analog communication lecture 07

8/14/2019 ANALOG COMMUNICATION Lecture 07

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Amplitude Modulation (Contd)

Lesson 07

EEE 352 Analog Communication SytemsMansoor Khan


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DSB-FC Full AMAM modulation is a fundamental modulation process incommunication system.Carrier frequency signal >> than modulating frequency signal.=> f c >> f m .Modulator is used to generate AM signal, am DSB-FC ( t). It is shown

in block diagram below.

t t v E t v cmc AM cos)()(v m (t )

v c(t )

AM Modulator

Modulatingsignal

Carrier signal

AM modulated signal f 2


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3

Let :

Therefore, am DSBFC signal can be expressed:

Given the modulation index :

am DSBFC can be deduced to:

From trigonometry identities:

Therefore:

c

m

E

E m

t t m E t v cmc AM coscos1)(

B A B A B A cos2

1cos

2

1)cos()cos(

t mE t mE t E

t t mE t E t v

mcc

mcc

cc

mcccc AM

cos2

cos2

cos

coscoscos)(

t t E E t v

t t v E t v

cmmc AM

cmc AM

coscos)(

cos)(

t E t vccc cos)(t E t v mmm cos)( and


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Signal frequency spectrum ; am DSBFC

t t mE t E t v mcmcccc AM coscos2

cos)(

Carrier signal Sidebands signal

)(V Amplitud

)( 1rads c mc mc 0

c E

2c

mE

2c

mE

m

22mc E mE

m E

where

LSB USB

Carrier bandModulating band


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One input to a conventional AM modulator is a 500kHzcarrier with an amplitude of 20Vp. The second input isa 10kHz modulating signal that is of sufficientamplitude to cause a change in the output wave of

7.5Vp. Determine

Upper and lower side frequencies.Modulation coefficient and percent modulationPeak amplitude of the modulated carrier and the upper and lower sidefrequency voltages.

Maximum and minimum amplitudes of the envelope.Expression for the modulated wave.Draw the output spectrum.Sketch the output envelope.

EXAMPLE


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Example

If the modulated wave has the equation,

find (a) the carrier freq (b) the usf and lsf (c) the modulating signal freq (d) the peak amplitude of the carrier signal (e) the upper and lower side signal peak amplitude (f) the change In peak amplitude of the modulated wave (g) the coefficient of modulation.

V t t t t vam )2182cos(60)2822cos(60)2502sin(150)(


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Full-Carrier AM: Time Domain Modulation Index - The ratio between the amplitudes

between the amplitudes of the modulating signal andcarrier, expressed by the equation:

c

m

E

E m =


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Modulation index from AM waveform

minmax

minmax

V V

V V

E

E

c

m

)(2

1minmax V V E m

)(2

1minmax

V V E c

)(4

1

2minmax V V

E E E

mlsf usf

E USF = PEAK AMPLITUDE OF THE UPPER SIDE FREQUENCYE LSF = PEAK AMPLITUDE OF THE LOWER SIDE FREQUENCY

mcmc E E V E E V minmax ;

ASSUMPTIONS: MODULATING SIGNAL IS A TONE MODULATING PROCESS IS SYMMETRICAL

(EQUAL + and ENVELOPE EXCURCIONS)


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AMPLITUDE MODULATION (DSB-FC)

Modulating Signal

Unmodulated Carrier

50% Modulation

100% Modulation


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Overmodulation and DistortionThe modulation index should be a number between 0 and 1.If the amplitude of the modulating voltage is higher than thecarrier voltage, m will be greater than 1, causing distortion.If the distortion is great enough, the intelligence signal becomesunintelligible.Distortion of voice transmissions produces garbled, harsh, orunnatural sounds in the speaker.

Distortion of video signals produces a scrambled and inaccuratepicture on a TV screen.


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Modulation Index for MultipleModulating Frequencies

Two or more sine waves of different, uncorrelatedfrequencies modulating a single carrier is calculated by theequation:


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P o w e r

Frequency

P c = 1000W

P lsb = 160W P usb = 160W

f lsb f usb f c

P o w e r

Frequency


f lsb f usb f c

No Carrier

Calculate total power

Conclusion ???


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DSBFC is wasteful of Power

75.6% of total transmitted power taken up by carrier.

P o w e r

Frequency

P c = 1000W


f lsb f usb f c

In transmitting 1320W of the total power, the carrier contains1000W and does not contain any information being transmitted. Theside freq each have 160W and each carries a copy of the same infosignal.

So, 1320W is being used in order to transmit only 160W.


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DSB is wasteful of Bandwidth

DSB has a Wide Bandwidthwasteful BW usage i.e info in USB = info in LSB

If so much of the transmitted wave is not required, then whytransmit it? any alternative? DSBSC?


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DSB Suppressed Carrier (DSBSC)

Generated by circuit called balanced modulator where it producessum ( f usb ) and difference ( f lsb ) freq but cancel or balance out thecarrier ( f c).

P o w e r

Frequency


f lsb f usb f c

The total power being transmitted is now reduced to 320W

No Carrier

DSBSC helps in reducing power but bandwidth still the same asDSBFC.


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Suppressing the carrierEliminating the carrier results in a double-sideband suppressed carrier (DSSC or DSB)signal shown below.

Suppressed carrier AM signal (DSB)

-2

-1.5

-1

-0.5

0

0.5

1

1.5

Time (sec)

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Time (sec)

V o

l t a g

e ( V )

Full carrier AM signal

Note the phase transitions


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DSBSC in frequency domainSuppressed carrier AM signal (DSB)Full carrier AM signal

0 1000 2000 3000 4000 5000 6000

0.1

0.2

0.3

0.4

0.5

Frequency (Hz)

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Time (sec)

V o

l t a g e

( V )

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Time (sec)

0 1000 2000 3000 4000 5000 6000

0.1

0.2

0.3

0.4

0.5

Frequency (Hz)

Frequency domain Frequency domain


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AM Power Distribution

The average power dissipated in a load by unmodulatedcarrier is equal to the rms carrier voltage, E c squareddivided by the load resistance, R.

Mathematically, power in unmodulated carrier, P c is:

R

E

R

E

R

E P ccrms

c

c 2)2()( 22

2)(


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The upper and lower sideband powers:

where mE c /2 is the peak voltage of usf and lsf . Then,

Total transmitted power in DSBFC AM envelope:

R

E m

R

mE PP cclsbusb

82)2( 222

cc

lsbusb Pm R E mPP

424

222

21

2

4422

22

mPP

mP

Pm

Pm

P

PPPP

ccc

ccc

lsbusbct


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21

244

2222 mPP

mPP

mP

mPPPPP cccccclsbusbct

Power Spectrum for

AM DSBFC wave

Note:Carrier power in the modulated signal is the same in the unmodulated signal i.e

carrier power is unaffected by the modulation process.The total power in an AM envelope increase with modulation (i.e as m , P t ).Major disadvantage of AM DSBFC is most of the power is wasted in the carrier. (It

does not contain info, info is contained in the sidebands).


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Sideband and Carrier Power (cont)

The sideband power is the useful power and the CarrierPower is the power wasted

We define the Power Efficiency as

%100*

)(

)(

)(21

2

)(2

1

~~~~~~~

22

~~~~~~~2

~~~~~~~

22

~~~~~~~

2

t m A

t m

t m A

t m

PP

P

TotalPower

r UsefulPowe

sc

s


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Sideband and Carrier Power (cont)

For the special case of tone modulation

then its power is

then

The max value when (100% modulation) is

t At m m cos)(

2)(

2~~~~~~~2 A

t m

%100*

2%100*

2

2%100*

221

2

22

1

2

2

22

2

22

2

A A

A

A A

A

1 %33


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ExampleDetermine the maximum sideband power if the carrieroutput is 1 kW and calculate the total maximumtransmitted power.

SinceE

SF =mE

c /2,It is obvious that the max SB power occurs when m = 1 or 100%, andalso when m = 1, each side freq is the carrier amplitude.Since power is proportional to the square of voltage, each SB has of the carrier power i.e x 1kW, or 250W.Therefore, total SB power is 250W x 2 = 500W.And the total transmitted power is 1kW + 500W = 1.5kW

f h


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Importance of High-percentageModulation

m P c P 1SB P SBs P T E1.0 1kW 250W 500W 1.5kW 0.3

0.5 1kW 62.5W 125W 1.125kW 0.1

NotesEven though the total transmitted power has only fallen from 1.5kW to

1.125kW, the effective transmission has only the strength at 50% modulation ascompared to 100%.

Because of these considerations, most AM transmitter attempts to maintainbetween 90 and 95 percent modulation as a compromise between efficiency andthe chance of drifting into overmodulation.

Table: Effective transmission at 50% versus 100% modulation


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Switching Modulator


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Switching Modulator (cont)

The input is with c>>m(t) so the switchingaction does not depends on m(t)

)(cos t mt c c

)()(cos' t wt mt cv cbb

.....5cos5

13cos

3

1cos

2

2

1)(cos' t t t t mt cv ccccbb

etct t mt cv ccbb cos)(2cos

2'


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Demodulation of AM Signals

We do not need a local generated carrier in this case

If we have undermodulation then we can use

1. Rectifier detection2. Envelope detection


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Rectifier detector


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Rectifier detector (cont)

If the AM wave is applied to diode and resistor circuit. Thenegative part of the AM is supressed. This is like saying thatwe have half wave rectified the AM

Mathematically

.....5cos5

13cos

3

1cos

2

2

1cos)(' t t t t t m Av cccc R

termsother t m Av R

)(1

'


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Rectifier detector (cont)

If we pass this voltage thru a LPF we get

If we use a capacitor, we block the DC and we obtain

)(1 t m Av filtered

)(1 t mvout


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Envelope detector


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Quadrature Amplitude Modulation(QAM)

DSB signals occupy twice the bandwidth required for the baseband.

This disadvantage can be overcome by transmitting two DSBsignals using carriers of the same frequency but in phasequadrature

The message signals m 1(t) & m 2(t) are in-phase & quadrature-phase components of QAM(t)

t t mt t m ccQAM sin)(cos)( 21


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Modulation and Demodulation of QAM


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QAM (cont) We can obtain both the signals by using two local carriers in

phase quadrature

Similarly the output of the lower branch can be shown as A slight error in phase leads to distortion and mixing of signals

t t t mt t mt t x ccccQAM cossin)(cos)(2cos2)( 211t t mt t mt mt x cc 2sin)(2cos)()()( 2111

t t t mt t mt t x ccccQAM cossin)(cos)(2cos2)( 211 sin)(cos)()( 211 t mt mt x

analog communication lecture 07

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