ANALOG MODULATION

PART I: AMPLITUDE MODULATION

WHAT IS ANALOG MODULATION?

Analog modulation consists of two parts Analog message Carrier

Analog message is then impressed upon the amplitude, frequency or phase of the carrier

We then say carrier is “modulated” by the message

Amplitude Modulation

Define the message by a baseband signal m(t).

Define the “carrier” by a sinusoidal signal c(t) with amplitude Ac and frequency fc

The AM signal is then given by

s t( )=Ac 1+kamt( )[ ]cos2πfct( )

AM Envelope

The “envelope” of an AM signal is the factor in front of the carrier,

Envelope is a baseband signal and contains all the information an AM signal carries

1+kamt( )[ ]

AM Example

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-1

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1

2

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AM BASEBAND

carrier

message

Requirements on the Envelope

To prevent “overmodulation”, the envelope must always be positive,

1+kam(t)≥0 ∀t

or

kam(t) ≤1

1+kam(t)≥0 1+kam(t)<0overmodulation

Finding AM Spectrum

Fourier transform modulation property says

Taking Fourier transform

s t( )=Ac 1+kamt( )[ ]cos2πfct( )=Ac cos2πfct( )+Ackamt( )cos2πfct( )

S f( )=Ac δ f −fc( )+δ f + fc( )[ ]+Acka

2M f − fc( )+M f +fc( )[ ]

Showing AM Spectrum

Baseband

AM

M(f)

upper sidebandlower sideband

fc

W

fc+Wfc-W-fc

AM bandwidth=2W

Tone Modulation

Tone modulation is when the message is a simple sinusoid

The corresponding AM signal ismt( )=Amcos2πfmt( )

s t( )=Ac 1+μcos2πfmt( )[ ]cos2πfct( )

with

μ =kaAm

Modulation Index

The quantity is called modulation index or modulation factor.

Modulation index must be less or equal to 1

Finding Modulation Index

Amax

Amin

Amax=Ac 1+μ[ ]

Amin=Ac 1−μ[ ]Ac 1+μcos2πfmt( )[ ]cos2πfct( )

μ =modulation index=Amax−Amin

Amax+Amin

What is Here?

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-3

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AM BASEBAND

AM Power

The power of AM signal is split between carrier and message

Ac 1+μcos2πfmt( )[ ]cos2πfct( ) =

Ac cos2πfct( )+Acμcos2πfmt( )cos2πfct( )

=Ac cos2πfct( )carrier

1 2 4 3 4 +

Acμ2

cos fc + fm( )t+Acμ2

cos fc −fm( )t

message1 2 4 4 4 4 4 4 3 4 4 4 4 4 4

Components of AM Power

Carrier power=Upper sideband power=Lower sideband=

Carrier power=Total message power=

12

Ac2

18

μ2Ac2

18

μ2Ac2

12

Ac2

14

μ2Ac2

Total AM Power

Total AM power is the sum of carrier and message power

PT =Pcarrier+Psideband=12

1+μ2

2⎡

⎣ ⎢ ⎤

⎦ ⎥ Ac

2

Distribution of AM Power

How much of the total power goes into carrier and how much into message?

Even under full modulation, only a third of the transmitted power goes into the message component

Psideband

Ptotal

=μ2

2+μ2 =0,μ =0

13,μ =1

⎧ ⎨ ⎩ ⎪

Example of Power Allocation

FCC power rating is based on average carrier power. If a transmitted is rated at 5KW and is connected to a 50 ohm load, find the total power and how it is split between carrier and message

Answers...

Carrier peak amplitude is given by

Peak sideband(message) power is

Pc =12

Ac2⎛

⎝ ⎞ ⎠ 50=5000⇒ Ac =707 volts

Ps =14

μ2Ac2 ⇒

14

⎛ ⎝

⎞ ⎠ 1( ) 707( )2 =2500W

AM Power Efficiency

Here is AM power distribution Carrier power=5000W Sideband power=2500W Total power=7500 Efficiency=(sideband power)/total power=

(2500)/(7500)=1/3.

Out of 3 watts of transmitted power, only 1 watt is going into the message

Variations of AM

Amplitude modulation has a number of flavors. They are Double sideband Double sideband-suppressed carrier (DSB-

SC) Single sideband (SSB) QAM

Double Sideband

Baseband

AM

M(f)

upper sidebandlower sideband

fc

W

fc+Wfc-W-fc

AM bandwidth=2Wcarrier

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-3

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-1

0

1

2

3

AM BASEBAND

Double Sideband-SC

To conserve power, we drop the carrier term from the AM expression

DSB-SC is then given by

s t( )=Ac 1+kamt( )[ ]cos2πfct( )=

Ac cos2πfct( )drop

1 2 4 3 4 +Ackamt( )cos2πfct( )

s t( )=carrier×message=Acmt( )cos2πfct( )

DSB-SC Spectrum

Spectrum is identical to before except for the removal of the carrier(two impulses)

upper sidebandlower sideband

fc fc+Wfc-W-fc

Single Sideband(SSB)

The two sidebands are identical. Only one is sufficient to carry all the message information. Block either the upper or lower sideband

Bandwidth=2W Bandwidth=W

Expression for SSB-AM

The SSB signal when the upper signal is kept is given by

Practice: show that S(f) is indeed an SSB signal(contains only the upper sidebands)

s t( )=

Ac

2mt( )cos2πfct−

Ac

2ˆ m t( )

HTofm(t){ sin2πfct

AM Signal Generation

Switching modulator

m(t)

~

R

c(t)=Accos2pifct

V2(t)=AMV1(t)

v1 t( )=m(t)+Ac cos2πfct

v2 t( )=v1 t( ),c t( )>0

0,c t( )<0⎧ ⎨ ⎩

Modeling Diode Effect

Diode acts as a switch. Its operation can be modeled by a gating operation

Fourier analysis shows v2 is AM

v2 t( )= Ac cos2πfct+m(t)[ ]×gTot( )

gTot( )

To/4-To/4To

SSB Generation

One way to generate an SSB signal is through selective filtering

DSB-AMf

Ideal highpass filter

Leaks some of the lower sideband

Issues in Sideband Suppresion

Consider a tone modulated signal with carrier frequency of 10MHz, and message frequency of 30Hz. Want to make a SSB signal out of it

60Hz

10MHz

carrier

10,000,030 Hz

Required Q’s

We are trying to separate a frequency of 10.000030 MHz from 9.999970 MHz at 10 MHz.

The required Q is Q=10MHz/60Hz=166,667. Too high

Solution: Translate the signal down to 100KHz. There, Q=100KHz/60Hz=1666. Acheivable

SSB transmitter

Take a 60Hz tone message.

Audio ampBalancedmodulator

Highpassfilter

Balancedmodulator

Power amp

Conversionoscillator

RF carrier

100 KHz 10 MHz

60Hz

99,940 Hz 100,060 Hz9,899,940 Hz

10,100,060 Hz

100,060 Hz

200.120K

Passes the higherterm

Power Distribution in SSB

If an SSB transmitter with total power of 10KW were to be replaced by a standard AM signal with the same total power, compare their respective carrier and sideband powers

PT =Pc +μ2Pc

4+

μ2Pc

4sidebands

1 2 4 3 4 =10,000

letμ =1

10,000=Pc +Pc

2⇒ Pc =666.67W

Sideband Powers

Subtracting carrier power from total power leaves us with the sideband power

The power in one sideband is half at 1,666.67W

Psb =PT −Pc =10,000−6,666.67=3,333.33W

Comparisons

A 10 KW AM transmitter ends up with 1,666 watts in each sideband

A 10 KW SSB transmitterends up with 10 KW per sideband. There is nowhere else for the power to go

AM Demodulation

There are two was to demodulate an AM signal coherent(synchronous) noncoherent(asynchronous)

Coherent detection is defined when the phase of the incoming signal is assumed known

Coherent Detection

∫s(t)

Ac' cos2πfct+φ( )

?

phase angle assumed known

local oscillator

Detector Output...

Let’s start with a DSB-SC signal s(t). Noting that integration is the same as low pass filtering

Remember this

v t( )=Ac'cos2πfct+φ( )s(t) =AcAc'mt( )cos2πfct( )cos2πfct+φ( )

2cosxcosy=cos(x+y)+cos(x−y)

…Detector Output

At the integrator input we have

Therefore, detector output is a scaled version of the message

v t( )=12

AcAc'm(t)cos4πfct+φ( )

blocked by lowpass filter1 2 4 4 4 4 3 4 4 4 4

+12

AcAc'm(t)cosφ( )

survives1 2 4 4 4 3 4 4 4

Effect of Phase Error

If phase error is known , or can be tracked, its effect can be reversed.

If not, the message m(t) is scaled by a factor, possibly time varying, causing sever distortion

v t( )=12

AcAc'm(t)cosφ( )

survives1 2 4 4 4 3 4 4 4

=

12

AcAc'm(t),φ =0

0,φ =90o

⎧ ⎨ ⎪

⎩ ⎪

Achieving Phase Lock:Phase Locked Loops

PLLs are precisely the kind of devices that are need in coherent detection

Their mission is to track the phase of the incoming signal and adjust the phase of the local oscillator accordingly

Costas Loop

X

X

-90 phaseshift

LPF

VCO LPF

LPF

Xs t( )=Acm(t)cos2πfct

Aocos2πfct+θe( )

v2 t( )=12

AoAcsinθe⎛ ⎝

⎞ ⎠ m(t)

v1 t( )=12

AoAccosθe⎛ ⎝

⎞ ⎠ m(t)

v3 t( )~m2(t)sin2θev4 t( )~sin2θe

Envelope(noncoherent) Detection

Envelope detection is the simplest form of AM demodulation but it requires a full DSB-AM signal. No carrier phase information is needed

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AM BASEBAND

Idea Behind E.D.

The message is in the envelope of AM signal. Envelope is a very slow changing signal relative to the carrier.

Envelope detector must not respond to fast carrier amplitude swings

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AM BASEBAND

E.D. Circuit Diagram

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-3

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-1

0

1

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AM BASEBAND

~

RsRlC

AM

1.Capacitor is charged up2.Carrier goes negative, diode opens, C discharges slowly3.Carrier goes positive, diode closes, C charges back up

How to Select RC?

Remember voltage across C of an RC circuit follows

Must avoid fast discharges to allow for the following of the envelope, ->large RC.

Must avoid slow discharges not to be left behind in tracking the envelope, ->small RC.

v t( )=Vo exp−t RC( )

Some Bounds on RC

To keep the output from following the carrier too closely

To make the output track the envelope

RC>>1fc

RC<<1fm

E.D. Output

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AM BASEBAND

ED output roughly follows the envelope through charge/dischargecycles~

RsRlC

AM

Quadrature Amplitude Modulation(QAM)

The AM signal corresponding to a message with bandwidth W takes up a bandwidth of 2W.

Therefore, it appears that in a bandwidth of 2W, only one AM signal can be transmitted.

There is, however, a way to double channel capacity

Transmit two AM signals in 2W bandwidth

First Try

m1 t( )

m2 t( )

cosωct

cosωct

+

m1 t( )cosωct+

m2 t( )cosωct

Quadrature Multiplexing

Clearly, the two signals cannot be separated by their frequency content alone

Instead, use two carriers that are in phase “quadrature” with respect to each other(QAM), i.e..

s t( ) =Acm1 t( )cosωct+ Acm1 t( )sinωct

Demodulating a QAM Signal

Ac′cos2πfct

Ac′sin2πfct

-90o

LPF

LPF

12AcAc

′m1 t( )

12AcAc

′m2 t( )

s t( )=Acm1 t( )cosωct+

Acm1 t( ) sinωct

X

X

Orthogonal signaling

QAM is a special case of a much broader class of signals that are “orthogonal” as defined by

For example, following signals are orthogonal

si∫ t( )sj t( )dt=0

cos2πfct,cos4πfct,cos6πfct,...

Lesson Learned

We can safely add up orthogonal signals and later recover each one individually by a quadrature receiver.

s1

s2

s3received signal has alarger projection on s2 than s1 or s3

-----> s2 was trans’d

Superheterodyne Receiver:The Problem

It is difficult to design highly selective RF stages at high carrier frequencies. Adjacent stations may leak

RF stage

f

Brf

fc Desired stationAdjacent stations

BT

IF Response

The IF stage is permanently centered at 445 KHz(for AM) and 10.7 MHz(for FM).

The IF bandwidth is just wide enough to let the modulated signal through (8~10 KHz for AM)

fIF fc

BIF ≅BT

blocked

Superhet Receiver Diagram

RF X

~

Demod

LO

fLO = fc ±fIF

BT <BRF <2fIF

IF amp

Baseband filter

fc +fLO =2fc ±fIFfc −fLO = fIF

⎧ ⎨ ⎩

Stops this

BIF ≅BT

Difference freq always remains thesame

Problem of “Image Frequency”

We are tuning in a station at fc. Let’s say there is another station at fc +2fIF

LO now runs at

Mixer’s output difference frequency is

fLO = fc −fIF = fc +2fIF( )−fIF = fc + fIF

fLO −fc = fc + fIF −fIF = fIF

Two stations, one at fc and the other at fc+2fif appear simultaneously at the IF input

Image Frequency Example

Take WIP at 610 KHz. The image station is at 610+2x455=1520 KHz

If this station were to pass through the RF stage, it would interfer with 610

However, RF stage is selective enough to suppress two frequencies that are 910 KHZ apart.Hence this problem does no arise.

Commercial AM

Here are some numbers Carrier frequencies: 540-1600 KHz Carrier spacing:10 KHz IF frequency:455 KHz IF bandwidth:6-10 KHz Audio bandwidth: 3-5KHz

Frequency Division MUX

Frequency division multiplexing (FDM) is a proven method for carrying simultaneous conversations

FDM allocates signals to nonoverlapping parts of the spectrum. Receiver then filters each out

f1 f2 f3 f4

Frequency Translation

Want to move a signal centered at f1 to f2

The way to it is through a mixing operation

X filter

f1 f2

f1

fLWhat should fL be?

LO and filter Characteristics?

Filter’s input will have two components, one at f1-fL and one at f1+fL

X filterf1

fL

f1f1-fL f1+fL

want this termto be at f2 fL=f2-f1

Example

Want to translate signal centered at 500 KHZ up to a center frequency of 750 KHz

X filterf1=500 KHz

fL=250 KHz

750 KHz250 KHz

750 KHz

750

Suggested Homeowork

3.73.83.15