ENEE 420 FALL 2010 COMMUNICATIONS SYSTEMS ANSWER KEY TO TEST # 2:

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1 ENEE /FALL ENEE FALL COMMUNICAIONS SYSEMS ANSWER KEY O ES # :..a Fix in R. I is plain ha z = Φy AM = aa c + k A m cos πf c + ba c + k A m cos πf c = aa c + k A m cos πf c + b A c + k Am + cos πf c = aa c + k A m cos πf c + b A c + k Am + b A c + ka m + k Am cos πf c...b Firs we feed he signal z : R R hrough a low-pass filer wih cu-off frequency B, namely if f B Hf = if f > B. Recall ha m is a low-pass signal cuoff frequency B, so ha he signal m is also a low-pass signal bu wih cuoff frequency B. herefore, provided he condiions w = h z = b A c + k Am, R B < f c B B < f c B hold. his amouns o 3B < f c.

2 ENEE /FALL Nex we use he square-rooer on w o obain w = b A c + k A m provided + k A m, R. Finally a use a dc-blocker o exrac k A m. Noe ha his approach has he advanage of avoiding muliplicaion by cos πf c in order o achieve demodulaion...a By sard properies of Fourier ransforms we readily find P f = amf + + K k= K k= b k j a k M M f k + M f + k f k M f + k, f R...b he required filer h : R R needs o saisfy he requiremen h p = y DSB SC, or equivalenly, Y DSB SC f = HfP f, f R in he frequency domain, wih Y DSB SC f = A c Mf f c + Mf + f c, f R. Upon comparing wih he expression for P f, we see ha h can be seleced as if f f c B Hf = if f f c > B provided he consrains f c = k for some k =,..., K such ha a k b k = B < hold..c Yes since he periodic carrier c : R R admis a Fourier represenaion provided similar condiions as above hold. More precisely, wih Fourier series expansion c = k c ke jπ k, R

3 ENEE /FALL we ge Fc mf = k c km f k, f R he needed condiions for y DSB SC = h p are now f c = k for some k =, ±,... such ha a k b k = B < a We have H VSB f = if f f c B B f fc + B if fc B f f c + B if f c + B f f c + B if f c + B < f. wih H VSB f = H VSB f, f. From his expression for H VSB or graphically, i is easy o check ha if f B H VSB f f c + H VSB f + f c = if f > B.3 he requisie condiion holds. 3.b he ransmission bwidh B is given by 3.c By consrucion so ha wih Here, B = B + B = 3B. y VSB = h VSB y DSB SC Y VSB f = H VSB fy DSB SC f, f R Y DSB SC f = A c Mf f c + Mf + f c, f R. Mf = A m δf f m + δf + f m, f R

4 ENEE /FALL so ha I is now plain ha Y VSB f Y DSB SC f = A ca m = H VSB fy DSB SC f = A ca m When + A ca m + A ca m δf f c f m + δf f c + f m δf + f c f m + δf + f c + f m.. H VSB f c + f m δf f c f m + H VSB f c f m δf f c + f m H VSB f c + f m δf + f c f m + H VSB f c f m δf + f c + f m. < c <, we noe ha herefore, we conclude ha Y VSB f = A ca m H VSB f c + f m = H VSB f c + f m = H VSB f c f m = H VSB f c f m =. δf f c + f m + δf + f c + f m, f R, whence y VSB = A ca m cos πf c + f m, R. 3.d When < c <, we now have H VSB f c + f m = H VSB f c + f m = f m + B = c + B H VSB f c f m = H VSB f c f m = B f m + B = c + whence Y VSB f = A ca m + A ca m I is now easy o see ha y VSB = A ca m c + δf f c f m + c + c + δf + f c f m + c + δf f c + f m δf + f c + f m. c + cos πf c + f m + c + cos πf c f m

5 ENEE /FALL for each in R...a he signal ε is periodic wih period, we have e jk P m d = d =. hus, he signal ε admis a Fourier series represenaion of he form ε = k c ke jπ k, R wih Fourier coefficiens c k = e jk P m e jπ k d, k =, ±, ±,.... Now, for a given k =, ±, ±,..., we ge For k =, c k = = = e jk P m e jπ k d + e jk P e jπ k d + e jk P e jπ k d + c = e jk P m e jπ k d e jk P e jπ k d e jk P + e jk P = cos kp. e jk P e jπ k d..5 For k = ±, ±,..., c k = e jk P e j + ejk P e j jπ k jπ k = e jk P e j + e jk P e j j j = e jk P k + e jk P k j j k e jk P e jk P = j k = sin k P.6

6 ENEE /FALL As a resul, ε = cos k P sin k P k k k = cos k P sin k P k= k = cos k P j sin k P k= = cos k P + j sin k P π l + sin e jπ k e jπ k e jπ k sin π k π l +..7.b he PM signal y PM : R R can be evaluaed as follows: For each in R, we ge y PM = A c cos πf c + k P m = A c Re e jπfc+k P m = A c Re e jπfc ε = A c Re e jπfc cos k P + j sin k P π = A c cos k P cos πf c A c π sin k P sin πf c = A c cos k P cos πf c A c π sin k P + A c π sin k P l + cos l + cos l + sin π l + l + sin π l + π f c l + π f c + l +..8.c he PM signal y PM : R R has frequency conen a he discree frequencies f = ±f c f = ± f c ± l +, l =,,... he power of he PM signal y PM : R R can now be compued in he usual manner: P ypm = lim A A = A c lim = A c A A y PM d A A A + lim A A A cos πf c + k P m d A A cos πf c + k P m d = A c.9

7 ENEE /FALL for i is easy o see ha lim A A cos πf c + k P m d = A A by a symmery argumen ha uses he form of m. Anoher approach is o use your answer in Par.c: A P ypm = lim y PM d A A A = A c cos k P + A c π sin k P + A c π sin k P l + = A c cos k P + A c π sin k P l + l + = A c.. his las fac follows from he well-known ideniy Indeed. holds if cos k P + 8 π sin k P his condiion can be rewrien as. validaes i! 8 π sin k P l + = π 8.. l + =, l + = cos k P = sin k P, his is also a consequence of Fourier series analysis wih he help of Parseval s heorem. You were no expeced o know i.