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examII_s03 - EE 350 EXAM II 3 March 2003 Last Name...

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Unformatted text preview: EE 350 EXAM II 3 March 2003 Last Name: Solui‘a on} First Name: ID number (Last 4 digits): Section: ‘DO‘ NOT TURN THIS PAGE UNTILYOU ARE TOLD TO DO SO Instructions. 1. You have two hours to complete this exam. 2. This is a closed-book exam. You are allowed to use both sides of a hand-written 8.5” by 11” note sheet. ~- 3. Calculators are not allowed. 4. Solve each part of the problem inthe 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. (5 points) The block diagram in Figure 1 contains three LTI systems whose impulse response functions are h1(t), h; (t), and h3(t). The input to the overall system is f(t), and the output is y(t). Represent the composite system using a single impulse response function h(t) so that 1105) = f (t) * h“): and express h(t) in terms of h1(t), hg(t), and 11.30.). f(t) y(t) §z (t) Figure 1: System comprised of several subsystems specified by their impulse response function. allot) : 1am aunt-e) u) gait) = We) at big) (z) #3 (+3 ; git-Q a< hJH'J. (,3) 44/50 mic. these, ace) Can be, expruyefi and We) = -0\4,H=3 +yzzt) egg—e) C1) Subs-éfévfil? QDVW-£(Ofls (I) throw]? [3) ”1.1.0 (Y) fil® ft) = — flasks» + ‘Fl'Q’kMLH'J — (Heywofi has) ’1 JFltHrE-k‘c—D 1412,04.) 4:1“) New} W hat) 2 2. (8 points) Consider a LTI system that has the impulse response h(t) = (2 — 25—5t) u(t). Is the system BIBO stable ? You must justify your answer to receive credit. jwlhwfloflfi = wlz-zéfiloQJc 6€c¢vse Z -- 2,2, 3. (12 points) It is known that a certain physical system with input f(t) and output y(t) can be repre- sented by an ODE of the form W) + 01W) = NW C 199 In order to determine the unknown parameters a and ,8, an engineer applies a unit-step input and observes the zero-state unit~step response. Using this data, the engineer determines that the impulse response of the physical system is 4t er ‘H: T 5 w n “CEVPMQ mg h(t)=3e "(t)- X ff, tho, stEWb 0!an So Determine the value of the parameters a and ,3. a; ___ q bfi I n Sfec'EIu/i _l \ SOL/Q, (A) Law 406) a «6-6) 01mg. g(o) =0] tkw‘E [SJ 'pmfl tl‘Q. taro—Ltw‘lz unit—5W feS/yonse. «ac-é: w 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) * h(t) for f(t) e—W2 h(t) —_- u(t + 2) — u.(t) In order to receive credit, clearly specify the regions of integration, and, for each region, provide a sketch of f and h. Summarize your results by providing an analytical expression for y(t) for the different regions of integration. Do not sketch y(t). .pc-Q) th) kHz-1') JOB $j £[t)k{-‘I:"C)CQ‘C +| I —t/Z —C t>o 4; 2+4: ?(‘C3;€ or 1+4: 2/ 2+i: -Z+-l> ’ (asj 4/; _: —22. z —.-_-—2<J_ L—ez') # i e a i < 2. (10 points) Suppose that both f(t) and h(t) are causal signals and that W) = f(t) * Mt)- Using the convolution integral, show that y(t) is also a causal signal. 3 0 Bacwvse, 9C?) I6 ox. Couvsvrg Sig/’V‘Q') ‘Céo. ' (SQC‘VR h(-E 'T) 75 av coat/SQ Jain/Q; )7[-[: ’C) 429,» 4:»: 40 or -c.>—t.. wmfl an; fiat \I (e) f 4 (I) h (.5 ~23th \ .F-( t) :_O ‘Ful‘ 7-}?75 ‘QLC'E WIIOWJ’ US in efiflrexs aC‘t’\ 05 CD 3%» =— J P cqh (t 'rMQt O [Q ~Fw— C4?) £(CBL6 :30 a l'vw) Problem 3: (25 points) 1. (5 points) A LTI system with input f(t) and output y(t) is represented by the frequency response function Jw2 Jw+1' Find the ODE representation of the system, and express your answer in standard form. HOW):- (he the Lora: (uni 4;” WWW wywen—em-émi £1 ] w"~. 014:" 5 (fl ) Y N wZ. .12.; fill/w): ’96::— M ’2. doc/F: \\ dwy'i‘}; 2. (10 points) Once again consider the LTI system with input f(t) and output y(t) that has the frequency response function _ _ jw2 Using phasor analysis techniques, determine the sinusoidal steady—state response of the system for the input f(t) = 5 + 3 cos(t + 20°). ’V’ 1‘ W "0 W ;‘ U510? the, rad/+2 SMVSOVOr-Q riacdyd—S‘iu‘i.‘ $2770 n52. L- TI: +149 = M w) a (t) = 9’60: (wot-re}, . ._ Q'Hya31505(%t+5 + 21 99%)) a [Via .5 ago, rpos z’iSluA 6 live) 640:3 : g leaQ‘CaSGJt-i-Oa‘ INN/0)) 3. (10 points) An engineer has generated the exact magnitude and phase response plots of a LTI system using the MATLAB command script w = logspace(—3,3,1000); bode([8,7,2,1,0], [4,0,7,2,0,1], w) Write a MATLAB script that calculates and plots the zero-state response of the same system to the input f(t) : 28in(5t) over the interval 0 S t g 10 using a time vector containing 800 points. = Qmsyaue. Co) IOJ 800); t- 4 = z =x .sm (ac-y '5)» e 2' E89742) loo—1) Q : EfiJOJ'Z/ZJO/ljJ j_ = QSM” (KQJ‘FJ‘EUJ plat: (4:,&) xlahe/i CHE/mo. (see) ’) flinch-oi C ’ccn//.£ufi,’) 10 Problem 4: (25 points) 1, (6 points) Show that the set of three real-valued basis functions {451(t), ¢2(t), 45305)} defined over the interval 0 g t S 1 in Figure 2 are mutually orthogonal. ¢1m gm gm 11 2. (6 points) Once again consider the three basis functions shown in Figure 3 and determine if they are orthonormal. #p) #1:) gm Figure 3: A set of there basis functions {¢1(t),¢2(t),¢3(t)}. | “Mm = jmfiwe =. 531+: ;/ / O I ($2) 4’2.) :3‘4931-fi5fi‘b ; Inf-I: ;) l/ \ '00 / «53,459 ;§¢3chx£t 7! t e I Seem-15¢ (l) tke. <25; (1:) ave, mu'me/(a of thcdoA‘Q DZ‘a C2) <G5£JdJL> >1 ‘Fo‘t‘ E;g 2, 3 a, ScynaQS Ed), (65 fifth 91+); ‘6 mvfiug “tanner/ml, 12 3. (13 points) The basis functionsconsidered in parts 1 and 2 are shown below in Figure 4, and are now used to represent the signal N) = St over the interval 0 S t S 1 as f(t) = c1‘131(t)+ CW2“) + 03¢3(t) + 63(15): Where e(t) is the approximation error. Determine the coefficients c1, ca, and cs so that the energy of the approximation error e(t) is minimized. gt) em +1 0.25 0.5 0.75 1.0 Figure 4: A set of there basis functions {¢1(t), ¢3(t), q53(t)} used to approximate f(t) = 8t over the interval 0 S t _<_ 1. CL 7' <44“) 2. <‘E 95D (beca‘be’ <¢3~J¢5> 3 1/ OWL <¢LJ¢J):O (327 3. a o ’/y 3/7 J.— 3 :mi" -a_:e:/“+m7~‘ 2 0 Z '/'I P 3/1 r-—~—--’\ ‘_'_______.__..1 .. r—‘f’i ‘— ,3... - 7....2’ +3... + £1 - .7,"— 37" 37,, 32 23 37- if C'— O CZ :1 O 13 ...
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