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HW1solution - ECEM301: Homework 1 Solutions Due January 26,...

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Unformatted text preview: ECEM301: Homework 1 Solutions Due January 26, 2009 in class SEE QVDENDUW HT EN D FDR WRTHEE EXVLANHW (a Linear, stable (b Memoryless, linear, causal, stable (d Linear, causal, stable ) ) ) ) (e) ) ) 2 causal, stable meme? lGSS or (a) $(t) periodic with period T; yl (t) periodic, period T/ 2. (b) ( ) (d) y2(t) periodic; period T; 13(25) periodic, period T/ 2. Time invariant ,. Linear, stable gag‘l’ @Wmiiha (g Time invariant, linear, causal fl nu consider fie w‘ 1.3 All statements are true. yl (t) periodic; period T; x(t) periodic, period 2T. 0 33(t) periodic with period T; y2(t) periodic, period 2T. 1.49 We have a +jb = re”, Where 7" = \/(12 + b2, and 6 = tan—1(b/a). (Be mindful when using tan‘1 of What quadrant a + jb is in). JV f ’4 (a) 1 + N3 = 2ej7r/3 , 1W “.9 ° (b) —5 = 569‘” r 5/2 9:1713 (d) 3 + 4j : 5ejtan—1(4/3) % 56309273 ( air/L1 __ ; j7T/3 1) 1+j¢§ ~— 26 I 353 4/4 '2: 2 z '/z fl, a+’ Hyfi I , = “ "3 :‘1[‘3-23fa+']=i,[wl3il‘ Mrs (ya—c) 3 -i ‘Ee -l‘ i \ J Eli" JWB N 6 o \hVolV€ 2 Sample‘i oi xl’c) ' —l*')£3 ‘7. Z. s 1 ‘J r twafilflaurfi axis Figure 1: Problem 1.49 1.51 Euler’s relation, 6” 2 005(6) +jsin(6) (1) = 005(6) — j sin(6) (2) (a) Adding (1) and (2), we get ej6 + e‘j6 = 2 008(6) 1 => 008(6) = -2—(ej9 + 6—”) (b) Subtracting (2) from (1), we get 619 — 6—39 = 23' sin(6) :> 5111(6) 2 TR") — 6—30). .7 (C) we have 6mm = we”, :S—O(tc}<l)?% + 3 9%wa *1 3m cos(6 + 925) + j sin(6 —|— $) = (cos 6 cos ([5 —— sin 6 sin ¢) + j(sin 6 COS ¢ + COS 9 Sin ¢) With 6 = qfi, we have (‘77 eiwfimfl “We {64‘ {was m ffif’aim abwe) cos 26 = cos.2 6 — sin2 6 With 6 = —¢, we have 1 = cos2 6 + Sin2 6 Adding these two equations and simplifying gives 1 cos2 6 = §(1 + cos 26) -2‘ WMM__WWNMM (d) Observing the real part of (3), we have 005(6 + (/5) cos 6 cos ([5 — Sin 0 sin <25 and c0s(6 — ¢) ‘= cos «9 cos ¢ + Sin (9 sin <15. M (29d mmufiis’t) . Subtractlng these equations, we obtain 1 ((50 (2M) sin6sin ¢ = §[cos(6’ — ng) — 008(0 + 915)] (e) Equating the imaginary parts of (3), we have sin(0 + ¢) 1.54 (a) For a = 1, it is obvious that sin 6 cos ¢ + cos 0 sin qfi t on QHQ‘HIQF {‘Qflmfi a? [si'Sum are Candid b7 WS 6F 2mm Thnhkof‘ Pi ‘HMS Way‘- Zan I N because \ 15 04023 N twmeS. ‘2‘ %. 4n - eaWCQuqJ“°m n=0 I If m “'0” " [9+ 9:“ i Z (A lasi‘term of _ h f 31.. For a aé 1, we have :' k. 2-“; summed 5 ' N—l 'N—1 f (1—a)Zo/‘ = a”— =1—cgN n20 n20 - ‘ M N\ T n _ 1 — a \ |si+erm 77.20 a Summa‘hon (b) For |a| < 1, we have lirnNnoo aN = 0. Therefore, from (a), we have 00 N—l 1 n = n = . n: (c) Differentiating (b), we obtain 1 | (d) We have for [ozl < 1. 1.55. a The desired sum is "’ 9 ~ 1— .. . ‘ Zen/2: ‘ —A ' (b) The desued sum IS “:0 , 7 9 Z 'ej’lm/Z m ej7rn/2 : n=—2 71:0 “I Prod ' 17a : PL; n_—.o ‘ ‘1 — (1/2 W (d) The desired sum is. i(l/2)nej1rn/2 :(1/2)nej7rn/2 = _ —0 n=2 : 44 n— (e) The desired sum is 9 9 9 i (member _ 1 - /2 l —]1m/2 __ _ _ _ Z 1 ‘ 3 wk?“ Iagdd’rg ECOSUm/m - 528]” 2:8 (1 +3) + 2(1 J) n ( a; E "=0 "z > ":30 n_ I ,wkeré’A-J/z (1 j ’4 : '2 )wlime d : 1‘ 3 (f) The desired sum is "z if ‘ 2“) 2 M 2 “A \ ‘ \ \ "" E 00 2. -" 'J +- 2 +i i 2(1/2)" cos(1rn/2) = 7-0 a) 0 i 4 , 2' 4 _ 2 . 4 ' —‘ 2’3 i ‘ "16+716 fi_7fi=§' _ 2y . q 2 I ‘ “:J‘ 4 3:3 I: " i ' b 5 5 1 a) W Eemuse oufiufi 409$ 'i‘o“ degeng 51mg»?! em WW“? a? ¥fesgmz.,§k‘ma 5 mean“ \3 X.lt):§(t)lom+?fi (s 5(t-2)+w ;zglt_2):\lllt) no’f Jrlmmnvamvfi because 0C mph Ks X’AHFX'Q‘Z); ou+ \fi rs (Spa—2) \mear Manse = 3(a) . M4) + a? #2939; 1. 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HW1solution - ECEM301: Homework 1 Solutions Due January 26,...

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