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A#08_Solutions - Assignment#8 Solutions- pl ECSE-2410...

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Unformatted text preview: Assignment #8 - Solutions -« pl ECSE-2410 Signais & Systems - Spring 2006 Due Tue 10/03/06 Using the properties and transform Tabies (no integrations of the definition) find the Fourier transform, X (ca) , for the foliewing signals: . ‘20 (21) x0) x u(t) - u(t — 2) (Express answer as a sine function.) wmw) -;. ~L {firng r jbo u) - «- are e“; M .. 1 M2: 5;, wwfifiej “we a 132m “jag “Lu “flat: 0:. W “310 ~ ' leg 3/9))“; 6 g 6 )2; 83“](83 m6 éjfia'giéw)= gaging) / m "i‘i i “a! “(t‘z'g Cb) W) ee""*"u(r 2) :- eabfi if )LLL'€~<1)""”& e. ate-z) W “i “:32” i “O‘U‘mi (Us) 1 H‘JLL} 3L6 - ‘ \ (0) x0) =5(t+1)—5(r—1) (Write asatrig function.) fiflwplé “:18 - J SM 00)] We. 2%) as MW muzfiw 1’9 3% WW? _ m, ‘ ‘rrt- (d) x(r)=cos(7rt+§~) =3 7'": 2.. Z . w: i? A + w my: ass/{firm r 5: gm r3 ) z /#r W“ “,7? r 2W i Assignment #8 - Solutions -~ p.2 ECSE‘24IO Signals & Systems - Spring 2006 Due Tue 10/03/06 (6) x0) = e" [u(z‘) - u( r~ 1)} (Express under a common denominator.)& I) J; --I M " ma 3.. m M g 6““W a e we) «a e W) W (t) x0) 2 e“ .31: h. A _.L_, {2+- w} .. e w.) 4—9 170122” W i w , M W _ / .L/ :2 Z ~—~_5 I fimm): ij +£+jw H'LUZ’ W" {we} «m: M—1)a———-» fi®==mda M z—z—w?‘ M FF...— #— Assignment #8 - Solutions ~— p.3 ECSE-24IO Signals & Systems n Spring 2006 Due Tue 10/03/06 I (g) x(t) = Isincfinflr (simplify answer) ww 1% W 245%}:J mm W Eta} 2% 31E} +m§m>fldd d1) MOW M0 =2 9m CM) 9133* M 4449‘” m9 2: ¥gk<€$§ A?» ‘“ a) E E M -.\AL COM/LVN +0 Mflaskcbfilm 331:?“ M Amid :wflu) my)“ SzficéflMTM flail-w in; “W. 13* { TL" nglw) Mid-[TM & m J14) 2’ ) ECO)" O ) else \ Assignment #8 - Solutions —— pA 13088-2410 Signais & Systems - Spring 2006 (11) Due Tue 10/03/06 Assignment #8 - Soiutions — p.5 ECSE-2410 Signals & Systems - Spring 2006 Due Tue 10/03/06 (j) One technique for finding Fourier transforms is to differentiate the signal} until it reduces to a waveform for which we have the Fourier transform in one of our Tables. Then the time differentiation property (or the integration property) can he used to find the transform of the original signal Use this technique to find the Fourier transform 0fx(z) glotth below. m) MEN) edit W (H Assignment #8 - Solutions — {1.6 ECSE-2410 Signals & Systems - Spring 2006 Due Tue 10/03/06 0) continued. J 2 £05 awn/{J Alfie? W W [muvwzm 7950M gala-93w) , Am WEI/w} {fPLHQmfi/w} ( fl : fibg 96* Lima 4: #2, wt t I” t ,"’*‘$’2__ Ft "35L y; 5' g mew ii? iii“? 72*"? f \ #2); 439g =.—> +24 59— ...
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