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A#22_Solutions-1

# A#22_Solutions-1 - Assignment#22 Solutions EECSE~2410...

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Unformatted text preview: Assignment #22 - Solutions EECSE~2410 gignalg & Systems — Spring 2007 Wed 05/02/07 1(14). Find Xiz), the: z—transform (in closed form) of the following repeating sequences: (axis). x122: o, »1, 2, 1, 0, -1, 2, 1, 0, -1, 2,...}. V «2. ~52. - ~ .. ~ 40 gig? {e Z \$3+2?~% 4264—21... .. ,_ ._ _ us~ I” £24 ﬁes-p z‘f(s~ LEW—Z 3+E "*3 WWW 35173) (W): 31:31; 2, 3, 1, 2, 3, 1, 2, 3, ...}. Cgmmim 4C,Cwlw§L/z,3/ 1,. 273).. l 2 «2.4m “like {ggjn 1‘ lam—2.1 M £53) == z”: jsllﬁ) ,4 -5 vi: ~l —2 4. Eiiédgw 1+ZSE+3£ +£ +28. ﬂimﬂz ) 4+ 4 ’ ~t~\$€3+w 2 03% 3(2) ﬁéﬁmw 2%) == 1+2~€i+ 32:2” 5% 4+ \$7" f: (+22” 3 l“ E a M‘ ESL 1+2él+3éz Sig “ l * ) Xm=£ior , 1 £ Mai A#22 ~— Solutions. p.1 El 1 ‘3} The? “10581 {0? *3 Pamwlar 33’5th is unknown, but we de have measurements available at the ‘3‘qu: Supper“ Weinputxlnl=u[nl and observe that yln] 2"» {Q,%7%a%,%,...}. FindH(z), the transfer fhﬁeiign for Eh i S system. "1 *2— I7 3 A: " Egrglw‘lwmwi my: e—‘ia + .473 +1.85%; 10% i" ‘l ‘2» v “3 1‘3; “‘3 was??? 3.. WM; 227%) 4.: pal ~ 4 ~ ~3 .22. “4' (iﬁgxia a: 14.363 + “9% ~+ ) ,. -—-3 7 “q is “S - ~ *3 LS “‘1‘ ...... 21.2 —-—~ Evy-E W 9': i31+2§€ +%Z‘ +7; +' é“? 4% 3 b "l f( §_.L ‘2" “5+(E 3“ 2+”, - 4 J, «4 ,1 xgé‘l—g—t}: 4493; +1493 + +1 g— a. ~ 2. "e ‘l -1 :: l+§£1+iE+éé+ﬁZ+ “ Jag) mcgfw WLWQMMISEM’Eﬁ W2 X, / ____ ,_ w If”. MZ‘ l 4213 g 2 gﬁto <: f.» «lei L n n ﬂ W «w» -l E] ._. w:1: M 20 i 2 E E j j: 9/?“ c Q ~G‘ jam] 3530': 2 274%; 8) lé) (2) m -l __ WM my ‘ 1 e “512%”: w '5“ { 2-. WT - v- m I- ——~ _ _ (—3,‘ lei-8‘ Ulla, [1-2 ) O_£LX:I_J£&‘) if“ "‘23" rel/T“ g Wager/m, 0%")(1-725; :1 3 “iii... 32%) A; E 1%.? if ; A#22 — Solutions. 13.2 3(18}, Find :Ew deﬁed form analytic expressions for the inverse z—transforms of the following. Use tables am} propertimé . -3, -q 1; z l M l Xglisj) .7: § ‘ 7-: ‘ ﬂ 6 9. ‘* ~ E,‘Z lmZ-E j’xaU‘L 2) l '5 m 1:“ {e as g" — H'zﬂ - 0+3}! \ I {’x‘g” ’ "923"" 2412 2" ’ l+2.-Z“ * H. a“ n w n—\ E ) mild : Kg} ,3“ Q4) MUaJ-t- {—2) {AU-4] 22 l MwL/W m A B {ﬁgs} .2: W :1 . L 4 “’74” M (l z‘vé-‘ézﬁ \+.%;“LJZ& (MEI/ll" 23’ l H?! (Jig! A m, wwiwm- 4/ —::. '"L‘" ‘" ~21 wig—13" ’ lug-1) Z; 3 l+€i= a s "1 A#22 —- Solutions. p.3 4(2()). The («iiifi’iﬁl‘éi‘géiiﬁ equation, y[n] “(1‘05))’[” " 1] = 00471]; represents a digital ﬁlter, where mi? iﬁifiiméiefi a , determines the “cutoff” frequency of the ﬁlter (ie, the bandwidth). (8X3) Find the traileﬂar function, H (z) , of this ﬁlter, egg.“ ‘1 ﬂex -— (We 3:72} :- o< ER) (h}(2}. P12}: the goies and zeros of H (Z) inthe Z_p1ane‘ 5 F _ (X % ~Lz ~04) / H1»qu )4 ewﬁeww = °( :L w- awz‘ Wadi) l- {Ir-o rt: :3 %< a o< ,4 A04) — -04) J3: W ‘I ~ _, i ,. 04 .. 1— e h 1 WP“ / / &’ U~°<l£ =0 51.1,. l~vk §_ n i a; M . ijmgihle M’Cn3~(i-oé)(i—o<) LL51 :2: (“0-09 Luhl é ' A#22 m Solutions. p.4 £(20). The {iifi'ereﬁeei equation, y[n] ~(1— a)y[n — l] = ax[n] , represents a digital ﬁlter, where {11¢ parameter? a , determines the “cutoff” frequency of the ﬁlter (1.6., the bandwidth). (r1301 Use MEX‘RAB and the “ﬁlter” function (type, “help ﬁlter”) to plot (i.e., stern(y) ) the step response for a r: 011 and {I :s: {3.9. A#22 — Solutions. 13.5 15 1pha=0. 1 ; p i a a=[1 ~(1~a1pha)]; b=[a1pha]; x=[1111111111111111]; k=[0:leng’th(x)—1]; y=ﬁ1ter(b,a,x); stem(k,y); grid hold on alpha=0.9; a=[1 —(1-alpha)]; b=[alpha]; x=[1111111111111111]; k=[0:length(x)-1]; y=ﬁlter(b,a,x); stem(k,y); hold off 5(22). FL}: {he discreic—tiﬂle, unity feedback system, (2:)(5). Use 1116- ﬁnal value theorem to ﬁnd the X(z) steady~3iato error, i.e., esx = e[n], whore é’if‘l] xiii] " ﬁn], fOT a UW‘ my”; (bias) 9? a; u-a“ma( Q ) Ewwﬁ? 2.?! 2}“?! £63 E 1% 11362.3) 1+}; 643) 1+ ~ ( '2. > S g?! z: a; x l—Jig‘ {173K Find 32-12:} when x[ﬂ] is a unit step. Find the closed-form equation for e[rz] . —\ ﬂamiﬂgfkngj I 2 {41—52 “26%) \:—.a (+3?” _ l A ' “3“?” 0% UM: 3) V?‘ ,, we :5 twé2= éke f J; 2; 53% m + M ' lei‘ [mrgz‘ WWW,“ , MW. 1. §E ‘ ” Z.Lﬁuaﬂ 5&3 :: guoﬂ +T§(é \ MW. «MW/va A#22 —— Solutions. p.6 5. (Continued) (mg). Uging gem equation for e[n] in part (b) evaluate the limit, lim e[n], directly in the n—domain. i h W c2511 3 i ' J. —— 3;; 11513 rug, 3 ha“, SMUO—P I§(é) j E) {13"}("13’1 US\$12: iviz’i‘lLAB to plot your equation in (b). Include your name in the title of your plot. figalm [:— ~+ Tag-(Zigjnjbﬂl Steady-state Error I r ~~~~~~~~ -------- ————————— u: .......... 4i ......... VVVVVVVV ‘ 1 r 1 1 1 1 1 . 1 I 1 1 1 1 1 1 1 1 1 I 1 1 1 1 1 1 1 1 1 I 1 1 1 x 1 1 1 , 1 1 1 1 1 1 1 1 1 ' 1 1 1 1 1 1 1 1 1 1 I 1 1 1 1 1 1 1 ‘ 1 I 1 1 1 1 1 1 1 a .—-1 1 1 I 1 1 1 1 1 1 1 ,V v .,.,..1 1 i 1 1 1 1 1 1 1 1 ' 1 l 1 1 1 1 1 1 1 . 1 ‘ 1 1 1 1 . 1 . 1 < 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 {3,3 0,357} 13.”? n=[0: 10]; e=0.2+(2/15).*(1/6).’\n; stem(n,e); grid title('Steady—state Error') A#22 -— Solutions. 1317 50(3). Use Zwii’aiwfbfif‘ls to solve the following difference equations; (mg), ﬁn} fig w 1] 2 %u[n] with initial condition, y[—l] = 3 . Siam—u» :EZT‘EH + M —;~ J M3 “31% g; «51% i3") 79%" 1&8 ‘ new; nglwinxw4l+éyW—Ql=%ﬂﬂamiyhﬂif;EE/”39' ?' ‘ =.#M» Z? J (’5’! 3/4+ 53? w ~ M :2 M \$93» “W .. .2, —1 / ”‘ ,Ll l"£ Cz~%€‘*%% ﬁg) 0 it” )0 “Ex ) C % 7%) ‘1’“ “ﬁx/5 + E —1 + Mfr vie! {~La leé - it 3:4 395w ..... ... 7:: A“ m E]; 14390-2“) 4 0—. {LE-ZJQ-QL) l—lie :b 6&2» 3 3/4.. 3 "Am—"a :2 1}“ m 0",; “ 1.5‘) Q— JiL4le”4) Fé-Z'l-ziﬁ Z‘Ltf A#22 - Solutions. 13.8 f,{1é}(§3). {{jg’gfﬁinuggﬁ,) Use z-transforms to solve y[n]--Z~ y[n —-1]+§y[n—2] = %u[n] and y[n] = O for :2 (3. A#22 — Solutions. p.9 ...
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A#22_Solutions-1 - Assignment#22 Solutions EECSE~2410...

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