S. Y. B.SC. SEM IV
ELECTIVE PAPER III UNIT III
(Solved Numerical)
Modulation techniques
PREPARED BY
PROF. VISHWAS DESHMUKH
ASSISTANT PROFESSOR
DEPT. OF PHYSICS
MODULATION TECHNIQUES 1
RIZVI COLLEGE OF ARTS, SCIENCE & COMMERCE
OFF CARTER ROAD, BANDRA WEST
MODULATION TECHNIQUES 2
1. If the maximum and minimum voltage of an AM wave are Vmax and Vmin
respectively. Show that the modulation factor is π =π½πππβ π½πππ
π½πππ+ π½πππ
Solution : The amplitude of the normal carrier wave is ;
π¬π =ππππ₯+ ππππ
2
If Es is signal amplitude then its magnitude
is given by ;
π¬π =ππππ₯β ππππ
2
According to definition ; Es = m Ec
π =πΈπ
πΈπ=
ππππ₯β ππππ
ππππ₯+ ππππ
MODULATION TECHNIQUES 3
2. The maximum peak-to-peak voltage of an AM wave is 12 mV and
the minimum voltage is 4mV. Calculate the modulation factor.
Solution : Maximum Voltage of AM wave is Vmax = 12/ 2 = 6 mV
Minimum voltage of AM wave is = Vmin = 4/2 = 2 mV.
Modulation factor π =ππππ₯β ππππ
ππππ₯+ ππππ=
6β2
6+2= 0.5
Percentage modulation is 50%
MODULATION TECHNIQUES 4
3. An AM wave is represented as ; π = π π + π. π ππππππππ ππππππ Γ ππππ
volts. ( a) what are the minimum and maximum amplitudes of AM wave? ( b)
What frequency components are contained in the modulated wave and what is the
amplitude of each component?
Solution: The Am wave equation; π = π π + π. π ππππππππ ππππππ Γ ππππ Volt
Comparing with standard equation π = π¬π π +πππππππ ππππππ
Ec = 5V , m = 0.6 , fs = ππ
ππ =
ππππ
ππ = π π²π―π , fc =
ππ
ππ =
πππΓπππ
ππ = πππ π²π―π
Minimum amplitude of Am wave = Ec β mEc = 5 β 0.6 x 5 = 2 V
Maximum amplitude of the AM wave = Ec + mEc = 5 + 0.6 x 5 = 8 V
β¦β¦ CNTD.
MODULATION TECHNIQUES 5
The Am wave will contain three frequencies
fc β fs fc fc + fs
336 - 1 336 336 + 1
= 335 KHz = 336KHz = 337 KHz
The amplitudes of the three components are
ππΈπ
2Ec
ππΈπ
20.6 Γ5
25
0.6 Γ5
2
1.5 V = 5V 1.5 V
MODULATION TECHNIQUES 6
4. A sinusoidal voltage of frequency 500 KHZ and amplitude 100V is AM
modulated by Sinusoidal Voltage of frequency 5 KHz producing 50% modulation.
Calculate the frequency and amplitude of LSB and USB.
Solution : Frequency of carrier fc = 500 KHz , Frequency of signal fs = 5 KHz
Modulation factor m = 0.5, Amplitude of carrier Ec = 100V
The lower side bands ( LSB) and upper side band ( USB) frequencies are;
fc β fs and fc + fs
ie 495 KHZ and 505 KHz
MODULATION TECHNIQUES 7
5. An audio signal of 1 KHz is used to modulate a carrier 500 KHZ.
Determine side band frequencies and band width required.
Carrier frequency = 500 KHz Signal frequency = 1 KHZ
LSB = 499 KHZ and USB = 501 KHz
Band width required = 501 β 499 = 2 KHz.
MODULATION TECHNIQUES 8
6. For an AM modulator carrier frequency is 100 KHz and maximum
modulating signal frequency is 5 KHz. Determine ( a) frequency limits for the
upper and lower side bands. ( b) Band width ( c) Upper and lower side
frequencies produced when the modulating signal is a single frequency 3Khz
tone. (d) draw the out put frequency spectrum
Solution : (a) the lower sideband extends from the lowest possible lower side
frequency to the carrier frequency , LSB = [ fc β fm(max)] to fc
= [ 100 β 5]KHz to 100 KHz
= 95 KHz to 100 KHz
the upper sideband extends from carrier frequency to the lowest possible upper
side frequency, USB = fc to [ fc + fm(max)]
= 100 KHz to [ 100 + 5] KHz
= 100 KHz to 105 Khz
MODULATION TECHNIQUES9
( b) the band width is equal to = 2 fm(max) = 2 ( 5KHZ) = 10KHz
( c) The upper side frequency is the sum of carrier and modulating frequency
= 100 KHz + 3 KHz = 103 KHZ
The lower side frequency is the difference between carrier and modulating
frequency,= 100 KHz β 3 KHz = 97 Khz
MODULATION TECHNIQUES 10
7. One input to AM modulator is 500 KHZ carrier with an amplitude 20Vp The
second input is a 10KHZ signal with amplitude to cause a change in out put wave
of 7.5Vp. Determine ( a) Upper and lower side frequencies ( b) Modulation
coefficient and percentage modulation ( c) peak amplitude of the modulated
carrier and upper and lower side frequency voltages. ( d) maximum and
minimum amplitude of the envelope.
Solution : ( a) The upper and lower side frequencies are simply the sum and difference frequencies
fusb = 500 + 10 = 510 KHz
f LSB = 500 β 10 = 490 KHz
( b) modulation coefficient , m = 7.5 / 20 = 0.375
Percentage modulation = 37.5%
( c) The peak amplitude of the carrier and the upper and lower side band frequencies is; EUSB =
ELSB = mEc / 2 = [ (0.375) ( 20) ]/ 2 = 3.75 V
(d) The maximum and minimum amplitudes of the envelop are ;
Vmax = Ec + Em = 20 + 7.5 = 27.5 V
Vmin = Ec - Em = 20 - 7.5 = 12.5 V
MODULATION TECHNIQUES 11
8. A modulating signal 10sin ( 2Ο x 103t) is used to modulate a carrier signal 10 sin(
2Ο x104 t). Find the modulation index, percentage modulation, frequencies of
sideband components and their amplitudes and also determine the band width of
the modulating signal.
Solution : The modulating signal is; 10sin ( 2Ο x 103t) therefore comparing with
the standard equation Em = 10V fm = I KHz , Ec = 10V , fc = 10Khz.
Modulation index m = Em / Ec = 10/10= 1 ; Percentage modulation = 100%
Frequencies of sideband components;
fUSB = 10 + 1 = 11 KHz
fLSB = 10 β 1 = 9 KHz.
Amplitude of side band frequencies = (mEc) / 2 = ( 0.5 x 10) / 2 = 2.5V
MODULATION TECHNIQUES 12
9. Consider the message signal X(t) = cos ( 2Οt) volt and carrier wave
c(t) = 50 cos ( 100Οt) Obtain an expression of AM for m = 0.75
Solution ; X( t) = 20 cos ( 2Οt) therefore fm = 1 HZ and Em = 20 V
C( t) = 50 cos ( 100 Οt) therefore fc = 50 Hz and Ec = 50 V
Modulation index m = 0.75
Expression for the Am wave is ;
s(t) = Ec [ 1 + m cos(2Οt)]cos(2Οfct)
= 50 [ 1 + 0.75 cos(2Οt)]cos(100Οt)
MODULATION TECHNIQUES 13
10. In Am modulating signal frequency is 100KHz and carrier
frequency is 1MHz. Determine the frequency components of side bands.
Solution: fc = 1 MHz = 1000 KHz , fm = 100 Kh
FUsb = 1MHZ + 100KHz = 1000 KHz + 100 KHz = 1100 Khz= 1.1 MHz
FLSB = 1MHZ - 100KHz = 1000 KHz -100 KHz = 900 Khz= 0.9 MHz
MODULATION TECHNIQUES 14
11. An AM wave is expressed as ; 10( 1+ 0.5cos 105 t + 0.2 cos 4000t)
cos 107 t. List the different frequency components
Solution: ( a) fm1 = 105/2Ο =15923.56 Hz
( b ) fm1 = 4000/2Ο =636.95 Hz
( c) carrier frequency = 107 / 2Ο = 1592356.68 Hz =15.92 MHz.
MODULATION TECHNIQUES 15
12.A carrier frequency of 10 MHz with peak amplitude 10 v is
amplitude modulated by a 10 KHz signal of 3 V. Determine the
modulation index.
Solution index m = Em / Ec = 3 /10 = 0.3
Amplitude of the side bands = (mEc) /2 = (0.3)(10) /2 = 1.5 V
( fc + fm) = 10000KHz + 10 Khz = 10010Khz
( fc -fm) = 10000Khz β 10 Khz = 9990 Khz.
MODULATION TECHNIQUES 16
REFERENCES
1. PRINCIPALS OF ELECTRONICS ----- V.K.MEHTA
2. PRINCIPALS OF ANLOG & DIGITAL COMMUNICATION --- J.S KATRE
3. ELECTRONIC COMMUNICATION SYSTEMS ---- W. TOMASI
THE END
MODULATION TECHNIQUES 17