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Solution of Communication Systems Test 01 Class 12 Notes | EduRev

Class 12 : Solution of Communication Systems Test 01 Class 12 Notes | EduRev

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CBSE TEST PAPER-01
CLASS - XII PHYSICS (Communication System)
Ans1: Digital communication.
Ans2: Modulation is done because low frequency signal cannot be transmitted to a longer
distance so in order to increase the range of transmission modulation is done.
Ans3: The transmitting and receiving antenna must be in the line of sight.
Ans4:  It means that signal jumps form one level to another in no time so its frequency will
be infinite.
Ans5:  Sky wave propagation is due to the reflection of radio waves by the ionosphere but
high frequency waves gets absorbed by the ionosphere and cannot be reflected by
the ionosphere.
Ans6: 2 d rh =
3
2 6250 50 10 d = × × ×
4
2.5 10 d m = ×
Area covered
2 4 2
3.14 (2.5 10 ) d p = = × ×
Area covered = 1963km
2

Ans7: Modulation index is the ratio of amplitude E m of caries wave to the amplitude E c of
carries (original) wave.
i.e.
m
c
E
E
µ =
Here Maximum Amplitude
c m
a E E = +
Minimum Amplitude
c m
b E E = - and
2 2
c m
a b a b
E E
+ - ? = =
?
a b
a b
µ
- =
+
Page 2

CBSE TEST PAPER-01
CLASS - XII PHYSICS (Communication System)
Ans1: Digital communication.
Ans2: Modulation is done because low frequency signal cannot be transmitted to a longer
distance so in order to increase the range of transmission modulation is done.
Ans3: The transmitting and receiving antenna must be in the line of sight.
Ans4:  It means that signal jumps form one level to another in no time so its frequency will
be infinite.
Ans5:  Sky wave propagation is due to the reflection of radio waves by the ionosphere but
high frequency waves gets absorbed by the ionosphere and cannot be reflected by
the ionosphere.
Ans6: 2 d rh =
3
2 6250 50 10 d = × × ×
4
2.5 10 d m = ×
Area covered
2 4 2
3.14 (2.5 10 ) d p = = × ×
Area covered = 1963km
2

Ans7: Modulation index is the ratio of amplitude E m of caries wave to the amplitude E c of
carries (original) wave.
i.e.
m
c
E
E
µ =
Here Maximum Amplitude
c m
a E E = +
Minimum Amplitude
c m
b E E = - and
2 2
c m
a b a b
E E
+ - ? = =
?
a b
a b
µ
- =
+
Ans8: Here h 1 = 80m
1 1
2 2 1
2
2 2 80 160
2 3
2 3 160
d h R R R
d h R d
h R R
= = × × =
= =
=
?
Ans9: E m = 0.5 E c
E max = E m + 0.5 E c = 1.5 E c
E min = E m - 0.5 E c = 0.5 E c
max min c c
max min c c
E - E 1.5 E - 0.5 E
E + E 1.5 E +0.5 E
µ = =
Ans10: 3.2 fm KHz =
34
96
fc MHz
KHz d
=
=
(a) Frequency modulated index
96
30
3.2
mf
fm
d
= = =
(b) Frequency range of the modulated wave
3
84 10 3.2
=83.997MHz to 84.003MHz
fc fm
KHz
= ±
= × ±
h
2
= 720m
0 .5 µ =
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