HW5+Solutions

# 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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