Problems on Modulation techniques
Transcript of 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.