Analog numerical

1. 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

2. 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 𝒎 =
𝑽𝒎𝒂𝒙− 𝑽𝒎𝒊𝒏
𝑽𝒎𝒂𝒙+ 𝑽𝒎𝒊𝒏
Solution : The amplitude of the normal carrier wave is ;
𝑬𝒄 =
𝑉𝑚𝑎𝑥+ 𝑉𝑚𝑖𝑛
2
If Es is signal amplitude then its magnitude
is given by ;
𝑬𝒔 =
𝑉𝑚𝑎𝑥− 𝑉𝑚𝑖𝑛
2
According to definition ; Es = m Ec
𝑚 =
𝐸𝑠
𝐸𝑐
=
𝑉𝑚𝑎𝑥− 𝑉𝑚𝑖𝑛
𝑉𝑚𝑎𝑥+ 𝑉𝑚𝑖𝑛

3. 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 𝑚 =
𝑉𝑚𝑎𝑥− 𝑉𝑚𝑖𝑛
𝑉𝑚𝑎𝑥+ 𝑉𝑚𝑖𝑛
=
6−2
6+2
= 0.5
Percentage modulation is 50%

4. MODULATION TECHNIQUES 4
3. An AM wave is represented as ; 𝒗 = 𝟓 𝟏 + 𝟎. 𝟔 𝒄𝒐𝒔𝟔𝟐𝟖𝟎𝒕 𝒔𝒊𝒏𝟐𝟏𝟏 × 𝟏𝟎𝟒𝒕
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; 𝒗 = 𝟓 𝟏 + 𝟎. 𝟔 𝒄𝒐𝒔𝟔𝟐𝟖𝟎𝒕 𝒔𝒊𝒏𝟐𝟏𝟏 × 𝟏𝟎𝟒
𝒕 Volt
Comparing with standard equation 𝒗 = 𝑬𝒄 𝟏 + 𝒎 𝒄𝒐𝒔 𝝎𝒔𝒕 𝒔𝒊𝒏𝝎𝒄𝒕
Ec = 5V , m = 0.6 , fs =
𝝎𝒔
𝟐𝝅
=
𝟔𝟐𝟖𝟎
𝟐𝝅
= 𝟏 𝑲𝑯𝒛 , fc =
𝝎𝒄
𝟐𝝅
=
𝟐𝟏𝟏×𝟏𝟎𝟒
𝟐𝝅
= 𝟑𝟑𝟔 𝑲𝑯𝒛
Minimum amplitude of Am wave = Ec – mEc = 5 – 0.6 x 5 = 2 V
Maximum amplitude of the AM wave = Ec + mEc = 5 + 0.6 x 5 = 8 V

5. …… CNTD.
MODULATION TECHNIQUES 5
The Am wave will contain three frequencies
fc – fs fc fc + fs
336 - 1 336 336 + 1
= 335 KHz = 336KHz = 337 KHz
The amplitudes of the three components are
𝑚𝐸𝑐
2
Ec
𝑚𝐸𝑐
2
0.6 ×5
2
5
0.6 ×5
2
1.5 V = 5V 1.5 V

6. 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 – fs and fc + fs
ie 495 KHZ and 505 KHz

7. 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 – 499 = 2 KHz.

8. 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 – fm(max)] to fc
= [ 100 – 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

9. MODULATION TECHNIQUES
9
( 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 – 3 KHz = 97 Khz

10. 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 – 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

11. MODULATION TECHNIQUES 11
8. A modulating signal 10sin ( 2π x 103t) is used to modulate a carrier signal 10 sin(
2π 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 ( 2π 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 – 1 = 9 KHz.
Amplitude of side band frequencies = (mEc) / 2 = ( 0.5 x 10) / 2 = 2.5V

12. MODULATION TECHNIQUES 12
9. Consider the message signal X(t) = cos ( 2πt) volt and carrier wave
c(t) = 50 cos ( 100πt) Obtain an expression of AM for m = 0.75
Solution ; X( t) = 20 cos ( 2πt) therefore fm = 1 HZ and Em = 20 V
C( t) = 50 cos ( 100 πt) 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(2πt)]cos(2πfct)
= 50 [ 1 + 0.75 cos(2πt)]cos(100πt)

13. 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

14. 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 = 105/2π =15923.56 Hz
( b ) fm1 = 4000/2π =636.95 Hz
( c) carrier frequency = 107 / 2π = 1592356.68 Hz =15.92 MHz.

15. 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 – 10 Khz = 9990 Khz.

16. 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

17. THE END
MODULATION TECHNIQUES 17