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Unformatted text preview: BB 350 EXAM I 2 October 2000 Last Name: First Name: ID#: 5Q); [L‘xggs Section: Test Form B This test consists of four problems. Answer each problem on the exam itself; if you use additional
paper, repeat the identifying information above, and staple it to the rest of your exam when you
hand it in. 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) In response to the input
f1“) = "(0: a LTI system produces the zero—state response
y1(t) = 2te"u(t). A new input, fﬁt), shown in Figure 1, is applied to the same system. Find an expression for the
resulting zerostate response y2(t). £20) I
1 0 1 Figure 1: The input f2(t) results in a. zerostate response 3;; (t). Express £1 ('9 055 (A... June % 5 CUJEQ dag
423mg .rhﬂiteg. In abs '9" HA 1?an = swam — Hutt)
ﬁat) 2' thaw: +0 — "i'FICbt) BQCawse. the. {QIS'bem is LTI MAL but) 3:. the. £210 ~5‘l3ufv3 Fawnﬁnk £0 "E. (’6‘) It 4:0”ij thw'é 2. (16 points) The zerostate responses of a certain system for three separate inputs are shown in Figure
2 on the next page. Using this information, answer each of the following questions in one or two
complete sentences. 0 (3 points) Is the system causal or noncausal ?
o (3 points) Is the system dynamic or instantaneous '?
o (4 points) Is the system timeinvariant or time varying ? o (6 points) Is the system zerostate linear or nonlinear ? ’ Tﬁe .sdrtem Is noncwgaﬂ. . The respoose. @300
b%‘ms bQCuen. the. 5th get) 1; “Mlle . ' 733 35122;!” 75 op namtc. beta2.03% the— WM
0042an Gr? (ltd/rod Value: % the, gnpu't. (see Qnrwer above.)  Tﬂe .s m is have—way}? . ObSeNQ, 5J6: '52} Ct) =' 'F; (t+Z)J however) at“? ("a 79;, {tIZD, 9’ 7J9, “gale”. 75 Dog/0156a)“ £39m the aﬂﬁév}
Quintin—y.) drug Fence, Saferﬂqsz‘ﬁmr} Jag; not boil. Observe, thOJE ﬂ (t) = 3 1C, (t3 + 3 4:364; but:
a; ct) :7.“ 35,616+ E/BHQ. 012345 012345 gm 012345 012345 Figure 2: The zerostate response of the system to the input f1(t) is y1(t). Similarly, the input
f2(t) produces the response y2(t) while f3(t) yields y3(t). Problem 2: (25 points) 1. (6 points) An inputoutput description of a certain system can be expressed in the form of an ordinary
differential equation as 11221 d3; __ df Determine if the system is asymptotically stable or unstable. Justify your answer. QC?) = 67sL+57s+1=o 2. (9 points) The output y(t) of a secondorder system is related to its input f(t) by
(D2 + a1) + 64)y(t) = (D — 4)f(t), where a. is an adjustable real parameter. For what ranges of the real parameter a is the system
underdamped, critically damped and overdamped ? Comfar.nd~
C905 : 31+ W!) 4— 6‘4 2.
to 62055 :bL—t— afwnb 1—Wn fax/254k:
WXT’ 6% =3) W0 =9
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21PM”) —' a” '7') F“: m : 7% e For “Life, ﬁber; ")3 be. crrétoag, germ/ﬂag 70:6 db 50 WQ Meg at. .= [6. ' For am aueFQuﬁI/IPQ rasﬂgnge, we, mar! (“Va f0>lj
N inlvqieﬁ'gi at > [6 3. ( 10 points) During an experiment the zerostate response y(t) of a LTI ﬁrstorder system to a unitstep input f(t) = u(t) was recorded at various time instances and the results are Ihown in Table 1. Using
these data1 ﬁnd 0 (3 points} the system rise time t,,
o (4 points) the time constant 1' of the system, and e (3 points) the root of the characteristic equation. Table 1: Recorded step response of a ﬁrstorder LTI system. From Tabie. I.) as; 2&9 (105:3 ‘—‘ ‘40 State 5° 707° 55' 75 36 win/Q. ’07:: 55 Is ‘I,
I": {Eta/s? {In}? 7/ ; USmdi the, rnwéronslnp I 7
”Fins'iT—ovpaer _ Sigemg) Problem 3: (25 points) Consider a secondorder system whose output y(t) relates to its input f(t) via the ordinary differential equation d’y dy _ 4 — 4 t = . a“, + d, + y() 3 ﬁt)
Suppose that the system is driven by the input f(t) = 8u(t), and that the initial conditions are y(0) = 2,
32(0) = 9 1. (6 points) Find the form of homogeneous solution yh (t) (do not solve for any undetermined coefﬁcients).
2. (5 points) Determine the particular solution yp(t). 3. (3 points) Evaluate the unknown constants in the homogeneous solution, and determine the total
response of the system. 4. (5 points) By directly solving the ODE, determine the zero—input response yn(t).
5. (6 points) By directly solving the ODE, determine the zerostate response y,,(t). l. 2. &cme, 4&3) is Coas'Eu/FE ‘L‘F 't 20 ghee .ﬂ __ ~zt: a ..
5108 ‘ 2Cle + Clea E  24: C20. 2t
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yCﬁ : "' 7% ‘" 52. 1‘1 = C2. .—_
#4:): —‘i€2’t+ +25% +6 t .20 Se}: ‘FGfT) : 0 "Fur t ZC’JJ “the my ér‘Q Cong/(£0005
own. d_(cb=z. 0090 9(0):? 60 191L411 BQ—Cocuse, ’dﬂe cLurchLercslstc. ang—é‘w, ,0,” mi
k
C/anggﬁ athft) =' C.e + Cz'lse, i320. ”Covet/QT J :15 40(7) :0 J gr) (1'33 = 0. This can be.
Seen 3 62.7%; a (4:) :1: ’4 then
p 2 ﬂ¢/*‘ ‘féé 1—~I%pr : 0 =9 H‘éO.
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—ee:“— mg” +6 tzo Problem 4: (25 points) 1. (8 points) Derive an ODE that relates the input voltage f(t) to the output voltage y(t) for the circuit
shown in Figure 3. If you introduce a node voltage (mesh current) to determine the ODE, clearly
indicate this variable in the circuit schematic and label its polarity (direction). Figure 3: Circuit with input f (t) and output y(t). KCL a£ :1er I9 ﬁnﬁej (,2 4.. W9 (H , (.
——— + 3,... yo
9. ) For 1h? Walcr" Ll {(blﬂntf '1‘" t
J 2.  2.0)4‘ czgwﬁmwﬂmcgc (a) «Subsitu'krﬁ (2) "7123 CI) ﬁlm t
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‘LQ 4“ L7, (vﬁrryﬁrﬂlw Kg: 4— 012— :0 C33 b1 ﬁeren'EKWLm bH‘ 5‘ (0&5 “Jar
I? i 2. (7 points) Once again consider the circuit from part 1, redrawn in Figure 4. Suppose that ﬁt) ~_— 6 u(—~t) + 3 an) and R1 : R; = 10 k9. Determine the initial conditions y(0+) = m5 V
d'y _ V
Et:o+ — 5 Figure 4: Circuit with input f(t) and output y(t). , Rt 17:0" the. mJuCEucs are. ~55de C‘I'tw'él' “’“QSO a,co)_. *CO‘) = .622... = 0.6,” My.
R‘ IOKJL
‘JHRL {JUI ‘ 6H: t =01} MOW”): 0.5m” amo ‘9 lath“): R. C109“): 6v? ' Eacavge, (3(0"): CL (0*) "' C, (0') = .6»: ﬂ") “’9' at” (gnu4.2 shook vkwr): e\(.= 319L030. ' USIﬂé v, :: L (01L _ %% 0mg We): Vﬂ'lﬁfy [“5 3. (10 points) A circuit, diﬁ'erent from the one considered in parts 1 and 2, is described by the ODE
3'] + 8g + 32 y(t) : 128 f(t). Given the input f(t) = u(t) and the initial conditions y(0) = 9(0) 2 0, ﬁnd y(t) for t 2 0. CumyarmU. Q55) =3 7&2'1'8? +32. to 'I. an» =— n1+szm +Wo reueqig
W76— 3: 32— % Wm = 432.. = .1 (—2—: _..L_.
ZFW 7" 3’ =7 19: £21: 5 anﬂ. so the, sdJExm is uanQhQumﬂQcQ arm/Q
#11 ('5) = e {Wt{E (AC3WOQA: + éfrané) “i320 WKWQ— Faun: ‘I amﬂ Woo—:Wndl“1°2'= "i. 8% Se. 43H) 1) COnJ‘an’E "Rr 'b 2.0 0mg 73, ¢ a gag.
Cﬂu
7‘), %0J a? (t) L: #. SubsL‘Evé,y y? {0% m 005. (7‘9‘05 .. A—
W + 3% 1— 3L/ 512.? 4 [Si:1 9mg. .50 at, Ct) = ‘1 t 2 o. The, 1’59"er Sea/Jam is g (t) = /h (1:) 47,93) 3 e“ Ht (9705 ‘115 + 8!"? V16) + 7.
Us“: '5“? tm‘EMQ ca L (on) ﬁycazo:n+y ==) 61:”?
jﬁb>~o :‘/lq+'4t3 =9 651927 9mg, 5 13 l4 ...
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