Uploaded by Prakash Das

AM Problems

advertisement
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 
Download