1. TE4201-Communication Electronics 1 10. FM Generation and Transmission Angle modulation Angle modulation FM analysis, Bessel function FM analysis, Bessel function Bandwidth of FM waves Bandwidth of FM waves Noise suppression Noise suppression Direct & Indirect FM generation Direct & Indirect FM generation Direct &Indirect FM generation Direct &Indirect FM generation

2. TE4201-Communication Electronics 2 FM Wave Modulating intelligence wave Difference between AM Wave and FM Wave Peak frequency of FM wave Peak amplitude of AM wave Peak of intelligence wave

3. TE4201-Communication Electronics 3 where : e= instantaneous voltage A= peak value of the carrier wave  c = carrier angular velocity (2  f c ) m p = maximum phase shift caused by intelligence signal amplitude  i = modulating(intelligence signal angular velocity (2  f i ) FM Analysis FMPM  = maximum frequency deviation (shift ) caused by intelligence signal amplitude m f = FM modulation index = ratio of maximum freq. deviation of the carrier to the intelligence frequency FM analysis, Bessel function

4. TE4201-Communication Electronics 4 Simple FM Generator C +  CLCLC -  CL FM wave varies from f L to f H with f O at the center (when there is no sound waves) Sound wave will cause capacitor microphone to change it’s capacitance by  C

5. TE4201-Communication Electronics 5 FM Equation FM where sine of sine Bassel function solution of the FM equation is

6. TE4201-Communication Electronics 6 Solution of Bassel function (chart)

7. TE4201-Communication Electronics 7 Solution of Bassel function (graph) 2f i fifififi fCfCfCfC

8. TE4201-Communication Electronics 8BW 4f i1 If  is fixed the m f x f i is constant. For small f i, m f will be large, and more side frequencies BW 10f i2 m f = 0.5 m f = 2.5 If maximum deviation of the FM wave is fixed at  20kHz.  Determine the BW required for an intelligence of fi = 10kHz. Example m f =  f i  20kHz/10kHz = 2 …..from table pair of important sideband is 4pairs (J 0 to J 4 ) or 8 side frequencies. BW required for an intelligence of f i = 10kHz is 8x10kHz = 80kHz If maximum deviation of the FM wave is fixed at  20kHz.  Determine the BW required if the intelligence is changed to fi = 5kHz. Example m f =  f i  20kHz/5kHz = 4 …..from table pair of important sideband is 7pairs (J 0 to J 7 ) or 14side frequencies. BW required for an intelligence of f i = 5kHz is 14x5kHz = 70kHz Bandwidth of FM waves

9. TE4201-Communication Electronics 9 Carson ’ s rule BW = 2 (  max + f i max ) Approximate BW prediction of a FM wave can be made by Carson ’ s rule ……. If maximum deviation of the FM wave is fixed at  20kHz.  Determine the BW required for an intelligence signal having a higher cutoff frequency at fi max = 10kHz. Example BW = 2 (  max + f i 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 J 0 = 1) then there will be 6 side frequencies (J 0 to J 3 ) as J 4 = 0.03 can be neglected. Then BW = 6x10kHz = 60kHz)

10. TE4201-Communication Electronics 10 Instantaneous voltage of the FM wave is e (t) = 2000 sin ( 2  x10 8 t+ 2 sin 2  x10 4 t ) It is applied to an antenna having an impedance of 50 . Determine: (a)Carrier frequency f C (b)Transmitted power P T (c)m f, (d) f i (e) BW (by two methods) (f)Power in the largest and the smallest sidebands predicted by the table of Bessel function. Example (a)Carrier frequency f C = 10 8 Hz =100MHz (b) Transmitted power P T = (E rms ) 2 / R = (2000/1.414 ) 2 / 50  = 40kW (c) m f = 2 (from 2 sin  x10 4 t ) (d)Information frequency (from 2 sin  x10 4 t ) f i = (10 4 /2)kHz = 5kHz (e)(Carson ’ s rule)  = m f x f i =2x5k=10kHz then BW = 2 (  max + f i max ) = 2 (10+ 10 )kHz = 40kHz  Bessel function) table, 4 pair (or 8 side frequencies, J 0 to J 4 ) then BW = 8x5kHz = 40kHz (f)(Bessel function) table largest SB2 is 58% of unmodulated carrier J 0 = 1, then P SB2 = (0.58E rms ) 2 / R = (0.58x2000/1.414 ) 2 / 50  = 13.5kW smallest SB4 is 3% of unmodulated carrier J 0 = 1,then P SB4 = (0.03E rms ) 2 / R = (0.03x2000/1.414 ) 2 / 50  = 0.036kW=36W

11. TE4201-Communication Electronics 11 Broadcast FM (entertainment) 200 kHz Channel width Carrier1 Maximum deviation  =  75kHz 200 kHz Channel width Carrier2 200 kHz Channel width 100MHz 200 kHz Channel width 2x25kHz guard band 0.1MHz0.1MHz 100.1MHz 100.2MHz Carrier accuracy  2kHz +75kHz-75kHz 100.4MHz 200 kHz Channel width  = 75kHz = m f x f i = fixed to a maximum value. Therefore m f decreases if f i increases (at high intelligent frequency no. of sideband is less because of small m f )

12. TE4201-Communication Electronics 12 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 f i = 3kHz. Narrowband FM has a (FCC) BW allocation of 10kHz to 30kHz. Narrowband FM has a modulation index m f (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

13. TE4201-Communication Electronics 13 (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. Example Example Determine the relative total power of the carrier and side frequencies when m f = 0.25 for a 10kW FM transmission

14. TE4201-Communication Electronics 14 FM limiter and detector AM limiter and detector External noise Noise at AM detector output Noise absent 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. Noise suppression

15. TE4201-Communication Electronics 15 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. FM noise analysis 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 :  =  x f i Where  = frequency deviation,  = phase shift (radians), f i = 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 “  ” is maximum when “ N ” is perpendicular to resultant vector “ R ”. Then  = sin -1 (N/S) = sin -1 (1/2) = 30deg = 0.52 rad. and maximum possible frequency deviation is  =  x f i = 0.52 x 15kHz = 7.5kHz R N S

16. TE4201-Communication Electronics 16 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 Example Determine the worst case output S/N for a narrowband FM communication that has a maximum deviation of  = 10kHz and maximum signal frequency of 3kHz. The input is S/N = 3 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 FM S/N is better than AM S/N

17. TE4201-Communication Electronics 17 Reversed bias capacitance of the Varactor Diode D 2 is in parallel with tuned capacitor C 1. C of D 2 change due to change of reverse bias by the modulating signal. Therefore output freq. will change due to change of C of D 2 Reversed bias capacitance of the Varactor Diode D 2 is in parallel with tuned capacitor C 1. C of D 2 change due to change of reverse bias by the modulating signal. Therefore output freq. will change due to change of C of D 2 Reactance Modulator Crosby Modulator VCO Modulator Varicap Modulator Varactor diode D 2 Varying reverse bias of Varactor changes the Varactor capacitance FM out Modulating intelligence signal Carrier f C oscillator L1L1L1L1 C1C1C1C1 Direct FM generation

18. TE4201-Communication Electronics 18 Reactance Modulator (equivalent capacitance) When modulating signal is connected to Gate of FET, V GS will change. This will make g m to change. Then C eq = CRg m will change. When modulating signal is connected to Gate of FET, V GS will change. This will make g m to change. Then C eq = CRg m 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. If terminal “e” is connected to the LC tuned circuit of the carrier frequency oscillator, then carrier frequency will change producing FM wave.

19. TE4201-Communication Electronics 19 Reactance Modulator (equivalent inductance) Example Show that the FET reactance circuit shown will produce an equivalent reactance of L eq =(CR/g m )

20. TE4201-Communication Electronics 20 Reactance FM modulator FET reactance circuit will provide changing equivalent inductance proportional to modulating signal amplitude. As it is in parallel with L 0 C 0 tuned oscillator (for carrier frequency), the carrier frequency will deviate according to dL produced by the FET reactance circuit, creating an FM wave. FET Reactance Modulator

21. TE4201-Communication Electronics 21 BJT Reactance Modulator When terminal “  C 0 =C eq ” is connected to the L 0 C 0 tuned circuit of the carrier frequency oscillator, then carrier frequency will change resulting an FM wave. When terminal “  C 0 =C eq ” is connected to the L 0 C 0 tuned circuit of the carrier frequency oscillator, then carrier frequency will change resulting an FM wave. When modulating signal is connected to Base of BJT, I B will change. This will make  i b to change. If current generator of BJT having an equivalent in such a way that  i b = g m e g where g m =(  i b )/e g =  r e =1/r e, then C eq = CRg m = CR/r e will change. Note that r e (=26mV/I C ) will change with I C (=  I B ) When modulating signal is connected to Base of BJT, I B will change. This will make  i b to change. If current generator of BJT having an equivalent in such a way that  i b = g m e g where g m =(  i b )/e g =  r e =1/r e, then C eq = CRg m = CR/r e will change. Note that r e (=26mV/I C ) will change with I C (=  I B )

22. TE4201-Communication Electronics 22 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

23. TE4201-Communication Electronics 23 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. Current source Schmitt trigger Buffer amplifier VCVC C1C1 R1R1 modulating input V+V+7 3 4 86 5 1 566 VCO

24. TE4201-Communication Electronics 24 CROSBY Modulator 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. f C =90 MHz  f = 75 kHz (max) f C =30 MHz  f = 25 kHz (max) 90 MHz 88 MHz 2 MHz f C =5 MHz  f = 4.167 kHz (max) Reactance modulator Primary oscillator 5MHz Frequency Tripler and Doublers Frequency Tripler Power Amplifier Frequency Converter (Mixer) To Antenna Discrimina tor Frequency Tripler and Doublers Crystal oscillator 14.67 MHz Audio input Frequency stabilization circuit

25. TE4201-Communication Electronics 25 Frequency Doublers High-distortion amplifier (class C) f C =5 MHz  f = 4.167 kHz (max) f C =10 MHz  f = 8.334 kHz (max) Tuned to 10MHz 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  f=4.167kHz enters the class C high distortion amplifier. Tank LC circuit of the amplifier output is tuned to 10MHz. Note that  f=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.

26. TE4201-Communication Electronics 26 f C =5 MHz  f = 0.1MHz Mixer f C =10 MHz  f = 0.1MHz Tuned to 10MHz Oscillator f o =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. frequency converter Frequency Converters

27. TE4201-Communication Electronics 27 Wideband Armstrong FM Mixer is responsible for achieving the final carrier from the basic crystal oscillator frequency 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 Frequency multipliers are responsible for achieving the final deviation frequency from the small deviation at the balanced modulator Indirect FM generation

28. TE4201-Communication Electronics 28 Design of mixer freq. and multiplying factor of freq. multiplier 1. Given: final= 88MHz,  75kHz, final multiplier= x64, before first xlier is carrier f C = 400k and f d =14.47Hz 2. then before multiplying f C =(88MHz/64) =1.375MHz, f d = (75kHz/64) =1.172kHz  3. before first xlier is f d =14.47Hz then first xlier is (1.172k/14.47)=81  4. before first xlier carrier f C =400k then after first xlier 400kx81=32.4MHz, then f 0 =32.4+1.375=33.81MHz 

29. TE4201-Communication Electronics 29 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?