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# hw3 - I i f i m.I ”m W PROBLEM 3.10 Ha F-wav‘e £3...

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Unformatted text preview: I i f i m .I ”m , W PROBLEM 3.10: Ha! F-wav‘e £3 Mme‘h‘ﬂ t (0.) xl’té): t 41\$ (b7 -I ’7 u, --' 4 E2, €JGA2)—w 8.36"wa / => 7541 _' 2n 022+ ax= 1-1; S xme“ m 4 => cup-o {2:1— keveM. McClellan, Schafer and Yoder, Signal Processing First, ISBN 0-13—065562-7. Pearson Prentice Hall, Inc. Upper Saddle Rive r, NJ 07458. © 2003 : -x(b\) = éwvmmu junk I { WWIIJIEIII’M;WTWW?! PROBLEM 3.19: Ckaraekra'ze each {mac Signal? (0.) 6 Féf‘l‘oés Fan“, t=~2 {'0 t: +3 =>T= 35%;ng DC value = 2 ém>0 =’> <P<° (b) 3 Perfoés From fr—o {-0 t= 2. => 7-=2. -sec. DC. value- = 0 90: 7r (0) 6F€WOJS ﬁrm». tam—~75 ta i= —.2 =9 T= %Sec DC value: 2 tm <0 2) cp>o (d) RENOA DC value = 0 ~ 3-1;, = g; 56:5 . Tim Freguenca'es (3) Z i>€ﬁ035 {rum é= -—2 iv; t=3~5 =>"H71?"?'575gcs TWO ﬁeﬁbevmies. ' DC. vaIUe? =0 (l) we: 31:0.2) => 'T'i'ﬂa = 5% s“ (p: 0.57:- >0 p; -= .2 ('2) “Jo: 3W(°3) because 0-3 Jun/mics 06%] 2:) T £1635: "3/5 ﬁccs. DC=O (3) Luke. (I). T: 5/4 Sec. DC=Z ‘- BU+ LP: ‘00ﬁﬁr< O (4) w = 2,4043 => ‘T' VOA‘: 2 §SCCSWIIMH 136— ~ 0 (5) <P=rr 'T'= ”.5— = ggsec.’ Thus) +ch mark/Ix is a _. : T (o) 4—; (3) I (c) H (x) f Esp“) (b) w9(5) (J>+—~<2) \$12,353:; 12233"? 1%;pifs’aééimm e in” \$251385 \$306556” “was“ ”'d"°"’”°"‘°d°’° “m““a‘wm mm d" “”3st “"5“ pennission hasbeen oblziin edfrom lhea ulhors December 30, 2003 ® I «9%? AZ: (*9) NM» Wdé> Qwﬁ ’ (Lu/2,3) ‘ ...
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hw3 - I i f i m.I ”m W PROBLEM 3.10 Ha F-wav‘e £3...

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