examII_f02 - EE 350 EXAM II 24 October 2002 Last Name...

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Unformatted text preview: EE 350 EXAM II 24 October 2002 Last Name: g (“)l GE 0 05 First Name: ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Weight Test Form A Instructions 1. 2. 3. You have two hours to complete this exam. This is a closed-book exam. You are allowed one 8.5” by 11” note sheet. Calculators are allowed. ' . 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.” N 0 credit will be given to a solution that does not meet this requirement. . Do not remove any pages from this exam. Loose papers will not be accepted and a grade of zero will be assigned. . 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. (15 points) A linear time-invariant system with input f(t) and output y(t) is described by the ODE 17+2§I+y= 2f(t). Determine the impulse response h(t) of the system. FImQ— the zero-simie. raspoare. &(£) ‘50 0" W’éfifiep mput -FL+,§ :Wl‘éflj then h(£\ = o’f/Jqéc 7“ 2 1:”, Gun: 7\1+Z?\+l = (n+0 .0 7\, 2L -4: —z+. 9”? (£3 ; 0" e + *tge t 20. 73:0 Fonchm‘L "PU” 1:20 once 2 23.10;. (0’7. ;f (t) : a ) a; {[t) L5 L3 I7"? *1; a £10; cFarmoflQf/fi}g 0 0L row 0 0 4 IO 2 ~ ._. 4.1) I + /I +/%7 2 WM.) :— z {:20 ,_._——- ~-l: 2 m JWHW a” C' 6 ”(Le + 3 t ‘ ._ C - '2— yCO) :o C) + Z .3 cl = -2_ ' (07 :0 1‘— ‘C1 + CZ : L g —'t’ We) - [Z — 20"“99- let) F——-" O .— 2. (10 points) Another LTI system is described by the impulse response function h(t) = eatu(t —- ,6), where a and fl are real-valued constants. o (5 points) For what range of values of a and fl is the system BIBO stable and causal ? In order to receive credit, you must justify your answer. 0 For 620 ana oC<o the 984:9.“ b 5/50 84:“ng 045 thmlé = fwemtié = tEe’a—tl: o (5 points) For what range of values of a and fl is the system not BIBO stable and noncausal ? In order to receive credit, you must justify your answer. . Pg, g; ‘0 ”(+3 is as. nonchdeQ ScyflccQ 0L3 t: Z-<). 77725 rhea/L5 file: Problem 2: (25 points) 1. (12 points) Use the graphical convolution method to determine an expression for the zero-state response y(t) = f(t) a: h(t) for 4e-2(t - 1)u(t — 1) 2[u(t+ 1) - u(t + 3)] W) N) 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 a sketch of y(t), as well as an analytical expression for y(t). {(T) hit) b ("E't’ t ”l ~C #1 ‘t’l -3 -I In this (0L5?) FCt3hIE’C) #0 4:0)" ‘39- t é"—1) (4an So I ’I «5 4:4 3 c f ;(e ’3 ¥It3h{t-t\c/C U31»? ”0. oan”/¢Elun Ya!» I new «OH/"145$ 4 IM'ECflYw‘LIUnc) r| -n gel-zflrnj‘ eZTJC g t‘t)gt : " gm L M _3 -I ~Zl‘E-(\ ‘Z -Z(£—D Z‘C _ — - 5 -9e [£2'3*- We [é 6 ~25 ~z(t+z) i > yét3=-HE€ -e ,o Summaw‘tmg 2- (7 points) Let 9(t) = f(-t) * f(t). and show that the energy E: of the signal f(t) is given by Er = 9(0)- ;(t) 2 ‘PC'fl‘kflfl zj‘ -F(- C) Hi: what 661(0) '—'— I 'Fét5-F(‘C)£C E j“, 1, wet. (a) -: E4: 3. (6 points) A system with input f(t) and output y(t) is comprised of three separate LTI systems whose impulse response functions are denoted as h1(t), h; (t), and ha (t), respectively. The zero-state response y(t) of the system to the input f(t) is given by W) = N) * [h1(t) - h2(t) * hs(t)] - Using separate blocks to represent the systems described by h1(t), h2(t), and h3(t), sketch a. block diagram of the overall system clearly showing where the input f(t) enters and the output y(t) exits. If a summer is needed, use + and — symbols to specify whether a given input is added to, or subtracted from, the summer output. ‘F-‘klh \L Ha Problem 3: (25 points) 1. (7 points) Consider a BIBO stable LTI system with input f(t) and output y(t) that is represented by the ODE fl+2y+2y=4f+2f+4fi o (3 points) Determine the frequency response function, H (1w) = f’/ 17‘, and express your answer in standard form How) = Z = W I“ 0w)“ +“n—1(Jw)"_1+a1(1w)+ a0 Raphae- ‘W n wad» (“an (0'0“)?- ‘Ar’ 4" 2%.») q + 2)? M VEC WYW— ltd») +2] == [ng)1+ 20 7 + <4] f: : "(Cam/)1]; 4 0'9“)"; + ”E e (2 points) Using the frequency response function, determine the DC gain of the system. (DC. am} 5 ”(£035 L. e (2 points) Using the frequency response function, determine the high frequency AC gain of the system. AC hfflk freauancd flat}; i'S gmfl ampmfl “0 - HC 3 — 2;,“ ”C >Z+im+~r ,gl/g—w d9“ wave {/W)Z " 1(zw) +13 9mm. can/MW“ *3 :Lm—sz (av/)7” 0° (007’ Hy». M5 6am =7 W V 2. (8 points) An engineer has written the following m—file to display the exact magnitude and phase plots of the frequency response function H (1w) for the circuit shown in Figure 1. omega = logspace(1,4,500); bode([8e—1], [lo-4,1], omega) Figure 1: Circuit with input voltage f(t) and output voltage y(t). o (2 points) By using only the arguments from the Bode command, determine the frequency response function H (10)) and express it the standard form m "1—1 I O Q H(Jw) Y _ 5111060) + b —1(Jw) +510“) +50~ ~ F _ (1w)" + an_1(Jw)"‘1---+ 01(Jw) + ao o (2 points) State the range of frequencies for which the exact magnitude an phase plots are shown, and specify how many points are contained in the vector omega. Omega, (“amid/rid 500 yolk}; r‘qnfllv from IO rag/[590 {:0 lo L! Faced/53¢“ o (4 points) Given that R1 = 200 D and R2 = 800 Q in the circuit shown in Figure 1, what must be the value of L ? Us!!!” Jol'lilAijL J‘QISIcJfl ’fl fie P5050), £001th (Na/ft Cacti/02101: (\a R2. F “a >’ Y = L "’ ~ = «+2.; .2. + RN“). *flw E I f“! “#97. ' 3. (10 points) An engineer located on a ship in the Pacific has been sipping coconut milk from his EE juice glass while monitoring a radio beacon. When he detects a weak oscillatory signal that cannot be attributed to known radio sources, he increases the volume level of the receiver but the signal decays away. In haste, to test his receiver, he tunes to a radio station located on a nearby island without first lowering the volume. He ends up spilling the coconut juice when the lyrics of Margaritaville blast through his headphones. He throws a switch that places a circuit, with the frequency response function given below, in the signal path to modify the bass and treble response of the receiver. ~ 2 2: =0.1'7w/10 +1 H : _._—_ (1w) F Jul/103 + 1 o (8 points) In order to understand how the circuit affects the sound, find the sinusoidal steady- state response y(t) of the circuit to the input f(t) = cos(10t) + cos(104t) u, - "‘ W figs): 0,! 004,032 ‘H ‘ ZLHVUQ ’- T‘qfl'fg}; _ "”7 I000 V(W/Io3)" ‘H fig?” 4, 4; ’________._. A, _ z ”(0071:; = ‘Hgmfl‘a *l {2411): (ogflo‘ta) =? F: -I. Y! d .’ 0° ’ (Tam... °-l— TW’.’ 0’0, +| 9% ’X,0 2‘5 0 :2 W3; ‘(0 0 0L refine"L t} =O“”") N 0:] e 1] 7- 01' you/Io“) z wwme ,, (W = ~ - _ -| 449° fit ‘02 4(an' 107'"““?0’f) ._ 1,06 re '0 fl a 44w" ’1’ ’ YL T. 0“ ~ ‘1 l: ; Cos (Io ‘1 t) 3;”) ; Ila. i V2. 6%") 3 7% F (t) 2 (as 001:) +- C05 (Io ) [-5) = o.\ (as (Io-t) 4- (o; (’0‘1 s; ) l y h‘yk F/aythlfl 5'?" flow £ré'fiV-eflg'bfl’ [791559) thrwyl') gym! a tum“ o (2 point) Will Jimmy Buffet be (1) upset because the engineer is attempting to drown out his vocals by amplifying the bass response of the guitar and drums, or (2) be pleased that the engineer is amplifying his high-pitched vocals ? SW” my w.” 99- “engage the}: filo-2, eny/néer )5 aWWVW‘E/y tk’ £9955 r95/0fi)24 {a ilflw‘t lot Ca), bt‘éL 6% MIA" file byter («away/,9, UO(CVIIS. 10 04 (0u¢$Q) rc 3mm} PTaC'blCOJ “,va he 3M7; “Q Wayloan it (“7Q {til/lef- w a; J Problem 4: (25 points) 1. (10 points) Consider a set of complex-valued functions 45"“) = e] n we t defined over the interval [0, To], where n is an integer and we 2 27r/To. Determine < d)”, ¢m > for the cases 72, = m and 71. 9e m. Case I 2 n=mT 1;, «my» = f (eflwfixeimwtfilt = S 0012 5 To. CM“ 7— n #m “a T5 anut *‘ E __ j (IF/”Qua, £ <d2n a > = (evmmt) (3* > 60 Se 1 2 m a To 0 - Jn~m)w9t — ,L..... [a ‘— am-nfiwo 6 j: ”WEE/r02 r] ._ .—L———- "l - flfmnfiu/o cl :6 BecOWSQ <2 film/”5L”- : (a5’(n-m)Z)T tjgmm'm) 2” PI) 44;", d‘Jm5 >0 4-1” Wm Using your results, justify whether or not the signals (1),. (t) are 0 mutually orthogonal, or o mutually orthonormal. T/Re 51 y (1 all! an), 3W lyecause. / (Cpnj¢».>= "PW ”#07. The are or; _ - 5T 3’. moba {1/ wtlmnanmufl, 11"- 70/2 I} {o 1:17} <95”, ($7 0 2. (15 points) Two real-valued functions, f(t) and g(t), are defined over an interval [—1, 1] and are exactly . represented by a set of three real-valued mutually orthonormal functions, ¢.-(t), also defined over the interval [-—-1, 1], as H W) W) ~2¢1(t) + 2¢a(t) «mm + Mt) +fl¢a(t) o (3 points) Determine the values of < f, 43,- > for i: 1, 2 and 3. CZz<F,d>g>/<¢L,¢;) : “MD as <e>i,¢z>:l I <‘FJ ¢z> 1C}, 20‘) (‘8 $37.:C3 = 2' t BQCWJSQ. Ee. =65 Parrot/a“: Theorem is a ”Malia 3 I E; z 3an <Mph) = CF+C§+ c3?- = 9 o (8 points) Determine the value of positive, real-valued parameters a and fl so that (1) the signals f(t) and g(t) are orthogonal, and (2) that g(t) has energy E ‘= 9. UJInfi_ paYYchKNS Theorem) .3 1 z 7— 2. 2. 7. -~ - :: icn<¢nn>ecl+‘1+‘3’0¢+’*é h:) E 3 a E; = “I => W C" <4“ 5‘) :0 g) Flt) and) 9(6) ave, orflogofli‘ a (pi?) ’- <'1¢5+Z¢3,00¢1+¢;+I$¢3> =0 0 O - 2.x. aft.) — 2 (MD - 2n <9£¢3> ' + 2x (16%!) + 2 «bi/<52) + 254%» 7—0 a 0 I ”Zen-LZIfi-zo => 0¢=8 LU 2— ea: 7. Svhst‘hkmfl CZ) Miro (I) gives 2c .., /\ L JO/lat \I’IW «’sz ”ml ‘° log"; 31) heard». we m“? stugmgn‘é Sieclnp 12 0" >0‘ ...
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