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Unformatted text preview: BB 350 EXAM III 15 November 2007 Last Name (Print): S plu‘é I o n S First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Problem Weight Test Form A w 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. 159°?” 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. (7 points) Consider the periodic signal A = I n =2
f'vﬂ ...————.
f(t) = a +ﬁcos(2 t + 1(1) + 'ycos(4 t + 45), where the parameters a, [3, 1/}, 'y, and 45 are realvalued and constant. Figure 1 shows the amplitude and phase
spectra of the periodic signal f(t). C n Figure 1: Amplitude and phase spectra of f (t). (a) (2 points) Determine the fundamental period of the signal f (t). {ICE}:— $C:E +70) Cebuth 11', = 1111‘ car.1 *4 To == 27"!” e ud/I'él‘eJ
To = "'0 5 Elm .) n: [J "’32. Smul‘lbf‘ urbtyed Sniff.59]? 6
To =T 19 No = E 1‘. 2.
To 2. (8 points) Find the trigonometric Fourier series coefﬁcients a0, an7 and bn of the periodic signal g(t) in Figure
2. Note that over the interval 0 S t S 277, the signal g(t) has the representation _67z —47z —27I 0 27: 47; 6,, Figure 2: Periodic signal g(t). 7; 2.77
‘4‘ ‘J‘ ‘ ‘ ———""—'  
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211 4—“ , S, c a) I}
1‘??— i’i (Cosérb) r M: 5m (nt)\o = 20,01. lacs/742w?» n ll ur (Q J use, 307—7
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C. \
L,— —_L os/Vr‘rn) E=,,‘ﬁ~ all!
“" nrr 3. (10 points) Figure 3 shows all the nonzero magnitude and phase spectra components of a periodic signal h(t)
whose fundamental period To is 7r sec. Find an expression for h(t) in terms of sinusoidal functions; do not
leave your answer as the sum of complex exponential terms. (I ID,, 4 D Figure 3: Magnitude and phase of all the nonzero exponential Fourier series coefﬁcients. ' Bemusz anl= lDnl anoL AD“ 3 — 4b,") “the 53mg bHD 15  From Dede.493 Do ﬂ 5’
m
9.. 9 uni '50 = — AB—
CD = 3 Dn‘ “n2. 6/) 2 A—DI) n F,.o,,‘ the. speaﬁm above. Co=ba= lDoleAbnz“! c, = 2.15).): '1 6,: lb.=1T/v
(LL: 2.11324 =7— 6;: lbL= 77/2, t 10.5 To = W"_’ (U0 3 277/70 :2”  usury these, fejults
q Co‘wao’t‘f‘ 9‘) 4 C2, CasCLle: +62) gCOQ :2 Co '1'
q + L: 0.05 CZt+U/~1) + zcosmt+ 77/2.) Problem 2: (25 points) 1. (6 points) Figure 4 shows the magnitude and phase spectra of a periodic signal f(t) whose fundamental
frequency wo is 10 rad/ sec. [D II Figure 4: All nonzero terms of the magnitude and phase spectra of f (a) (2 points) What is the average value of f(t)? 300 0°
[The «warm value. JP {(0 (5 Do’ mole} = 2.9, :2“, (b) (2 points) Is the signal f(t) real, imaginary, or complex valued? To receive credit7 you must justify your ansgfrc'ausz lbnl= lb—ol +(t) 3.5 feeQ VGA/99. Q01 3D“ = ' LD'FJ (c) (2 points) As a function of time, is the signal f(t) even, odd, or neither? To receive credit, you must
justify your answer. Not; that bl“ [Dth 3 l9] = Baggy f}; the, ho are, [lather pvr¢£i reoQ 0r qu mar/U
‘§:('b) 78 {)9th am am of Own 'Fvnc‘éwn inme 2. (12 points) Once again consider the signal h(t) with magnitude and phase spectra shown in Figure 4. The
periodic signal h(t) is passed through a LTIC system with the frequency response function w
—Jﬁﬁ H(jw) 2 we 5 g w g 15
0 otherwise to produce a steady—state response y(t). The ﬁlter parameters a and ﬁ are realvalued and satisfy a > 0 and 0 S ﬂ S 7T.
(80 (6 points) Specify all nonzero exponential Fourier series coefﬁcients D3 of y(t) in terms of the parameters
a and 5.
H pale} 1s onl} non zero gun n = IJ 9.5 we 1:: Lo raga/Sec,
_ g 4.17/1 gar/we)
U — ' ‘F'  c. a l ' e ‘—‘ 0(— 9
nm.  Home» 13, _ o e 
 n ' r “TN—6)
+ r; 9 /“l 3
 a; e i’ l . a ace I)if —— tic—04w) bf * h='l (b) (6 points) If the signal y(t) is real—valued and an even function of time, and the power Py of y(t) is 32,
what must be the numeric value of the parameters a and ﬂ? m&4Juaa dug, an even ‘RM0 00 % ‘b‘m‘J the” 1? éCD 3:.
we, mast ham. 5?; = 03;) ana. we, bﬁ‘ Must be. reg.
7M: rorvmms the}:
out = anew/1'9) = bitI =wéécw/PG)
in mg ma 5? so mﬁ
Dné‘. —v Dnél =09
: 9 on: —I 5°C) 3 _
as t)‘e. on}? nonzero Dn are. Oh ,i ‘9:
Be cwusﬁ— 09 70) 11)::ch im‘ = Zoo ’ 3. (7 points) A student generates a function g(t) that approximates a periodic signal f(t) using the m—ﬁle in
Figure 5. g = 3; x = 5*pi; t = linspace(1e1, 1e1, 2e4);
for n = 1 : 50 g = g  (2 / n) * (—1)“n * cos(n*x*t);
end; Figure 5: MatLab m—ﬁle for generating a function g(t) that approximates a periodic signal f(t). (a) (2 points) What is the average value of g(t)? W0
Nut: thatl; 50 r__'
Lt): '3 + 2. (13)(7 I)" (as (511' 0 —£ D
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T’RQ “VET device. Ct) 75 3. (b) (2 points) What is the fundamental period of g(t)? From pﬁf“: cubed?) 600: {ITO “ma 60 To: A" = 355%
“la O (c) (3 points) As a function of time, is the signal g(t) odd, even7 or neither? To receive credit, you must
justify your answer. From Park Coo) abwa note, thaé amfl 5° g Problem 3: (25 points) 1. (10 points) By direct integration determine the Fourier transform of the aperiodic signal
f(t) : 2 [u(t + 4) — u(t — 2)]. Express your answer in terms of the sinc function. {L457
i
4:
1 7.
2.
Z —J,\JL
w — wt " “I: — _.E—[
F092] +0966 4:1: = 23:1} ii: ‘ otw e’ —1
—w   3"“ Sm 3W
W .—
—_. "led .sm3u/ —— Lie. 3 /’ W 2. (8 points) Using the Fourier transform pair 1 e_atu(t) H ,
a + ]w a>07 and appropriate Fourier transform properties, determine the Fourier transform of g(t) = 36—2t [u(t — 1) — u(t — 4)].  z.£
3H) = 39 “1 bLL‘t‘D  3e web—w) 3. (7 points) Figure 6 shows the Fourier transform of an aperiodic signal f(t) in terms of a positive real—valued
parameter a. Determine the value of a so that the energy of the signal f (t) is 20. F(w) Figure 6: Fourier transform of an aperiodic signal f 6) E4: = i7;— chmILlw ’W Pa») 75 an 9w“ Rno‘lﬁm % W) Because,
2. a) 7"
54 = 275 Ile‘ﬂw = #F‘lea: z‘ﬂw
o o co
200
OC’ _‘j__ { cc. Vac ‘53:
: w W := _... 'l' ﬂ — M
:ln"f lo + 77 06 TV 77 W
= 5:5:
E4 77. "'77 1 5
oar—HF 10 Problem 4: (25 points) Figure 7 shows a squarelaw modulator. The input to the modulation system is a signal f (t) whose spectra is band
limited to B as shown in Figure 8. The signal 5(t) applied to the square—law device is 5(t) ——— f(t) + cos(wct), where the carrier frequency we is signiﬁcantly larger than the bandwidth 5 of the signal The output of the
square—law device is passed through a bandpass ﬁlter with center frequency wc, bandwidth B, and amplitude A as
shown in Figure 7 to produce the output signal Bandpass Filter F 2
m “3:2? m
a)
cos(wk.t) —ﬂ )8
Figure 7: Square—law modulator. Figure 8: Fourier transform of 1. (6 points) Sketch the Fourier transform of the signal 3(t) in Figure 9.
3 (+3 s 4H.) + Cos (Wot) = 1.10;» + 1’” + 71' Figure 9: Sketch of .7"{s(t)}. 11 2. (13 points) Sketch the Fourier transform of the signal 32 (t) in Figure 10.
2
320;) : f F L13 + cos (weDJ = 43711:) + 2 +55) coslwoLj + 008‘ prt) {acn +— zFtaciouwot) + i + 45¢“sz \l QviSZHOB —_, 3: Flow) + F(w hug} +FCwWc.) + $34". 3' Lou) 2 TT'
4' E 4"ch) 4’ £22, g’cw— 2.
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F cw) PM 86! 20. 749 Figure 10: Sketch of .7'{s2 12 3. (6 points) Specify the ﬁlter amplitude A and the smallest value of the ﬁlter bandwidth B so that y(t) = f(t) cos(wct). F‘V‘J‘t DehrmmOa Yb”) +rcm the. “have. l‘e{u:Eam)1tf) > Fat! +Wo) + PlW' We)
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 Fall '07
 SCHIANO,JEFFREYLDAS,ARNAB

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