dm t t A cos 2 10 t 10

Transcription

1 T.C. OKAN ÜNİVERSİTESİ Fauly o Engineering and Arhieure Elerial and Eleroni Engineering Program EEE 3 Analog Communiaions Fall 23 In Class Work Par 4 Soluions:. Skeh he FM and PM modulaed waveorms or he modulaing signal m() shown below a. = MHz, k = 2π 5, k p = π. + 2x -4 SOLUTION: FOR Frequeny Modulaion: m FM 8 5 A os 2 2 mxdx FOR Phase Modulaion: 6 2 m d 6 d 2 m d d 8 A os 2 m PM dm d PM + 2x -4

2 m + 2x -4 FM b. = MHz, k = 2π 5, k p = π/2. + SOLUTION: FOR Frequeny Modulaion: FOR Phase Modulaion: m 6 2 m FM A os 2 2 mxdx d 6 d 2 m d 2 d 8 PM A os 2 m 2 Noie ha m 2 is a phase hange o +π only a τ 2 2. A phase hange o π is hal a wavelengh.

3 dm d τ PM 2δ( τ ) 2δ( τ 3 ) τ 2 τ 3 2δ( τ 2 ) 2δ( τ 4 ) τ 4 m + FM 2. Deermine he bandwidh o he ollowing signal or PM and FM modulaion: a. = MHz, k = 2π 5 radians/se/v, k p = π. radians/v, + 2x -4 SOLUTION: FOR Frequeny Modulaion: m FM 8 5 A os 2 2 mxdx FOR Phase Modulaion: 6 2 m d 6 d 2 m d d 8 A os 2 m PM

4 For requeny modulaion, he peak ampliude o m() is, so m p =. This means ha he maximum requeny deviaion is ω = k m p =2π 5 =2π 5, and = ω/(2π) = 5 Hz = Hz = khz. Now deermine he bandwidh B o m(). For a periodi signal he Fourier series expansion is: os m Cn n n where ω is he undamenal requeny, alulaed as: The Fourier oeiiens or a periodi sawooh wave (above union) are: C n 8 n odd 2 2 n n even The harmoni ampliude dereases very quikly wih n. See he ollowing Malab ode: >> n = [:2:5] n = >> Cn = 8/pi^2./(n.^2) Cn = >> Cn./Cn() ans = Aording o he deiniion in your slides, we ould deine he bandwidh o he signal as he poin a whih he signal power drops o % o he power o he irs harmoni. So i we square hese oeiiens (o ind he power in ha harmoni). We an see ha he 3 rd index is abou 4% o he irs index. I we square his o ind heir relaive powers, we ind ha he 3 rd harmoni power is only (.4) 2 =.6 or.6 peren o he power o he irs harmoni, whih means i an be negleed. We an say ha he bandwidh o he signal m() is abou 2 imes is undamenal requeny ω = π. This means ha he bandwidh is abou 2 π. radians, whih orresponds o B = (2 π) / (2 π) = Hz = khz. This means ha he bandwidh o he FREQUENCY modulaed signal is B 2 B 2 khz khz 2 khz 22kHz. FM

5 For phase modulaion, we need o look a he bandwidh o he derivaive o he sawooh wave, whih is a square wave. The peak ampliude or he derivaive is ( ) 4 slope 2. So m 4 p = 2 4. This means ha he maximum requeny deviaion is ω = k m p =π 2 4 = 2π 5, and = ω/(2π) = 5 Hz = Hz = khz. The Fourier series or a square wave is: d m D n os n d n where ω is he undamenal requeny, is alulaed he same way as beore: The Fourier oeiiens or a periodi square wave (above union) are: D n 4 n odd n n even The harmoni ampliude dereases very quikly wih n. See he ollowing Malab ode: >> n = [:2:5] n = >> Dn = 4/pi./n Dn = >> Dn./Dn() ans = This ime he 6 h harmoni erm, is he irs erm smaller smaller han / h o he undamenal erm, whih means i is he irs erm suh ha he power o he harmonis is less han % o he irs harmoni. Based on he deiniion we alulae he bandwidh o be 5 imes he undamenal, whih means ha B = (5 π) / (2 π) = 25Hz = 25 khz. So sine neiher B nor hange rom he previous example, hen he bandwidh o he PHASE modulaed signal is B 2 B 2 25kHz khz 2 25kHz 25kHz. PM

6 3. A periodi square wave m() below (a) requeny modulaes a arrier o requeny = khz wih =khz. The arrier ampliude is A. The resuling FM signal is demodulaed, as shown in he igure (b) below. Skeh he waveorms a poins b,, d, and e. m() + T o ime a b d e FM modulaor d d Envelope deeor DC bloking Demodulaor For FM modulaion, our phase is equal o he inegral o he message. So i he message waveorm is : i.5ms i.5ms ms m i ms.5ms hen i.5ms 2ms i.5ms.5ms.5ms i.5ms ms M ms i ms.5ms.5ms.5ms i.5ms 2ms sfm A os 2 2 k m d A os 2 2 k M is skehed below.

7 Malab ode whih will skeh he waveorms: =:/e3:2e-3; m = [- -ones(,5) ones(, 5) -ones(,5) ones(, 5)] = (:5); M = [ - -.5e e-3+]; sfm=os(2*pi*e3*+2*pi*4e3*m); subplo(3,,) plo(,m) legend('m()') subplo(3,,2) plo(,m) legend('m()') subplo(3,,3) plo(,sfm) legend('s_{fm}()')

8 m() x -3 x -4 M() x -3 s FM () x -3 Waveorm a (b) The waveorm a () is ds FM d A 2 2 k msin 2 2 k m d A 2 2 k msin 2 2 k M

9 Malab ode whih will skeh he waveorms: d_sfm=-(2*pi*e3+2*pi*4e3*m).*sin(2*pi*e3*+2*pi*4e3*m); e = (2*pi*e3+2*pi*4e3*m); e_minus_d = 2*pi*4e3*m; subplo(5,,) plo(,m) legend('m()') axis([ 2e ]) ylabel('(a)') subplo(5,,2) plo(,sfm) legend('s_{fm}()') ylabel('(b)') subplo(5,,3) plo(,d_sfm) legend('ds_{fm}() / d') ylabel('()') subplo(5,,4) plo(,e) legend('envelope o ds_{fm}() / d') ylabel('(d)') subplo(5,,5) plo(,e_minus_d) legend('oupu o d bloker') ylabel('(e)')

10 (e) (d) () (b) (a) m() x -3 s FM () x 5 x -3 ds FM () / d - x x -3 envelope o ds FM () / d x 4 x -3 oupu o d bloker x -3

11 4. Le s() be an angle-modulaed signal ha a reeiver obains, 2os 2sin os s a. Find he bandwidh o his FM signal. b. I s() is sen o an (ideal) envelope deeor, ind he deeor oupu signal.. I s() is irs diereniaed beore he envelope deeor, ind he deeor oupu signal. d. Explain whih deeor oupu an be proessed o yield he message signal m() and ind he message signal m() i k = 2π radians/v/s. Soluion: a. Firs we need o ind he maximum requeny deviaion or any value o o ind ha irs alulae he insananious requeny: d i d 2sin os 2 2 os sin Looking a his, we an assume ha he maximum value will our when sin() = and sin(2π+.3π) =, in whih ase he angular requeny inreases by (4 π +3π)= 43π. So = 25Hz. The bandwidh o he message signal is aken o be he requeny o he larges signal here, so his is : 2 B max, khz 2 2 Finally inding he bandwidh : B 2 B 2 2.5kHz khz 2 3.5kHz 6.3kHz. PM b. I he oupu was sen o an ideal envelope deeor, sine he envelope o he signal is onsan a 2, he oupu would be 2 all he ime. The requeny holds he message, bu he envelope deeor will remove his.

12 . Diereniaing he signal gives us: ds 2 os 2sin os d d 2sin os d 2 os 2sin os So he oupu o he envelope deeor will be 2 2 os sin 2 22 os sin e d. Clearly he envelope o he diereniaed signal will give us inormaion abou he original message signal. Sine his was an FM signal, we know ha he inegral o he message signal was: s A os 2 2 k m d 2 os 2sin os FM Diereniaing boh sides o he phase gives us: 2 k m d 2sin os 2 k m 2 2 os sin So he message signal is: Noe: he unis o k, are saed o be radians/v, so we need o solve aordingly. k m 22 os sin 2 m 22 os sin m 2 os 2.3.5sin m 2os 2.3.5sin From par () he envelope o he diereniaed signal was 2 22 os sin e Thereore: m e 2 2 os sin 2

13 5. Consider an FM signal: sfm A os 2 2 k m d Le and 2 ( 2 > ) denoe he imes assoiaed wih wo adjaen zero rossings o x FM (). I 2 m d m hen show ha 2km where = Beween wo adjaen zero rossings, he phase should hange by π, so k m d 2 2 k m d 2 2 k m d 2 2 k m k m 2k m 2 2

14 6. The resul o he previous problem indiaes ha m() an be reovered by ouning he zero rossings in x FM (). Le N denoe he number o zero rossings in ime T. Show ha i T saisies he ondiion T where M is he bandwidh o m() in H (ω M =2π M ), hen N km 2T M Suppose ha m() is onsan over some inerval T (his requres he ondiion ha M << T. sfm A os 2 2 k m d Sine a zero rossing ours every ime he phase hanges by pi, he number o zero rossings over he inerval rom o 2 will be: 2 N k m d 2 2 k m d k m d k m 2 2 k m 2 k m2 k m N 2