HW2solution

HW2solution - ECE 301: Homework 2 Solutions Due January 30,...

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Unformatted text preview: ECE 301: Homework 2 Solutions Due January 30, 2009 in class 2.21 (a) The desired convolution is (b) Similar to (a), but now replace 0 with a: n = a” = (n + 1)a”u[n] k=0 (0) Case 1, n g 6: )([r\]= 1‘2)“ HUME! j . 9° 00 m: i mum-ta = z (-1 k it"s“ (Db-00 2’ fie ‘ k “‘k //ﬂ k2 : Z (*2) ‘i as long/2&3 neg k3” ﬁas/(apéﬁimh if: at? lea? as {@S‘im‘iwe 5‘5 “iii?” l W] = 19:4 ‘ 00 n lsﬂermmsum -0 2 4” —1 8 k : M- g / > I_ (d? _ 4n _ 1~(—1/8) 1 TL _ Case 2, n > 6: 00 n 3.0 -1 k- lsﬁﬁm go Mn] 2 Z (_1/2)k4n—k : Z < _ msum k=n—2 tun-'2 1' (.;g> __ 4n (—1/8)n—2 _ 1— (—1/8) H (512/9)(—1/2)” We have then that: \$4", n S 6 (512/9)(—1/2)n, n > 6 (EWIer ﬂags graﬂeméé Mn] (d) The convolution is plotted below: I, £3. 7 2.22 (a) The desired convolution is as — m ‘Blbﬂ 7 7777777720); 7’: xrioﬁéd 35c}; ¥ L} “(£0377 9 :“V i = Ate—“‘e‘mt—Wn t2 0 q: t to IN 5?), J(\nen erej‘M 6 once?) ZZSZJCE K ‘ 92PM] WC) 43H): Jgéﬁt MB] : {git—LEE] “(J0 "NWfTMMWWW.ommeiwwwmmw I V Figure 1: Problem 2.21(d) (b) The desired convolution is This may be written as 5- 31(75): 2135| (11-2 Therefore, , as if, 3 ) 3’;th y(t) = /°° “aha—am = /2 h(t — 7')d7' — /5 h(t — 7')d7'. 0 2 2M NH“): 2 f: ext—ﬂak — 25 ext—ﬂak, t < 1 ft: 62(t‘7)d'r - 25 raga—ﬂak, 1 < t < 3 _ 1:1 ext—och, 3 < t < 6 0, 6<t (1/2)[82t _ 26203—2) + 8205—5)], t S 1 '1 Ewan.wa Secure puke. moves “awrbr W“ are r1a1g41'""¥&'2“}‘;51;;2r;r"‘acug'rggggggggngW ‘L' as 1C tmcreaseg W See wdure wt +9 39+ “mfg efgs 0) ff! m ’83” ...t:1 , i‘L :14 - 05.111110113141533, (Z/W)[1_COS{W(t_1m, 19\$ 3 X 3111.; 2 d 3355 = 31(1): = 23351:: X mm) (V’ <2/vr>[cos{vr<t—3>}— 1], 3 st 3 5 262335”1 , J‘ﬁ 4;? Reid 0 5—4: 0, 5 < t (7;, 5 ) New) @Jcro it? '{3 ‘____> (d) Let \ x‘“ W) = W) — —6<t—2>, = 3.3%) ‘gﬁit-z) b . where T /3 0 <1: < 1 4 1 __ _ bl“) : { 0, otherwise ' NOW, \ l W) = Mt) * \$05) = [h1(t) * 96W — g3?“ — 2)- °° , r» 0+ 1.1! 01; to ﬂ, We have 1‘ 8mg th@fy 170:“; 5?; a? F (L §7 > t “bf "("5 "‘t/*ka+{91‘tzq’2t-’or .I/ijav “(J _ f1 _é12_1 _ 2 _ _ t-Ii’tet' 3 ‘2 M4 —._@1@_§,§_£tl:1/?;3,,— 3 [2m 2a(t 1) +191 b(t 1)]: V I '1 {on ,‘v ~ Therefore, k W 4 1 2 1 » 2 1 Maris an V mieg‘ml Come From IPV kli-ﬂgvo ,(t—O ‘ “EIMWOSJ‘e 90f imam uiar «\$52 is ~—7 buried at ﬁe g ? Ii (e) x(t) periodic implies y(t) periodic. determine 1 period only. We have _l ft_;(t—T—1)d»r+ff%(1-—t+T)dT=§+t—t2, —§ <t< W) = i The period of y(t) is 2. ié’Rmeg'i’ is Eacaieéé fir“! _‘/Z t ‘3 (“HEM ﬁgfﬂf] 4. i_—Hﬁ“+’§' ,éqd/Z M- ): \-1;+T’) a 2 1: =t-l T: A12 0 elsewiiore't \12 'L' ~Z 1 4? (Ft)? +1}? + ( v‘)?-\i ) ££ /l as s‘nowm chc‘g‘ure 2 'I'=t~( T—“lz 7 \ M r - ,sz [tvlthL-iirl : 7—3t*’ '2 TWHM Li + 1 8 ‘II A . L31(1—t+T)d¢+f§(t—1—T)dr=t2—3t+g, §<t< l 2 _¢vrw.-w._a —-F_\_.¥L M40 @ ":0 1' “[v1] —H[h—23 'Moves mam >7 n as h‘mcreasej 2.24. (a) We are given that h2[n] = 6[n] + 6[n — 1]. Therefore, __ m h 1“) H31: Mn] * h2[n] = 5[n] + 26[n —-1]+ 6[n — 2]. 2 g i i f i E E E i I I 5 I Since _ hln] = Mn] * [h2[n] * hzlnﬂy h a3 We get n (meré‘assgg'i h[n] =h1[n]+2h1[n— 1]+h1[n—2]. ' f Therefore, I= htol=hd01 => wold, mm M ' ‘5 : h[1] = h1[1]+ 2h1[0} => mm = 3, W ‘ \0: I42] = h1[2]+ 2h1[11+ IMO] => m2] = 3, WW J m = Sum» Lam : 3 n = h.[3] = h1[3]+ 2h1[2]+ hlill # M3] = 2" many 0+ Mn] 2 N.‘ (g: h[4] = mm + 2h1[3] + h1[2] é h1[4] = 1 ah f a ‘Le W = 3+Nr‘ q : M5] = h1[5]+ 2h1[4]+ m3] => h1[5] = 0. _ ﬂew 9 MW (0 ﬂ 7 1C ' " (OW g h1[n]=0forn<0andn25. ' it? N‘:S Fwﬁue (b) In this case, (n 890‘ (Degwwvﬁs y[n] = \$[n] * h[n] = h[n] — h[n —— 1]. ~ -__ [d 2_r‘vl )MDJ) 05kg") 2.40. .(a) Note that t V T}. _ F2 09 , I y(t) = / e‘(t*7')\$('r — 2)d’r =/ e_(t—2_TI)IE(TI)dTI. :B 6- (tdlz-T )u (f_r2_1u’>)( (TV/)3? ' ~00 —00 ~ W 3 Therefore, _ ‘ , 2 g H\\' “ 12“) = e_(t_2)U(t — 2)- 1,12-6Tt. zo‘hr u[~):\ ‘ =9 _~’ : - I‘L'Z—T”) (b) We have. > I} Hi ()‘ e M {ii-24],) y(t) = 00Image: ~ 7W r? k H2) ‘ > [00: ‘ . XHJ‘V) @130 = A e‘(7_2)[u(t —— T +1) — u(t — 7 _ 2) I . __> moves N3“? (15 h(r) and :1:(t — T) are as shown in the ﬁgure below, t mcreases Using this ﬁgure, we may write ( Egmgﬂjjl V 0’ {+1 2 t < 1 ' _ (w-zjéh = Mt) = f2”: (T- M7 ‘=: 1 _ ((t—l)’ 1 < t < 4 —e Weft/Q /t-2 e—(T—2)d7. '= F—(t_4)[1 _ 53], t > 4 2.46. Note th-at . dw) “at I I dt = —6e u(t —— 1) + 26(t — 1) : —3.'z:(t) + 26(t — 1). Given that \$(t) = 2e—3tu(t — 1) ——) / we know that % = ~—3:1:(t) +26(.t — 1) m ' 9M». the given information we may conclu - tat 2h(t — we have (‘QSVOINSQ Jro c1152 ;5 —3y{+) +67%) : afﬁrm» 35> 2H1Evt)=€’7tu(+) or Maxie ...
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HW2solution - ECE 301: Homework 2 Solutions Due January 30,...

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