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 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

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%

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 × 𝟏𝟎𝟒𝒕 Volt

Comparing with standard equation 𝒗 = 𝑬𝒄 𝟏 +𝒎𝒄𝒐𝒔𝝎𝒔𝒕 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

…… 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

2Ec

𝑚𝐸𝑐

20.6 ×5

25

0.6 ×5

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 – 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 – 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 – 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

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 – 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 – 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 ( 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

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)

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 = 105/2π =15923.56 Hz

( b ) fm1 = 4000/2π =636.95 Hz

( c) carrier frequency = 107 / 2π = 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 – 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