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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 = 2fct 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 .2m 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 .3m t .sin c 2m t .sin c m t J 2 (m) sin c .J 3 (m) sin c .J1 (m) sin c .sin c 3m 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 .