ φ′ ′+ − + + −

11( ) (1 ( )).

2QAMAg t s t

φ φ′== +

22/ 2

( ) (1 ( )).2QAMAg t s t

φ φ π′= −= +

( )QAM tβ 1.


Slide 35 I. Willms

• QAM may suffer from cross talk (in case of phase distortions)• QAM usable for data transmission (less suitable for speech, partly for audio stereo)

If the receiver exhibits a frequency shift of , described byFrequency shift distortions

ω∆cos(( ) ) it followsc tω ω φ∆+ +

( ) ( ) cos(( ) )DAM AM cz t y t tω ω φ∆= + +

Conclusions:

(1 ( )) cos( ) cos(( ) )c cA s t t tω φ ω ω φ∆= + + + +

(1 ( ))(cos( ) cos(2( ) )).2 cA s t t t tω ω φ ω∆ ∆= + + + +

1( ) ( ) ( ) (1 ( )) cos( )2DAM DAM LPAg t z t h t s t tω∆= ∗ = +


Slide 36 I. Willms

Con. Type 2 (Nonlinear Demodulation)- Very simple DRC-demodulator circuit is used- Simplicity is reasonable as cheap receivers for many consumers

were needed (Transmitter can be costly)Details are discussed based on an excitation with a rect-signal and based on a DSBAMwC and a DSBAMsCsignal:


Slide 37 I. WillmsDRC-demodulator

( )outu t( )inu t

DRC-demodulator output in case of rectangular excitation and ideal diode

0 0/ 2( ) ( ), 0 .in

t Tu t u rect uT

−= >

( )outu t

( )inu t0u


Slide 38 I. WillmsDRC-demodulation

( ) ( ) ( ) (1 ( )) cos( ) ( ),in AM cu t y t t A s t t tε ω φ ε= = + +

RCτ =The choice of is essential for the performance.

The DRC-demodulator output in case of

( )AMy t( )DRCg t 1

0 .11

212

c

s

A M

H zH z

A

m

ωω

πφ

=

==

=

=

10 .sτ =


Slide 39 I. Willms


DRC-demodulation

( )AMy t

( )DRCg t10.1

1

212

c

s

AM

H zH z

A

m

ωω

πφ

===

=

=

50 1/ .ssτ ω= >


Slide 40 I. Willms


DRC-demodulation

( )AMy t

( )DRCg t10 . 1

1

212

c

s

A M

H zH z

A

m

ωω

πφ

=

==

=

=

5 1/ .ssτ ω= <


Slide 41 I. Willms


DRC-demodulation

( )AMy t

( )DRCg t 10 . 1

1

212

c

s

A M

H zH z

A

m

ωω

πφ

=

==

=

=

1.5.AMm =


Slide 42 I. Willms

DRC-enhancement „Absolute value demodulator (AVD)“

( )i t

( )outu t

10 .1

1

21 .5 .

c

s

A M

H zH z

A

m

ωω

πφ

===

=

=


Slide 43 I. Willms

Output signal of DRC-enhancement AVD

( ) ( )AVD AMz t y t=

(1 ( )) cos( )cA s t tω φ= + + (1 ( )) cos( ) .cA s t tω φ= + +

Hence it results for positive 1 + s(t) :

( )AMy t ( )AVDz t ( )AVDg t

12 ( )

2

2 ( 1)cos( ) .4 1

c

nj n t

cn

t en

ω φω φπ

+∞+

=−∞

−+ =

−∑

12 ( )

2

2 ( 1)( ) (1 ( )) .4 1

c

nj n t

A V Dn

Az t s t e

nω φ

π

+∞+

= −∞

−= +

−∑


Slide 44 I. WillmsDRC-enhancement AVD

{ }1

2 ( )2

2 ( 1)( ) (1 ( ))4 1

c

nj n t

AVDn

AZ s t e

nω φω

π

+∞+

=−∞

−= +

−∑ F

{ } { }1

2 ( ) 2 ( )2

2 ( 1) ( ( ) )4 1

c c

nj n t j n t

n

Ae s t e

nω φ ω φ

π

+∞+ +

=−∞

−= +

−∑ F F

12

2

2 ( 1)( ) (2 ( 2 ) ( 2 )).4 1

nj n

AVD c cn

AZ e n S n

nφω πδ ω ω ω ω

π

+∞

= −∞

−= − + −

−∑

, ( )AVD rZ ω2

( ) (1 ( ))A V D

Ag t s t

π= +

( )LPH ω

( )S ω( 2 )cS ω ω−

(4 / 3)A (4 / 3)A

2 cω− 2 cωsωsω−


Slide 45 I. WillmsDRC-enhancement SLD

( )AMy t ( )SLDz t

Usage of a square law demodulator (SLD)

( )SLDg t

2( ) ( )SLD AMz t y t=2 2 2(1 ( )) cos ( )cA s t tω φ= + +2

2(1 2 ( ) ( ))(1 cos(2 2 )).2 cA s t s t tω φ= + + + +

2Assumption of ( ) 2 ( ) or s(t) 2 gives:s t s t<< <<2

( ) 2

2

( ) (1 2 ( ))(1 cos(2 2 ))2

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

SLD cs t

c c

Az t s t t

A t s t s t t

ω φ

ω φ ω φ

<< ≈ + + +

= + + + + +


Slide 46 I. Willms

Spektrum of SLD demodulated signal

, ( )SLD rZ ω

2 22 ( ) ( 2 ) ( 2 )).j jc cS e S e Sφ φω ω ω ω ω−+ + − + +

22 2

( ) 2( ) (2 ( ) ( 2 ) ( 2 )2

j jSLD c cs t

AZ e eφ φω πδ ω π δ ω ω π δ ω ω−<< ≈ + − + +

( )LPH ω

( )S ω( 2 )cS ω ω−

2( / 2)A π− 2( / 2)A π

2 cω− 2 cωsωsω−

2( )A π

( ) 2

2

It follows: ( ) (1 2 ( ))2s tSLDAg t s t

<<≈ +

Disadvantage of SLD demodulation: Low allowed modulation degree leads to low power efficiency of the transmission.


Slide 47 I. Willms

SSB-Modulation and Demodulation• Reducing bandwidth of DSBAM transmission methods leads to SSB

techniques• Problem: Low frequency components (about 20 Hz e.g.) in source

signals• SSB modulation needs suitable, (nearly ideal) LP or HP filtering

device with high filter selectivity


Slide 48 I. WillmsSSB-Generation

, ( )AM rX ω

s cω ω+

Problem: Suitable fast transition of frequency response of the filter (frompassband to stopband) requiring almost perfect rectangular frequency response. Possible solution: Use of Hilbert transform circuits and analytic signal

upper sidebands

lower sidebands

( )LSBH ω

( )USBH ω

s cω ω− − cω− cω

2A

Generation of an SSBAM-signal


Slide 49 I. WillmsSSBAM-Spectrum

, ( )SSB rX ω

s cω ω− − s cω ω+

2A

cω− cω( )rS ω

( ) ( )rS signω ωsω− sω

sω−

sω( ( ) ( ) ( )) ( )

4 r r cA S S signω ω ω δ ω ω+ ∗ −

( ( ) ( ) ( )) ( )4 r r cA S S signω ω ω δ ω ω+ − ∗ +

2A

s cω ω− − s cω ω+cω− cω

( ) ( ( ) ( ) ( )) ( ) ( ( ) ( ) ( )) ( ) .4

j jSSB c c

AX S S sign e S S sign eφ φω ω ω ω δ ω ω ω ω ω δ ω ω −⎡ ⎤= + ∗ − + − ∗ +⎣ ⎦


Slide 50 I. Willms

• In the following we will make use of the analytic signal and theequivalent LP signal

• The relation of the analytic signal and its Fourier transform to the normal signal and its Fourier transform is as follows:

0

0

( ) 2 ( ) ( ) (1 ( )) ( ) (1 ( )) ( )ˆˆ( ) ( ) ( ) with X( ) ( ) ( )

X X sign X j jsign X

x t x t jx t jsign X

ω ε ω ω ω ω ω ω

ω ω ω

= = + = − ⋅

= + = −


Slide 51 I. Willms

The analytic signal is also given by hence

Analytic signal

{ }0 1 0( ) ( )x t X ω−= F

{ }1( ) ( )ELP ELPx t X ω−= F

0( )0 ( ) ( ) cj tELPx t x t e ω φ+=

The analytic signal and the equivalent

low pass signal are related by:

where is an arbitrary phase.00 ( ) ( ) ,jELP cX X e φω ω ω= − 0φ

0 ˆ( ) ( ) ( ),x t x t jx t= +

{ }0( ) Re ( )x t x t=

{ }0( )Re ( ) cj tELPx t e ω φ+=

{ }0( )Re ( ) c ELPj t x tELPx t e withω φ+ += ( )( ) ( ) ELPj x t

ELP ELPx t x t e=

0( ) cos( ( ) ).ELP c ELPx t t x tω φ= + +


Slide 52 I. Willms

• Generation of SSB filtering can be realized as follows:

( ) ( )2 ( ) ( )2 ( )4 4

( )(1 ( )) ( )(1 ( ))4 4

( )(1 ( )) ( )(1 ( ))4 4

( )(1 ( )) ( ) ( )(1 ( )) (4 4

SSB c c c c

c c c c

c c c c

c

A AX S S

A AS sign S sign

A AS j jsign S j jsign

A AS j jsign S j jsign

ω ω ω ε ω ω ω ω ε ω ω

ω ω ω ω ω ω ω ω

ω ω ω ω ω ω ω ω

ω ω δ ω ω ω ω δ ω

= − − + + − −

= − + − + + − +

= − − ⋅ − + + + ⋅ +

= − ⋅ ∗ − + + ⋅ ∗ )cω+

• This formula then can be rewritten using convolution operations, 2 Hilbert transforms and by extending it with the zero phase φ of the carrier.


Slide 53 I. Willms

By use of it follows

Alternative generationof SSBAM

( ) ( ( ) ( ) ( )) ( )4

jSSB HILBERT c

AX S jS H e φω ω ω ω δ ω ω⎡= + ∗ −⎣

( ( ) ( ) ( )) ( ) jHILBERT cS jS H e φω ω ω δ ω ω − ⎤+ − ∗ + ⎦

( ) ( ),HILBERTH jsignω ω= −

( ) ( ( ) ( ( ) ( ) )4

j jSSB c c

AX S e eφ φω ω π δ ω ω δ ω ωπ

−⎡= ∗ − + +⎣

( ) ( ) ( ( ( ) ( ) )j jHILBERT c cS H j e eφ φω ω π δ ω ω δ ω ω − ⎤+ ∗ − − + ⎦

ˆThis gives: ( ) ( ( ) cos( ) ( )sin( ))2SSB c cAx t s t t s t tω φ ω φ= + − +

{ }ˆwith ( ) ( ) .s t s t=H


Slide 54 I. WillmsGeneration of SSBAM

( )SSBx t

ˆ( ) ( ( ) cos( ) ( ) sin ( )),2SSB c cAx t s t t s t tω φ ω φ= + − +

cos( )ctω φ+

Alternative circuit for generating an SSBAM-signal

sin( )ctω φ+

( )s t{ }H ...

{ }H ...


Slide 55 I. WillmsAnalytic signal for SSB

( )ELPx t{ } { }0 0 ˆ( ) ( ) ( ) ( )SSB SSB SSB SSBX x t x t jx tω = = +F F

We will try to figure out in case of SSBAM.

From this we obtain the equivalent lowpass spectrum:

( )(1 ( ))SSBX signω ω= +

2 ( ) ( ) with

( ) ( )2 ( ) ( )2 ( )4 4

SSB

j jSSB c c c c

XA AX S e S eφ φ

ω ε ω

ω ω ω ε ω ω ω ω ε ω ω −

=

= − − + + − −

0 ( ) ( ) ( ) .jSSB c cX AS e φω ω ω ε ω ω= − −

0 0( )0, ( ) ( ) ( ) ( )j j

ELP SSB SSB cX X e AS eφ φ φω ω ω ω ε ω− −= + =0( )( )(1 ( )) .

2jA S sign e φ φω ω −= +

0( ),

0

ˆHence we get: ( ) ( ( ) ( ))2

As is arbitrary we can set it to zero without problems.

jELP SSB

Ax t s t js t e φ φ

φ

−= +


Slide 56 I. WillmsAnalytic signal for SSB

, ,( ) ( ) cos( ( ) )SSB ELP SSB c ELP SSBx t x t t x tω φ= + +

2 2

0 0

ˆ( )ˆ( ) ( ) cos( arctan( ) )2 ( )

( ) cos( ( ) )2

c

c

A s ts t s t ts t

A s t t s t

ω φ

ω φ

= + + +

= + +

Amplitude and angle modulation of SSB signal (not always coinciding zeros)

0 2 12 0.6

2 2.40.750

sf

c

HzHz

Hza

ω πω π

ω π

φ

=

=

===

2( )SSBc sf

x t πω ω

−+

( )SSBx t


Slide 57 I. WillmsPower of SSB

/ 222

/ 2/ 22

2

/ 2

1(1 lim 2 ( ) ( ) )2

1lim ( )4

T

TTA M

TSSB

TT

A s t s t d tTP

P A s t d tT

→ ∞−

→ ∞−

+ +=

∫

∫

/ 2 / 222 2

/ 2 / 2

1 1lim ( ) lim ( ) .4

T T

SSB SSBT TT T

AP x t d t s t d tT T→∞ → ∞

− −

= =∫ ∫

/ 22

/ 2/ 2

2

/ 2

12(1 lim 2 ( ) ( ) )6 in case of sinusoidal ( ).

1lim ( )

T

TT

T

TT

s t s t dtT

s ts t dt

T

→∞−

→∞−

+ += =

∫

∫

For the power of a SSB signal holds (due to exercise 10):


Slide 58 I. Willms

Thus envelope demodulation will not work!

In case of SSB reception for sinusoidal signal ( ) holds:( ) cos( ) which shows a constant envelopeSSB c S

s tx t A t tω ω= +

Although this introduces also disadvantages concerning power efficiency now an additive carrier is applied according to:

( ) cos( ) cos( ) SSB c S cx t A t t A tω ω ω φ= + + +

Next question is how SSB signals can be demodulated


Slide 59 I. WillmsDemodulation of SSBAM

ˆ((2 ( )) cos( ) ( )sin( ))2SSBWC c cAx s t t s t tω φ ω φ= + + − +

It follows for the envelope:

( ) ( ) cos( ) gives with

ˆ ( ) ( ( ) cos( ) ( )sin( )) :2

WCSSB SSB c

SSB c c

x t x t A t

Ax t s t t s t t

ω φ

ω φ ω φ

= + +

= + − +

2 2 ˆ( )ˆ(2 ( )) ( ) cos( arctan( ) )2 2 ( )cA s ts t s t t

s tω φ= + + + +

+

2 2ˆ ˆ( ) ( ) ( )1 ( ) cos( arctan( ) ).4 4 2 ( )c

s t s t s tA s t ts t

ω φ= + + + + ++

2 2ˆ( ) ( )( ) 1 ( ) .4 4W CSSB

s t s tg t A s t= + + +

Hence if ( ) 1 holds it follows:s t << ( ) 1 ( ).WCSSBg t A s t≈ +


Slide 60 I. WillmsDemodulation of SSBAM

11 1 ,2xxx+ = +

( )( ) (1 ).2W CSSB

s tg t A≈ +

( ) ( ) cos(( ) ) with ( ) ( )SSB SSB c SSB SSBz t y t t t y t x tω ω φ∆ ′= + + =

In addition, by use of

we finally obtain

In order to avoid the additive carrier now synchronous demodulation with unknown carrier frequency (or little frequency error) and phase is considered:

ˆ( ) ( ( ) cos( ) ( )sin( )) cos(( ) )2SSB c c cAz t s t t s t t t tω φ ω φ ω ω φ∆ ′= + − + + +

( ( )(cos( ) cos((2 ) ))4 cA s t t tω φ φ ω ω φ φ∆ ∆′ ′= + − + + + +

ˆ( )( sin( ) sin((2 ) )).cs t t tω φ φ ω ω φ φ∆ ∆′ ′− − + − + + + +


Slide 61 I. WillmsUnknown phase & carrier

( ) ( ) ( )SSB SSB LPg t z t h t= ∗

ˆ( ( ) cos( ) ( )sin( )).4A s t t s t tω φ φ ω φ φ∆ ∆′ ′= + − + + −

This shown an additional phase shift and an additional modulation of both s(t) and its Hilbert transform compared with the situation without frequency error.

The interpretation using Fourier transforms (in frequeny domain) is as follows:

( ) ( )1( ) ( ( ) ( ) )4 2

j jSSB

AG S e S eφ φ φ φω ω ω ω ω′ ′− − −∆ ∆

⎡= − + +⎢⎣( ) ( )1 ˆ ˆ( ( ) ( ) ) .

2j jS e S e

jφ φ φ φω ω ω ω′ ′− − −

∆ ∆

⎤+ − − + ⎥

⎦

ˆWith ( ) ( )( ( )) it follows:S S jsignω ω ω= −



( ) ( )( ) ( ) ( )8

j jSSB

AG S e S eφ φ φ φω ω ω ω ω′ ′− − −∆ ∆⎡= − + +⎣

( ) ( )( ) ( ) ( ) ( ) )j jS sign e S sign eφ φ φ φω ω ω ω ω ω ω ω′ ′− − −∆ ∆ ∆ ∆ ⎤− − − + + + ⎦

( ) ( )[ ( ) (1 ( ) ( ) (1 ( ))]8

j jA S e sign S e signφ φ φ φω ω ω ω ω ω ω ω′ ′− − −∆ ∆ ∆ ∆= − − − + + + −

We define and as the lower and the upper edge frequencies of respectively. Let us sketch

( ) ( )( ) 2 ( ) ( ) 2 ( )8

j jA S e S eφ φ φ φω ω ε ω ω ω ω ε ω ω′ ′− − −∆ ∆ ∆ ∆⎡ ⎤= − − + + +⎣ ⎦

( ) ( )( ) ( ) ( ) ( ) .4

j jA S e S eφ φ φ φω ω ε ω ω ω ω ε ω ω′ ′− − −∆ ∆ ∆ ∆⎡ ⎤= − − + + +⎣ ⎦

slωsuω ( )s t

( ).SSBG ω



suωsuω− slω−slω

( )ε ω ω∆ − ( )S ω ω∆−

slω ω∆− +suω ω∆+

( )S ω

( )S ω ω∆+ ( )ε ω ω∆ + ( )( ) ( ) jS e φ φω ω ε ω ω ′−∆ ∆− −

( )( ) ( ) jS e φ φω ω ε ω ω ′− −∆ ∆+ + ( )

slSSBG ω ωω∆<

suω ω∆− suω ω∆−


Slide 64 I. Willms

If the frequency error can be neglected, it holds:

Unknown phase

( ) ( )0( ) ( ) ( ) ( ) .

4j j

SSBAG S e eφ φ φ φ

ωω ω ε ω ε ω∆

′ ′− − −= ⎡ ⎤= − +⎣ ⎦

• The human ear can be modelled as a bank of filters, whichare insensitive to phase pertubations

• SSBAM receives much attention for transmission ofspeech (due to little low frequency content). For transmission of data QAM would be better.

( ).φ φ′± −

The human ear is not thus sensitive for frequency shifts of less than 10Hz. But such a low frequency error can only be achieved with a PLL and a highly stable quartz reference clock


Slide 65 I. Willms

• Important application of VSBAM is analog transmission of videosignals with 5,5 MHz bandwidth

• Allows reduced bandwidth and modest demands on the transmitterVSB filter

• Band spread is well below DSBAM• Thus nearly a double number of TV channels can be transmitted

• Transmission of additional (low power) pilot carrier of nearly arbitraryfrequency allows synchronisation of receiver with transmitter

• Condition of the transmitter VSB filter (applied to the DAM signal) forperfect demodulation (as shown in the following slide) is:

( ) ( ) constant withinThis can be accomplished by a VSB BP filter with symmetry around

/ 2For the band spreading factor holds:

VSB C VSB C S S

C

S VSBVSBAM

S

H Hω ω ω ω ω ω ωω

ω ωβω

− + + = − ≤ ≤ +

+=

Vestigial side-band AM

VSBAM aspects and applications


Slide 66 I. Willms

• Using a AMsC signal and a suitable VSB band pass filter it results:

1

VSBAMsC AMsC

( ) ( ) cos( )

( ) ( ) ( ) with ( ) { ( )} givingX ( ) X ( ) ( )

( ( ) ( ) ) ( )2

AMsC C

VSB AMsC VSB VSB VSB

VSB

j jC C VSB

x t As t t

x t x t h t h t F HH

A S e S e Hφ φ

ω φ

ωω ω ω

ω ω ω ω ω

−

−

= +

= ∗ ==

= − + +

• A synchronous demodulation is applied here as described in the following relations:

( ) ( ) cos( )1( ) ( ( ) ( ))2

( ( 2 ) ( ) ) ( )4

( ( ) ( 2 ) ) ( ) giving finally4

= ( )[ ( ) ( )]4

VSBAMsC VSBAMsC C

VSBAMsC VSBAMsC C VSBAMsC C

j jC VSB C

j jC VSB C

j jVSB C VSB C

z t x t t

Z X X

A S e S e H

A S e S e H

A S e H e H

φ φ

φ φ

φ φ

ω φ

ω ω ω ω ω

ω ω ω ω ω

ω ω ω ω ω

ω ω ω ω ω

−

−

−

= +

= − + +

= − + −

+ + + +

− + +


Slide 67 I. WillmsThe VSBAMsC method

( )rS ω

sω− sω

ω

cω−, ( )AM rX ω

ω

( )VSBH ω

cω− cω

ω

s cω ω− − s cω ω+cωcω−

ω

2A

cω− cωs cω ω− −s cω ω+

s cω ω+s cω ω− −

VSBω

, ( )VSB rX ω

, ( )VSB rG ω

ω

cωcω− sω− sω

cω


Slide 68 I. WillmsAn overview on AM methods

Method TX Power TX costs RX costs

Purpose

DSBAMwCenv.dem.DSBAMwCsync.dem.

QAM

DSBAMsCSSBAMsCSSBAMwCVSBAMsCVSBAMwC

2 high low low

2 low moderate high1 very low moderate/high high

>1 very low moderate/high high1 high moderate low

2 moderate low high Data-, stereo-sig.

1 Low/moderate high high speech, audio-sig.

>1 high moderate low TV-signals

β


Slide 69 I. Willms

Chapter 1.2 Angle modulation

• Basics of FM and PM• Spectrum analysis of narrow/wide band FM• Demodulation• Comparison of FM to AM


Slide 70 I. Willms

1.2 Angle Modulation• For FM & PM only variations of carrier phase / frequency are needed:

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

FM FM

PM C PM

x t A f s t tx t A t f s t

φω

= += +

Angle Modulation

( ) cos( ( ) )

( ) cos( ( ) )

t

FM C

PM C

x t A t s d

x t A t s t

ω τ τ φ

ω φ−∞

= + +

= + +

∫

• Typical signals:


Slide 71 I. Willms

0 0 0

1 0

0

1 0 0 0 1

Now a comparison is made for the signal ( ) cos( (t)t+ )and the signal ( ) cos( ( )) at a specific time instant .For this specific time instant it is required:

d( ) ( ) and dt

x t Ax t A t t

t

x t x t x

ω ϕϕ

=

=

=0 00

d( ) ( ) dtt t t tt x t= ==

0

0

0 0 0 0 0 0 0 0

0 0 0 0

0 0 0

This leads to cos( ) cos( ( )) or sin( ) sin( ( ))

dsin( (t)t+ ) ( ) sin( ( ))dt

dsin( ( )) ( ) sin( ( ))dt

and finally gives the definition of the instantaneous f

t t

t t

t t t t

t t

t t t

ω ϕ ϕ ω ϕ ϕ

ω ω ϕ ϕ ϕ

ω ϕ ϕ ϕ

=

=

+ = + =

⋅ = ⋅

⇒ ⋅ = ⋅

requency ( )( )id tt

dtϕω =


Slide 72 I. WillmsAngle Modulation

( ) 0.d tdtϕ

>

Usually, is assumed to be a strictly monotonically increasingfunction of time. Therefore,

( )tϕ

( )tϕ

Now the spectrum of a simple FM signal is analysed. Analysis cannot be made for arbitrary s(t) due to nonlinear properties.


Slide 73 I. Willms

Angle Modulation

• Example of an FM-signal with a sinusoidal source signal s(t) :

( sin( ) ) ( )

( ) cos( cos( ) ) with ( ) cos( ) ( )

cos( sin( ) ) and (t)= ( )

Re{ } Re{ ( ) }I

C I S C

t

FM C FM S FM S

tFM

C S IS

m

j t m t j t

x t A t d s t t t

A t t s s d

A e A z t eω ω φ ω φ

ω ω ω τ τ φ ω ω ε

ωω ω φ τ τω

−∞

−∞

+ + +

= + ∆ + = ∆

∆= + +

= =

∫

∫

sin( )with ( ) I S Sjm t jn tn

nz t e z eω ω

∞

=−∞

= = ∑

0

0

!

sin( )0

and the complex Fourier coefficients (as its a periodic signal)

1 ( ) with

1 1( / ) with set to - / 22 2

1 [cos( sin( ) ) sin( si2

S

S

I

t Tjn t

n SS t

jmjn jnS S

I I

z z t e dt tT

z e d e e d t T

m n j m

ω

π παα α

π π

ω α

α ω α απ π

α απ

+−

− −

− −

= =

= =

= − +

∫

∫ ∫

n( ) )] ( )

due to the zero value of sin( sin( ) )

n I

I

n d J m

m n d

π

π

π

π

α α α

α α α

−

−

− =

−

∫

∫


Slide 74 I. WillmsBessel functions

2 4 6 8 10

1

We obtain the first kind n-th order of the Bessel-function for whichonly a Taylor-series can be derived:

( )2

0

( 1)( ) ( ) , 0.! ! 2

kn kI

n Ik

mJ m nk n k

∞+

=

−= ≥

+∑0 ( )IJ m

1( )IJ m2 ( )IJ m

3 ( )IJ m4 ( )IJ m

5 ( )IJ m

Im


Slide 75 I. WillmsFM signal

( )

( )

( )

( ) Re{ ( ) },

Re ,

Re{ ( ) },

( ) cos(( ) ).

C

s C

s C

j tFM

jn t j tn

n

jn t j tn I

n

n I c sn

x t A z t e

A z e e

A J m e e

A J m n t

ω φ

ω ω φ

ω ω φ

ω ω φ

+

∞+

=−∞

∞+

=−∞

∞

=−∞

=

⎧ ⎫= ⎨ ⎬⎩ ⎭

=

= + +

∑

∑

∑

0( ) ( ) cos( ) with ( ) ( 1) ( )nFM I C n I n Ix t AJ m J m J mω φ −= + + = −

1

( ) cos(( ) ) ( 1) cos(( ) ) .nn I C S C S

n

A J m n t n tω ω φ ω ω φ∞

=

⎡ ⎤+ + + − − +⎣ ⎦∑

A new writing of xFM(t) now gives:

0

01

( ) ( )

( ) ( )( ( ) ( ) )

( ) ( ( )) ( ( ))

( ( )) ( ( )) .

j jFM I C C

j jI C S C S

n

j n j nC S C S

X A J m e e

A J m n e n e

n e n e

φ φ

φ φ

φ π φ π

ω π δ ω ω δ ω ω

π δ ω ω ω δ ω ω ω

δ ω ω ω δ ω ω ω

−

∞−

=

+ − −

= − + + +

⎡ − + + + +⎣

⎤+ − − + + − ⎦

∑


Slide 76 I. Willms

Here we see: • x(t) only contains cosine expressions• Amplitude is given by the Bessel function values• Frequency values are given by:• The phase is not changed

C Snω ω+

This corresponds to a line spectrum with lines at the carrier frequencyand at n times the source frequency.

Thus the FM spectrum is in principle extended to an infinetely wide band around the carrier frequency.

This wide-spread frequency band gives the desired „threshold effect“ in case of modest noise power as receiver only cares about frequency changes and not for changes in the amplitude.

The FM spectrum is in detail determined in the next slides.

Due to properties of the Bessel function essentially only the lines corresponding to the following equation need to be taken into account.

( ) ( ) cos(( ) )I

I

N m

FM n I c sn N m

x t A J m n tω ω φ≈

=− ≈

= + +∑


Slide 77 I. WillmsFM spectrum

,

The case 1 refers to the well known Carson bandwidth: 2( ).FM Carson FM sB

αω ω

== ∆ +

A more precise consideration requires a summation within the limit

2 2( ) with 1 2.S

I

Sinus I s

N m

B N mω

α

α ω α

± = ± +

≈ ≈ + ≤ ≤

It is now assumed that s(t) consists out of not only one but of several cosinesas follows:

, ,

,

,1

, , ,1 ,

( sin( ) )

1

( ),

( ) cos( ),

( ) sin( ) ,

( ) ( ) , ( )

( ) ( ) .

I l s l l

l s l l

l

L

l s l ll

Ll

I I l s l l I ll s l

Lj m t

l ll

jn tl nl I l

n

s t t

s t m t m

z t z t z t e

z t J m e

ω φ

ω φ

ω ω φ

ωω φω

=

=

+

=

∞+

=−∞

= ∆ +

∆= + =

= =

=

∑

∑

∏

∑


Slide 78 I. WillmsFM spectrum

max ( ) 1Is t and 1ImFor narrowband FM holds with:

( ) 1( ) cos( ( ) )

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

IFM c Is t

c I c I

c c I

x t A t s t

A t s t A t s tA t A t s t

ω φ

ω φ ω φω φ ω φ

= + +

= + − +

≈ + − +

This corresponds to an AM signal with carrier!


Slide 79 I. Willms

( ) 1( ) ( ( ) ( )

1 ( ( ) ( ) ) ( )2

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

( ( ) ( ) )

( ) ( )(2

I

j jFM C Cs t

j jC C I

j jC C

j jC C

j jC C

j jC C

C

X A e e

A e e Sj

A e e

A e e Sj j

A e e

S e S eA

φ φ

φ φ

φ φ

φ φ

φ φ

φ φ

ω π δ ω ω δ ω ω

π δ ω ω δ ω ω ωπ

π δ ω ω δ ω ω

δ ω ω δ ω ω ωω

π δ ω ω δ ω ω

ω ω ω ωω ω ω

−

− +

−

− +

−

− −

= − + +

− + − − ∗

= − + +

− + − − ∗

= − + +

+ ++ −

+)

Cω+

This shows that a narrowband FM signal essentially has the same band spreading factor as a DSBAMwC signal.

For the spectrum holds:


Slide 80 I. WillmsFM demodulation

( ) cos( ( ) ).FM c Ix t A t s tω φ= + +

• Demodulation of FM( )( ) ,Ids ts t

dt=

( ) ( )FM FMy t x t= with ( ) sin( )I I ss t m tω=

( ( ))( ) sin( ( ) )c IFMc I

d t s tdy t A t s tdt dt

ω ω φ+= + +

( ( )) sin( ( ) ).c c IA s t t s tω ω φ= + + +

Derivation in required frequency range can be accomplished by

system with e.g. ( ) ( )C

FM

H j rectB

ω ωω ω

−= ⋅

Onset: Derivation of received signal and subsequent envelope demodulation


Slide 81 I. Willms

AM vs FM• FM receiver is very robust against additive noise (e.g. rapid

amplitude fluctuations)• SNR of about 10dB is the treshhold• Such noise is often found in practise• Carson bandwidth is easily scalable• Doubling bandwidth gives doubled SNR• Constant envelope ensures power efficient transmitter


Slide 82 I. WillmsAM versus FM

• In broadcast radios FM is the most popular analog modulationtechnique. Do you agree?

• Envelope detection is possible in AM as well as FM. Comment.

• Envelope detection shows better performance in FM than AM.Do you agree, why?

• AM bandwidth is not adjustable as FM bandwidth. Give tworeasons.


Slide 83 I. WillmsChapter 2Discrete Signal Transmission

Chapter 2

Fundamentals of Discrete Signal Transmission

2.1 A short review on information theory2.2 A short review on detection theory


Slide 84 I. Willms

We assume the alphabet X is formed by L different symbols xl

Information theory

2.1 Information TheoryConsider a discrete source, which emits messages x(k)

We assume a stationary source (a shift of messages doesn‘tchange probabilities) or the source properties never change!

One can regard as a random process

in form of where each random variable takes

values of the alphabet with the probability

( ), (1)x k k = −∞ ∞

( )X k ( )X k

lx χ .lp


Slide 85 I. Willms

Please note:• The alphabet is represented by a caligraphic letter X• It contains the L symbols xl

• A sample (a real message sequence) by x(k)• The whole process (set of message sequences) is

represented by X(k)


Slide 86 I. WillmsInformation theory

( ) ( )l mI x I x> if l mp p<

If the events of the two symbols xl and xm are independent with

for all k and each l and m we require

How to define „information“ ? (Measure of uncertainty)

{ } { } { }1 2 1 2( ) , ( ) ( ) ( )l m l mP X k x X k x P X k x P X k x= = = = =

,

( , )l mx x independent

l mI x x = ( ) ( )l mI x I x= +

Therefore,we define „information“ as

21 1( ) log ( ) ( ).

def

ll l

I x ldp p

= =


Slide 87 I. WillmsEntropy

1 1

1( ) ( ) ( ).L L

l l ll ll

H p ld p ld pp

χ= =

= = −∑ ∑

( )H χ1

( )L

l ll

I x p=∑

What is „entropy“ ?

Defined as average information content

Hence,

E.g. for 2 symbols with and it follows1p p= 2 1p p= −

( ) ( ) (1 ) (1 ) ( ).def

H pld p p ld p H pχ = − − − − =


Slide 88 I. WillmsEntropy

Bounds on „entropy“ [bits/symbol] of an alphabet

The equal sign is valid only if all symbol show the same probability 1/L

( )Example:2 symbols have equal probabalities giving:

2 ( ) 2 1 /

H ldL

L H ld bit symbol

χ

χ

≤

= ⇒ = =


Slide 89 I. WillmsBerger‘s channel diagram

+

A memoryless communication channel with additive noise:Often additive white Gaussian noise is assumed (AWGN channel)


Slide 90 I. Willms

• In the following: Analysis of a communication channel fordifferent power levels of noise

• For noise with zero mean the variance describes thepower level

• In general Gaussian distributed noise is assumed• This is not always true (impulsive noise might be

present)

Note: The entropy can be specified for an alphabet of a source or a sink (thus indirectly for a source and sink)

Berger‘s channel diagram


Slide 91 I. WillmsBerger‘s diagram

( , )T yχ( )H y

( | )H yχ

SourceSink

is denoted as equivocation

is called irrelevance(„undesired“ info, e.g. noise)

is called transinformation

(„Lost“ info, specifies averageinfo needed to specify theinput symbol in case of knownoutput symbol)

( )H χ

( | )H y χ


Slide 92 I. Willms

Berger‘s diagram

( , ) ( ) ( | ),T y H H yχ χ χ= − ( , ) ( ) ( | ),T y H y H yχ χ= −

For noiseless channel:

For useless channel:

Equivocation is zero, no infois lost!

Equivocation is the source entropyand all info is lost (no communicationat all)!

Thus we obtain:

The diagram shows clearly the following:

( , ) ( , ).T y T yχ χ=


Slide 93 I. Willms

Berger‘s diagram

, ,( , ) ( , ) ( ( , )) .def

X Y X YH y p x y ld p x y dxdyχ∞ ∞

−∞ −∞

= ∫ ∫

The following can be shown:( , ) ( ) ( | ) ( ) ( | )H X Y H Y H X Y H X H Y X= + = +

(Joint entropy)


Slide 94 I. WillmsChannel capacity

( )(max ( )) ( | ).

Xp xC H y H y χ= −

, ( , ) ( ) ( ),X N X Np x n p x p n=

( ) ( ) ( ) ,Y X Np y p x p y x dx∞

−∞

= −∫

The channel capacity is a function of pX(x) and is dependant on channel properties. It is defined as

which leads to

(due to independant noise)

(as sum of x and n gives y)

C gives upper bound of transferabel info!

Thus we obtain:

Example: A channel with additive noise and y = x + n:

( )max ( , ),

Xp xC T yχ=

,( ) ( , ) ,Y X Yp y p x y dx∞

−∞

= ∫ , ( , ) ( ) ( ).X Y X Np x y p x p y x= −


Slide 95 I. Willms

Channel capacity

, ,( , ) ( , ) ( ( , ))

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

( ) ( ) (without proof)

X Y X Y

X X X

H y p x y ld p x y dxdy

p x ld p x dx p x H N dx

H X H N

χ∞ ∞

−∞ −∞

∞ ∞

−∞ −∞

= −

= − − −

= +

∫ ∫

∫ ∫

Now the joint entropy is rewritten based on the last relation:

Comparing the last result with( , ) ( ) ( | ) ( ) ( | )

shows: ( | ) ( ) as already indicated in Berger's diagramH X Y H Y H X Y H X H Y X

H Y X H N= + = +

=

(Conclusion: Noise is the „undesired“ info)


Slide 96 I. WillmsChannel capacity

2

2( )

21( ) ,2

N

N

n

NN

p n eµσ

π σ

− −

=

2

2

1 (1 ).2

XGAUSS

N

C ld σσ

= +

Another example: Additive noise is GAUSSIAN distributed with

If the signal is bandlimited with cut-off frequency

If

it follows (see the corresponding exercise)

[bits/symbol]

[bits/s]

Each symbol needs 1/2fco time due to 1. Nyquist criterion

(for 37 db S/N ratio and 3100 Hz bandwidth of a memoryless telephone line)

2

2(1 )XS GAUSS co

N

C f ld σσ

= +

23.7

2 10 ,X

N

σσ

= 3.73100 (1 10 ) 38130S GAUSSbit bitC lds s

= + =


Slide 97 I. WillmsDetection theory

The received signal is assumed to be

Given are M deterministic source signals xm(t) with a

finite duration T and a finite energy Ex .These „waveforms“ and

their duration must be known to the receiver.

( ) ( ) ( ).my t x t n t= +

where the noise is n(t) which is modeled as a random process.

Detection problem:Given is y(t), the sequence of source signals xm(t) has to be determined.

2.2 Detection theory


Slide 98 I. Willms

Let us consider M =2 and in case H0 of a logical zero x0(t) = 0 and in case H1 of a one x1(t) = x(t) described by:

Sufficient statistic

0 : ( ) ( )H y t n t=1 : ( ) ( ) ( ).H y t x t n t= +

The detection is based on minimization of probabililty of a

wrong decision Pe . To consider that y(t) is decomposed as

0( ) ( ),n n

ny t Y f t

∞

=

= ∑

0

where ( ) is the set of orthonormal functions defined

in 0 t T with ( ) ( ) / .n

x

f t

f t x t E≤ ≤ =

0 0 0 00

0

( ) and the set ( ) should be selected such that it holds:

( ) ( ) ( ) ( ) due to

1Example: ( ) ( 0.5) with and ( ) ( 0.5)

n

n n x xn

x

x t f t

x t X f t X f t E f t X E

t tx t rect E T f t rectT T T

∞

=

= = = =

= − = = −

∑


Slide 99 I. WillmsSufficient statistic

0( ) ( ),n n

nn t N f t

∞

=

=∑0

( ) ( ) .T

n nt

N n t f t dt=

= ∫Moreover, it holds:

It can be shown that Nl and Nm are uncorrelated and (as Gaussian random variables) also statistically independent.

The set {Nn } fully describes the noise.

Moreover it can be show that this enables the reformulationof the detection problem to:

0 : n nH Y N= 1 : n n nH Y X N= +

0

with ( ) ( )T

n nt

Y y t f t dt=

= ∫


Slide 100 I. WillmsSufficient statistic

0 00 0

1( ) ( ) ( ) ( ) .T T

t tx

Y y t f t dt y t x t dtE= =

= =∫ ∫

0 0 0:H Y N=

Therefore, the whole decision should be solely based on

Due to earlier assumptions (Gaussian noise), the detectionproblem further simplifies to:

Thus Y0 gives (or is called) a sufficient statistic as it is theonly data useful for decision-making.

It includes the whole relevant information.

1 0 0 0 0: .xH Y X N E N= + = +


Slide 101 I. Willms

Error probability

0 0 0assume : if YH R∈

By dividing the range of real numbers R into two seperated subsets R1 and R0 we can state generally with:

{ } { }0 1 0 00 1

1 1 0 0( ) ( )e Y H Y HR RP P H p y H dy P H p y H dy= +∫ ∫

{ } { }0 1 0 01 1

1 1 0 0( ) ( )Y H Y HR R RP H p y H dy P H p y H dy

−= +∫ ∫

{ } { } { }0 1 0 01

1 1 1 0 0( ) ( ) .Y H Y HRP H P H p y H P H p y H dy⎡ ⎤= − −⎣ ⎦∫

0 1 0 1

0 1

1 1due to 1 ( ) ( )Y H Y HR R

p y H dy p y H dy= +∫ ∫

0 1 0 0 1 0 Example: : 0 and : 0 R R R R Y R Y∪ = < >

1 0 1assume : if YH R∈


Slide 102 I. WillmsLikelihood ratio

2

2

0 0

20

1( )2

N

y

Y HN

p y H e σ

πσ

−

=

{ }{ }

01 0

1

Assume case : ( ) .P H

H if YP H

Λ >

{ }{ }

00 0

1

Assume case : ( )P H

H if YP H

Λ ≤

For GAUSSIAN noise holds:

Problem:

Both pdfs and the 2 probabilities need to beknown

{ } { }0 1 0 01 1 0 0Hence ( ) ( ) is chosen

as it maximizes the integral in the last equationY H Y HP H p y H P H p y H>

0 1

0 0

1

0

( )By introducing the likelihood ratio ( ) it follows:

( )Y H

Y H

p y Hy

p y HΛ =

2

2

0 1

( )2

01( )

2

x

N

y E

Y HN

p y H e σ

πσ

−−

=



22 2

0 0

1 1( ) ( ) ( )2 .

T T

N Ny t x t dt x t dt

eσ σ−∫ ∫

=

20

2 02

202

( )22

20

2

( )

xx x

NN

N

y Ey E E

y

eY e

e

σσ

σ

−− −

−Λ = = 0 0

1with Y ( ) ( )T

x

y t x t dtE

= ∫

Consequently, we can rewrite as

Since the likelihood is often met in exponentional form, it is customary to define the loglikelihood ratio as

0( )YΛ

20 0 2 2

0 0

1 1( ) ln ( ) ( ) ( ) ( ) .2

T T

N N

Y Y y t x t dt x t dtλσ σ

= Λ = −∫ ∫

20

1 1( ( ) ( ) )2

T

xN

y t x t dt Eσ

= −∫



{ }{ }

00 0

1

: ( ) lnP H

H if YP H

λ ≤

{ }{ }

01 0

1

: ( ) ln .P H

H if YP H

λ >

{ } { }0 1P H P H=

Then, the decision rules becomes

accept

accept

(ML-rule)

In case the decision rules could be written as

accept

accept

0 0: ( ) 0H if Yλ ≤

1 0: ( ) 0.H if Yλ >


Slide 105 I. WillmsDetection of M Signals

{ } { } { }0 1 1... .MP H P H P H −= = =

0, 0,0 0

1( ) ( ) ( ) ( ) .m

T T

m m mx

Y y t f t dt y t x t dtE

= =∫ ∫

For simplicity we assume

At the receiver, we have first to calculate

Now, we consider M signals xm(t)

0,0

1 ( ) ( ) .m

T

m mx

Y y t x t dtE

= ∫


Slide 106 I. WillmsDetection of M Signals

( )y t

1( )x t

2 ( )x t

Optimum receiver for memoryless modulationin AWGN channel

Choosethe

Largest

( )Mx t

1

12 xE−

2

12 xE−

12 MxE−

( )truemx t

0

...T

dt∫

0

. . .T

d t∫

0

. . .T

d t∫


Slide 107 I. WillmsMatched filter concept

00

1 ( ) ( ) .T

x

Y y t x t dtE

= ∫

00

1 ( ) ( )T

x

Y y x dE

τ τ τ= ∫( ) ( )

0

1 ( ) ( )Tx h T

x

y h T dE

τ τ

τ τ τ= −

= −∫

Let us re-consider the fundamental detector equation.

This processing can be realised using the convolution integral:

Thus a „matched filter“ with impulse responsecan be applied.

00

1( ) ( ) ( )t

t T t Tx

Y g t y h T dE

τ τ τ= == = −∫

( ) ( ).h t x T t= −


Slide 108 I. Willms

Matched filter concept

( )y t ( )truemx t

Optimum receiver for memoryless modulationin AWGN channel (Matched Filter Implementation)

Choosethe

Largest2 ( )x T t−

( )Mx T t−

1

12 xE−

2

12 xE−

12 MxE−

1( )x T t−


Slide 109 I. Willms

Some reasons for analog-to-digital conversion:

Introduction

• Precision and Dynamic

• Component Tolerancy

• Temperature behaviour, Aging

• Integration

• Frequency regions

• Multiplexing

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