Communication Systems Practice Questions - DPP for JEE

 Page 1


1. (d) The maximum line-of-sight distance between the transmitting and
receiving antennas is
d
M
 = 
where h
T
 and h
R
 are the heights of transmitting and receiving antennas
respectively.
? d
M
 = 
( Q h
T
 = h
R
 = h)
2. (b) The frequencies present in amplitude modulated wave are :
Carrier frequency = ?
c
Upper side band frequency = ?
c
 + ?
m
Lower side band frequency = ?
c
 – ?
m
.
3. (b) p ? ( l/?)
2
4. (d)
5. (c)
6. (d)
7. (b) For given transmission band 88-108 MHz (?f)
max
= 75 kHz
given 
 % modulation 
8. (a) Comparing (x 
AM
)t = 100 [1 + 0.5 t] cos?
c
t for 0<t<1
with standard AM signal x 
AM
 = E
c
 [1+m
a
 cos ?
m
t] cos ?
c
t
We have modulating signal t and m
a
 = 0.5.
9. (d)
Page 2


1. (d) The maximum line-of-sight distance between the transmitting and
receiving antennas is
d
M
 = 
where h
T
 and h
R
 are the heights of transmitting and receiving antennas
respectively.
? d
M
 = 
( Q h
T
 = h
R
 = h)
2. (b) The frequencies present in amplitude modulated wave are :
Carrier frequency = ?
c
Upper side band frequency = ?
c
 + ?
m
Lower side band frequency = ?
c
 – ?
m
.
3. (b) p ? ( l/?)
2
4. (d)
5. (c)
6. (d)
7. (b) For given transmission band 88-108 MHz (?f)
max
= 75 kHz
given 
 % modulation 
8. (a) Comparing (x 
AM
)t = 100 [1 + 0.5 t] cos?
c
t for 0<t<1
with standard AM signal x 
AM
 = E
c
 [1+m
a
 cos ?
m
t] cos ?
c
t
We have modulating signal t and m
a
 = 0.5.
9. (d)
10. (d) For good demodulation,
 or, 
11. (b) Modulation index 
B = 25, A = 60
M.I.  m% 
12. (d) Frequency of EM wave ? = 830 kHz
= 830 × 10
3
? 
Hz.
Magnetic field, B = 4.82 × 10
–11
 T
As we know, frequency, ? = 
or ? = 
?  360 m
And, E = BC = 4.82 × 10
–11
 × 3 × 10
8
= 0.014 N/C
13. (a) Here, f
c
 = 1.5 MHz = 1500 kHz,   f
m 
= 10 kHz
? Low side band frequency
= f
c  
– f
m 
= 1500 kHz – 10 kHz = 1490 kHz
Upper side band frequency
= f
c  
+ f
m 
= 1500 kHz + 10 kHz = 1510 kHz
14. (b) The frequency of AM channel is 1020 kHz whereas for the FM it
is 89.5 MHz (given). For higher frequencies (MHz), space wave
communication is needed. Very tall towers are used as antennas.
15. (a) The critical frequency of a sky wave for reflection from a layer of
atmosphere is given by f
c
 = 9(N
max
)
1/2
Page 3


1. (d) The maximum line-of-sight distance between the transmitting and
receiving antennas is
d
M
 = 
where h
T
 and h
R
 are the heights of transmitting and receiving antennas
respectively.
? d
M
 = 
( Q h
T
 = h
R
 = h)
2. (b) The frequencies present in amplitude modulated wave are :
Carrier frequency = ?
c
Upper side band frequency = ?
c
 + ?
m
Lower side band frequency = ?
c
 – ?
m
.
3. (b) p ? ( l/?)
2
4. (d)
5. (c)
6. (d)
7. (b) For given transmission band 88-108 MHz (?f)
max
= 75 kHz
given 
 % modulation 
8. (a) Comparing (x 
AM
)t = 100 [1 + 0.5 t] cos?
c
t for 0<t<1
with standard AM signal x 
AM
 = E
c
 [1+m
a
 cos ?
m
t] cos ?
c
t
We have modulating signal t and m
a
 = 0.5.
9. (d)
10. (d) For good demodulation,
 or, 
11. (b) Modulation index 
B = 25, A = 60
M.I.  m% 
12. (d) Frequency of EM wave ? = 830 kHz
= 830 × 10
3
? 
Hz.
Magnetic field, B = 4.82 × 10
–11
 T
As we know, frequency, ? = 
or ? = 
?  360 m
And, E = BC = 4.82 × 10
–11
 × 3 × 10
8
= 0.014 N/C
13. (a) Here, f
c
 = 1.5 MHz = 1500 kHz,   f
m 
= 10 kHz
? Low side band frequency
= f
c  
– f
m 
= 1500 kHz – 10 kHz = 1490 kHz
Upper side band frequency
= f
c  
+ f
m 
= 1500 kHz + 10 kHz = 1510 kHz
14. (b) The frequency of AM channel is 1020 kHz whereas for the FM it
is 89.5 MHz (given). For higher frequencies (MHz), space wave
communication is needed. Very tall towers are used as antennas.
15. (a) The critical frequency of a sky wave for reflection from a layer of
atmosphere is given by f
c
 = 9(N
max
)
1/2
? 10 × 10
6
 = 9(N
max
)
1/2
16. (c) Above critical frequency (f
c
), an  electromagnetic wave penetrates
the ionosphere and is not reflected by it.
17. (c) Modulation index
Equation of modulated signal [C
m
(t)]
= E
(C)
 + m
a
E
(C)
 sin ?
m
t
= A (1+ sin ?
m
t) sin ?
C
t
(As E
(C)
 = A sin ?
C
t)
18. (b) E
c
 = 100 V,  m
a
 = 0.4,  R = 100 ?,
19. (c) Average side-band power 
Here m
a
 = 0.5
P
c
 = 10
20. (d)
Power gain
Page 4


1. (d) The maximum line-of-sight distance between the transmitting and
receiving antennas is
d
M
 = 
where h
T
 and h
R
 are the heights of transmitting and receiving antennas
respectively.
? d
M
 = 
( Q h
T
 = h
R
 = h)
2. (b) The frequencies present in amplitude modulated wave are :
Carrier frequency = ?
c
Upper side band frequency = ?
c
 + ?
m
Lower side band frequency = ?
c
 – ?
m
.
3. (b) p ? ( l/?)
2
4. (d)
5. (c)
6. (d)
7. (b) For given transmission band 88-108 MHz (?f)
max
= 75 kHz
given 
 % modulation 
8. (a) Comparing (x 
AM
)t = 100 [1 + 0.5 t] cos?
c
t for 0<t<1
with standard AM signal x 
AM
 = E
c
 [1+m
a
 cos ?
m
t] cos ?
c
t
We have modulating signal t and m
a
 = 0.5.
9. (d)
10. (d) For good demodulation,
 or, 
11. (b) Modulation index 
B = 25, A = 60
M.I.  m% 
12. (d) Frequency of EM wave ? = 830 kHz
= 830 × 10
3
? 
Hz.
Magnetic field, B = 4.82 × 10
–11
 T
As we know, frequency, ? = 
or ? = 
?  360 m
And, E = BC = 4.82 × 10
–11
 × 3 × 10
8
= 0.014 N/C
13. (a) Here, f
c
 = 1.5 MHz = 1500 kHz,   f
m 
= 10 kHz
? Low side band frequency
= f
c  
– f
m 
= 1500 kHz – 10 kHz = 1490 kHz
Upper side band frequency
= f
c  
+ f
m 
= 1500 kHz + 10 kHz = 1510 kHz
14. (b) The frequency of AM channel is 1020 kHz whereas for the FM it
is 89.5 MHz (given). For higher frequencies (MHz), space wave
communication is needed. Very tall towers are used as antennas.
15. (a) The critical frequency of a sky wave for reflection from a layer of
atmosphere is given by f
c
 = 9(N
max
)
1/2
? 10 × 10
6
 = 9(N
max
)
1/2
16. (c) Above critical frequency (f
c
), an  electromagnetic wave penetrates
the ionosphere and is not reflected by it.
17. (c) Modulation index
Equation of modulated signal [C
m
(t)]
= E
(C)
 + m
a
E
(C)
 sin ?
m
t
= A (1+ sin ?
m
t) sin ?
C
t
(As E
(C)
 = A sin ?
C
t)
18. (b) E
c
 = 100 V,  m
a
 = 0.4,  R = 100 ?,
19. (c) Average side-band power 
Here m
a
 = 0.5
P
c
 = 10
20. (d)
Power gain
directive gain 
21. (10) Band width
= 2 × frequency
= 2 × 5000
= 10 kHz
22. (380) For x(t), BW = 2(?? + ?)
?? is deviation and ? is the band width of modulating signal.
? BW = 2(90 + 5) = 190
For  x
2
 (t),  BW = 2 × 190 = 380
23. (10) Comparing given expression with
(e)
AM
 = E
c
? (1 + 
m
a
 cos ?
m
t) cos ?
c
t
peak value of carrier wave, E
c
= 10V.
24. (50) Here and 
Now, 
25. (9.6)