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EE 313 Exam _2 from Fall 2009

# EE 313 Exam _2 from Fall 2009 - EUAL EE 313 Exam#2 from...

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Unformatted text preview: EUAL EE 313 Exam #2 from fall 2009 Attached are four problems, and solutions, from Exam #2, fall 2009. There was a problem (5), but since it is not relevant to our forthcoming Exam #2, it has been omitted. i_....__.__..4 A W .P‘ l l Part (a) For an LTIC system described by the following diﬁerential equation, ﬁnd the system transfer function H(s). dJth) dzya) 4110‘) dsz) 6175(1) 6 — 1—— 6 r=3 7—— 5 r at:3 + dr’ 1 dt- + yo dtz + dt + x0 Part (b) For an LTIC system speciﬁed by the following transfer function H(s), ﬁnd the diiferential equation relating the output y(z‘) to the input x(t). ' :2 +3.9 + 5 HS =————— 0 53+8S2+55+7 ‘qu w»; cm "32 i1“ 4— 453" 10% —H sits) ~1.— (9 its}: \$ SL‘IlsX 4—?5 3731+ 5 [83.x— GSLw—‘i‘g. + (511(33: E3331+~73 +53 31(53 ﬁg.“ mi _ 1(s\ _ 3 37-4—734— S S‘N—ésL—us +.La ....-——.—....__..._.._.__ ...__. _,__...,._._._._..._._...._. (—l ts): 033 H (s\ — 1‘“ __. 5:33 3 '3 M...“ 34-“ s54— Es"-+€3 +7!— . ﬁst msﬂswrﬂirsi =7 (SL433 +53Ztsx Car-msna‘uagamﬂ armarﬁwgtut eyuaA—tom :s . M If it y 1 i i i : a T ﬁbﬁéﬁxfz 7' ﬂ 7 h w” W" '7 W '7 7 if v 7 7 a v W Consider the convolution c(t) = x,(t)*x2(z‘), where 361(1) =,u(t) and x20) = e'3’,u(t). Evaluate the convolution integral C(z‘) using 1? the Laplace transform table and a fundamental property of Laplace transforms. (No credit for directly evaluating the , convolution integral in the time domain, you must use a frequency domain approach based on Laplace transform concepts.) I l l -—3 1LT : : C(H: Xruanexlte\:,u(%\ % a ,m-L—vl ! Zlo— A—m-mafrcar‘ldk L HS Max @H-Q C ,{D *3 i:- Lava—(22. we, U..£a,L_Q cakao‘Ucl-‘FOM 'v‘k 44‘4— I C3} = iii/(1H1 fie “At-a I ‘ ‘ +th£ akDLUDJl/t Co”—"-~‘"P”"“‘“ 4'9 I ‘1 ; I . k ‘ (L an. '2 2 C (EX =~ “I; x ml“ : ' mu LHMN‘L'HO‘L l“ W“ m? “m 3 q. -- x . . ‘ 5+5 8‘ (8 4a» atom To CCA:\ k” ‘3‘ \ﬂﬁQa—Q "\‘rD eat-(‘4 cut-3&— (J—I ‘(DaL'F‘l‘l‘K‘aiv-k @114 VS‘WLDM 'QJMPQLMSLKGM (:1. I CCS i :a -~--- :1. wlm “SELL. 3 3+3} 5 g .: um I K i l. 61" i ' i '31 M“ .‘ E€ 3+3) S, p" :1 D l f " ' anus-1.1.90 (1 ’ ’ %I~~ ~ ~~C~~ ¢~~e_ir~~~~~~ ~~~~~ ~~ ~~ ‘ ‘2 ” ‘Ezarg'g 3 i VM'W. '5‘ :2 ~ , /‘ ' 3 z : tailarLuVA 1 . C(ss: ;,(,I_,\_L .L I 5 v 5 3 5+5 J \\\DL§J 464%" h.‘ 9.0 L. {+3. act/«:9» i 6‘. t 1 3“M['E)“§€MZ~Z:) 2:83;! -—C“L AHA _ I *3*E Sail-alive low-5 '“ly‘m 1' ~l-~n <. Problem. 3 . Using the Laplace transforfn, solve the following diﬁ‘erential equation for y(t) subject to the following initial conditions y(o‘) = 2 and y(0‘) = 1; and When x(t) = e‘4’/.z(t). (Note x(0‘) = Q) j l i WW) Mil Jxﬂ ’ ‘3 , ‘ I _ dtz + dt +6y(t)— dt +x(t) . Tam“; W DE knollwm as; Lu—s mw— M4 war Gai- 3mm“ 3 we”) Milo-3 4L 5( silo—woﬂnt (91(8) 1 S 2153 ~ xzm 4, ZS) we ‘K‘kk I“ acﬁ‘ \.‘(~I"r:>//L # ; ' guh's‘LtLuAﬂmq— i.” .Hw, Va-qu‘ m-gﬁ +LLL‘L—er/kl'li‘ 4% I am“ “ME-“i Woo“ Elsxi" Xe” MHAJ: slink—Sm —{«_3 + Balm — 5(2) +4 im \ i ll _ ‘— — o + —- ” S~<\$+£§ ‘ 3+4 7 J < _, R.A_-.._ A 4 WW... 1 '37: ~» \f' _ ,_,_ .__ I: _ , ll WHEN-9,) 43),,”7237 7 r ‘9, W4), "Wigiiﬁiww . i "me, xx 4; * 1*{*5*)***:*(234—14541"\$44 ~ W 7 ~ 9 *7 WV ﬂ " (3 +5 3 (a) . S+4_ "' ._.__ L§l+sg+ey i(g\,—— SM +ZS+H _ s—HA- (ZS-HD(S++\ ~ ‘ s++ . . . lj . LS’W‘FB - LSL+S\$+L) i(\$\: W lsl+gpg+£§5 ‘ (5+4) Lad—4) . ZSL+1DS+45 cu Q,_ (13 “ab—L ‘._. ‘1(3\_ ZSM-zos +45 M A r —-. L\$+‘F)é.s “in-5.3%) 7 ' (3+4x7(s+z\<'s+33 “54—43 LS+L3 @z—s) (Pact—*‘imk. 'c‘l‘aa—K-tr‘Jl/L ®>9l3°°mgtom .;‘ L Q ch : “23 +.2’9___...— 4‘45 \7— Zf—L'rELJI—zazj 43 4:45 3. 32—90 +45; _ (quay—441% Fiﬁ-'3 \Q‘ﬂlcSJr 136 s-vaw “ " 5’ . S = — 4. ' ‘ q x 2. 5.5 + 2 0 s ~l— 45 x I L - ‘————-——-————. W 7 (i344) £~si+~15ljtrs+83 Cougr—up ‘ L ZL-n"+2w(—_z_)+45 L1)(4\—dl-D~I—4S_ é 5‘ I .V g,_l #214.» (I243) (2)03 ' ‘ (‘13,: 25L+205 4.45. \ _ 2é~531+loﬁ~§l +49 _ 2(4N—60M55, m3 .. 7 _ , _ _ m" -‘.“““".‘"”“T"‘.""“""" V , l .... ASH l. _ _ .,~ ‘ ...._J\...-.L§... _ .. A ’ ' t " ‘ 3 “mufﬁn S_ 3 c. +)( 3+2) Ll z 7 7' ﬂ" 7 "7" W" ""7" WAWWW WWW iiiriiiiiﬁauﬂFr—UP’ _ ' Are —2 E —.3' \_L(&3= -— 1.3.. +_ 9.5 __ 3.0 :> %.U:\=Elvﬁ€ 4—-é.5€, ~32rlruﬁf3 4/5 I _ m C\$+43 Girl-L) 34:3 ’ PI‘OD‘IEIIII‘IW r v v V 7 ~ 7 L w V 77" 7 Consider the following diﬁ‘erence equation for an LTID system. I y[n+2]+6y[n+1]+9y[n] = x[n +2]+3x[n +1] 1 2 :; 'E T Find the zero~input solution yo[n], subject to the initial conditions y [-1] =__ and y [—2] = __. : ‘ o 3 o 9 f i Aclosed form solution is required. I \ t 3 l / i! i I ' i s 1 ‘ E 1 iz¢ra wi‘meri— 'x LM1:,-o ' ‘lh . CMurt¢9\-J—m'.t+.‘c~ agucc'vfai/K {S I. +=( 10 bv. O Qemumxmaesn mes—2 ’xL—u—z é “Ea”: “"F “‘3‘ Sﬂuﬂvcmm ‘35 {M3 :2. 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