examII_f03 - EE 350 EXAM II 20 October 2003 Last Name...

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Unformatted text preview: EE 350 EXAM II 20 October 2003 Last Name: Solutions First Name: ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Score 2 25 100 Test Form A Instructions 1. You have two hours to complete this exam. . This is a closed-book exam. You are allowed to use both sides of a 8.5” by 11” note sheet. 2 3. Calculators are not allowed. 4 . Solve each part of the problem in the space following the question. If you need more space, continue your solution on the reverse side labeling the page with the question number, for example, “Problem 2.1) Continued.” No credit will be given to a solution that does not meet this requirement. 5. Do not remove any pages from this exam. Loose papers will not be accepted and a grade of zero will be assigned. 6. The quality of your analysis and evaluation is as important as your answers. Your reasoning must be precise and clear; your complete English sentences should convey what you are doing. To receive credit, you must show your work. Problem 1: (25 Points) 1. (9 points) A LTI system with input f(t) and response y(t) is represented by the ODE jj—7y+6y(t) :30 fit) Find the impulse response representation of the system and, if your solution contains a term with a delta functional, simplify that term. F'mJ. the Ezra-shafie. unrlz—s E-ep response. alt» thQn h ”fl = «lg/904:. ——_—. 62(2): 11-724 6 = C’A-IMJrQ =0 7w = 4", 32..“ gbca= c.et+c2.e‘t tzé fiecwse ‘Fi'C'kl‘l For 172.0 4-2an 0 76 ”5+3“ Char“‘w’5£'c |"col-7) a? (4:3 := 0C, SAW? “FD" OC- yiveJ " ' _,, _—_-_ 7. = 5‘ t 2 . a? -7gPl—GJF" 30 ) grad 6(— O t’ 6— 9ce‘, = 35%) 4. (,(t) = C,e + CLe, +§ 1320. 069. the Inolzmg— conootbms {:0 ’(mi Cl afloQ C2,. {seem/.32. w?— aurQ, J¢£ermlrnq thg ZQra —S‘£‘u.é<.. (“OJ’qonse anoQ ékb .Sdséem 1.5 S trio'téi proper) 51 (0) :y0 (05 =0 : as; 5‘ 5"> 3(a);o:c,+cl_+§ }ec‘ $— 30)} >0 = C, +6cz g) QL-s —2‘-C. .. Uqct) = LvéeJc + 86f: 4-5:]wct) 2.0 912,. £40 " 6i: r’_l tea-z 03H) = L'Geti-Cth-Juct) + [—Go'bl—Q +57$w 2. (8 points) A LTI system, different from the one considered in part 1, has the impulse response h(t) = [Ze—e + 1] 'u.(t). Is this system BIBO stable ? Show work to justify your answer. LJthHoQt :: [OOHZe't-r ‘)WL€)lcQ‘{: “'09 \1 I N N co 't-l't/ :00 O BQC‘WJQ- f [h Lvtn‘flfi is 0.56 meéel tl‘Q' Sastszm 13 not (51/30 “Laue. 3. (8 points) Derive the identity 6 (2) = 0 am where a > 0 is a. nonzero, positive, real-valued constant parameter. That is, by directly evaluating the appropriate integral expressions, show that the generalized functions 6(t/a) and a 6 (t) have the same effect on a. given function f(t). 03 Q) I HQ “,8“ ’TVQWE = 0* F(7)j‘ S—(t Wk“: 5 0&0) =0 unless is =1- Qeé (L: t’T E=M+T fin) s(¢:;:5)&£ l 0,“, g 2,; = arc/r 1C CM ”)flfiJW -Oo/Ov =0 aneSS u,—_—.o \l =‘ a. PCT) [3% Cause. r {(+3 «gas—flow.— = SHJGWCtz-gbié ‘ 0”me tl-L yanerqllwi Lnoélbns SCt/o‘) aan, 0" 8'6“” “'9’ flan-(swag, hecawée/ bk? flew; the, Jame. etcéo‘é 0A dn arbitruy' quno‘lla‘tan ‘F[t). Problem 2: (25 points) 1. (15 points) Use the graphical convolution method to determine an expression for the zero-state response y(t) = f(t) Ir h(t) for W) W) [u(t) —- u(t — 2)] + 2[u.(t — 3) + u(t — 4)] 3 e'"(t _ 2) u.(t — 2) In order to receive credit: 0 Specify regions of integration by writing the convolution integral with appropriate limits of integration and specifying the range of t for which the integral is valid. 0 For each convolution integral, substitute for f and h using the expressions given above and simplify the integrand by eliminating unnecessary unit-step functions. 0 For each region of integration, provide a sketch showing the position of f and h. 0 Do not evaluate the convolution integral for each region of integration. Hz) b (u h (r + a W: -t) 2. 3 3’ I: 44? ‘ ‘C 12 '2. hacue. not) Era overlap (Faggot/é) FW 0 (7: < t—z when 24 tcw, WE 2 T. “ME—w when o<t <2. L 30:) = 34961:)th ._— I‘l ‘V—Mfit 3G or 74% froflob’e +Lt3\n(b" 75 1.5 non um au-Zr CC 1: L1 “"9— 3<t<-£—2_ TF2. ”cowl: FL'C)VsCt"C) Is flan-zom oztcz mi 34:01. OU—zw -(-l:-"C—z) ¢ L s '50 2. (10 points) Using the definition of the convolution integral, show that z(t) * h(t) = h(t) * z(t) for all values of )3. Show all steps and clearly define any change of variables. on lei: =-l-:"C 'c=Jc—p x(—L-.3-)<hl4=3 = £:(t)h(~L—-t\ai:c k oLszv-OL'C) " j x(t-F)h(p3 ('31?) oa 1' Sm htp) XC-L- —p3JZP : l1 (4:) 5X X (t)- Problem 3: (25 points) 1. (9 points) The circuit shown in Figure 1 that has input voltage f(t) and output voltage y(t) is represented by the frequency response function 1" —10 F _ H0”) — W' Assume that the operational-amplifier is ideal and R3 = 1 k9. Determine the numeric value of R1 and C. In order to receive credit you must show work. If you introduce a. node voltage (mesh current) in your analysis, clearly indicate this variable in the circuit schematic and label its polarity (direction). Figure 1: Active RC operational amplifier circuit. R .L. V = ‘_ 22- ~ — 7. (WC. Rzi’J" V F' 2 l __._#_w° F' Pu @ -._«;- z to or a: E: - m Rw— ‘ IO )0 .. J— l l ._ @ CQZ— loo or C: IoogL 2:. 7—0? C -lo ll _—________—-.———o fl W/loo 1' l i 2. (9 points) A LTI system, different from the one considered in part 1, has the frequency response function whose magnitude and phase plots are shown in Figure 2. In response to the input f(t) = 4 cos(a. t + 90°) + cos(b t + 90°) the sinusoidal steady-state response is y(t) = z cos(a. t + 140°) + 0.01 cos(b t + 9). Find the numeric value of a, b, 2:, and 9 (assuming we know that |9| S 90°). In order to receive partial credit, clearly mark the points of interest on the Bode magnitude and phase plots in Figure 2. H6056“: +7o°> H INC; MICosGd: +700 + AWCJ-a” x cos (we +l‘10°) ll fie» QGS'EI Gauqhé/ rev Viv-e5 A. H (6/0") 7- +S-Oo. FYam F‘dvm’ ’2.J A Mala.) = 4-50" when w t: 0b:- SOEE' Pow WNW... Cos(bt + 90°) ——>-IH'l01b)lcoa(b'l: +‘/0° +r 23$”ij —_-.- 0.01 cast'L' +6) T/y‘er 905*: efiuqlt'ég Peéulms [Hf/lg), =o.old OY‘ " L/oJB. From 5;,qu 2.) li+(d.w)l: "104,6 when OUT—b: (Grad/Jen. DY“ ‘100 ‘."4/59"' 082. the, 'Cum'l: the; [9| 4... ?’0° '60 clfiooSe tit comm/’17 Lreveenca: 10 Phase (deg); Magnitude (dB) Bode Diagrams = 5'0 2:. \100 Frequency (rad/sec) Figure 2: Bode magnitude (in dB) and phase (in degrees) plots as function of frequency ( in rad/sec). 11 3. (7 points) An EE 403W student has designed an active filter circuit with input voltage f (t) and output voltage y(t) that is represented by the following ODE, that is not necessarily written in standard form 0232(0'1' a1 3]“) + Go 1105) = 52);“) + 51 f“) + be N)- The student generates the Bode magnitude and phase plots over two decades starting at 100 rad / sec using the MATLAB commands omega = logspace(c1, c2, 100); bode( [1,-5,P3], 4,—2, 9 , omega); Specify the numeric valifias of the parameters a9, a1, a0, b3, b1, (20, c1 and c2. T"- ev—e, are two IOCI : loo => CI :2. “leaflet/e) b&41M3-9" lOCZ : lo“i loz’ omeg— ‘0‘. “Hi/39° ill 0 P H .c 12 Problem 4: (25 points) 1. (10 points) Consider a set of three functions ¢1(t) = 1 ¢2(t) = t m) = tug defined over the interval [—1, 1]. Are these functions either mutually orthogonal or mutually orthonor- mal ? Justify your result by showing the appropriate calculations. I :2 O l <d>.(t7,¢,m> =-' £50Ct)<’l4= = 5:: Z ( <d>.m,4>3cto>= J cow—file = [ +3 -5 'I 13 2. (8 points) Consider a. set of three mutually orthonormal signals 1/11(t), ¢2(t), $30) that are defined over the interval [0, 1]. It is know that a certain function f(t) can be represented by these three functions f(t) = 2 1/11(t) — 2 1/13(t) + 3 $303) + e(t) with the approximation error e(t) = 6t over the interval [0, 1]. Determine the numeric value of < f, f >. BCCausQ elk.) =0 ‘filr fl {— 5, E0313 we. Canno'b 05$ Vowseuwl‘) Theomm, rnyéeaflj use, tko. redo/'6 =1 because th. 3 F. = <¥F>’ 2d <41 #2 “mism— @— 3 M. ” ' D ”.3545, ”may”! Ee- = “J9 = jechWE s J36’Lzl‘é = 32-53) =12. O O 3 0 «£0 -— (27.Z + (-2? + mi) N H (fit—5 : 12 4 t7 14 3. (7 points) Consider a. set of two-hundred mutually orthonormal complex-valued functions {¢i(t)}, i: 1, . . .200, defined over the interval [t1' t9]. Find the numeric value of the integral t; 10 10 ft (Z¢n(t))( ¢;,(t)) dt. In order to receive credit, justify your answer by showing each step of your analysis. it o o L“ él¢n1e))( 23,463 mij’: to ‘0 ti. n M e 3 E o 31: E if n :l m =1 4:, [0 lo — n :l m :1 <¢fl3 (bin > ' W o n it: In because. the, 2 l 8 lg nods are. N 5"“ 15...
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