UNIT-5 ANGLE MODULATION (FM) I

Transcription

1 UNIT-5 ANGLE MODULATION (FM) I Topis: Basi deinitions, FM, narrow band FM, wide band FM, transission bandwidth o FM waves, and generation o FM waves: indiret FM and diret FM. Angle odulation is a ethod o analog odulation in whih either the phase or requeny o the arrier wave is varied aording to the essage signal. In this ethod o odulation the aplitude o the arrier wave is aintained onstant. Angle Modulation is a ethod o odulation in whih either Frequeny or Phase o the arrier wave is varied aording to the essage signal. In general or, an angle odulated signal an represented as s( = A os[ θ ( ]...( 5.) Where A is the aplitude o the arrier wave and θ( is the angle o the odulated arrier and also the untion o the essage signal. The instantaneous requeny o the angle odulated signal, s( is given by i dθ ( ( = 2π dt...( 5.2) The odulated signal, s( is norally onsidered as a rotating phasor o length A and angle θ(. The angular veloity o suh a phasor is dθ(/dt, easured in radians per seond. An un-odulated arrier has the angle θ( deined as θ ( t ) = 2 π t + φ...( 5.3) Where is the arrier signal requeny and ϕ is the value o θ( at t = 0. The angle odulated signal has the angle, θ( deined by θ ( = 2π t + φ(...( 5.4) There are two oonly used ethods o angle odulation:. Frequeny Modulation, and 2. Phase Modulation. Analog Couniation 0EC53 Page

2 Phase Modulation (PM): In phase odulation the angle is varied linearly with the essage signal ( as : where k p is the phase sensitivity o the odulator in radians per volt. Thus the phase odulated signal is deined as Frequeny Modulation (FM): θ ( = 2π t k ( + [ 2 t k ( ) ]...( 5.6) s( = A os π t + p In requeny odulation the instantaneous requeny i ( is varied linearly with essage signal, ( as: ( = k ( i +...( 5.7) where k is the requeny sensitivity o the odulator in hertz per volt. The instantaneous angle an now be deined as p...( 5.5) θ ( = 2π t + 2π k t 0 ( dt...( 5.8) and thus the requeny odulated signal is given by s( = A os ( ) t π t π k t dt 0...( 5.9) The PM and FM waveors or the sinusoidal essage signal are shown in the ig-5.. Fig: 5. PM and FM Waveors with a essage signal Analog Couniation 0EC53 Page 2

3 Exaple 5.: Find the instantaneous requeny o the ollowing waveors: (a) S ( = A Cos [00π t π ] (b) S 2 ( = A Cos [00π t + sin ( 20 π ] () S 3 ( = A Cos [00π t + ( π t 2 ) ] Solution: Using equations (5.) and (5.2): (a) i ( = 50 Hz; Instantaneous requeny is onstant. (b) i ( = os( 20 π ; Maxiu value is 60 Hz and iniu value is 40 Hz. Hene, instantaneous requeny osillates between 40 Hz and 60 Hz. () i ( = (50 + The instantaneous requeny is 50 Hz at t=0 and varies linearly at Hz/se. Relation between Frequeny Modulation and Phase Modulation: A requeny odulated signal an be generated using a phase odulator by irst integrating ( and using it as an input to a phase odulator. This is possible by onsidering FM signal as phase odulated signal in whih the odulating wave is integral o ( in plae o (. This is shown in the ig-5.2(a). Siilarly, a PM signal an be generated by irst dierentiating ( and then using the resultant signal as the input to a FM odulator, as shown in ig-5.2(b). Fig: 5.2 Shee or generation o FM and PM Waveors Analog Couniation 0EC53 Page 3

4 Single-Tone Frequeny Modulation: Consider a sinusoidal odulating signal deined as: ( = A Cos( 2π. (5.0) Substituting or ( in equation (5.9), the instantaneous requeny o the FM signal is ( = + k A os(2π = + os(2π i where is alled the requeny deviation given by = k A... (5.a) and the instantaneous angle is θ ( = 2π = 2π t + where β = t 0 ( dt i = 2π t + β sin( 2π sin( 2π ; odulation index... ( 5.b) The resultant FM signal is [ 2π t + β sin(2π )]...( 5.2) s( = A os t The requeny deviation ator indiates the aount o requeny hange in the FM signal ro the arrier requeny on either side o it. Thus FM signal will have the requeny oponents between ( - ) to ( + ). The odulation index, β represents the phase deviation o the FM signal and is easured in radians. Depending on the value o β, FM signal an be lassiied into two types:. Narrow band FM (β << ) and 2. Wide band FM (β >> ). Exaple-5.2: A sinusoidal wave o aplitude 0volts and requeny o khz is applied to an FM generator that has a requeny sensitivity onstant o 40 Hz/volt. Deterine the requeny deviation and odulating index. Solution: Message signal aplitude, A = 0 volts, Frequeny = 000 Hz and the requeny sensitivity, k = 40 Hz/volt. Frequeny deviation, = k A = 400 Hz Modulation index, β = / = 0.4, (indiates a narrow band FM). Analog Couniation 0EC53 Page 4

5 Exaple-5.3: A odulating signal ( =0 Cos(0000π odulates a arrier signal, A Cos(2π. Find the requeny deviation and odulation index o the resulting FM signal. Use k = 5kHz/volt. Solution: Message signal aplitude, A = 0 volts, Frequeny = 5000 Hz and the requeny sensitivity, k = 5 khz/volt. Frequeny deviation, = k A = 50 khz Modulation index, β = / = 0, (indiates a wide band FM). Frequeny Doain Representation o Narrow Band FM signal: Expanding the equation (5.2) using trigonoetri identities, s( = A = A os os For NBFM, (β << ), we an approxiate, os [ 2π t + β sin(2π ] ( 2π os[ β sin(2π ] A sin( 2π sin[ β sin(2π ] [ β sin(2π ] and sin[ β sin(2π ] β sin(2π Hene, s( = A os(2π A β sin(2π sin(2π...( 5.3) Using trigonoetri relations; A β s( = A os(2π + π t 2 [ os(2π ( + ) os(2 ( ) ]..( 5.4) The above equation represents the NBFM signal. This representation is siilar to an AM signal, exept that the lower side requeny has negative sign. The agnitude spetru o NBFM signal is shown in ig-5.3, whih is siilar to AM signal spetru. The bandwidth o the NBFM signal is 2, whih is sae as AM signal. S() + + Fig: Magnitude Spetru o NBFM Waveor. Analog Couniation 0EC53 Page 5

6 Frequeny Doain Representation o Wide-Band FM signals: The FM wave or sinusoidal odulation is given by s( = A = A os os [ 2π t + β sin(2π ] ( 2π os[ β sin(2π ] A sin( 2π sin[ β sin(2π ] The FM wave an be expressed in ters o oplex envelope as: s( = Re The oplex envelope o the FM wave [ A exp( j2π t + jβ sin( 2π )] = Re[ ~ s ( exp( j2π ]... ( 5.5) ~ s ( = A exp π [ jβ sin( 2 ] and ~ s ( : periodi untion with The oplex envelope is a periodi untion o tie, with a undaental requeny equal to the odulation requeny. The oplex envelope an be expanded in the or o oplex series: ~ s ( = exp[ j2π n t]... ( 5.6) n n= The oplex Fourier oeiient, n equals, n = = 2 2 A 2 2 ~ s ( exp exp ( j2πn dt [ jβ sin( 2π j2πn t] dt... ( 5.7) Substituting x = (2π t ), in the above equation we an rewrite A π n = exp 2 π π ( j( β sin x nx) ) dx...( 5.8) The n th order Bessel untion o the irst kind is deined as J n ( β ) = 2π π π exp ( j( β sin x nx) ) dx...( 5.9) Coparing equations (5.8) and (5.9), we get C n = A J n (β) Substituting in (5.6), the oplex envelope is ~ s( n n = - = A J ( β )exp t ( j2πn )...( 5.20) Analog Couniation 0EC53 Page 6

7 Substituting in (5.5), the FM signal an be written as s( = A Re s( = A n n= n n= J ( β )exp J ( β )os [ j2π ( + n ) t] [ 2π ( + n ) t]...( 5.2) The above equation is the Fourier series representation o the single tone FM wave. Applying the Fourier transor to (5.2), A S ( ) = J n ( β ) + 2 n= [ δ ( n ) + δ ( + n )]...( 5.22) The spetru S() is shown in ig-5.4. The above equation indiates the ollowing: (i) FM signal has ininite nuber o side bands at requenies ( + n ). (ii) Relative aplitudes o all the spetral lines depends on the value o J n (β). (iii) The nuber o signiiant side bands depends on the odulation index (β). With (β<<), only J 0 (β) and J (β) are signiiant. But or (β>>), any sidebands exists. (iv) The average power o an FM wave is P = 0.5A 2 (based on Bessel untion property). J 0 (β) J (β) J (β) + S() J 0 (β) J 2 (β) J (β) J β) (β) J 2 ( Bessel s Funtion: Fig: Magnitude Spetru o Wide Band FM Wave. Bessel untion is an useul untion to represent the FM wave spetru. The general plots o Bessel untions are shown in ig-5.5 and table (5.) gives the values or Bessel untion oeiients. Soe o the useul properties o Bessel untions are given below: (a) n J ( β ) = ( ) J ( β ) n n or all n (5.23a) (b) 2n J n+ ( β ) + J n ( β ) = J n( β ) β (5.23b) () 2 n n= J ( β ) = (5.23) Analog Couniation 0EC53 Page 7

8 β (d) For saller values o β, J0 ( β ), J( β ) and Jn ( β ) 0, or n > 2 2 Fig: 5.5 Plots o Bessel untions Table: 5. Analog Couniation 0EC53 Page 8

9 The Spetru o FM signals or three dierent values o β are shown in the ig-5.6. In this spetru the aplitude o the arrier oponent is kept as a unity onstant. The variation in the aplitudes o all the requeny oponents is indiated. For β =, the aplitude o the arrier oponent is ore than the side band requenies as shown in ig-5.6a. The aplitude level o the side band requenies is dereasing. The doinant oponents are ( + ) and ( + 2 ). The aplitude o the requeny oponents ( + n ) or n>2 are negligible. For β = 2, the aplitude o the arrier oponent is onsidered as unity. The spetru is shown in ig-5.6b. The aplitude level o the side band requenies is varying. The aplitude levels o the oponents ( + ) and ( + 2 ) are ore than arrier requeny oponent; whereas the aplitude o the oponent ( + 3 ) is lower than the arrier aplitude. The aplitude o requeny oponents ( + n ) or n>3 are negligible. The spetru or β = 5, is shown in ig-5.6. The aplitude o the arrier oponent is onsidered as unity. The aplitude level o the side band requenies is varying. The aplitude levels o the oponents ( + ), ( + 3 ), ( + 4 ) and ( + 5 ), are ore than arrier requeny oponent; whereas the aplitude o the oponent ( + 2 ) is lower than the arrier aplitude. The aplitude o requeny oponents ( + n ) or n>8 are negligible. Fig: 5.6 Plots o Spetru or dierent values o odulation index. (Aplitude o arrier oponent is onstant at unity) Analog Couniation 0EC53 Page 9

10 Exaple-5.4: An FM transitter has a power output o 0 W. I the index o odulation is.0, deterine the power in the various requeny oponents o the signal. Solution: The various requeny oponents o the FM signal are, ( + ), ( + 2 ), ( + 3 ), and so on. The power assoiated with the above requeny oponents are: (Reer (5.2)) (J 0 ) 2, (J ) 2, (J 2 ) 2, and (J 3 ) 2 respetively. Fro the Bessel untion Table, or β = ; J 0 = 0.77, J = 0.44, J 2 = 0., and J 3 = 0.02 Let P = 0.5(A ) 2 = 0 W. Power assoiated with oponent is P 0 = P (J 0 ) 2 = 0 (0.77) 2 = W. Siilarly, P = P (J ) 2 = 0 (0.44) 2 =.936 W. P 2 = P (J 2 ) 2 = 0 (0.) 2 = 0.2 W. P 3 = P (J 3 ) 2 = 0 (0.02) 2 = W. Note: Total power in the FM wave, Exaple-5.5: P total = P 0 + 2P + 2P 2 + 2P 3 = (.936) + 2(.2) + 2(.004) = 0.05 W A 00 MHz un-odulated arrier delivers 00 Watts o power to a load. The arrier is requeny odulated by a 2 khz odulating signal ausing a axiu requeny deviation o 8 khz. This FM signal is oupled to a load through an ideal Band Pass ilter with 00MHz as enter requeny and a variable bandwidth. Deterine the power delivered to the load when the ilter bandwidth is: (a) 2.2 khz (b) 0.5 khz () 5 khz (d) 2 khz Ans: Modulation index, β = 8 k / 2 k = 4; Fro the Bessel untion Table- 5.; or β = 4; J 0 = -0.4, J = , J 2 = 0.36, J 3 = 0.43, J 4 = 0.28, J 5 = 0.3, J 6 = 0.05, J 7 = 0.02 Analog Couniation 0EC53 Page 0

11 Let P = 0.5(A ) 2 = 00 W and P 0 = P (J 0 ) 2 = 00 (-0.4) 2 = 6 Watts. P = P (J ) 2 = 00 (-0.07) 2 = W. P 2 = P (J 2 ) 2 = 00 (0.36) 2 = W. P 3 = P (J 3 ) 2 = 00 (0.43) 2 = W. P 4 = P (J 4 ) 2 = 00 (0.28) 2 = W. P 5 = P (J 5 ) 2 = 00 (0.3) 2 =.690 W. P 6 = P (J 6 ) 2 = 00 (0.05) 2 = W. (a) Filter Bandwidth = 2.2 khz The output o band pass ilter will ontain only one requeny oponent. Power delivered to the load, P d = P 0 = 6 Watts. (b) Filter Bandwidth = 0.5 khz The output o band pass ilter will ontain the ollowing requeny oponents:, ( + ), and ( + 2 ) Power delivered to the load, P d = P 0 + 2P + 2P 2 = 42.9 Watts. () Filter Bandwidth = 5 khz The output o band pass ilter will ontain the ollowing requeny oponents:, ( + ), ( + 2 ), and ( + 3 ), Power delivered to the load, P d = P 0 + 2P + 2P 2 + 2P 3 = 79.9 Watts. (d) Filter Bandwidth = 2 khz The output o band pass ilter will ontain the ollowing requeny oponents:, ( + ), ( + 2 ), ( + 3 ), ( + 4 ), and ( + 5 ), Power delivered to the load, P d = P 0 + 2P + 2P 2 + 2P 3 + 2P 4 + 2P 5 = Watts. Exaple-5.6: A arrier wave is requeny odulated using a sinusoidal signal o requeny and aplitude A. In a ertain experient onduted with = khz and inreasing A, starting ro zero, it is ound that the arrier oponent o the FM wave is redued to Analog Couniation 0EC53 Page

12 zero or the irst tie when A =2 volts. What is the requeny sensitivity o the odulator? What is the value o A or whih the arrier oponent is redued to zero or the seond tie? Ans: The arrier oponent will be zero when its oeiient, J 0 (β) is zero. Fro Table 5.: J 0 (x) = 0 or x= 2.44, 5.53, β = / = k A / and k = β /A = (2.40)(000) / 2 =.22 khz/v Frequeny Sensitivity, k =.22 khz/v The arrier oponent will beoe zero or seond tie when β = Thereore, A = β / k = 5.53 (000) / 220 = 4.53 volts Transission Bandwidth o FM waves: An FM wave onsists o ininite nuber o side bands so that the bandwidth is theoretially ininite. But, in pratie, the FM wave is eetively liited to a inite nuber o side band requenies opatible with a sall aount o distortion. There are any ways to ind the bandwidth o the FM wave.. Carson s Rule: In single tone odulation, or the saller values o odulation index the bandwidth is approxiated as 2. For the higher values o odulation index, the bandwidth is onsidered as slightly greater than the total deviation 2. Thus the Bandwidth or sinusoidal odulation is deined as: B T = 2( β + ) = 2 + β ( 5.24) For non-sinusoidal odulation, a ator alled Deviation ratio (D) is onsidered. The deviation ratio is deined as the ratio o axiu requeny deviation to the bandwidth o essage signal. Deviation ratio, D = ( / W ), where W is the bandwidth o the essage signal and the orresponding bandwidth o the FM signal is, B T = 2(D + ) W... (5.25) Analog Couniation 0EC53 Page 2

13 2. Universal Curve : An aurate ethod o bandwidth assessent is done by retaining the axiu nuber o signiiant side requenies with aplitudes greater than % o the unodulated arrier wave. Thus the bandwidth is deined as the 99 perent bandwidth o an FM wave as the separation between the two requenies beyond whih none o the side-band requenies is greater than % o the arrier aplitude obtained when the odulation is reoved. Transission Bandwidth - BW = 2 n ax, (5.26) where is the odulation requeny and n is the nuber o pairs o side-requenies suh that J n (β) > 0.0. The value o n ax varies with odulation index and an be deterined ro the Bessel oeiients. The table 5.2 shows the nuber o signiiant side requenies or dierent values o odulation index. The transission bandwidth alulated using this ethod an be expressed in the or o a universal urve whih is noralised with respet to the requeny deviation and plotted it versus the odulation index. (Reer ig-5.7). Table 5.2 Fro the universal urve, or a given essage signal requeny and odulation index the ratio (B/ ) is obtained ro the urve. Then the bandwidth is alulated as: BT BT BT = ( ) = β ( )...( 5.27) Analog Couniation 0EC53 Page 3

14 Fig: 5.7 Universal Curve Exaple-5.7: Find the bandwidth o a single tone odulated FM signal desribed by S(=0 os[2π0 8 t + 6 sin(2π0 3 ]. Solution: Coparing the given s( with equation-(5.2) we get Modulation index, β = 6 and Message signal requeny, = 000 Hz. By Carson s rule (equation ), Transission Bandwidth, B T = 2(β + ) B T = 2(7)000 = 4000 Hz = 4 khz Exaple-5.8: Q. A arrier wave o requeny 9 MHz is requeny odulated by a sine wave o aplitude 0 Volts and 5 khz. The requeny sensitivity o the odulator is 3 khz/v. (a) Deterine the approxiate bandwidth o FM wave using Carson s Rule. (b) Repeat part (a), assuing that the aplitude o the odulating wave is doubled. () Repeat part (a), assuing that the requeny o the odulating wave is doubled. Solution: (a) Modulation Index, β = / = k A / = 3x0/5 = 2 Analog Couniation 0EC53 Page 4

15 By Carson s rule; Bandwidth, B T = 2(β + ) = 90 khz (b) When the aplitude, A is doubled, New Modulation Index, β = / = k A / = 3x20/5 = 4 Bandwidth, B T = 2(β+) = 50 khz () when the requeny o the essage signal, is doubled New Modulation Index, β = 3x0/30 = Bandwidth, B T = 2(β+) = 20 khz. Exaple-5.9: Q. Deterine the bandwidth o an FM signal, i the axiu value o the requeny deviation is ixed at 75kHz or oerial FM broadasting by radio and odulation requeny is W= 5 khz. Solution: Frequeny deviation, D = ( / W ) = 5 Transission Bandwidth, B T = 2(D + ) W = 2x5 khz = 80 khz Exaple-5.0: Q. Consider an FM signal obtained ro a odulating signal requeny o 2000 Hz and axiu Aplitude o 5 volts. The requeny sensitivity o odulator is 2 khz/v. Find the bandwidth o the FM signal onsidering only the signiiant side band requenies. Solution: Frequeny Deviation = 0 khz Modulation Index, β = / = k A / = 5; Fro table (5.2) ; 2n ax = 6 or β =5, Bandwidth, B T = 2 n ax = 6x2 khz = 32 khz. Exaple-5.: A arrier wave o requeny 9 MHz is requeny odulated by a sine wave o aplitude 0 Volts and 5 khz. The requeny sensitivity o the odulator is 3 khz/v. Deterine the bandwidth by transitting only those side requenies with aplitudes that exeed % o the unodulated arrier wave aplitude. Use universal urve or this alulation. Analog Couniation 0EC53 Page 5

16 Solution: Frequeny Deviation, = 30 khz Modulation Index, β = 3x0/5 = 2 Fro the Universal urve; or β = 2; (B / ) = 4.3 Bandwidth, B = 4.3 = 29 khz Generation o FM Waves: There are two basi ethods o generating FM waves: indiret ethod and diret ethod. In indiret ethod a NBFM wave is generated irst and requeny ultipliation is next used to inrease the requeny deviation to the desired level. In diret ethod, the arrier requeny is diretly varied in aordane with the essage signal. To understand the indiret ethod it is required to know the generation o NBFM waves and the working o requeny ultipliers. Generation o NBFM wave: A requeny odulated wave is deined as: (ro equation 5.9) [ 2π t + φ ( )]...( 5.28) s( = AC os C t Where φ ( = 2π k ( dt t 0 s ( = AC os(2π C os[ φ( ] - AC sin (2π C sin[ φ( ] Assuing ϕ ( is sall, then using os[ϕ (] = and sin[ϕ ( ] = ϕ (. s ( = A s ( = A C C os(2π os(2π C C - A sin (2π C - 2π k A C C sin (2π.[ φ ( ] C. t 0 ( dt...( 5.29) The above equation deines a narrow band FM wave. The generation shee o suh a narrow band FM wave is shown in the ig.(5.8). The saling ator, (2πk ) is taken are o by the produt odulator. The part o the FM odulator shown inside the dotted lines represents a narrow-band phase odulator. The narrow band FM wave, thus generated will have soe higher order haroni distortions. This distortions an be liited to negligible levels by restriting the odulation index to β < 0.5 radians. Analog Couniation 0EC53 Page 6

17 Frequeny Multiplier: Fig: 5.8 Shee to generate a NBFM Waveor. The requeny ultiplier onsists o a nonlinear devie ollowed by a band-pass ilter. The nonlinear devie used is a eory less devie. I the input to the nonlinear devie is an FM wave with requeny, and deviation, then its output v( will onsist o d oponent and n requeny odulated waves with arrier requenies,, 2, 3, n and requeny deviations a, 2, 3,... n respetively. The band pass ilter is designed in suh a way that it passes the FM wave entered at the requeny, n with requeny deviation n and to suppress all other FM oponents. Thus the requeny ultiplier an be used to generate a wide band FM wave ro a narrow band FM wave. Fig: 5.9 Frequeny Multiplier Generation o WBFM using Indiret Method: In indiret ethod a NBFM wave is generated irst and requeny ultipliation is next used to inrease the requeny deviation to the desired level. The narrow band FM wave is generated using a narrow band phase odulator and an osillator. The narrow band FM wave is then passed through a requeny ultiplier to obtain the wide band FM wave, as shown in the ig:(5.9). The rystal ontrolled osillator provides good requeny stability. But this shee does not provide both the desired requeny deviation and arrier requeny at the sae tie. This proble an be solved by using ultiple stages o requeny ultiplier and a ixer stage. Analog Couniation 0EC53 Page 7

18 Fig: 5.9 Generation o WBFM wave Generation o WBFM by Arstrong s Method: Arstrong ethod is an indiret ethod o FM generation. It is used to generate FM signal having both the desired requeny deviation and the arrier requeny. In this ethod, two-stage requeny ultiplier and an interediate stage o requeny translator is used, as shown in the ig:(5.0). The irst ultiplier onverts a narrow band FM signal into a wide band signal. The requeny translator, onsisting o a ixer and a rystal ontrolled osillator shits the wide band signal to higher or lower requeny band. The seond ultiplier then inreases the requeny deviation and at the sae tie inreases the enter requeny also. The ain design riteria in this ethod are the seletion o ultiplier gains and osillator requenies. This is explained in the ollowing steps. Fig: 5.0 Generation o WBFM wave by Arstrong ethod Analog Couniation 0EC53 Page 8

19 Design Steps: Q: How to hoose n and n 2 or the given speiiations?. Selet the value o β < 0.5 or the narrow band phase odulator. This value liits the haroni distortion by NBPM to iniu. 2. The requireent is that the requeny deviation produed by the lowest odulation requenies is raised to required. So hoose the requeny deviation o NBFM, by seleting the iniu value o. = β (in) (a) 3. Frequeny Multipliers hange the requeny deviation. Hene the total hange in the requeny deviation is produt o the two deviations: n.n 2 = / (b) 4. Frequeny Translator (ixer & osillator) will not hange the requeny deviation, it only shits the FM signal to either upwards and downwards in the spetru. The output o ixer is For down ward translation: = n 2 ( 2 - n. ) ---- () and or upward translation: = n 2 (n. - 2 ). 5. Choose suitable value or 2 and solve the equations (b) and () siultaneously to ind the ultiplying ators n and n 2. Exaple 5.2: Design Arstrong FM generator or the generation o WBFM signal with = 75 khz and = 00 MHz, using the narrow band arrier as 00 khz and seond arrier as 9.5 MHz. Find the suitable ultiplying ators. Assue the essage signal is deined in the range, 00Hz ~ 5KHz. Solution:. Let phase deviation be, β = Frequeny deviation o NBFM, = β (in) = 0.2x 00 = 20 Hz 3. n.n 2 = / = 75000/20 = 3750 (A) 4. Let and 2 be 0.MHz and 9.5 MHz and is given as 00 MHz. = ( 2 - n. )n 2 ; 00 M = (9.5 M - n. 0.M )n 2 (B) Solving the equations (A) and (B) siultaneously we get n = 75 and n 2 = 50. Analog Couniation 0EC53 Page 9

20 Generation o WBFM using Diret Method: In diret ethod o FM generation, the instantaneous requeny o the arrier wave is diretly varied in aordane with the essage signal by eans o an voltage ontrolled osillator. The requeny deterining network in the osillator is hosen with high quality ator (Q-ator) and the osillator is ontrolled by the inreental variation o the reative oponents in the tank iruit o the osillator. A Hartley Osillator an be used or this purpose. Fig: 5. Hartley Osillator (tank irui or generation o WBFM wave. The portion o the tank iruit in the osillator is shown in ig:5.. The apaitive oponent o the tank iruit onsists o a ixed apaitor shunted by a voltage-variable apaitor. The resulting apaitane is represented by C( in the igure. The voltage variable apaitor oonly alled as varator or variap, is one whose apaitane depends on the voltage applied aross its eletrodes. The varator diode in the reverse bias ondition an be used as a voltage variable apaitor. The larger the voltage applied aross the diode, the saller the transition apaitane o the diode. The requeny o osillation o the Hartley osillator is given by: i ( = 2 π + ( L L ) ( ) 2 t...( 5.30) Where the L and L 2 are the indutanes in the tank iruit and the total apaitane, ( is the ixed apaitor and voltage variable apaitor and given by: ( = 0 + os t ( 2π )...( 5.3) Let the un-odulated requeny o osillation be 0. The instantaneous requeny i ( is deined as: i ( = 0 + os π t 0 2 ( 2 )...( 5.32) Analog Couniation 0EC53 Page 20

21 where 0 = 2 π ( L + L ) ( 5.33) ( = i 0 + os 0 0 os 20 Thus the instantaneous requeny i ( is deined as: ( 0 + os t i ( 2π ( 2π 2 ( 2π )...( 5.34) The ter, represents the requeny deviation and the relation with is given by: = ( 5.35) Thus the output o the osillator will be an FM wave. But the diret ethod o generation has the disadvantage that the arrier requeny will not be stable as it is not generated ro a highly stable osillator. Generally, in FM transitter the requeny stability o the odulator is ahieved by the use o an auxiliary stabilization iruit as shown in the ig.(5.2). Fig: 5.2 Frequeny stabilized FM odulator. The output o the FM generator is applied to a ixer together with the output o rystal ontrolled osillator and the dierene is obtained. The ixer output is applied to a requeny disriinator, whih gives an output voltage proportional to the instantaneous requeny o the FM wave applied to its input. The disriinator is iltered by a low pass ilter and then apliied to provide a d voltage. This d voltage is applied to a voltage ontrolled osillator (VCO) to odiy the requeny o the osillator o the FM generator. The deviations in the transitter arrier requeny ro its assigned value will ause a hange in the d voltage in a way suh that it restores the arrier requeny to its required value. Analog Couniation 0EC53 Page 2

22 Advantages and disadvantages o FM over AM: Advantages o FM over AM are: Disadvantages o FM:. Less radiated power. 2. Low distortion due to proved signal to noise ratio (about 25dB) w.r.t. to an ade intererene. 3. Saller geographial intererene between neighbouring stations. 4. Well deined servie areas or given transitter power.. Muh ore Bandwidth (as uh as 20 ties as uh). 2. More opliated reeiver and transitter. Appliations: Soe o the appliations o the FM odulation are listed below: I. FM Radio,(88-08 MHz band, 75 khz, ) II. III. TV sound broadast, 25 khz, 2-way obile radio, 5 khz / 2.5 khz. Additional Exaples: Exaple 5.3: An FM wave is deined below. S( = 2 sin(6x0 8 π t + 5 sin250 π Find the arrier and odulating requenies, the odulating index, and the axiu deviation o the FM wave. Also ind the bandwidth o the FM wave. What power will the FM wave dissipate in a 0 oh resistor? Solution: Fro equation 5.2, we have [ 2π t + β sin(2π )] s( = A os t Coparing with the given FM wave, Carrier requeny = 3x0 8 Hz = 300 MHz Analog Couniation 0EC53 Page 22

23 Modulating signal requeny, = 625 Hz Modulation Index, β = 5 ; Maxiu requeny deviation, = β = 325 Hz. Using Carson s rule, Bandwidth = 2( ) = 7500 Hz Power dissipated aross resistor = P, ( A ) 2 44 P = = = 7. 2W 2R 20 Exaple 5.4: Consider an FM signal with : = 0 khz, = 0 khz, A = 0 V, = 500 khz Copute and draw the spetru or FM signal. Solution: Modulation index, β = 0 k / 0 k = ; Fro Bessel untion Table- 5.; or β = ; the oeiients are J 0 = 0.77, J = 0.44, J 2 = 0., J 3 = The spetru is deined as A [ δ ( n ) + δ ( + n )] S = ( ) J n + 2 n= The single sided spetru is shown in ig:ex-5.4. Fig: Ex-5.4 Frequeny Spetru (or exaple 5.4) Analog Couniation 0EC53 Page 23