ANALOG COMMUNICATION Lecture 07

Amplitude Modulation (Contd)

Lesson 07 EEE 352 Analog Communication Sytems Mansoor Khan

DSB-FC – Full AM   

AM modulation is a fundamental modulation process in communication system. Carrier frequency signal >> than modulating frequency signal. => fc >> fm. Modulator is used to generate AM signal, amDSB-FC(t). It is shown in block diagram below.

vm(t)

AM Modulator

Modulating signal

vc(t) Carrier signal

v AM (t )  Ec  vm (t )  cos ct AM modulated signal

  2f

vm (t )  Em cos mt

vc (t )  Ec cos ct



Let :



Therefore, amDSBFC signal can be expressed:

and

v AM (t )  Ec  vm t  cos c t

v AM (t )  Ec  Em cos mt  cos c t 

Given the modulation index :

Em m Ec

v AM (t )  Ec 1  m cos mt cos ct 



amDSBFC can be deduced to:

1 1 cos( A) cos( B)  cos A  B   cos A  B  2 2 From trigonometry identities:

v AM (t )  Ec cos  ct  mEc cos  ct cos  mt 

mEc mEc  Ec cos  ct  cos c   m t  cos c   m t 2 23

Therefore:



Signal frequency spectrum ; amDSBFC

v AM (t )  Ec cos ct 

mEc cosc  m t  cosc  m t  2

Carrier signal

Sidebands signal

Amplitud (V ) Modulating band

0

where

Carrier band

Em

mEc 2

m

c  m LSB

Ec

c

mEc 2

c  m USB

mEc Em  2 2

 (rads 1 )

EXAMPLE 

One input to a conventional AM modulator is a 500kHz carrier with an amplitude of 20Vp. The second input is a 10kHz modulating signal that is of sufficient amplitude to cause a change in the output wave of 7.5Vp. Determine  Upper and lower side frequencies.  Modulation coefficient and percent modulation  Peak amplitude of the modulated carrier and the upper and lower side  

 

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

Example 

If the modulated wave has the equation, vam (t )  150 sin(2 250t )  60 cos(2 282t )  60 cos(2 218t )V – – 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.

Full-Carrier AM: Time Domain • Modulation Index - The ratio between the amplitudes between the amplitudes of the modulating signal and carrier, expressed by the equation:

Em m= Ec

Modulation index from AM waveform

V

max

 Ec  Em ;

1 Em  (V max  V min ) 2 1 Ec  (V max  V min ) 2

Eusf  Elsf

min

 Ec  Em

ASSUMPTIONS: • MODULATING SIGNAL IS A TONE • MODULATING PROCESS IS SYMMETRICAL (EQUAL + and – ENVELOPE EXCURCIONS)

Em V   Ec V

Em 1   (V 2 4

V

V max  V max

max

min min

V

min

)

EUSF = PEAK AMPLITUDE OF THE UPPER SIDE FREQUENCY ELSF = PEAK AMPLITUDE OF THE LOWER SIDE FREQUENCY

AMPLITUDE MODULATION (DSB-FC) Modulating Signal

Unmodulated Carrier

50% Modulation

100% Modulation

Overmodulation and Distortion  The modulation index should be a number between 0 and 1.  If the amplitude of the modulating voltage is higher than the carrier voltage, m will be greater than 1, causing distortion.  If the distortion is great enough, the intelligence signal becomes unintelligible.  Distortion of voice transmissions produces garbled, harsh, or unnatural sounds in the speaker.  Distortion of video signals produces a scrambled and inaccurate picture on a TV screen.

Modulation Index for Multiple Modulating Frequencies • Two or more sine waves of different, uncorrelated frequencies modulating a single carrier is calculated by the equation:

Power

Pc = 1000W Plsb = 160W

flsb

• Calculate total power • Conclusion ???

Pusb = 160W

fc

fusb

Frequency

Power

The total power being transmitted is (1000).(1 + 0.82) = 1320W 2 No Carrier Plsb = 160W

flsb

Pusb = 160W

fc

fusb

Frequency

The total power being transmitted is now reduced to 320W

DSBFC is wasteful of Power 

75.6% of total transmitted power taken up by carrier.

Power

Pc = 1000W Plsb = 160W

flsb

Pusb = 160W

fc

fusb

Frequency

The total power being transmitted is (1000).(1 + 0.82) = 1320W 2 In

transmitting 1320W of the total power, the carrier contains 1000W and does not contain any information being transmitted. The side freq each have 160W and each carries a copy of the same info signal. So,

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

DSB is wasteful of Bandwidth 

DSB has a Wide Bandwidth  wasteful BW usage i.e info in USB = info in LSB

If

so much of the transmitted wave is not required, then why transmit it? – any alternative? – DSBSC?

DSB Suppressed Carrier (DSBSC) Generated by circuit called balanced modulator where it produces sum (fusb) and difference (flsb) freq but cancel or balance out the carrier (fc).

Power



No Carrier Plsb = 160W

flsb

Pusb = 160W

fc

fusb

Frequency

The total power being transmitted is now reduced to 320W DSBSC

DSBFC.

helps in reducing power but bandwidth still the same as

Suppressing the carrier 

Eliminating the carrier results in a doublesideband suppressed carrier (DSSC or DSB) signal shown below. Suppressed carrier AM signal (DSB)

2

2

1.5

1.5

1

1

0.5

0.5

Voltage (V)

Voltage (V)

Full carrier AM signal

0

0

-0.5

-0.5

-1

-1

-1.5

-1.5

-2 T ime (sec)

-2 T ime (sec)

Note the phase transitions

DSBSC in frequency domain Suppressed carrier AM signal (DSB)

2

2

1.5

1.5

1

1

0.5

0.5

Voltage (V)

Voltage (V)

Full carrier AM signal

0 -0.5

0 -0.5

-1

-1

-1.5

-1.5 -2

-2

T ime (sec)

T ime (sec)

0.5

0.5

0.4

Voltage (V)

Voltage (V)

0.4

Frequency domain 0.3

0.2

0.2

0.1

0.1

0

Frequency domain

0.3

0 0

1000

2000

3000 Frequency (Hz)

4000

5000

6000

0

1000

2000

3000 Frequency (Hz)

4000

5000

6000

AM Power Distribution 

The average power dissipated in a load by unmodulated carrier is equal to the rms carrier voltage, Ec squared divided by the load resistance, R.



Mathematically, power in unmodulated carrier, Pc is:

Pc 

( Ec ( rms) ) 2 R



2 )2

( Ec R

2

Ec  2R

AM Power Distribution 

•

The upper and lower sideband powers: (mEc 2) 2 m 2 Ec Pusb  Plsb   2R 8R

2

– where mEc/2 is the peak voltage of usf and lsf. 2 – Then, m2  E  m2

Pusb  Plsb 



c Pc   4  2R  4

Total transmitted power in DSBFC AM envelope: Pt  Pc  Pusb  Plsb

m2 m2  Pc  Pc  Pc 4 4  m2  m2   Pc  Pc  Pc 1  2 2  

AM Power Distribution  m2  m2 m2 m2  Pt  Pc  Pusb  Plsb  Pc  Pc  Pc  Pc  Pc  Pc 1  4 4 2 2  

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 , Pt ). 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).

Sideband and Carrier Power (cont) • The sideband power is the useful power and the Carrier Power is the power wasted • We define the Power Efficiency as ~~~~~~~

1 2 m (t ) 2

~~~~~~~ 2

UsefulPowe r Ps m (t )     *100% ~~~~~~~ ~~~~~~~ TotalPower Pc  Ps A2 1 2 2 A  m (t ) 2  m (t ) 2 2

Sideband and Carrier Power (cont) For the special case of tone modulation

m(t )  A cos m t then its power is

 A2 (t ) 

~~~~~~~ 2

m

then

2

A 1 A 2  2   2 2 2 2 *100% *100%  *100% 2 2 2  A 1 A A 2  A  2 2 2 2 2

2

The max value when   1(100% modulation) is   33%

Example 

Determine the maximum sideband power if the carrier output is 1 kW and calculate the total maximum transmitted power.

Since ESF = mEc/2, It is obvious that the max SB power occurs when m = 1 or 100%, and also when m = 1, each side freq is ½ the carrier amplitude. Since power is proportional to the square of voltage, each SB has ¼ of the carrier power i.e ¼ x 1kW, or 250W. Therefore, total SB power is 250W x 2 = 500W. And the total transmitted power is 1kW + 500W = 1.5kW

Importance of High-percentage Modulation m

Pc

P1SB

PSBs

PT

E

1.0

1kW

250W

500W

1.5kW

0.3

0.5

1kW

62.5W

125W

1.125kW

0.1

Table: Effective transmission at 50% versus 100% modulation Notes Even though the total transmitted power has only fallen from 1.5kW to 1.125kW, the effective transmission has only ¼ the strength at 50% modulation as compared to 100%. Because of these considerations, most AM transmitter attempts to maintain between 90 and 95 percent modulation as a compromise between efficiency and the chance of drifting into overmodulation.

Generation of AM Signals • Any DSB-SC modulators are valid if the modulating signal is A  m(t ) • Because the carrier does not need to be suppressed, we do not need balanced circuits

• The modulators are then very simple

Switching Modulator

Switching Modulator (cont) • The input is c cos ct  m(t ) with c>>m(t) so the switching action does not depends on m(t)

vbb'  c cos c t  m(t )w(t ) vbb'

1 2  1 1   c cos c t  m(t )   cos  c t  cos 3 c t  cos 5 c t..... 3 5  2  

vbb'

2 c    cos  c t  m(t ) cos c t   etc  2 

Demodulation of AM Signals •

We do not need a local generated carrier in this case

•

If we have undermodulation then we can use 1. 2.

Rectifier detection Envelope detection

Rectifier detector

Rectifier detector (cont) • If the AM wave is applied to diode and resistor circuit. The negative part of the AM is supressed. This is like saying that we have half wave rectified the AM Mathematically 1 2  1 1  v R'  A  m(t )cos c t    cos  c t  cos 3c t  cos 5c t..... 3 5  2  

vR ' 

1



A  m(t )  other terms

Rectifier detector (cont) • If we pass this voltage thru a LPF we get

v filtered 

1



A  m(t )

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

vout 

1



m(t )

Envelope detector

Quadrature Amplitude Modulation (QAM) • DSB signals occupy twice the bandwidth required for the base band. • This disadvantage can be overcome by transmitting two DSB signals using carriers of the same frequency but in phase quadrature

 QAM  m1 (t ) cos c t  m2 (t ) sin c t • The message signals m1(t) & m2(t) are in-phase & quadraturephase components of φQAM(t)

Modulation and Demodulation of QAM

QAM (cont) • We can obtain both the signals by using two local carriers in phase quadrature x1 (t )  2QAM cos c t  2m1 (t ) cos c t  m2 (t ) sin c t cos c t

x1 (t )  m1 (t )  m1 (t ) cos 2c t  m2 (t ) sin 2c t • Similarly the output of the lower branch can be shown as • A slight error in phase leads to distortion and mixing of signals

x1 (t )  2QAM cos c t  2m1 (t ) cos c t  m2 (t ) sin c t cosc t    x1 (t )  m1 (t ) cos   m2 (t ) sin 