examII_s01 - EXAM II EE 350 Continuous Time Linear Systems...

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Unformatted text preview: EXAM II EE 350 Continuous Time Linear Systems February 26, 2001 Last Name (Print): II" First Name (Print): ID. number (Last 4 digits) : Official Section number : Do not turn this page until xou are told to do so u— II— Total Test form B Instructions 1. You have 2 hours to complete this exam. 2. This is a closed book exam. You are allowed one sheet of 8.5" x 11" of paper for notes as mentioned in the syllabus. 3. Solve each part of the problem in the space following the question. Place your final answer in the box when provided. If you need more space, continue your solution on the reverse side labeling the page with the question number; for example, “Question 4.2) Continued”. NO credit will be given to solution that does not meet this requirement. 4. DO NOT REMOVE ANY PAGES FROM THIS EXAM. Loose papers will not be accepted and a. grade of ZERO will be assigned. 5. The quality of your analysis and evaluatiori is as important as your answers. Your reasoning must be precise and clear; your complete English sentences shall convey what you are doing. To receive credit, you must show work. Problem 1: (25 Points) 1.1 When we apply an input 1’ (t) = 2u(t - 3) to a certain LTI system, the zero—state response y; (t) is 3:1 (t) = 4e‘(“3)u(t , 4) - 6u(t — 1) o (5 points) Determine the zero-state unit—step response y(t) of the system. Simcc Wme 4'4 m , wt) -——» we) = wimp/3 MP, 2u(£—~3) ——-9 2‘11Hr3) caged) VH2) -"-' .21. 131(‘l:+3) = fi-(4l. u({+3—4)_ 511({34'3'0 o (4 points) Determine the impulse response h(t) of the system. . = 0L . 3mm, htt) a: Smut—mp AMPW; -t -t g: d‘é = ~21 MIL-IHZL 5tL—1)-35(£+2) a: t but): —2[ wt—n- 2,1"5ll:~I)—35(£+2) 1.2 (8 points) The system shown in Figure 1 is composed of a difierentiator and two LTI systems whose impulse response functions are h1(t) 2 411(1) *- 3) and 1120?) = 2e‘tu(t). This composite system can be represented by a single block with an impulse response h(t) so that y(t) = f (t) at h(t). Find 11(3); do not leave your final answer in terms of h;(t) and h2 (t). sum as: m) “t m Differentiator Figure 1: System Diagram for Problem 1.2 _...__ FROM W cling/1m , NJ ({7} == dX ‘3‘ L72 (.H cit sutsli’mlr out}: ¥(’c)*ln1(’cl => gttki {-P(lc)*ljl(i:)i¥ll(t) d’c A‘Yld. so, “3%) : 'Fl'lil #[dlfl‘iitl * L, [H] OH: 1 o g: — " WV di: hf)" ‘ 450° 3) but) [dead-3) as him] a; We) «:3 G‘— H Fun—.m- 1.3 A certain LTI system has an impulse response h(t) = e't(sin(t) — cos(t))u(t + 2). o (2 points) Is the system causal or noncausal? Explain in one sentence. "Wu Wit/m Aid 4100066!me bxcmx 1% MFA/0&4, Marco/144 is momcwsafi £42,, Wt) 940 , Vt <0, 0 (6 points) Determine if the system is bounded-input, bounded-output (BIBO) stable. Justify your jlhufl (it = 5, |£_t(sf'Int-cos1;)‘ uh”) <13 _°° —OD °° J; r. p. [gtmt—COStl (it —2 00 .— "’T. 00 S f 12.1: J3 0LT. = -_9-__ (J3) *2 ‘1 -2 2 = 2. J3 < 00 (-Fl'lmfllg) Simu. S \hmlolt is ln'mfh, Wu W’rm {s BIBO slam Nob. | gimlt) wont” = N2 cosft+5§ll é Ji 4 Problem 2: (25 Points) 2.1 (15 points) The input f (t) shown in Figure 2 is applied to a LTI system whose impulse response is h(t). Determine the zero-state response y(t) = f (t) * h(t) for all values of it using the graphical approach. Do not sketch y(t). In order to receive credit, clearly specify the regions of integration and, for each region, provide a sketch of f and h. f(t)=e '“+2’u(1+2) h(t) Figure 2: The graphs of the input f (t) and the impulse response h.(t). o Skai’ch in) mid. mic-1:). «FLU (continued) Amd m, «3801- 2!- ’ -2 2.2 (10 points) Find the impulse respouse function h(t) for the circuit shown in Figure 3, with R = 5M] and C = 1011.1”. Assume that the operational amplifier is ideal. (39-0an is iolwdl =3 -l:(’c) ___ cdvc R dt + ‘50:) - Figure 3: The CiI‘Cult for Problem 2.2 = ‘ci‘é dt Problem 3: (25 Points) 3.1 Consider an orthogonal signal set {1131 (t),¢2(t)} over the interval t E [0, 00) where «2516) = (3e—2‘ — 2e“)u(t), mu) : e—‘um, (¢1(t)’¢1(t)> = la and (¢2(t)a¢~2(t)) = g o (5 points) Show that the signal set {4516), ¢2(t)} is orthogonal but not orthonormal. If 00 -t -—t -'t <¢1,¢1> 5(312-11 )1 0H: 0 = (0+0)-—(-1+1) = O acaifiogomwe {451 3952 3 1'3 000+ MlfiM/Im/maf £2ch <¢,,¢,> $1 o (10 points) Find the best approximation of the signal f (t) = 206—3tu(t) in the form, 1‘“) :3 6194910) + 02%“), Where e(t) = f (t) — c1961 (t) — 62452 (t) is an approximation error and c1, (32 are chosen so that the energy EC of the approximation error e(t) is minimized. 1% but appnowirmot’n'o'nfsohimd 1913 choosfmg co .. -21: «t C __ <¥3¢,> __ 20,1},t (3.2. ~24, )0“? 1 <¢ (12> - ° __._________h__ 1’ 1 1/4 ‘0 -gt ~4t = 4 60-9. -40x cit 0 DO -51: = 4 ( £191 -— 40 -4’c) = 9 ,5 -.__. 1. [ t *4- o $ <1> > °° t = < —3 _ 02 ___:_i = 2 20,2(1, )dt “52,959 0 ~4t oo :- 40 JL 3.2 (10 points) Consider a complete orthonormal signal set {x,:(t)}§§1 = {2:1 (t),1'2(t), . . . ,mn(t), . . .}. If a signal f (t) is represented in terms of 1.:(t),z' = 1,2,... as f(t) = 9:1(t) + $2226) + (32 3:3(t)+(«21-)3 me) + by using Parseval’s theorem, compute the energy E; of the signal f (t). Hint: You may need the following identity °° 1 Ea”: , |a|<1. l—a n:0 simu $0“ = 6,961+ 029624- csxa +,,_ y 1 (11-1 5 C’- 2 ’ C3 (2.) 3'”: C1.” ( And so, big szvafl ’5' WW, 1 (01M +0 MffiOmmafi'm) 00 E = Z 0.: <96,1 96,” -F m=1 °° TM 2. oo 71-! = elm = 2, [2H °° m = Z .1. _ ‘1 Yn=o (4) " 1‘ V4 = 3:}- 10 Problem 4: (25 Points) 4.1 (6 points) An input f (t) = 2 — 3sin(10t —— 45°) + 4sin(30t — 30°) is applied to the lowpass filter whose frequency response is given in Figure 4. Determine the sinusoidal steady-state response y(t). |H(jw)| { H0“) 4 -20 -10 O 10 20 Figure 4: The magnitude and phase of the frequency response function of the lowpass filter ¥ m = 2,- 3 Sim (wt-45°) + 4 sim (sot~3o°). U94. sowfmf’rl'w 4. gm ) = ‘3 [I 2 helm 31:10:): 2- H '0 mm = (-3) (magnet 1‘31"“ = '3 3“” Hot—45°) -135°+Qo°) = —3 cost1ot~135°) laid} = —$ nos(1ot-¢5°) “Cam = 49"” (3°t'300) ham yam = o bum Hrjso)=o_ ‘90“ = \é'tt) +\d,L’c)-\~133L’c) Kath) 2 9 — b coshot—45’) 11 4.2 (7 points) Using phasor analysis techniques, find the frequency response function H (jw) of the circuit in Figure 5. Put your final solution in the form bm (Mm + bm_1(jw)""_1+...+ b1(jw)+ be H(jw) = Uni)“ +a.n—1(J'W)fl_1 +_..+a1(jw}+a0 L—ngL Figure 5: The circuit for Problem 4.2 Us; W1” OUVIISI'O’n a} mod; A : g = E. a + 'wL+4—- R J me a; : E--—1_. = FR 1 5"” R+'uL+I . T“ J .- W1 JHC _—-—-—-—--—"‘""——-.___. jLJRC + (J'U)ZLC + 1 II 12 4.3 (12 points) Consider a causal LTI system whose input f(t) and output y(t) are related by dzy dy _ df F '1' 6% + 531“) — 2f(t) +43}. Write an m—file that o in Figure 1, plots the zero-state response y(t),0 _<_ t S 4 due to the input f(t) = sin(3t)u(t). Your plot should contain at least 200 data points, and a in Figure 2, plots the Bode magnitude and the phase plots of the system from 0.01 rad/sec to 100 rad/sec using 100 logarithmically equally spaced points. 3) limsPau (0 , 4 ,200); 'l: = .9: Shani-t); P = [4,2]; (9. = [1,6,5]; w =. logqucx, (—2,2,100) lll'gwu. Ln lSI'rm( 19,51, lift) $1'WUL) looolL ( PHI, w) 13 ...
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