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Unformatted text preview: EE 350 FINAL EXAM 5 May 2000 b Name: 9 l n3 ID#: Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Test Form A Important Guidelines You have 1 hour and 50 minutes to complete the exam. This test consists of six problems. The exam score is calculated by adding the ﬁve
highest problem scores. This is a closed book exam. You are allowed one sheet of 8.5” x 11” paper for notes
as stated in the syllabus. Answer each problem on the exam itself. The quality of your written solutions is as important as your answers. Your reasoning
must be precise and clear. Your grade will reﬂect the clarity of your presentation. V Problem 1: (20 points)
A linear timeinvariant system with input f(t) and output y(t) is described by the transfer
function
u “— — 3 _,__S....—_—
_ F(s) '— 33 + K s + 4 = (5+MS'H)
K=s Where K is a real number. 1. (5 points) Determine the zero and pole locations of the tranfer function assuming that K = 5.
29m: Sachsé; 4+0) :0 0’?— IS = O I 22m)  roles Poles Salas£d_ HLS) =°’ 0?— (S+:)LS+~¢):0 cm 2. (5 points) For the value of K in part 1, is the system asymptotically stable ? 82.01052, b°’l.'J p.90 Rave. nOaanan. (‘qu Part5) "the...
.SﬂﬂQm ‘o Stable. 3. (5 points) Determine the DC gain of the system. DC 60cm = Melba =0, 714.. mac"? :5 are, 4. (5 points) Determine the value of K so that the response of the system to a unit step
input is critically damped. For as ch't‘tqug (app/25¢. ﬂ“?— muﬁtf 5Q
fleaabNQg [Gag cm (Jeni?ch ﬁéwnlmhla In this ca» .32" 1 K5 +7 :— 5?— ... 2fwn5 +y/n?_' ﬂaﬂ sa
cu}: Li ml 21%“ = K, sol/:2} Q» K (7M2: 2mm 2‘1,
Nat thﬂi: The. Sashn TS cation"
5 J1 9Q 312cm cl: ha; K T— SL+ : ﬁn a, dc, “209/
‘f ioqﬂnttc% / Problem 2: (20 Points) b 1. (10 points) Given a LTI system with impulse response
Mt) "—— edt u(t) and input
f0?) = 6—3' “(0 ﬁnd an expression for the zero—state response y(t) using the method of Laplace trans forms.
. _ t I .
U5.8 the transLorna p44» ea' “(5) é—a ﬂ: dew} I f Ugly ﬁn, com/wa out) 5 4643* Mt)
l ..
YLSE =1 PC9N—5) : m "' 5+2. 5+3 A (“‘9 3 are, fun—Jen}: 2. (10 points) Given a LTI system with impulse response
h(t) = Eng: 11.“) and input
f0) = 6'“ RU)
ﬁnd an expression for the zerostate reaponse y(t) using graphical convolution.
In order to receive credit, clearly specify the regions of integration and, for each region, provide a sketch of f and h. No credit will be given if you quote
a solution from the Convolution Table (Table 2.1) provided in the text. 016:3 = FH:):a~h(t) : jsngpG—ﬁcgt ~ZCtr) Problem 3: (20 points) 1. (10 points) The operational ampliﬁer in Figure 1 is ideal. Determine the transfer L function and express it in the form
_ Y(s) _ 1313+!)D
H(s)— F(s) _ 3+0”.
The coefﬁcients an, 60 and bi must be speciﬁed in terms of the component values R1,
R2, R3, L, and Figure 1: ActiVe circuit with input f(t) and output y(t). L/ The trdnfggr ‘pv‘ﬂc'blon ‘Fﬁm #(t) 1b 2.66..) is oé‘éa‘moog ()5: n& 001+?“ IV “Stun Fun «9%) a woe/h ECS) = I PC)
YB) = (SL—t R4422.
————~R‘ > 13(5)J Subs‘l'l'bvﬁon a (£385 2. (10 points) The actiVe circuit shown in Figure 1 is driven by a. periodic signal with
period To = 1 and f(t) = 3—” for 0 S t S To. As a ﬁrst tep in calculating the steadystate response y(t), determine the coefﬁcents
D“ of the complex exponential Fourier series representation of  cool:
Dn =' 2? Heed" OQi: 3, we :. z'n‘ Problem 4: (20 points) 1. (10 points) Find the transfer function of the closedloop system hown in Figure 2 and
show that it can be represented as Y(3) bm3m+b 1Sm"1+"'+bl 8+bo H(3) = Fm: s"+an_1s“'1+~+a1s+ao ' 51 4» 3 q“; __———— 5 _, “15+?—
57 «s + 3(1511.) 2. (10 points) For a. certain closedloop system the transfer function from the reference
input r(t) to closedloop error e(t) is E(s)_ 32+403+4
R(s) _ W' For a. unit step input r(t) = u(t), choose the proportional gain K so that the absolute
value of the steadystate error is less than 0.1 e“ = e(t) < 0.1. 05.? “Ll2. ﬁtﬂﬂ «9,1,2, fjeofem
855 3 SECS) = aﬂsﬂiew '5 5° 530 £65) II
Q €55 ‘— ——~" 40.; :9 1+K > 70 =>
34—: [Va[2 thj; 4;»— ﬂo. Lanna Value. theorem £2) £0. «fling/9) 55(5) cunno'E ,QdUQ. any [9on in tie ry‘léﬂJLIg/Me or M
a“ iw aux—LS. ' 10 Problem 5: (20 points) 1. (10 points) A LTI system has the frequency response function .. 3"“ + 1
HUw) = : = 10 100
F .271 + 1
1000
Construct the Bode magnitude and phase plots using the semilog grids provided in Figure 3. In order to receive credit: 0 In both your magnitude and phase plots, indicate each term separately using
dashed lines. 0 Indicate the slope of the straightline segments and corner frequencies of the ﬁnal
magnitude and phase plots. 0 Do not show the 3 dB corrections in the magnitude plot. I DO NOT COPY THE DISPLAY FROM YOUR CALCULATOR !! 11 ' '=s;p~+++2+eéé;;»swe«éevaee:
Was; L 504: . . . . . . . i . . I . . . . , . _ . _ . ‘ . _ . . . . . . . . . . I . . . . . . _ . , $ ._ Phase deg I
4:.
01 C _90 . . . . ‘ 10° 10‘ 102 103 1o4 10 Frequency (rad/sec) Figure 3: Blank graphs for sketching the magnitude and phase response of the frequency
reponse function. 2. (10 points) A secondorder linear timeinvariant system has a. frequency response func
tion H ( 3w) with the magnitude and phase plots shown in Figure 4. The input to the
sytem is 15" 9/ f(t)=100+cos(700t+45°). =3 AF: —. I Find the steadystate reponse y(t) of the system. U From tlo, 15°94, maaﬁcbﬂI‘L ﬁloE 4:
A 5') II
“b +3
+ —> 1.3""
a E)
(EC
+
“f
E T‘ T
DC ‘Uavm SJGuSotcﬂoa Dc {awn $.0v5b1ﬂt—Q tw
4:2!”00
ﬂ 0‘ 90mm
g! 2' ‘F. 2: ’00 {0— ____ (o
1H“ 740) ; V2 2‘ F1. l 7°01” Cg / _. o " 70° ‘ I eat 75 Joe a “(5‘0
: [0 e at £063 : [o + [o €9.5(700'I': ’75“) 13 Phase (deg); Magnitude (dB) Bode Diagrams 3 1 02 70 o 10
Frequency (rad/sec) Figure 4: Bode magnitude and phase plots for the LTI system considered in part 2. 14 0 Problem 6: (20 points) A double sideband supressed carrier (DSB—SC) modulation system produces a signal 3(t)
Whose spectrum is 5(w) = M(w +wo) + M(w — we). This signal is demodulated by the system shown in Figure 5 where H (w) an ideal lowpaas
ﬁlter. Ideal Lowpass Filter oos(coot) Figure 5: Demodulator for the DSBSC' signal 3(t). 15 1. (10 points) Assume that «1., = 100 rad/sec and M(w) = 34”” 10. Sketch the spectrum on the graph provided in Figure 6. 2 6;, 25 CE) COS «uon :. 2L1? {SiB335" Q’ECo; «NoL]
7, 4? 5909 3k 1; E Svfw 1way 4i Sifwawa']
2,! = JESLVMQ + Aida/Mo) guLs‘L'Ev'émﬂ 5(w\ :1 M (weouch I~ M (w we) E5“) =‘ JgMCV’rz'UcBF f’”@ 4— Jimfw) +— fMCw—Ywu)
E (U?) = fMLwFLWQ 4— Mfw) + tMCW—Zwa) 1 a 00° : \oo) Mfuh: a 9"1/! Figure 6: Blank graph for sketching 16 2. (10 points) If possible, determine the cutoff frequency we of the ideal lowpass ﬁlter so that the output of the demodulator matches the original message m(t). If we
cannot be found to satisfy this condition, expalin why in one or two sentences. 14: “L5 065 FOSSIL‘Q '50 EIIMIkK/g {‘9‘ Com/moné! .1; M [w fwo) dnj/ J}; M (w, W‘s ova—t1) (Do—E J¢S£Wg,y M (w). 7}“) 75 bQCMLSQ M(oc) :6
“at 0.1 ba JWAGQ' 17 ...
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 Fall '07
 SCHIANO,JEFFREYLDAS,ARNAB

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