Note 14. Worked-Out Fourier Transform Pairs-2

Note 14. Worked-Out Fourier Transform Pairs-2 - l‘f Note...

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Unformatted text preview: l‘f Note 5. Worked-Out Fourier Transform Pairs ' ECSE-2410 Signals & Systems (Wozny) Spﬁﬂg 2000 .e Fourier transfonn of a signal 25(1), is defined as X(w) m fx(t)e'j“’dt. Also written as X(w) x F{x(t)}. an The inverse Fourier transform is deﬁned as x(t) = 3‘; {X{w)e”"dw. Aso written as 1(1) : F'1{X(a3)}. v —oo The signal x(t) and its transform X(a2) represent a Fonn'er transform pair, MI} (—9 X (to). (See Text, p.301) Ali transform pairs can be found by direct integration of the defining integrals. In many cases, however, the integration can be simplifed and often avoided by the use of transform properties. First we deveiop a basic set of four transform pairs and then use the transform properties in text, Table 4.1 to ﬁnd transforms and/or inverse transforms of new signals. Basic Set of Transform Pairs. x(t) «a X{w) 1. Find the Foutier pansfonn of x{t) : (“14(1), ﬂew} > 0, by direct integration. (See text, Example 4.1.) 2. Find the Fourier transform of x(t) : 5(1), by direct integration. (See text, Example 4.3.) 3. Find the Foun'er transform of x(t) 2: um. Unfortunately this integral does not converge. We can force convergence by deﬁning the step function as limemuv). Thus we transform of e""'u(t) and then take the 0—H) .limit a —> 0. (See these notes, p2.) . Find the inverse Fourier transform of X(w) :2 2715(0)) by direet integration. Summary of basic transform pairs. 1(1) X(w) Wh 13m) 1 imam“ 1+?» awn-mitt) 1. x(t)me““u(t), 92g (1 > 0 : Mm); _ wt { } I (1+th} w at J 2.- ‘x([) a 2. x0) :2 Xuﬂ) =1 r f 3 co X”) - 1 3. W) a W) , ’- X(a)} ..—. 73+ wmnjf "h w . I J a) ﬂu?“ . 1(1) :1 <= X(w):2n6(w) s w W i— 5. Find th) for x(:) :2 11(1) by direct integration. V m an no - 3 . . Xiw) = IMUE—jmd! : féw’w'dt = "i—em’w' . But upper limit is indetermrnatei Can’t evaiuate Integral. —«> 0 1w 0 7. Use convergence factor. Let 5%(2) 2 e”“'u(t), where real number a > 0. Thus 14(1) = 5333(1), and . ~ ~ ~ ~ a) X023) = imgXuo), where X.(w) w XR(w)+jX,(a)) = 1 z a wj a+jw c22+w2 erg-ta)? Note first term. '- Trcglw) & I ! ‘19.?63. w ‘39 J. 213. . tiotw) I ? => iiﬂBXR(w)==7E5(w) => ' a—n-b " ~ ” a _ a) . - . “ , where the area, IXR(w)da) = {We} : Ian {M} =2t,1s constant and 1ndependent of a’. ma +a)2 0 m Therefore, X(w) = iim)?(w) : 7:5(w)- j—1—, or Mr) <-—-> #+m§(w) . a-w-JO a; la) Linearity Property. ax](t)+ bxzu) («-9 aXﬁw) +bX2(a)}, where 0,!) are constants. Time Shifting Property. x(t-to)+—-:~e"“"”X(ar). a T 1 1 6. Find the Fourier transform X(w) ofa single pulse x(t) = {1’ I *l < . 0, else “T. T‘ t Let x(t) = u(t + T, ) -— u(t - 7]), then trﬁnsform the shifted step functions using 0 linearin property. X(w) : F{tt(t «t» 71—— F{u(t ~— T, - timeshifting property. F{u(t + Ti : em“ F{tt(t)}, F{u(l — 32)} 2 e'W‘F{u(t)} ' 1 - transform 3 from basic set. F u(t) = -;-—+ 7:6(w) { } 1w Ebb) ‘ (HT ZT‘ Ans. X(w) = 2 5"“ t) a zrjsincmn), i.e., («a 27]sinc(coT,) . ‘ co T 1: Example 4.4 in text ﬁnds this transform by direct integration of IO). Time Reversal Property. x(~—I) <-~> X(-—w). i 5'45“) 7. Find rhe Fourier transform of x(r) = 6““, a > 0. At . Express x(z) z x§{t)+x1{*-t), where 36(1):: e‘a'um, and use - linearity property. X(w) = F{x1(r)} + F{I§ (“0} - time reversai property.‘ If F{x] (1’)} 2 X1 {w}, then F{x1 (-0} 2 X1(—-a)) 1 file a+jm' 4 J. 202 2:2 "‘ " & Ans. X(m) : 2 2, i.e., x(t)= e'al‘l, a > 0 (w) 2 2 . ' i a +0) (1 +0) a) 4 Example 4.2 in text ﬁnds the transform by direct integration of x(t). "6" Frequency Shifting Property. e’w“'x(r)<——> X(w~»wn). fir) 8. Find rhe inverse Fourier transform x0) of X(a)) m 27r5£w~ we). Use w Id - frequency shifting property. F”1{27I5(w n 600)} = e’w°‘F"l {2%5(w)}. v inverse transform 4. F’]{27r5(co) = I Ans. x0) = em“, a compiex signai, or, em“ 4-) 27:6(a) «— wﬁ) . 9. Exercise. Find X(w) for 25(1) m e“i"cos(101). .Ans. X(a))=——~l-m~5+«m~m}~m—m§. Euro—10) i+(w+]0) 10. Find the inverse Fourier transform x(1) of any weighted sum X (w) z 27: Z a*5(co -— kwn) of equally spaced W 31.9) impulses. Use i M‘ I 27! zak5(w—kwg)} m XakF"{2n5(w-mkwo)}. ‘U ‘ linearity property. F ”‘{ -wo a we we - transform 1. F {x1 (1)} = tbs-m tram - frequency shifting property. F”'{2n5(w -— kw“ m e’*”“'F'1{2K6(w)} '- transform pair 4. F"l{2n5(a))} = 1. Answer is 1(1): F"1{X(w)} = Zak}:-l {235(a) — kw” m ayejm‘“ , rhe complex form of the Fourier k=‘°° kz—oe 1 1. Exercise. Find X (co) for 25(1) periodic, with period 4 sec. Use 1 «likew‘g'dr :lsinc[k£) 44 2 2 ' for periodic signals. ah :— -I- J. x(I)e""‘°’°'dr = (TD . - X(a)) = 27: iakcﬁww kwe) = 27: k=v~m yer) 12. Exercise. Find X(a)) for 1(1) = 25(I“”T)- n9.» f: 1 2a -T 0 ‘3“ 1“" Ans. X(a)) 2 2:: —5(co— fawn), cos n m. (See IEXE, Example 4.8.) - kzm T T 300) 2* .‘ 1* *5 T (U 13. 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Linea]in ax(I) + [3320) CIXCCO) + bY(w) Time Shiﬁing ‘ m — :0) e'jw'°X{co) Frequency Shifting eimu'xm X(a3 m (06) Conjugation x‘ (I) _ X’ (—00) Time Reversal x(——z) Xl-GJ) Time Scaling Mat) l0] 0 Convolution in Time x(z) * y(r) X(CD)Y(£0) Mulziplieation in Time x(I)}‘(I) :21- XW) * 17(0) yr Differentiation in Time 5x0) faJX(w) r Integration in Time unmr; -,—}-X(a)) + xX(0)5(co) Differentiation in Frequency :x(:) ji X03» dw X (co) = X '(—-w) . 932{X(w)} :2 9ie{X(~a))} 3m{X(w)} : -Sm{X(—m)} Conjugate Symmetry for x0) i‘eal Real Signals [X (03)} : lX (—60)! ZXUO) m —./_’X(-—c0) Symmetry for I , Rea] & Even Signals xU) red} end even X(a)) real and even wig??? 35d Signals 1(1) real and odd X(a)) purely imaginaiy & odd EvenuOdd Decomposition for x3”) 2 5143(0)} [xm real] SEE-{XWH Real Signals x00) = Od{x(t)} (m) real] jSm{X(co)} Duality gU) +9 f (w) ﬂabiﬂwe dv f(r)<-+27rg(ww) Paiseval’s Relation for Apen'odic Signals f]x(t)l2dt a a}; JIX (@1202!) -Zg- 'f‘ ﬂ> 3.". L aﬁﬂavx; «W n ~35va N? 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