This tutorial session enables students to solve numerical problems AM

1. 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

2. Important formulae to solve AM problems
c
m
E
E
m 
c
c
c
c f
t
E
t
e 

 2
where
)
cos(
)
( 

t
E
t
e m
m
m 
cos
)
( 
m
c E
E
E 

max m
c E
E
E 

min
min
max
min
max
E
E
E
E
m




c
c
total
t P
m
m
E
P 



















2
1
2
1
2
2
2
2
)
( c
c
USB
LSB P
m
E
m
P
P
4
8
2
2
2



 
  2
2
2
2
2
2
1
2
/
efficiency
on
transmissi
m
m
P
m
P
m
c
c




m
f
2
Bandwidth  1
R
for
2
2
2
2
2
2




 c
c
c
E
c
E
P
R
E
R
P
c

3. For an AM, amplitude of modulating signal is 0.5 V and
carrier amplitude is 1V.Find Modulation Index.
c
m
E
E
m 

Solution: Ec = 1 V and Em = 0.5 V
We know that
1
5
.
0

m
5
.
0

m

4. 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
c
total
t P
m
P 









2
1
2
)
(
We know that











2
1
2
)
(
m
P
P total
t
c
W
Pc
3
10
8
.
7 














2
75
.
0
1
10
10
2
3
c
P
KW
Pc 8
.
7



5. 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
c
total
t P
m
P 









2
1
2
)
(











2
1
2
)
(
m
P
P total
t
c
W
Pc
3
10
064
.
16 














2
7
.
0
1
10
20
2
3
c
P
KW
Pc 064
.
16



6. 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
c
total
t P
m
P 









2
1
2
)
(
We know that











2
1
2
)
(
m
P
P total
t
c
W
Pc 67
.
666














2
1
1
10
1
2
3
c
P
c
LSB
USB P
m
P
P 









4
2
67
.
666
4
1








 LSB
USB P
P
W
Pc 67
.
166



7. 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
c
total
t P
m
P 









2
1
2
)
( 500
2
6
.
0
1
2
)
( 









 total
t
P
W
P total
t 590
)
( 

m
c
USB f
f
f 

 m
c
LSB f
f
f 


We know that
KHz
fUSB 101

 KHz
fLSB 99



8. 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
c
total
t P
m
P 









2
1
2
)
( 400
2
75
.
0
1
2
)
( 









 total
t
P
W
P total
t 5
.
512
)
( 

m
c
USB f
f
f 

 m
c
LSB f
f
f 


We know that
KHz
fUSB 5
.
1002

 KHz
fLSB 5
.
997


c
LSB
USB P
m
P
P 









4
2
400
4
75
.
0 2









 LSB
USB P
P W
25
.
56

m
f
BW 2


KHz
KHz
BW 5
5
.
2
2 




9. 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
c
total
t P
m
P 









2
1
2
)
( 750
2
5
.
0
1
2
)
( 









 total
t
P
W
P total
t 75
.
843
)
( 

m
c
USB f
f
f 

 m
c
LSB f
f
f 


We know that
KHz
fUSB 1002

 KHz
fLSB 998


c
LSB
USB P
m
P
P 









4
2
750
4
5
.
0 2









 LSB
USB P
P W
875
.
46

KHz
f
BW m 4
2 


 KHz
f
f
BW LSB
USB 4





10. 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)
c
c
total
t P
P
m
P
2
3
2
1
2
)
( 










c
c
P
P
2
3
4
5
saved
power
of
Amount 
LSB
c P
P
P 

suppressed
12
10
 833
.
0
 %
3
.
83

Solution: m = 1
We know that
c
c P
m
P
P 










4
2
suppressed c
c
c P
P
P
P
4
5
4
1
suppressed 









total
P
Psuppressed
saved
power
of
Amount 

11. Calculate the percentage power saving when one side
band and carrier is suppressed in an AM signal if
percentage of modulation is 50%.
c
c
total
t P
P
m
P
8
9
2
1
2
)
( 










c
c
P
P
16
17
8
9
saved
power
of
Amount 
LSB
c P
P
P 

suppressed
17
8
16
9


 944
.
0
 %
4
.
94

Solution: m = 0.5
We know that
c
c P
m
P
P 










4
2
suppressed c
c
c P
P
P
P
16
17
16
1
suppressed 









total
P
Psuppressed
saved
power
of
Amount 

12. 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
m
c
USB f
f
f 

 m
c
LSB f
f
f 


KHz
fUSB 1220

 KHz
fLSB 1180


min
max
min
max
also
E
E
E
E
m



2
min
max E
E
Ec


2
& min
max E
E
Em


We also know that
2
90
110

c
E V
Ec 100

2
90
110

 m
E V
Em 10


90
110
90
110



m 1
.
0

m

13. 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 m
c
USB f
f
f 

 m
c
LSB f
f
f 


KHz
fUSB 5
.
100

 KHz
fLSB 5
.
99


R
E
P c


2
power
carrier
also
2
c
c
m
E
E
m 
2
band
side
of
Amplitude c
mE

We also know that
50
10

m %
50
2
.
0 

m
2
50
2
.
0 
 V
5

600
2
2500


c
P 08
.
2

I.frequency of side bands
II. Bandwidth
III. Modulation index
IV. Amplitude of each side band.
V. Transmission efficiency
VI. Total power delivered to a load of 600Ω.
m
f
BW 2


KHz
Hz
BW 1
500
2 



c
P
m
P 









2
1
power
total
and
2
t
125
.
2

 t
P
2
2
2
eff.
m
m


Transmission
Efficiency
2
2
2
.
0
2
2
.
0
eff.


%
96
.
1
196
.
0
eff. 
