Problems on Modulation techniques

Problems on Modulation techniques

Quote of the day

“Any man who reads too much and uses hisown brain too little falls into lazy habits ofthinking.”

― Albert Einstein

Important formulae to solve AM problems

c

m

E

Em

cccc ftEte 2 where)cos()(

tEte mmm cos)(

mc EEE max mc EEE minminmax

minmax

EE

EEm

cc

totalt PmmE

P

21

21

2

222

)( cc

USBLSB PmEm

PP48

222

2

2

2

2

221

2/efficiencyon transmissi

m

m

Pm

Pm

c

c

mf2Bandwidth 1Rfor 22

22

2

2

cc

cE

c

EP

R

E

RP

c

For an AM, amplitude of modulating signal is 0.5 V andcarrier amplitude is 1V.Find Modulation Index.

c

m

E

Em

Solution: Ec = 1 V and Em = 0.5 V

We know that

1

5.0m

5.0m

When the modulation percentage is 75%, an AM transmitter radiates 10KW Power. How much of this is carrier Power?

Solution: Pt = 10 KW and m=0.75

ctotalt Pm

P

21

2

)(

We know that

21

2

)(

m

PP

totalt

c

WPc

3108.7

2

75.01

10102

3

cP

KWPc 8.7

An AM transmitter radiates 20KW. If the modulation Index is 0.7. Find the carrier Power.(June-July2009(2002 scheme))

Solution: Pt = 20 KW and m=0.7

We know that

ctotalt Pm

P

21

2

)(

21

2

)(

m

PP

totalt

c

WPc

310064.16

2

7.01

10202

3

cP

KWPc 064.16

The total Power content of an AM signal is 1000W. Determine the power being transmitted at carrier frequency and at each side bands when modulation percentage is 100%. .(Dec 2014-Jan 2015)

Solution: Pt = 1000 W and m=1

ctotalt Pm

P

21

2

)(

We know that

21

2

)(

m

PP

totalt

c

WPc 67.666

2

11

1012

3

cP

cLSBUSB Pm

PP

4

2 67.6664

1

LSBUSB PP

WPc 67.166

A500W, 100KHz carrier is modulated to a depth of 60% by modulating frequency of 1KHz. Calculate the total power transmitted. What are the sideband components of AM Wave?(Dec –Jan 2010(2006 scheme))

Solution: Pc = 500 W, fc =100 KHz, m = 60% = .6 and fm =1 KHz

We know that

ctotalt Pm

P

21

2

)( 5002

6.01

2

)(

totaltP

WP totalt 590)(

mcUSB fff mcLSB fff

We know that

KHzfUSB 101 KHzfLSB 99

A400W, 1MHz carrier is amplitude-modulated with a sinusoidal signal 0f 2500Hz. The depth of modulation is 75%. Calculate the sideband frequencies, bandwidth, and power in sidebands and the total power in modulated wave. (June-July2008(2006 scheme))

Solution: Pc = 400 W, fc =1 MHz, m = 75% = .75 and fm =2.5 KHz

We know that

ctotalt Pm

P

21

2

)( 4002

75.01

2

)(

totaltP

WP totalt 5.512)(

mcUSB fff mcLSB fff

We know that

KHzfUSB 5.1002 KHzfLSB 5.997

cLSBUSB Pm

PP

4

2

4004

75.0 2

LSBUSB PP W25.56

mfBW 2

KHzKHzBW 55.22

A Carrier of 750 W,1MHz is amplitude modulated by sinusoidal signal of 2 KHz to a depth of 50%. Calculate Bandwidth, Power in side band and total power transmitted.

Solution: Pc = 750 W, fc =1 MHz, m = 50% = .5 and fm =2 KHz

We know that

ctotalt Pm

P

21

2

)( 7502

5.01

2

)(

totaltP

WP totalt 75.843)(

mcUSB fff mcLSB fff

We know that

KHzfUSB 1002 KHzfLSB 998

cLSBUSB Pm

PP

4

2

7504

5.0 2

LSBUSB PP W875.46

KHzfBW m 42 KHzffBW LSBUSB 4

Calculate the percentage power saving when one side band and carrier is suppressed in an AM signal with modulation index equal to 1.(VTU Model QP-2014)

cctotalt PPm

P2

3

21

2

)(

c

c

P

P

23

45

savedpower ofAmount

LSBc PPP suppressed

12

10 833.0 %3.83

Solution: m = 1

We know that

cc Pm

PP

4

2

suppressed ccc PPPP4

5

4

1suppressed

totalP

Psuppressedsavedpower ofAmount

Calculate the percentage power saving when one side band and carrier is suppressed in an AM signal if percentage of modulation is 50%.

cctotalt PPm

P8

9

21

2

)(

c

c

P

P

1617

89

savedpower ofAmount

LSBc PPP suppressed

178

169

944.0 %4.94

Solution: m = 0.5

We know that

cc Pm

PP

4

2

suppressed ccc PPPP16

17

16

1suppressed

totalP

Psuppressedsavedpower ofAmount

A Sinusoidal carrier frequency of 1.2MHz is amplitude modulated by a sinusoidal voltage of frequency 20KHz resulting in maximum and minimum modulated carrier amplitude of 110V & 90V respectively.

Calculate

I. frequency of lower and upper side bands

II. unmodulated carrier amplitude

III. Modulation index IV. Amplitude of each side band.

Solution: fc =1.2 MHz, Emax = 110V, Emin = 90V and fm =20 KHz

We know that

mcUSB fff mcLSB fff

KHzfUSB 1220 KHzfLSB 1180

minmax

minmax alsoEE

EEm

2

minmax EEEc

2& minmax EE

Em

We also know that

2

90110cE VEc 100

2

90110 mE VEm 10

90110

90110

m 1.0m

An audio frequency signal 10 sin(2π×500t) is used to amplitude modulate a carrier of 50 sin(2π×105t).Calculate. (JAN2015)

Solution: fm =500 Hz , Em = 10V, fc =100 KHz and Ec = 50V.

We know that mcUSB fff mcLSB fff

KHzfUSB 5.100 KHzfLSB 5.99

R

EP c

2power carrier also

2

c

c

m

E

Em

2band side of Amplitude cmE

We also know that50

10m %502.0 m

2

502.0 V5

6002

2500

cP 08.2

I.frequency of side bandsII. BandwidthIII. Modulation index

IV. Amplitude of each side band.V. Transmission efficiencyVI. Total power delivered to a load of 600Ω.

mfBW 2

KHzHzBW 15002

cPm

P

21power totaland

2

t125.2 tP

2

2

2eff.

m

m

Transmission Efficiency

2

2

2.02

2.0eff.

%96.1196.0eff.