final_f07 - EE 350 EXAM IV S olu‘LCo AS 19 December 2007...

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Unformatted text preview: EE 350 EXAM IV S olu‘LCo AS 19 December 2007 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Test Form A INSTRUCTIONS 1. You have 2 hours to complete this exam. . This is a closed book exam. You may use one 8.5” X 11” note sheet. Calculators are not allowed. aspen: . 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 1.2 Continued. NO credit will be given to solutions that do 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. (10 points) Determine the transfer function for the circuit in Figure 1 and express your answer in the standard form Y(s) bmsm + bra—13'”—1 + - - - + bls + b0 F(s) sn+an_1s"_1 +~'+013+ao C4»): + P65) (/3 fa) a R R yap—am) Figure 1: Passive circuit with input voltage f (t) and output voltage y(t). Begun we out. Jail-ermmm} 5‘0- ‘Q/no’EIOAJ 59ft am cm‘r-Euns to 2mm. 08!? u oHIILyq, 9' w LS] on Y“) = gL/IR Fco .{LIIR + 31;: Y“) 5:”! Sc (5L+ a.) fig; . “‘R + J— SCCs pr 8-) sud“ so SLLLIL “Etc—i : ' -1... SLLcK-r SL 1— K La. “3 d- ckecl‘q 9’ a) m “(Er '5 gt“ 0/)? P do a b . g HC$)\ ’ 0 9'5 afq'fufi b thllé do. m! C. g'am 5:0 or. short filer m [Nab ' = =3 0, Ca- : HIE/h {wovme 3am , “(fitsw ls we") Mn so Y F 2. (8 points) The input current f (t) and the stored energy determine the total output current y(t) for the network shown in Figure 2. Given that C = 0.5F, R = 19, L 2 5H, 110(0‘) 2 2V, and y(0_) = 2A, determine just the zero-input response yz¢(t) of the circuit for t Z 0. V00) + _ C f(t) R L Figure 2: Passive circuit with input current f(t) and output current y(t). To JEmJL the aeruvmflil“ (“PM 5‘1: Fee) = ° (ret’h‘w m (WW/M: Swag 51. R L ;,, LL; s’jmu‘n JED.— wn'bh an open circa-b3 : + “2(5) ’11.. (0') __dl R ‘1' '3-(0‘) '- a (.Lco') = 709’) From thq, crr‘wa'L' gin-gram YCE— ’fiw’: S— Sh‘l'fl 5+"? 3. (7 points) An engineer seeks your help in determining the impulse response h(t) of a LTI system, and provides the MATLAB data in Figure 3. >> [r,p,k] = residue(num,den) r: 8 2 P: -2 -1 k: 4 Figure 3: Expanding H (s) using MatLab. o (3 points) What is the impulse response function h(t)? o (4 points) Specify the transfer function, and express your answer in the standard form bmsm + bm-lsm—l + "'+ bis + 50 H3 2 () sn+an_1s"_1+-~+als+a0 Us"? the rOSul't‘ ‘Fram MATLAB, rm HZ) a ‘_—"_-_. .P M," H“) “ K + s« pm sweat) ,. 8 ,_?‘.—/ ‘ i *— 7: i" w u: ’tM-e) bL-l-A = wear 94— M55 *‘ 1" 900‘ = (3+2) (5-H) Problem 2: (25 points) 1. (10 points) Figure 4 shows a feedback control system with reference input 1'(t) and output y(t). Determine the closed-loop transfer function and express your answer in the standard form Y(s) bmsm+I)m_1sm—1 +--'+bls+b0 R(s) s"+an_1s"_1+---+als+a0 Figure 4: Feedback control system. more, and: (5(5) = «La- ECs\ " "L— EC‘B “ml 5° 5+L 5+3 \ - \ 8L5) _ l .. _L_.. z. W =- 31.”? 5-5 +6 ER) " 5+; 5+3 Lyn-365+» L £4 yum Pecan/cos 1‘70 03m; thus Magi-l: tbO, bloc» any 2. (15 points) Figure 5 shows another closed—loop system where the parameters a and K are real—valued constants. Figure 5: Feedback control system with real—valued constant parameters K and a. (a) (7 points) Given that the parameter a is chosen so that the closed-loop system is BIBO stable, specify the value of K so that the DC gain of the closed—loop system is four. I M Y- Muse») 3 Ad’— 9‘ ‘- ___L__...—- s"+-">¢$+¥‘ ‘4‘ K $(5-r‘306) TN— D0 841"" air w 6(05ag—Qoor (d .SlQm IS Lest k 51; u 7‘l’ l .i (b) (8 points) Set K = 1 and Choose the value of a so that the unit-step response of the closed-loop system is under-damped with C = 0.6. I Y _______I______... : fi/j. (L s $z+3co>+| 52+ 29W!” Hy" Problem 3: (25 points) 1. (13 points) An engineer determines the frequency response function of a LTI system by applying the input f (t) = cos(w0t) and recording the sinusoidal steady—state response y(t) = Acos(wot + 9) for a range of frequencies wo. Table 1 shows the experimental data. Determine the frequency response function of the system and express your answer in the standard form, for example, (JW/wl + 1) mi How) = Carefully read note 6 on page 1 of the exam. In order to receive partial credit, clearly describe your approach. ’/L Dc mugmbvfll- (6Jb Jaw») :25 e— 70‘ 00” than pm gig“ S fact» 0' loo Jet/r0052— ”‘ m nfirvfl. (of H043) “was one flecwfiz Table 1: Frequency dependence of the sinusoidal steady-state response. twl pbufe, get/mew?) 9 15’0". ’° ° '1‘" t k 'F .. ('05 o F m 9 w My 071 9.4% w - 70&8/fle° . 7"?» 1" aflflc‘bw) ‘the, maimdufl, rolls 12% 4.1.4“, “Data two 90b5, Nih- ‘thué at 5.0 raj-[mo : LC 7'th main . a.) U4va, L' 6013 900w“) ain’t {10; p v.61. «mac-Ea) {hi bo‘li earner Mflodéu am. {swa- 7.5 ow—l‘d‘c ’2’” Do ‘ 700 ThIJ 2. (12 points) Consider the system whose transfer function representation is H _100(s—1) c lea LED—r. |o ill-4‘ (S)— s+10 ’ 1° (lilo-(5 5/10" The zero of the transfer function H (s) is called an unstable zero because it resides within the right—half of the s-plane. (a) (1 point) What is the magnitude and sign of the DC gain? EC apu'u : HID): -Io (Hto)l= ZOJBJ 2‘- Héfi 7' —,8o° (b) (1 point) What is the magnitude and sign of the high frequency gain? "'3" frefigam == H'Coo) == loo lH—(~)l=‘{oJfl) ‘LW‘fllm" (c) (2 points) Provide a hand sketch of the magnitude and phase reponse of the zero term yea/1 — 1. Carefully label the slope of the magnitude and phase plots and indicate the location of the corner frequency. ld.‘~/t —1 [5 060734533} :- ldW/l +1): V(I)"*(-r 7’ -—§ The, magm’bvvfiz 4 d-W/I ‘I all are (flefl‘é‘la-Q. (4.: 2; (d—w/l '13 = To"!~| (T I"! “V (In. ‘ e ld-W/l ’ll + 20431399 046 l w FVI‘ an vns'bq $2r‘3 i) — “IS—“flack (d) (8 points) Sketch the straight-line approximation of the magnitude and phase response of H (s) in Figure 6 (page 10) and Figure 7 (page 11), respectively. Label a_ll relevant features of the graphs. H9”): lo fili— d,u./,o +1 9 Magnitude [dB] 10¢ 10" Magnitude construction plot. 10‘ 10° 10‘ 1oz- Magnitude plot. 10" 10° 10 ‘ 10 *- Frequency [rad/sec] Figure 6: Graphs for constructing the Bode magnitude response. 10 Phase [Degrees] Phase [Degrees] 10"- Phase construction plot. Phase plot. 10° 10' 10’- Frequency [rad/sec] Figure 7: Graphs for constructing the Bode phase response. 11 10° Problem 4: (25 points) 1. (13 points) A LTI system has the transfer function representation l3s+45 32 + 103+9' (a) (3 points) Specify the system pole(s) and zero(s). V {II} zerocs) sort-:59, I35 1- qr = 0 :=> '29“? 5’ polocg sucks“;— 5L+lo$ + cl = (5+q)C.S+¢\ :0 =5 pole) we: S:-C{ elm—a ss-l (b) (1 point) Is the system BIBO stable? Be. (also. be H1 polo) ham. noaa‘liwe mag «om/'ng tk, 5:51;,” 13 13:60 s-thle. H(s) = (C) (2 130111138) What is the DC gain of the system? D ‘ x n a C we 3¢C4vft the, rSas'bcm I; filéo sfil. gel/0 1.5 fl Hme = 15— = 5‘ Cf (3 points) Is the zero-state unit-step response under—damped, over-damped, or critically damped? Because, the. {,9ch ? H'CQ are, facing) Mga‘bueJ czan J‘Jél'x‘é ‘5“. 2‘: Ora-5434.432— um —s '50? Ya)” one, B we‘rqfla acad— (e) (4 points) Find the impulse response representation of the system. __ (gs—wo— ’ I35+yg _ ,4 + L "‘5" .2. * ’ 5+: 5+? 5 1403+? CS+1)(.S+V) A: (32% 35+"? ii _ l1 C5 (5+9) __ 3 '— ‘8 -7 +155 ~1I7+~4r ,. 1.73: = B =“Z{ l35+ ‘15‘ l = (‘9‘) = ’8 — ,3 5 v9 12 2. (12 points) A LTI system with input f(t) and output y(t) has the ODE representation dzy dy __ 5 _ - 2 . (a) (3 points) Find the transfer function representation of the system and express your answer in the standard form Sn+an—18n_1+---+als+a0 . ' .éfi Fina. the. Laplu4¢ transfer-m a, botk 9.94.; 7 tie/ODE- m‘H" In! ConJL'kwm Sa-E be zero :1 rm + $5 ch\ 4— 6 Us) = F0) [1(3) 2 (b) (5 points) Given the initial states y(0”) = 0 and y(0-) = 1, find the Laplace transform Y(s) of the unit-step response, and express your result in the standard form bmsm + bra—181n—1 + ' ' ' + bls + ()0 s" +an_1s"_1 + ~v+als+a0 ‘ 3 .L. Raped? garb («4-) 5J5 [helm/(b the "HEWQ CUth‘Ltu/zs C(un 56$ Fm 5 Y(s) = .L $2 Y(s)-53(o")-;;(o‘) + 55YC9-53(J) + 6”” _,_ Biz): 0 l 0 Y6) {SI-#55 4-43 '3 ~1— -r +JS/Oi) +6’y/Zga) .1. Ya» -_-. _,_$_1_____. 57-4-55 +6 (0) (4 points) Using Y(s), determine the steady—state value of the unit—step response. Nah. and: sch) =. ._§_'L‘.——v— '— ___$_;*_L.— ha; Ins-EL polo; S '- + :5 +6 c s +2\(5+3\ m flu. Lil-WJ 4mg 5° W and ualve. Uaeorem 75 apphcaye‘ . 5+] L a ' M :— 955 ‘ S W“ ' 0mm) 6 13 ...
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This note was uploaded on 07/23/2008 for the course EE 350 taught by Professor Schiano,jeffreyldas,arnab during the Fall '07 term at Pennsylvania State University, University Park.

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final_f07 - EE 350 EXAM IV S olu‘LCo AS 19 December 2007...

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