Uploaded by Prakash Das

AM Problems

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Problems on Modulation
techniques
Quote of the day
“Any man who reads too much and uses his
own brain too little falls into lazy habits of
thinking.”
― Albert Einstein
Important formulae to solve AM problems
e(t )  Ec cos(c t ) where c  2f c
Em
m 
Ec
em (t )  Em cosm t
Emax  Ec  Em
Bandwidth  2 f m
Pt (total )
E 2c

2
Emin  Ec  Em
Emax  Emin
m 
Emax  Emin
E 2c
2
2
E
E
Pc  2  c  Pc  c
R
2R
2
 m2   m2 
1 
  1 
 Pc
2  
2 

m
2
PLSB  PUSB

for R  1
m 2 E 2c m 2


Pc
8
4
/ 2 Pc
m2
transmission efficiency 

2
1  m 2 Pc 2  m 2


For an AM, amplitude of modulating signal is 0.5 V and
carrier amplitude is 1V.Find Modulation Index.
Solution: Ec = 1 V and Em = 0.5 V
We know that
Em
m 
Ec
0.5
m 
1
m  0.5
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
We know that
 m
 1 
2

2
Pt (total )

 Pc

10 103
 Pc 
 0.752 
1 

2 

 Pc 
Pt (total )
 m2 
1 

2 

 Pc  7.8 103W
 Pc  7.8KW
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
 m
 1 
2

2
Pt (total )

 Pc

20 103
 Pc 
 0.7 2 
1 

2 

 Pc 
Pt (total )
 m2 
1 

2 

 Pc  16.064103W
 Pc  16.064KW
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
We know that
Pt (total )
 m2 
 Pc
 1 
2 

1103
 Pc 
 12 
1  
2

PUSB  PLSB
 m2 
 Pc
 
 4 
 Pc 
Pt (total )
 m2 
1 

2 

 Pc  666.67W
1
PUSB  PLSB     666.67
4
 Pc  166.67W
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
Pt (total )
 m2 
 Pc
 1 
2 

 Pt (total )
 0.6 2 
500
 1 
2 

 Pt (total )  590 W
We know that
 fUSB  f c  f m
 f LSB  f c  f m
 fUSB  101 KHz
 f LSB  99 KHz
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
 fUSB
We know that
 fc  fm
f
LSB
 fc  fm
 BW  2 f m
 fUSB  1002.5 KHz  f LSB  997.5 KHz  BW  2  2.5KHz  5KHz
We know that
Pt (total )
 m2 
 Pc
 1 
2 

 Pt (total )
 0.752 
400
 1 
2 

 Pt (total )  512.5 W
PUSB  PLSB
 m2 
 Pc
 
 4 
PUSB  PLSB
 0.752 
400  56.25 W
 
 4 
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
 fUSB  f c  f m
 f LSB  f c  f m
 fUSB  1002 KHz  f LSB  998 KHz  BW
 BW
 fUSB
 2f LSB
f m 44 KHz
KHz
We know that
Pt (total )
 m2 
 Pc
 1 
2 

 Pt (total )
 0.52 
750
 1 
2 

 Pt (total )  843.75 W
PUSB  PLSB
 m2 
 Pc
 
 4 
PUSB  PLSB
 0.52 
750  46.875 W
 
 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)
Solution: m = 1
We know that
Pt (total )
 Psuppressed
 m2 
3
 Pc  Pc
 1 
2 
2

 m2
 Pc  
 4
Psuppressed  Pc  PLSB
5
1

 Pc  Psuppressed  Pc    Pc  Pc
4
4


Amount of power saved 
5
4
3
2
Psuppressed
Ptotal
Pc
10
Amount of power saved 

 0.833  83.3%
Pc
12
Calculate the percentage power saving when one side
band and carrier is suppressed in an AM signal if
percentage of modulation is 50%.
Solution: m = 0.5
We know that
Pt (total )
 Psuppressed
 m2
 1 
2


9
 Pc  Pc
8

 m2
 Pc  
 4
Psuppressed  Pc  PLSB
17
1
 P
Pc
 Pc 
 Pc
suppressed  Pc  
16 
16


Amount of power saved 
Amount of power saved 
9
8
17
16
Psuppressed
Ptotal
Pc
9  16

 0.944  94.4%
Pc
8  17
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
 fUSB  f c  f m
 fUSB  1220 KHz
We also know that
 f LSB  f c  f m
 f LSB  1180 KHz
E max  E min
Emax  Emin
Emax  Emin
also m 
& Em 
Ec 
E max  E min
2
2
110  90
110  90
110
.90
m

0
1
E  10V
 E
EE
m 
m m
c c  100V
2
2
110  90
An audio frequency signal 10 sin(2π×500t) is used to amplitude
modulate a carrier of 50 sin(2π×105t).Calculate.
(JAN2015)
I.frequency of side bands IV. Amplitude of each side band.
II. Bandwidth
V. Transmission efficiency
III. Modulation index
VI. Total power delivered to a load of 600Ω.
Solution: fm =500 Hz , Em = 10V, fc =100 KHz and Ec = 50V.
We know that  fUSB  f c  f m  f LSB  f c  f m  BW  2 f m
 f LSB  99.5 KHz  BW  2  500Hz  1KHz
E
10
m  0.2  50%
We also know that m  m m 
50
Ec
Transmission
mEc 0.2  50  5V
Efficiency
Amplitude of side band 

2
2
m2
eff. 
2
2
Ec
2

m
2500
also carrier power Pc 
 2.08
Pc 
2
0
.
2
2 R
2  600
eff. 
2
2  0.2 2
 m 
 Pc  Pt  2.125
and total power Pt  1 
eff.  0.196  1.96%
 fUSB  100.5 KHz


2 
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