AM Modulation Problem1 - Analog Modulation Chapter 2...

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Unformatted text preview: Analog Modulation Chapter 2 Example 2.4 For an AM modulator with a carrier frequency of 150 kHz and a modulating signal frequency of lOkHz, determine the : i) Frequency for the upper and lower sideband . ii) Bandwidth. Sketch the output frequency spectrum. Solution i ) The lower and upper side band frequency fLSB = fc - fm' = 150kHz — IOkHz = 140 kHz fUSB = fc + fm = 150kHz +10kHz = 160kHz ii ) Bandwidth BW = 2fm = 2 ( 10 ) kHz = 20 kHz The output frequency spectrum ”is as shown , Amplitude Vc BW = 20kHz ~31 Analog Modulation Chapter 2 e) Power content in AM The power in the AM signal is the total power in the carrier and the power in the sidebands, PT = Carrier power + LSB power + USB power PT = PC + PLSB + PUSB VLSB = VUSB = maVc (peak value ) Note that , 2 = maVc (r.m.s value ) 2x12 Therefore, PT = VC2/2R + meCZ/sR +ma2vC2/8R = Vc2/2R + manc2/4R = ch/2R(l+maZ/2) where ch / 2R =Pc, giving, PT = PC(1+ma2/2) (2.11) So, the total power in the AM wave depends on the carrier power and modulation index. Example 2.5 For an AM wave with a peak unmodulated carrier voltage Vc =20 V, a load resistance RL = 20 Q and a modulation index, ma— — 0. 2, determine; a) Power contained in the carrier and the upper and lower sidebands. b) Total sideband power 0) Total power of the modulated wave 32 Analog Modulation Solution ,- a) The carrier power is: Pc = (V0)2 The upper and lower sideband power is ; PUSB =PLSB : Ina2 ch = ma2 Pc 8R 4 = (0.2)2 (10) 4 = 0.1 b) Total sideband power, PSB = maZPC/z = (0.2)2(10)/2 = 0.2w 0f, PSB = PUSB + PLSB = 0.1 + 0.1 = 0.2 W e) The total power in the modulated wave , PT = Pc[1+ maZ/Z] =10[1+(0.2)2/2] = 10.2w 01', PT =Pc + P513 = 10 + 0.2 = 10.2w 33 Chapter 2 ...
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  • Summer '16
  • Hemant Patil

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