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10. FM Generation and Transmission Angle modulation FM analysis, Bessel function Bandwidth of FM waves Noise suppression Direct & Indirect FM generation TE4201-Communication Electronics 1 FM Wave Peak of intelligence wave Peak amplitude of AM wave Modulating intelligence wave Difference between AM Wave and FM Wave Peak frequency of FM wave TE4201-Communication Electronics 2 FM analysis, Bessel function FM Analysis PM e A sin wct mp sin wi t FM e A sin wct mf sin wit where : e= instantaneous voltage A= peak value of the carrier wave wc= carrier angular velocity (2pfc ) mp= maximum phase shift caused by intelligence signal amplitude wi= modulating(intelligence signal angular velocity (2pfi ) d = maximum frequency deviation (shift ) caused by intelligence signal amplitude mf = FM modulation index = ratio of maximum freq. deviation of the carrier to the intelligence frequency mf TE4201-Communication Electronics d fi 3 Simple FM Generator capacitor microphone FM radio waves sound waves Oscillator C C + DC C - DC 1 L fO L fL L fH 2 p LC 1 2 p L( C DC ) 1 2 p L( C DC ) Sound wave will cause capacitor microphone to change it’s capacitance by DC FM wave varies from fL to fH with fO at the center (when there is no sound waves) TE4201-Communication Electronics 4 FM Equation FM e A sin wct mf sin wit d where mf f i sine of sine Bassel function solution of the FM equation is fc ( t ) J0 ( mf ) cos wct J1 ( mf ) cos wc wi t cos wc wi t J2 ( mf ) cos wc 2 wi t cos wc 2 wi t J3 ( mf ) cos wc 3 wi t cos wc 3 wi t fc ( t ) FM wave component J0 ( mf ) cos wct carrier component J1 ( mf ) cos wc wi t cos wc wi t first sideband fi component J2 ( mf ) cos wc 2 wi t cos wc 2 wi t sec ond sideband 2 fi component 2 4 6 mf mf mf 2 2 2 mf 1 where JN ( mf ) n ............... n ! 1 ! n 1 ! 2 ! n 2 ! 3 ! n 3 ! 2 TE4201-Communication Electronics 5 Solution of Bassel function (chart) TE4201-Communication Electronics 6 Solution of Bassel function (graph) 2fi fC fi 2fi TE4201-Communication Electronics 7 Bandwidth of FM waves mf = 0.5 sin ce mf mf = 2.5 BW 4fi1 d fi then mf fi d BW 10fi2 If d is fixed the mf x fi is constant . For small fi , mf will be large, and more side frequencies Example If maximum deviation of the FM wave is fixed at d 20kHz. Determine the BW required for an intelligence of fi = 10kHz. mf = d / fi 20kHz/10kHz = 2 …..from table pair of important sideband is 4pairs (J0 to J4) or 8 side frequencies. BW required for an intelligence of fi = 10kHz is 8x10kHz = 80kHz Example If maximum deviation of the FM wave is fixed at d 20kHz. Determine the BW required if the intelligence is changed to fi = 5kHz. mf = d / fi 20kHz/5kHz = 4 …..from table pair of important sideband is 7pairs (J0 to J7) or 14side frequencies. BW required for an intelligence of fi = 5kHz is 14x5kHz = 70kHz TE4201-Communication Electronics 8 Carson’s rule Approximate BW prediction of a FM wave can be made by Carson’s rule……. BW = 2 (dmax + fi max ) Example If maximum deviation of the FM wave is fixed at d 20kHz. Determine the BW required for an intelligence signal having a higher cutoff frequency at fi max = 10kHz. BW = 2 (dmax + fi max ) = 2 (20+ 10 )kHz = 60kHz Note from important side frequency of Bessel function is BW= 80kHz. ( If neglecting sidebands having an amplitude less than 10% of maximum value of fundamental amplitude component (unmodulated carrier J0 = 1) then there will be 6 side frequencies (J0 to J3 ) as J4 = 0.03 can be neglected. Then BW = 6x10kHz = 60kHz) TE4201-Communication Electronics 9 Example Instantaneous voltage of the FM wave is e (t) = 2000 sin ( 2p x108 t+ 2 sin 2p x104t ) It is applied to an antenna having an impedance of 50W. Determine: (a) Carrier frequency fC (b) Transmitted power PT (c) mf , (d) fi (e) BW (by two methods) (f) Power in the largest and the smallest sidebands predicted by the table of Bessel function . (a) Carrier frequency fC = 108Hz =100MHz (b) Transmitted power PT = (E (c) rms )2/ R = (2000/1.414 )2/ 50W = 40kW mf = 2 (from 2 sin p x104t ) (d) Information frequency (from 2 sin p x104t ) fi = (104/2)kHz = 5kHz (e) (Carson’s rule) d = mf x fi =2x5k=10kHz then BW = 2 (dmax + fi max ) = 2 (10+ 10 )kHz = 40kHz Bessel function) table, 4 pair (or 8 side frequencies, J0 to J4 ) then BW = 8x5kHz = 40kHz (f) (Bessel function) table largest SB2 is 58% of unmodulated carrier J0 = 1, then PSB2 = (0.58E rms )2/ R = (0.58x2000/1.414 )2/ 50W = 13.5kW smallest SB4 is 3% of unmodulated carrier J0 = 1,then PSB4 = (0.03E (0.03x2000/1.414 )2/ 50W = 0.036kW=36W TE4201-Communication Electronics rms )2/ R = 10 Broadcast FM (entertainment) Channel width 200 kHz Channel width 200 kHz Channel width 200 kHz Carrier1 Carrier2 100MHz +75kHz Maximum deviation d = 75kHz -75kHz 2x25kHz guard band Carrier accuracy 2kHz Channel width 200 kHz 0.1MHz 0.1MHz 100.1MHz 100.2MHz Channel width 200 kHz 100.4MHz d= 75kHz = mf x fi = fixed to a maximum value. Therefore mf decreases if fi increases (at high intelligent frequency no. of sideband is less because of small mf ) TE4201-Communication Electronics 11 Narrowband FM (communications) Narrowband FM is widely used in communications such as police, aircraft, taxicabs, whether service, and private industry networks. Often voice transmission whose highest frequency allowed is fi = 3kHz. Narrowband FM has a (FCC) BW allocation of 10kHz to 30kHz. Narrowband FM has a modulation index mf (or a deviation ratio) of (10kHz/3kHz)=3.3 to (30kHz/3kHz)= 10 compared to (75kHz/15kHz) = 5 in Broadcast FM In FM, the waveform amplitude never varies just frequency. Therefore the transmitted power must remain constant regardless of level of modulation. It is thus seen that whatever the energy is contained in the side frequencies has been obtained from the carrier. No additional energy is added during the modulation process. In addition amplitude of the carrier in FM depends upon intelligence signal TE4201-Communication Electronics 12 Example (a) Determine the permissible range in maximum modulation index for commercial FM that has 30Hz to 15kHz modulating frequencies. (b) Repeat for a narrowband system that allows a maximum deviation of 10kHz and 100Hz to 3kHz modulating frequencies. d 75 kHz d 75 kHz ( a ) mf (max) 2500 to mf (min) 5 fi 30 Hz fi 15 kHz d 10 kHz d 10 kHz ( b ) mf (max) 100 to mf (min) 3.3 fi 100 Hz fi 3 kHz Example Determine the relative total power of the carrier and side frequencies when mf = 0.25 for a 10kW FM transmission at mf 0.25 , J0 0.98 , J1 0.12 and for 10 kW transmitting power , Pcarrier ( 0.98 ) 2 10 kW 9.604 kW and Psideband ( 0.12 ) 2 10 kW 0.144 kW Ptotal Pcarrier Psideand 9.604 0.144 9.892 kW 10 kW TE4201-Communication Electronics 13 Noise suppression FM limiter and detector Noise absent at AM detector output External noise AM limiter and detector Noise at AM detector output The most important advantage of FM over AM is the superior noise characteristics. Static noise is rarely heard on FM although it is quite common in AM. In FM intelligence is not carried by amplitude changes but instead by frequency changes. The spikes of external noise picked up during transmission are clipped off by a limiter circuit and through detector circuits insensitive to amplitude changes. Unfortunately the noise spike still causes an undesired phase shift of the FM signal, and this frequency shift cannot be removed. TE4201-Communication Electronics 14 FM noise analysis If FM receiver is tuned to 96MHz, the receiver provide gain only for frequencies near 96MHz. The noise also should be around this frequency since all other frequencies are greatly attenuated. Adding the noise to the desired signals will give a resultant signal with a different phase angle then the desired FM signal alone. Since the noise can cause phase modulation, it indirectly causes an undesired FM. Value of frequency deviation caused by phase modulation is : d = f x fi Where d = frequency deviation, f = phase shift (radians), fi = intelligence frequency Consider the noise signal to be one-half of the amplitude of the desired signal. S/N = 2 intolerable in AM. But in FM it is not so bad as shown by the calculations below. Since the noise “N” and signal “S” are at different frequencies (but in the same range, as dictated by receiver tuned circuits) the noise is a rotating vector with “S” as reference. The resultant phase “ f ” is maximum when “N” is perpendicular to resultant vector “R”. Then f = sin-1 (N/S) = sin-1 (1/2) = 30deg = 0.52 rad. and maximum possible frequency deviation is d = f x fi = 0.52 x 15kHz = 7.5kHzR N f S TE4201-Communication Electronics 15 FM S/N is better than AM S/N In standard broadcast,maximum deviation is 75kHz at full intelligence signal amplitude. Now the deviation from noise is 7.5kHz then S/N of the FM detector will be (75kHz/7.5kHz) = 10 = S/N which is better than AM whose S/N = 2 Example Determine the worst case output S/N for a broadcast FM program that has a maximum signal frequency of 5kHz. The input is S/N = 2 f sin -1 (N/S) sin -1 (1/2) 30deg 0.52 rad d f x fi 0.52 x 5kHz 2.5kHz S 75 kHz S ( FM out ) 30 ( FM in ) 2 N 2.5 kHz N Example Determine the worst case output S/N for a narrowband FM communication that has a maximum deviation of d = 10kHz and maximum signal frequency of 3kHz. The input is S/N = 3 f sin -1 (N/S) sin -1 (1/3) 19.5deg 0.34 rad d f x fi 0.34 x 3kHz 1 kHz S 10 kHz S ( FM out ) 10 ( FM in ) 3 N 1 kHz N TE4201-Communication Electronics 16 Direct FM generation Reactance Modulator Varicap Modulator • Reversed bias capacitance of the Varactor Diode D2 is in parallel with tuned capacitor C1 . C of D2 change due to change of reverse bias by the modulating signal. Therefore output freq. will change due to change of C of D2 TE4201-Communication Electronics VCO Modulator FM out Crosby Modulator FM modulating signal C1 L1 Modulating intelligence signal Varying reverse bias of Varactor changes the Varactor capacitance Carrier fC oscillator Varactor diode D2 17 Reactance Modulator (equivalent capacitance) C R e e g i1 R but i1 R jXc iD i1 eg e e R R jXc g e R id gme g m R jXc Ceq= CRgm Z V gm gm0 1 GS VP eg e R jXc jXc e 1 id gm e R gm gm R If R is chosen to be R Xc jXc 1 1 Then Z j Then Ceq CRgm gm R wCRgm jwCeq • When modulating signal is connected to Gate of FET, VGS will change. This will make gm to change. Then Ceq = CRgm will change. • If terminal “e” is connected to the LC tuned circuit of the carrier frequency oscillator, then carrier frequency will change producing FM wave. TE4201-Communication Electronics 18 Reactance Modulator (equivalent inductance) Example R Show that the FET reactance circuit shown will produce an equivalent reactance of Leq =(CR/gm) e g i1 ( jXC ) C i1 If e eg e R jXC id gm e g z iD i1 eg Leq=(CR/gm) e ( jXC ) R jXC jXC ( gm e ) R jXC R jXC jwCR e e 1 jXC ( gme ) id jXC gm gm gm R jXC jwCR jwCR 1 CR then z jwLeq Where Leq gm gm gm gm TE4201-Communication Electronics 19 FET Reactance Modulator Reactance FM modulator VDD RFC FM C modulating signal C0 i1 iD L0 R DC0 modulating signal FET reactance circuit will provide changing equivalent inductance proportional to modulating signal amplitude. As it is in parallel with L0C0 tuned oscillator (for carrier frequency), the carrier frequency will deviate according to dL produced by the FET reactance circuit, creating an FM wave. TE4201-Communication Electronics 20 FM output BJT Reactance Modulator R6 Master oscillator R2 FM C C0 modulating signal • When modulating signal is connected to Base of BJT, IB will change. This will make bib to change. If current generator of BJT having an equivalent in such a way that bib = gmeg where gm=(bib )/eg = b/ bre =1/re , then Ceq = CRgm = CR/re will change. Note that re (=26mV/IC) will change with IC (= bIB) TE4201-Communication Electronics R5 BJT Ractance modulator L0 R R3 R4 DC0 VCC C3 modulating signal • When terminal “DC0=Ceq” is connected to the L0C0 tuned circuit of the carrier frequency oscillator, then carrier frequency will change resulting an FM wave. 21 Linear IC ,Voltage Controlled Oscillator (VCO) FM generation) A voltage controlled oscillator (VCO) produces an output frequency that is directly proportional to a control voltage level. The circuitry necessary to produce such an oscillation with a high degree of linearity between control voltage and frequency was formerly productive on a discrete component bias. But now low cost monolithic LIC VCO are available, they make FM generation extremely simple. Voltage Controlled Oscillator TE4201-Communication Electronics 22 V+ R1 modulating input 6 Schmitt trigger 566 VCO C1 3 V VC V 4 8 Current source VC 5 2 V VC f0 Note R1 C1V 7 Buffer amplifier 3 Buffer amplifier 4 1 The figure shown is a 566VCO IC. It provide a high-quality FM generator with the modulating voltage applied at VC terminal. The FM output can be taken as a square or a triangular waveform from the IC. Feeding of any of these two outputs into an LC tank circuit resonant at the center frequency of the VCO ( = carrier frequency of FM) subsequently provide standard FM signal by flywheel effect. TE4201-Communication Electronics 23 CROSBY Modulator Audio input Reactance modulator fC =5 MHz fC =30 MHz fC =90 MHz Df = 4.167 Df = 25 kHz Df = 75 kHz kHz (max) (max) (max) Primary Frequency Frequency oscillator Tripler and Tripler 5MHz Doublers Discrimina 2 MHz tor Frequency stabilization circuit Crystal oscillator 14.67 MHz Frequency Converter (Mixer) 88 MHz Frequency Tripler and Doublers Power Amplifier 90 MHz To Antenna The mixer shown has the 90MHz carrier and 88MHz crystal oscillator signal as inputs. The mixer output only accepts the difference component of 2MHz. Which is fed to the discriminator. A Discriminator is the opposite of a VCO, in that it provides a dc level output based upon the frequency input. Discriminator output is zero dc if it has an input of exactly 2MHz which occurs when the transmitter is at precisely 90MHz. Any carrier drift up or down causes the discriminator output to go positive or negative, resulting in the appropriate primary oscillator readjustment. TE4201-Communication Electronics 24 Frequency Doublers fC =5 MHz Df = 4.167 kHz (max) High-distortion amplifier (class C) fC =10 MHz Df = 8.334 kHz (max) 1 Tuned to f C 10MHz 2 p LC Frequency Multipliers (doublers,triplers, etc.) are high distortion amplifier which will produce high harmonic content at the output. Then the output is tuned to the desired frequency multiple of the input frequency. In the above example shown above, a 5MHz carrier having a frequency deviation of Df=4.167kHz enters the class C high distortion amplifier. Tank LC circuit of the amplifier output is tuned to 10MHz. Note that Df=4.167kHz is also doubled to 8.334kHz. The reason is as follows: Suppose a 1 MHz deviating +0.1MHz enters the amplifier. The output will double of 10.1 x2 = 20.2 = a carrier of 2MHz and a double deviation of 0.2MHz. TE4201-Communication Electronics 25 Frequency Converters fC =5 MHz Df = 0.1MHz frequency converter Mixer fC =10 MHz Df = 0.1MHz 1 Tuned to f C 10MHz 2 p LC Oscillator fo =5 MHz It is very important to note that in the case of a frequency converter only the center frequency is changed, not the side frequency. When 5.1MHz is mixed with 5MHz oscillator, the output will be 5.1+5=10.1MHz meaning that the deviated frequency remains the same after the frequency conversion. TE4201-Communication Electronics 26 Indirect FM generation Wideband Armstrong FM FM output Audio input 400kHz Crystal Oscillator fC =400kHz RF Power amplifier Preemphasis Balanced modulator 90deg phase shifter fC =90.1MHz fd =75kHz + RF amplifier x81 first Freq. multipliers fC =400kHz fd =14.47Hz fC =32.4MHz fd =1.172Hz x64 final Freq. multipliers fC =1.41MHz fd =1.172Hz Mixer f0=33.81MHz Oscillator Wideband Armstrong FM system • Mixer is responsible for achieving the final carrier from the basic crystal oscillator frequency • Frequency multipliers are responsible for achieving the final deviation frequency from the small deviation at the balanced modulator 27 TE4201-Communication Electronics Design of mixer freq. and multiplying factor of freq. multiplier 1. Given: final= 88MHz , 75kHz, final multiplier= x64, before first xlier is carrier fC = 400k and fd=14.47Hz 2. then before multiplying fC =(88MHz/64) =1.375MHz, fd = (75kHz/64) =1.172kHz 3. before first xlier is fd=14.47Hz then first xlier is (1.172k/14.47)=81 4. before first xlier carrier fC =400k then after first xlier 400kx81=32.4MHz, then f0 =32.4+1.375=33.81MHz TE4201-Communication Electronics 28 HW: A Wideband Armstrong FM system is to be transmitted at 88MHz. with maximum deviation of 75kHz. A 400kHz crystal oscillator is used with a small deviation of 15Hz.at balanced modulator output. The final multiplier has a multiplication factor of 64. (a) Design the freq. of Oscillator and the freq. multiplying factor of the multiplier before the mixer. Note that a mixer will not change the deviation while the carrier is changed and that a multiplier will change both carrier and deviation freq. (b) If the crystal available is 410kHz, what changes will you make to the above FM system? TE4201-Communication Electronics 29