Solved numericals on Am of the robaotics and artificial intelligence subject which is a popular subject nowdays.pptx

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1. RAI , SEM V (Solved Numerical) Modulation techniques Prepared By: Dikshant Sharma MO D U LAT I ON T E C H N I QUE S 1

2. MO D U LAT I ON T E C H N I QUE S 2 1. If the maximum and minimum voltage of an AM wave are V max and V min respectively. Show that the modulation factor is 𝒎 = 𝑽 𝒎𝒂𝒙 − 𝑽 𝒎𝒊𝒏 𝑽 𝒎𝒂𝒙 + 𝑽 𝒎𝒊𝒏 Solution : The amplitude of the normal carrier wave is ; 𝑬 𝒄 = 𝑉 𝑚𝑎𝑥 + 𝑉 𝑚𝑖𝑛 2 If E s is signal amplitude then its magnitude is given by ; 𝑬 𝒔 = 𝑉 𝑚𝑎𝑥 − 𝑉 𝑚𝑖𝑛 2 According to definition ; E s = m E c 𝑚 = 𝐸 𝑐 𝐸 𝑠 = 𝑉 𝑚𝑎𝑥 − 𝑉 𝑚𝑖𝑛 𝑉 𝑚𝑎𝑥 + 𝑉 𝑚𝑖𝑛

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 V max = 12/ 2 = 6 mV Minimum voltage of AM wave is = V min = 4/2 = 2 mV. 6 + 2 MO D U LAT I ON T E C H N I QUE S 3 Modulation factor 𝑚 = 𝑉 𝑚𝑎𝑥 − 𝑉 𝑚𝑖𝑛 = 6−2 = 0.5 𝑉 𝑚𝑎𝑥 + 𝑉 𝑚𝑖𝑛 Percentage modulation is 50%

4. 3. An AM wave is represented as ; 𝒗 = 𝟓 𝟏 + 𝟎. 𝟔 𝒄𝒐𝒔𝟔𝟐𝟖𝟎𝒕 𝒔𝒊𝒏𝟐𝟏𝟏 × 𝟏𝟎 𝟒 𝒕 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? V olt S o lutio n : The Am wa v e equ a ti o n; 𝒗 = 𝟓 𝟏 + 𝟎 . 𝟔 𝒄𝒐𝒔𝟔 𝟐 𝟖 𝟎 𝒕 𝒔𝒊𝒏𝟐𝟏𝟏 × 𝟏𝟎 𝟒 𝒕 Comp a ring with stand a rd equ a ti o n 𝒗 = 𝑬 𝒄 𝟏 + 𝒎 𝒄𝒐𝒔 𝑚 𝒔 𝒕 𝒔𝒊 𝒏 𝑚 𝒄 𝒕 E c = 5V , m = 0.6 , f s = 𝒔 𝑚 𝟔𝟐𝟖𝟎 c = = 𝟏 𝑲𝑯 𝒛 , f = 𝑚 𝒄 𝟐 𝝅 𝟐 𝝅 𝟐 𝝅 𝟐𝝅 MO D U LAT I ON T E C H N I QUE S 4 = 𝟐𝟏𝟏×𝟏𝟎 𝟒 = 𝟑𝟑𝟔 𝑲𝑯𝒛 Minimum amplitude of Am wave = E c – mE c = 5 – 0.6 x 5 = 2 V Maximum amplitude of the AM wave = E c + mE c = 5 + 0.6 x 5 = 8 V

5. …… CNTD. MO D U LAT I ON T E C H N I QUE S 5 The Am wave will con t ain th r ee f requenc i es f c – f s f c f c + f s 336 - 1 336 336 + 1 = 335 KHz = 336KHz = 337 KHz The amplitudes of the three components are 𝑚𝐸 𝑐 2 E c 𝑚𝐸 𝑐 2 0.6 ×5 2 5 0.6 ×5 2 1.5 V = 5V 1.5 V

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. MO D U LAT I ON T E C H N I QUE S 6 Solution : Frequency of carrier f c = 500 KHz , Frequency of signal f s = 5 KHz Modulation factor m = 0.5, Amplitude of carrier E c = 100V The lower side bands ( LSB) and upper side band ( USB) frequencies are; f c – f s and f c + f s ie 495 KHZ and 505 KHz

7. 5. An audio signal of 1 KHz is used to modulate a carrier 500 KHZ. Determine side band frequencies and band width required. MO D U LAT I ON T E C H N I QUE S 7 Carrier frequency = 500 KHz Signal frequency = 1 KHZ LSB = 499 KHZ and USB = 501 KHz Band width required = 501 – 499 = 2 KHz.

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 = [ f c – f m(max) ] to f c = [ 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 = f c to [ f c + f m(max) ] = 100 KHz to [ 100 + 5] KHz = 100 KHz to 105 Khz MODULATION TECHNIQUES 8

9. MO D U LAT I ON T E C H N I QUE S 9 ( b) the band width is equal to = 2 f m(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. MO D U LAT I ON T E C H N I QUE S 10 7. One input to AM modulator is 500 KHZ carrier with an amplitude 20V p The second input is a 10KHZ signal with amplitude to cause a change in out put wave of 7.5V p . 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 f usb = 500 + 10 = 510 KHz f LSB = 500 – 10 = 490 KHz ( b) modulation coefficient , m = 7.5 / 20 = 0.375 Percentage modulation = 37.5% E USB = ( c) The peak amplitude of the carrier and the upper and lower side band frequencies is; E LSB = mE c / 2 = [ (0.375) ( 20) ]/ 2 = 3.75 V (d) The maximum and minimum amplitudes of the envelop are ; V max = E c + E m = 20 + 7.5 = 27.5 V V min = E c - E m = 20 - 7.5 = 12.5 V

11. 8. A modulating signal 10sin ( 2π x 10 3 t) is used to modulate a carrier signal 10 sin( 2π x10 4 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 10 3 t) therefore comparing with the standard equation E m = 10V f m = I KHz , E c = 10V , f c = 10Khz. Modulation index m = E m / E c = 10/10= 1 ; Percentage modulation = 100% Frequencies of sideband components; f USB = 10 + 1 = 11 KHz f LSB = 10 – 1 = 9 KHz. Amplitude of side band frequencies = (mE c ) / 2 = ( 0.5 x 10) / 2 = 2.5V 11 MO D U LAT I ON T E C H N I QUE S

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 12 MO D U LAT I ON T E C H N I QUE S Solution ; X( t) = 20 cos ( 2πt) therefore f m = 1 HZ and E m = 20 V C( t) = 50 cos ( 100 πt) therefore f c = 50 Hz and E c = 50 V Modulation index m = 0.75 Expression for the Am wave is ; s(t) = E c [ 1 + m cos(2πt)]cos(2πf c t) = 50 [ 1 + 0.75 cos(2πt)]cos(100πt)

13. 10. In Am modulating signal frequency is 100KHz and carrier frequency is 1MHz. Determine the frequency components of side bands. MO D U LAT I ON T E C H N I QUE S 13 Solution: f c = 1 MHz = 1000 KHz , f m = 100 Kh F Usb = 1MHZ + 100KHz = 1000 KHz + 100 KHz = 1100 Khz= 1.1 MHz F LSB = 1MHZ - 100KHz = 1000 KHz -100 KHz = 900 Khz= 0.9 MHz

14. 11. An AM wave is expressed as ; 10( 1+ 0.5cos 10 5 t + 0.2 cos 4000t) cos 10 7 t. List the different frequency components MO D U LAT I ON T E C H N I QUE S 14 Solution: ( a) f m1 = 10 5 /2π =15923.56 Hz ( b ) f m1 = 4000/2π =636.95 Hz ( c) carrier frequency = 10 7 / 2π = 1592356.68 Hz =15.92 MHz.

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. MO D U LAT I ON T E C H N I QUE S 15 Solution index m = E m / E c = 3 /10 = 0.3 Amplitude of the side bands = (mE c ) /2 = (0.3)(10) /2 = 1.5 V ( f c + f m ) = 10000KHz + 10 Khz = 10010Khz ( f c -f m ) = 10000Khz – 10 Khz = 9990 Khz.

16. THE END MO D U LAT I ON T E C H N I QUE S 16