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# Angle Mo dulation In this yp of mo dulation the frequency or phase of carrier is aried in prop ortion to the amplitude of the mo dulating signal

ct Figure 1 An angle mo dulated signal If cos is an angle mo dulated signal then 1 Phase mo dulation brPage 2br where 2 requency Mo dulation dt dt dt Phase Mo dulation If cos2 is the message signal then the phase mo dulated signal is giv en brPage

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## Angle Mo dulation In this yp of mo dulation the frequency or phase of carrier is aried in prop ortion to the amplitude of the mo dulating signal

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## Presentation on theme: "Angle Mo dulation In this yp of mo dulation the frequency or phase of carrier is aried in prop ortion to the amplitude of the mo dulating signal"â€” Presentation transcript:

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Angle Mo dulation In this yp of mo dulation, the frequency or phase of carrier is aried in prop ortion to the amplitude of the mo dulating signal. c(t) Figure 1: An angle mo dulated signal If cos )) is an angle mo dulated signal, then 1. Phase mo dulation:
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where 2. requency Mo dulation: dt dt dt Phase Mo dulation If cos(2 is the message signal, then the phase mo dulated signal is giv en
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cos )) Here, is phase sensitivit or phase mo dulation index. requency Mo dulation If cos(2 is the message signal, then the requency mo dulated signal is giv en

cos(2 sin (2 here, is called fr quency deviation ( and is called mo dulation index ). The requency mo dulated signal is giv en
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cos(2 sin(2 )) Dep ending on ho small is FM is either Narr owb and FM( << or Wideb and FM( Narro w-Band FM (NBFM) In NBFM << 1, therefor reduces as follo ws: cos(2 sin(2 )) cos(2 cos sin(2 )) sin(2 sin sin (2 )) Since, is ery small, the ab equation reduces to cos(2 sin (2 sin (2
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The ab equation is similar to AM. Hence, for NBFM the bandwidth is same as that of AM i.e., messag bandw idth (2 ). NBFM signal is generated as sho wn in

Figure ?? DSB−SC oscillator m(t) A cos( t) −Asin( t) NBFM signal /2 Phase shifter Figure 2: Generation of NBFM signal
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Wide-Band FM (WBFM) WBFM signal has theoritically innite bandwidth. Sp ectrum calculation of WBFM signal is tedious pro cess. or, practical applications ho ev er the Bandwidth of WBFM signal is calculated as follo ws: Let bandlimited to Hz and sampled adequately at Hz. If time erio is to small, the signal can appro ximated sequence of pulses as sho wn in Figure ??
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Figure 3: Appro ximation of message signal If tone mo

dulation is considered, and the eak amplitude of the sin usoid is the minim um and maxim um frequency deviations will and resp ectiv ely The spread of pulses in frequency domain will
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as sho wn in Figure ?? w - w + Figure 4: Bandwidth calculation of WBFM signal Therefore, total BW is and if frequency deviation is considered (2 2(
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The bandwidth obtained is higher than the actual alue. This is due to the staircase appro ximation of ). The bandwidth needs to readjusted. or NBFM, is ery small an hence is ery small compared to This implies But the bandwidth for NBFM

is the same as that of AM whic is etter bandwidth estimate is therefore: 2( 2( This is also called Carson ’s ule
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Demo dulation of FM signals Let an FM signal. cos d This signal is passed through dieren tiator to get )) sin d If observ the ab equation carefully it is oth amplitude and frequency mo dulated. Hence, to reco er the original signal bac an en elop detector can used. The en elop tak es the form (see Figure ?? ):
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nv el ope )) FM signal Envelope of FM signal Figure 5: FM signal oth Amplitude and requency Mo dulation The blo diagram of the demo

dulator is sho wn in Figure ??
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t ( ) F ( t) fm fm d/dt Detector Envelope A( + k m(t)) Figure 6: Demo dulation of an FM signal The analysis for Phase Mo dulation is iden tical. Analysis of bandwidth in PM
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)] max pm 2( pm 2( The dierence et een FM and PM is that the bandwidth is indep enden of signal bandwidth in FM while it is strongly dep enden on signal bandwidth in PM. wing to the bandwidth eing dep enden on the eak of the derivative of m(t) rather than itself
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Angle Mo dulation: An Example An angle-mo dulated signal with carrier

frequency 10 is describ ed the equation: 12 cos sin 1500 10 sin 2000 1. Determine the er of the mo dulating signal. 2. What is 3. What is 4. Determine the phase deviation. 5. Estimate the bandwidth of )? 1. 12 72 units 2. requency deviation need to estimate the instan taneous frequency:
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dt 500 cos 1500 20 000 The deviation of the carrier is 500 cos 1500 20 000 When the sin usoids add in phase, the maxim um alue will 500 20 000 Hence 11 193 66 3. 11 193 66 1000 11 193 4. The angle sin 1500 10 sin 2000 ). The maxim um angle deviation is 15, whic is the phase deviation. 5. 2( 24

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