HW5+Solutions - 1 2 0 1 l M l | i x“ = t be a periodic...

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Unformatted text preview: 1 2 0 1 l/\ M l/\ |/\ i x“) = t, be a periodic signal with fundamental period T = 2 and Fourier coefficients ak. (a) Determine the value of a0. . (b) Determine the Fourier series representation of dx(t)/dt. ‘ b (c) Use the result of part (b) and the differentiation property of the continuous-time Fourier series to help determine the Fourier series coefficrents of x(t). t H Consider the following three continuous-time signals with a fundamental period of T = 1/2: x(t) = cos(47rt), y(t) = sin(47rt), ztt) = x(t)y(t)- (3) Determine the Fourier series coefficients of x(t)‘. (b) Determine the Fourier series coefficients of y(t). (c) Use the results of parts (a) and (b), along with the multiplication property of the continuous-time Fourier series, to determine the Fourier series coefficients of r 20) = x(t)y(t)- (d) Determine the Fourier series coefficients of z(t) through direct expansion of z(t) in trigonometric form, and compare your result with that of part (c). H @Let x(t) be a periodic signal whose Fourier series coefficients are H Use Fourier series properties to answer the following questions: (a) Is x(t) real? (b) Is x(t) even? (c) Is dx(t)/dt even? i’i A discrete—time periodic signal ic[n] is real valued and has a fundamental period N = 5. The nonzero Fourier series coefficients for x[n] are 2, k = 0 j(%)lki, otherwise” ~1r a0 = 2, a2 = aiz = Zejfllé, a4 = (1:4 = e13. Express x[n] in the form x[n] = A0 + ZAk sin(wkn + (pk). k=1 H @Consider the following three discrete~time signals with a fundamental period of 6: 271' 7r x[n] = 1 + cos y[n] = sin + Z), z[n] = x[n]y[n]. (a) Determine the Fourier series coefficients of x[n]. (b) Determine the Fourier series coefficients of y[n]. (c) Use the results of parts (a) and (b), along with the multiplication property of the discrete—time Fourier series, to determine the Fourier series coefficients of z[n] = x[n]y[n]. ((1) Determine the Fourier series coefficients of z[n] through direct evaluation, and compare your result with that of part (c). X65): C44“), 3(JC):S(V\(LWJQ/ 96—);M433g07 294-22“; 3&07‘373t m 7 N40): MUM) > e, / Jr :57: 6/ 67 WW; w m av» #va 2n 9'4 *2/31; ‘(4 23%; (b) Ewe); gimw’mt) ; 6’ m; “3:” 2/) {L 5 fl 7 A y 14 ii. -;) L31: :: AWA, O) ?l: (C) I , 4”” ‘ Sum fit); XLfc)‘ Qt) & firm 2 L 0w; W \ 2 1L: ’0“ a +00 ‘ Wm > Cr 39w"; 2 m «a my 3% b490, Hi1, mam? [Mt ~ We Cw ‘NMW Mfkfl V117 MM, 3% «My ‘Qwr'a 12, QLK,WW+ _ A 4 \z_”1; EV" M \44/("3 i7 / A V _ \L: O) 2, yr“ '" J’— J ,, , ‘ M 5,,{L A, 5,30 1% Lo) an: MN at-qu 1 (30+ L <3) ' , \a ,, A, L; L h" XQ‘N‘LW 44"":5 i AM“ 3“ 5, (*3 :v m L, , 4, A, 1.- .L L 17 Q1” al'k’fr 4/1 ’1 ~ 1 < _ PM ‘Lz’flk , 3r N X ~\l XY‘1912/ZAC ’ in} @1 j 21110, 3“ j M , vL’EflM , M “Ip‘L : d Raiakflg NJ: a? N4 ‘ fikfilfl‘m : 2% low , }{ New “Fifi. , . : :24 MA @(kximfil) 31—: K 1 A w gznpflfw M; Ibo ~ I N" : [$4.3m (314+ f L ‘4” Sm [Lzfi’bh'l'cflb] ' Wig? WM; W’Offllil mm)” “at” éflmm 4) A“ 6vaL AF: / A012/ Lffilzti All” W15 A1111 LIV 17:er R510) Lfk‘g ‘ TY“. Min (W11; 3: ‘ 3 «am ...
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