Exponential Sawtooth

Transcription

1 ECPE 36 HOMEWORK 3: PROPERTIES OF THE FOURIER TRANSFORM SOLUTION. Exponenial Sawooh: The eaie way o do hi problem i o look a he Fourier ranform of a ingle exponenial funcion, () = exp( )u(). From he able in Appendix II of he Roden book, we know ha he following are Fourier ranform pair: exp( )u() ψ! +jßf I hen become much eaier o calculae he Fourier erie coefficien for he awooh waveform: y() = X n= exp( [ n]) u( n) The funcion () i replicaed and hifed over ineger period of econd o form he periodic funcion y(). Since he period of y() i econd, he fundamenal frequency f i equal o = Hz. Thu, he nh complex Fourier coefficien igiven by he following formula developed in Lecure 5: c n = S(nf) = S(n=) = +jßn Thi i he final anwer.. Malab Analyi: The iniial wo graph of he exercie were plo of he periodic exponenial awooh and i Fourier ranform:

2 SOLUTION. Exponenial Sawooh. ignal (econd) 7 Fourier Tranform of Exponenial Sawooh 6 5 pecrum f (Hz) Since he waveform i periodic, he pecrum coni of pecral line. Ploing a ingle exponenial waveform in he ime domain produce he following graph:

3 ECPE 36 HOMEWORK 3: PROPERTIES OF THE FOURIER TRANSFORM 3 Exponenial ignal (econd) The following Malab command were ued o conruc hi graph: yy =.5*(+ign(-.)).*exp(-); plo(,yy); e(gca,'fonsize',); xlabel(' (econd)'); ylabel('ignal'); ile('exponenial'); The Fourier ranform of he exponenial wa graphed wih he following command: plo(f,ab(ffhif(ff(yy)))); e(gca,'fonsize',); xlabel('f (Hz)'); ylabel('pecrum'); ile('fourier Tranform of Exponenial'); Thi produce he following graph in he frequency domain:

4 SOLUTION Fourier Tranform of Exponenial pecrum f (Hz) Noe ha he hape of he periodic exponenial Fourier ranform and he hape of he ingle exponenial Fourier ranform are idenical. The key difference i ha he periodic exponenial Fourier ranform coni of pecral line inead of a coninuou pecrum. 3. Pareval' Theorem: a. Fir, we recognize ha if we have a ignal (x) = +ß x, hen we can apply Pareval' heorem a follow: + R ( + ß x ) dx = = + R j(x)j dx + R js(f)j df where (x) and S(f) are Fourier ranform pair. (Don' be confued by he ue of x inead of f i' ju a variable name and doe no affec he mahemaic. If i ge oo confuing, ju ubiue everywhere you ee x) The acual Fourier ranform pair ha we ue o evaluae hi inegral may be found in App. II: +ß x ψ! exp( jfj)

5 ECPE 36 HOMEWORK 3: PROPERTIES OF THE FOURIER TRANSFORM 5 Subiuing he frequency domain funcion exp( jf j)ino he above equaion produce + R ( + ß x ) dx = = = = + R j exp( jfj)j df + R + R + R exp( jfj) df exp( f) df + exp( f) df = exp( f) fi fifi + R exp(+f) df = b. Recognize ha if we have a ignal ( ) = a follow: + R + R in ß d = = in ß, hen we can apply Pareval' heorem + R j( )j d + R js(f)j df where ( ) and S(f) are he Fourier ranform pair (from App. II): jfj in ß ψ! ßu Subiuing he frequency domain funcion ßu jfj a box funcion from f = = o f = += ino he above equaion produce in ß d = + R fi fi jfj fi ßu fi df uni ep quared i ill a uni ep + R = ß jfj u df inegral i zero for f< = and f>= = ß += R df over =» f» += uni ep i = = ß Noe ha he clever ue of Pareval' Theorem urn nearly-impoible inegral ino much eaier inegral.. Omied hi week ince we did no cover he maerial in ime.

6 6 SOLUTION 5. Convoluion: The wo ignal, () and (), are convolved ogeher. The ignal are lied below: () = ( jj)u( jj) = () = u( jj) = 8 >< >: +»»»» elewhere (»» elewhere Thee wo ignal are very imple o he graphical convoluion example in he book on page 7. The mo difficul par of a convoluion problem i eing up he inegral ued o calculae he convoluion reul. A face value, i doe no look like a difficul problem ju plug he wo ignal ino he following convoluion inegral and you're done: Anwer = Z+ ( )( ) d The difficul par abou evaluaing hi inegral i ha here are wo variable in he inegrand, and. The variable i he variable of inegraion, while may be reaed a a conan. However, for differen value of, he folding-and-muliplicaion of and look very differen and require differen ype of piece-wie evaluaion of he convoluion inegral. Noe: we could have choen he convoluion inegrand o be ( )( ) and we would ge he ame anwer. Baically, a increae, ignal lide ino he non-zero range of ignal. Every ime a diconinuiy in ignal croe a diconinuiy in ignal, a new e of piece-wie inegraion mu be wrien o olve he convoluion inegral. The oluion for hi problem i wrien below. Follow he graph a he end o underand why he range and piece-wie inegraion keep changing. I. In hi region ( < ), he non-zero region of he funcion ( ) and ( ) do no overlap a all. Therefore, when hey are muliplied ogeher, he inegrand of he convoluion (and, of coure, he reul of he inegraion ielf) i zero. II. In hi region (» < ), he fir half of he riangular region of ( ) creep ino he non-zero inerval of ( ). The reuling convoluion inegrand i non-zero over he inerval [ ;+ ]. The inegral for hi region i given by Z+ ( )( ) d = Z + [+( )][] d = ( + ) fi fi fi + = III. In hi region (» < ), he fir half of he riangular region of ( ) i compleely underneah he non-zero inerval of ( ); he la half of he riangular funcion i alo enering hi region. The reuling convoluion inegrand i non-zero over he inerval [ ; + ]. For he ake of eaier calculaion, he inegraion will

7 ECPE 36 HOMEWORK 3: PROPERTIES OF THE FOURIER TRANSFORM 7 be performed in wo piece-wie egmen from [ ;] and from [; + ]. Thu, he inegral for hi region i given by Z+ ( )( ) d = Z + [+( )][] d + Z [ ( )][] d fi = (+) fi fi + fi +( ) + fi fi = IV. In hi region (» <), he la half of he riangular region of ( ) i compleely underneah he non-zero inerval of ( ); i i he fir half ha i only parially conained, runcaed a =. The reuling convoluion inegrand i non-zero over he inerval [ ; ]. The reul, like region III, mu be expreed a wo piece-wie inegraion from [ ;] and from [; ]. The reuling inegral for hi region i Z+ Z Z ( )( ) d = [+( )][] d + [ ( )][] d fi = (+) fi fi fi +( ) + fi fi = V. Thi region (» < ) ee only he la half of he riangular region of ( ) underneah he non-zero porion of ( ). The reuling convoluion inegrand i nonzero over he inerval [ ; ]. A a reul, he inegraion i Z+ Z ( )( ) d = [ ( )][] d = ( ) + fi fi fi = 8 +8 VI. The region for > i imilar o region I in ha no non-zero inerval of he funcion ( ) and ( ) line up. Thu, he convoluion inegral i equal o zero for all >. Phew! Tha wa a lo of work. The mo difficul par of convoluion problem are defining he range of inegraion for he differen inerval of ime,. Remember, when liding one funcion pa anoher along he -axi, you mu re-wrie he inegral every ime a diconinuiy in one funcion croe a diconinuiy in anoher funcion. To ummarize he oluion we ju found, we may wrie he convolved funcion a () Ω () = 8 >< >: < +8 +8» <» < 8 +8» < >

8 8 SOLUTION I. <- - ( ) (- ) Convoluion Inegrand ( ) (- ) II. -<<- - + (- ) III. -<< (- ) IV.<< (- ) V. << (- ) VI. > (- )