6mmmm

March 26, 2018 | Author: HusamHasko | Category: Frequency Modulation, Modulation, Phase (Waves), Bandwidth (Signal Processing), Hertz


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Angle Modulation1 Module outcome After this module, you should be able to: Describe and explain the differences between AM and angle modulation schemes and advantages and disadvantages of each. Describe and explain the differences between frequency and phase modulation and show the relationship between the two. 2 Introduction Three parameters can modulated with information signal  Amplitude  Frequency  Phase Phase and frequency changes are related, so we group them together in term angle modulation 3 Introduction In FM, frequency of modulated signal varies with amplitude of baseband signal. In PM, phase of modulated signal varies with amplitude of baseband signal. In angle modulation amplitude of carrier signal doesn’t change with modulation. 4 Introduction FM is more familiar in our daily life: Radio broadcast Sound signal in TV Mobile radio system Satellite communications Cellular telephone systems. PM is less familiar and mostly used in data communication 5 . There are two methods to vary the angle linearly with message signal  Frequency modulation (FM)  Phase modulation (PM) 6 . s (t )  Ec sin  2 f c t   0   Ec sin  (t )  (t )  2 f c t   0 for simplicity we consider 0 = 0.Basic Definitions Let the unmodulated carrier a sinusoidal waveform. f sig (t )  f c  k f em (t )  f c  f fsig(t) = signal frequency as a function of time fc = unmodulated carrier frequency kf = frequency sensitivity of modulator in hertz per volt em(t) = modulating signal f = instantaneous frequency deviation 7 . frequency of modulated signal varies with amplitude of baseband signal.Frequency Modulation (FM) In FM. frequency deviation is f sig (t )  f c   sin m t 8 .Frequency Modulation (FM) If the modulating signal is a sine wave em (t )  Em sin m t then f sig (t )  f c  k f Em sin m t The peak frequency deviation (Hz) will be   k f Em Thus. The frequency deviation formula can be rewritten as f sig (t )  f c  m f f m sin m t 9 .Frequency Modulation (FM) Modulation Index Modulation index mf for sine wave is mf  k f Em fm   fm mf has no theoretical limits and can exceeds one. Frequency Modulation (FM) Waveform f sig (t )  f c  k f Em sin m t - Maximum + deviation 10 . 94 MHz 11 .Frequency Modulation (FM) Example An FM modulator has kf = 30 kHz/V and operates at a carrier frequency of 175 MHz. Find the output frequency for an instantaneous value of the modulating signal equal to: a) 150 mV b) –2V (a) f sig   175  106 Hz    30  103 Hz / V   150 10 -3V   175.94  106 Hz  174.0045MHz (b) f sig   175 106 Hz    30 103 Hz / V   -2V   174.0045 106 Hz  175. Frequency Modulation (FM) Example The same FM modulator as in previous example is modulated with 3 V sine wave.24V  127. Calculate the peak frequency deviation Solution: ac voltages are assumed RMS unless otherwise stated Em  3 2  4.2kHz 12 .24V   k f Em  30kHz / V  4. Find the modulation index for sinusoidal modulating signal with frequency of : (a) 15 kHz (b) 50 Hz mf   75kHz   5.00 f m 15kHz  75  103 Hz mf    1500 fm 50 Hz 13 .Frequency Modulation (FM) Example An FM broadcast transmitter operates at its maximum deviation of 75 kHz.  (t) is varied linearly with the message signal m(t)  (t )  2 f c t  k p em (t )   c  k p em (t ) c = 2fct is the angle of the unmodulated carrier. kp is the phase sensitivity of the modulator.Phase Modulation (PM) In PM. expressed in radian per volt. PM signal in the time domain is s(t )  Ec cos  2 f c t  k p em (t )  14 . Phase Modulation If modulation signal is a sinusoidal one. phase of the modulated carrier is given by  (t )   c  k p Em sin m t = c   (t )  (t) is the phase deviation in radian The peak phase deviation is defined as phase modulation index. given as m p  k p Em (rad) 15 . Phase Modulation Example A phase modulator has kp = 2 rad/V. What RMS voltage of a sine wave would cause a peak phase deviation of 600? m p  k p Em Em  mp kp VRMS   V peak 2   / 3 rad 2rad / V  0.524V  0.37V 16 .524 2  0. the instantaneous frequency is f sig (t )  f c  k f em (t ) Since t  (t )  2  f sig (t )dt 0 t t 0 0  (t )  2   f c  k f em (t ) dt 2 f c t  2 k f  em ( )d The FM signal is therefore   s (t )  Ec sin  2 f c t  2 k f  em ( )d 0   t 17 .Relationships between FM and PM With FM. FM signal may be regarded as a PM signal in which t em (t )   em ( )d 0 18 .Relationships between FM and PM The FM signal in time domain is given by   s (t )  Ec sin  2 f c t  2 k f  em ( )d 0   t = Ec sin  2 f c t  k p em (t ) Recalling PM signal in time domain s (t )  Ec sin  2 f c t  k p em (t )  Thus. Relationships between FM and PM Conversely. f sig (t )  f c  k f dem (t ) dt t  (t )  2  f sig (t )dt  2 f c t  2 k f em (t ) 0  2 f c t  k p em (t ) 19 . a PM signal can be generated by first differentiating em(t) and then using the result as the input to a frequency modulator. we concentrate our attention on FM signals.Relationships between PM and FM We may thus deduce all the properties of PM signals from those of FM signals and vice versa. Henceforth. 20 . mp or mf represent peak phase deviation from the phase of unmodulated carrier. For angle-modulated signal with sine wave. 21 .Relationship between mf and mp Either FM or PM results in changes in both frequency and phase of modulated waveform. What is the maximum phase shift that it produces?  mf  fm max  m f   5000   16.7rad fm 300 22 .Relationship between mf and mp Example An FM communications transmitter has maximum frequency deviation of 5 kHz and a range of modulating frequencies from 300 Hz to 3 kHz. Find the maximum frequency deviation it produces with a sine-wave input of 2 V peak at frequency of 1 kHz?  (t )  k p Em sin m t m p  max  k p Em  3rad / V  2V  6rad  max  m p  m f     m f f m  6  1kHz  6kHz fm 23 .Relationship between mf and mp Example A phase modulator has a sensitivity of kp= 3 rad/V. we consider single-tone modulation produces a narrow band FM signal. we consider single-tone modulation produces wideband FM signal. Next.FM Spectrum FM is nonlinear modulation so its spectrum is not related in a simple manner to that of modulating signal. To simplify FM spectral analysis we proceed in the following manner: First. that that 24 . frequency deviation is f sig (t )  f c   sin m t 25 .FM Spectrum If the modulating signal is a sine wave em (t )  Em sin m t then f sig (t )  f c  k f Em sin m t The peak frequency deviation (Hz) will be   k f Em Thus. the instantaneous frequency is f sig (t )  f c   sin m t Since t  (t )  2  f sig (t )dt 0 t  (t )  2   f c   sin(2 f m t )  dt 2 f c t  0  cos(2 f m t ) fm  2 f c t  m f cos(2 f m t ) 26 .FM Spectrum With FM. 27 . the FM signal is given by s (t )  Ec sin  2 f ct  m f cos(2 f mt ) Depending on the value of mf. for which mf is large compared to one. for which mf is small compared to one ( m f  1). Wideband FM. we may distinguish two cases: Narrowband FM.FM Spectrum Thus. we get  m f sin(2 f m t ) s (t )  Ec sin  2 f ct  cos  m f sin(2 f mt  Ec cos  2 f c t  sin Assuming mf is small compared to one radian.FM Spectrum Narrowband FM sin( x  y )  sin x cos y  cos x sin y The FM signal is given by s (t )  Ec sin  2 f ct  m f cos(2 f mt ) By expanding this relation. cos  m f sin  2 f mt   1 sin  m f sin  2 f mt   m f sin(2 f mt ) s (t )  Ec sin  2 f c t   m f Ec cos  2 f c t  sin(2 f mt ) 28 . FM Spectrum Narrowband FM Because s (t )  Ec sin  2 f c t   m f Ec cos  2 f ct  sin(2 f mt ) 1 cos x sin y   sin  x  y   sin  x  y  2 s (t )  Ec sin(2 f ct )  1 m f Ec  sin  2 ( f c  f m )t   sin  2 ( f c  f m )t   2 This expression is similar to AM signal. Thus for m f a1narrowband FM signal requires the same transmission bandwidth as the AM signal (2 fm) 29 . Sideband are separated from carrier by multiple of fm. Amplitude of side bands tends to decrease with their distance from carrier. Sidebands with amplitudes less than 1% of total signal voltage can be ignored.FM Spectrum Wideband FM For mf >> 1. FM produces infinite sidebands even for single tone. 30 . FM Spectrum Wideband FM FM Spectrum in this case can be expressed as a series of sinusoids using Bessel function of first kind. 31 . m  t .2m  t .FM Spectrum Wideband FM Based on Bessel functions s(t) can be rewritten as s (t )  A sin  c t  m sin m t   A{J 0 (m) sin c t .3m  t .sin  c  2m  t .sin  c  m  t  J 2 (m)  sin  c .J 3 (m)  sin  c .J1 (m)  sin  c .sin  c  3m  t L } 32 . FM Spectrum Wideband FM 33 . FM Spectrum Wideband FM 34 . power in each of sidebands are Power in whole signal will be A2 2 PT  J 0  2 J12  2 J 22 L  RL PSB1 J12 A2  RL  35 . Power at carrier frequency is J 02 A2 Pc  RL Similarly.FM Spectrum Wideband FM Let unmodulated carrier have a voltage of A volts RMS across a resistance of RL . power in each of sidebands are PSB1 J12 A2  2 RL Power in whole signal will be A2 PT  RL  J 02  2 2  J1  J 2  L   2   36 . Power at carrier frequency is J 02 A2 Pc  2 RL Similarly.FM Spectrum Wideband FM Let unmodulated carrier have a peak voltage of A volts across a resistance of RL . Carrier frequency is 160 MHz. Calculate the RMS signal voltage VT Calculate RMS voltage at carrier and first three sets of sidebands Calculate power at carrier and each of first three sidebands 37 . Its total power PT is 5 W.FM Spectrum Wideband FM An FM signal has deviation of 3 kHz and modulating frequency of 1 kHz. developed across a 50 Ω load. 49. J 3  0. J1  0. J 2  0.34.8V ( RMS ) RL Modulation index mf   3 fm From Bessel table we have J o  -0.FM Spectrum Wideband FM Signal power is constant with modulation.31 38 . thus VT2 PT   VT  PT RL  15.26. 85W 39 .9V .11V .74V .  P1   0.  P2  V3  J 3 VT  4.37V .8dBm RL PT  Pc  2( P1  P2  P3 )  4.2W  30.8dBm RL V32  0.FM Spectrum Wideband FM Vc  J 0 VT  4.338W  25.3dBm RL V12 V1  J1 VT  5.48W  26.  P3  V22  1.3dBm RL V2  J 2 VT  7.576W  27.  Pc  Vc2  0. For FM. 40 . The above two effects work in opposite directions making FM bandwidth to some extend constant.FM Spectrum Bandwidth For FM. situation is complicated by the fact  mf  fm Increase in fm will reduce mf and thus number of sidebands. bandwidth varies directly with fm as well as with mf. Increase in fm means further apart sidebands in frequency. FM Spectrum Bandwidth 41 . Using Cason’s rule calculate the bandwidth.FM Spectrum Bandwidth Carson’s Rule: The bandwidth of FM signal is given by  B  2   f m  max   Example: An FM signal has a deviation of 3 kHz and a modulating frequency of 1 kHz. B  2    fm   2  3kHz  1kHz   8kHz 42 . BW of an FM signal is generally limited by government regulations that specify: Maximum frequency deviation Maximum modulating frequency 43 .FM Specifications No theoretical limits to modulation index or frequency deviation of an FM signal. In general larger values for deviation result in increased S/N. FM transmitter 44 .
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