problems on modulation techniques
TRANSCRIPT
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.