Fall 2009 Exam 2 Solutions

Fall 2009 Exam 2 Solutions - ESE 3%7 Exam I]: FO-M 7—005...

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Unformatted text preview: ESE 3%7 Exam I]: FO-M 7—005 Sou/films 1 I Compute the impulse response of a DT system described by 3/[11] — y[n k 1] : 2am] — 51471 — 1] + 7U]n — 2] M -’1u]n — 3] by direct substitution. Is it FIR or HR? (10 points) ® Repeat. the problem for 1;]71] A 1.23;]71- 1] : 'u,['n, 7 1] (10 points) Sow-DIM: (D [OCH] : lfitm—O + ZlLEhJ —5U\U\—t1+‘]m'_lr\—7_1 —4—U\D\—-33 To 392% l'vau/lse vesanse , \eyt- Mm}: 504], Jrlwn m oM‘PvaS wI’ck “st—k] :0 M2: nzo, Iqu1=M~lJ+25Lo1—Sb[—fl +75C-21—4SE-3] = o+7_—o +o—o :2 h—_\ , V):\1—_rqcp3+sz:3-s$tol +15E—u]~<_SE—zl : 1+0 —S+o~o =—-3 nzz, VJE21=10D1+28E21 ‘551'43 +78E01-4SC“J = ~5+o—o +7-o :4 V\-—3 , %L31=%Ez]+2§[31—55E21+7SEQ—48b] = 4+O—o—ko—4—zo and VJEM=0, ff” m7x4. TIM/LS 171m. veep/Lee is hCo]=2, \ACIlz-K, kcz1=4 Me]. \nan =0 {W n25. The Slflé‘l‘alm has mahg 3 WM emtIn‘es Ems impulse meme mi; w, is a FIR Jim. min] = L1 Win-t] + UL'Em—x] To 39* {Mu/he Vesanse , \eyt- Mn}: 51;“! Jrkevx “cl/UL oM‘Pvdrs wT’cln VHZ—fl :0 Me: n=o , V3120] = LL‘OE-Vj + 5[—\] = 0+ 0 = o I’V—M WCI1=\~ZV)L°3+ 8L0] = 0+ \ =\ V\=Z, WEI]: \-Z"Jl'_l3 T SD] : \.z+o=_(.7_ \n=3, V333 -—\.zv)c21 'r gm 2 l.z7‘+o : [.21 MA So fwth. ln gene/PM , we, We Win]: l-zIrH , new mac. TIM/b6 flap. (Mm/{6e IEJSP‘UMSQ. Ls VEM=LZVH "S‘UYVIZO. Lt- MM LHImi-Febfl wwwa erko Mfies . There-Ewe Jul/st (s an 11R :erm. —~. 2. Repeat Problem 1 using the z-transform. Sch/mums: (D As (3va WEnl-‘fim-fl = ZlAChl'SULCIA—I] +7mCm-21—4uln—3] ApPl/Iohj 14M. 2- Wfiwm mews T02) — z“ Tm = zUcz) ~52" u(z)+72'20czu—4z‘30ca> u—z"> Tm : (2—52 “Hz-Z— 42—3) Uta) TIM, TM Ws+e+ {Mam is HQ): cz> 2 2—52"+72'Z—42-5 ULZ) \..2‘\ ZZS—Sz"+7~z — 4 2 23— 21 No-fe Z —32—‘+4Z—L 2%‘21 jig—521+7i —4 7.23—2.21 “7221+7Z —4{- —3zz+%i 42 —-+ +2 —+ 0 “(Ms He) 2 z~3z“ +42‘Z/ _ ,3. .i —. —- z + 21 NW6 H- MS FDKQS ONLY 00(— 0/ hence +0415 (:5 a, QB §‘I\Te/r. @As STVGAA [GEM- LL‘jEn-I] 1 Win—I] APPl/Ig“ m3 ‘HAL 2- Wfiwm UJTe/HS T03)— 12-2“ W72) 2 2“ 0(2) (l— 1.22“) YE) = z" 0sz TIAMA TM Ws+e+~ {mam is H_C_Z) : TC‘Z) 2 ‘Z -| = l Uta) l—l.22“ 2—1.7. 1-1.- kus WPDKe 06(- l-Z omd deswflaes an 11R {TH—e» I _—— —. ® ® 3. Compute the impulse and step responses of a DT system with transfer function _ 22%;") ” z2+1.2zm'1.6 Note that. it, has two real poles. (10 points each) HM So[ m’o‘ums : Q 1m M52 res e'vxse : P F HC‘Z) we 6K1)on E 0.5 Has) 7i '3 _ i; a _ ECZZ—H—zz —l.6) ~Z(~Z+2)C~Z—o.8) k0 + kl k7. : Z 2+2 + ‘Z-°~‘3 wH-L‘ k0: 12-3 \ = —3 = E 2+2) (2—0.3) i=0 Z'(-°-8) 3 k 7.2—5 _ ‘4 '5 _43 I: ZCZ—ag) \Z="Z \ -Z (-2 S) — ZS z~2~3 : l 6‘5 _“ 83 k1: 2—Ci+2) l2=0-8 0 3 28 56 W S = 3:5 _ Q. i __ _8_§_ _ FL UL Hg) 8 18' “24-2 3’6 3.43.3 ‘Z Z Z We. @3135 1 Md bn C—=> 240 w; (av? ImpuJSe V2.9[30‘m98 S h cm] = $5433 a 43% (—2) — 37 0.8n'rfw-hao 6"“ = 3.\7—S 8313/0 - l~6o7| (—1)“ ._ \.S-l79 O'Bfl (Farnao (9 9g: raspmse: Z Z 19(- ZMPM— [92 $96? qe'RMmce 1 14mm um) @Ucayz A I -Z—\ we expcmd Jig—Z) (1,5 112) _ch>Uc:) a A _ 4 ? -——z— _ 21+‘.zz —\6 2“ 2.2-5 : (2+23c2—o.%><2—n k. k; + k3 2 8+1 + 2—08 Wm k: “"5 \ “1:95 \ (z—o mtg—n ~2=-z —z.3(-3) [tr k 21-3—5 |6_S so 7—— Cz‘erz) (2—H \‘3—0 3 _ z 8C—o 7.) h 7 2.3—5 \ ,_ "5 _ k3: (2+L)(‘Z—o_8) ‘Z:\ 3-O~Z 3 z FIT/HAS = —[_s:- z z - - Y“; 14 in + 7 3“”8 5 2“ Wae )oh<’“—— Zib , We ‘3“ SJ‘Q’P VQSFW‘SQ m5 ~ 30 h (GEM = «if; (—2)“ + '7‘Q)-?3)n “5'1,’§°YV\Z° w w] = - 1.07:4 L—z>"+ («m-0.2" — 5-1”, ch nae 4. Consider a DT system with transfer function H z : ~——.——u~¢m—— ( l (2: + 0.5)3(z ~ 0.8)(22 + z + 0.5) where N z) is a polynomial of degree «1 with leading coefficient. 2 and NH) 2 5. Viral, is Ihe general form of its step response yln]? (12 points) ® What is yln] as n A 00? (4 pointsglliat are r n, for n, :7 0 : 2? (4 points) SDlUL'bitm : z 2 Q lel— ium be sl-e? finance, The» um) => Uta»: , ‘Z—\ we [Am/Q. \fCZ) 02$: N (‘3) Z Y€3>=HCZ)U(1) 2~5_—z—— . — (21—0-5) C-Z-o.%) (a +‘Z+o.g) ~75 —\ on _ -‘i ‘ (73“ gel‘rllfifil—Eelfl) NC-Z)--Z m) _ = + k° T 2-H 5 + (2-1-0.sz + (‘3 +05)3 2—03 k5 Z + ké Z kn‘Z ’——_3—’ 4- + _. IE MTV - 3 2—4 L Q 2 - L: em“ (were -. k5: E , m awe owl/quote. MGMOWPLQ a»: low [in FreSlnlawk Sol/Wt‘fievx Pwlo 1.1% ) we z—er/xsfwm pairs, M olej- fie? reeFmee In general farm ‘jthl = koédtn] + k, (-03)” + k2 n (‘0-S>h+k3HZC—o£)v‘ 1-K; (M3)n +125 Cg)” STn(—3T_-7Tn+kg) 4- EACH!” NO) 2 05 Do, CIA]le =HU): O “a '0 M (Ho-5)" CH .3) (MM—r) 5 80 = —’—— = ~ m» 2.34 l—Sg- 0.2-7.5 Z7 5 ® ‘f'lMl-hwmer‘oofw 0'8: HQ?) lAaf: clessvu 4/ dub Hm. ale/mm“.wa [A05 Agava 6. Tke cirf'fe/rwce [5 M21, hence ffiEol=o , ‘flEflr—o, Mu: law“? “Wilt Laws? Cé-o.%)C£Z+~Z+°-S) - 2 ’— I ...
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Fall 2009 Exam 2 Solutions - ESE 3%7 Exam I]: FO-M 7—005...

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