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Unformatted text preview: BB 350 Last Name: EXAM III 18 November 2002 Solutions First Name: ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Score



 0  Total 10 Instruct ions 1.
2.
3. You have two hours to complete this exam. This is a closedbook exam. You are allowed one 8.5” by 11” note sheet. Calculators are not allowed. . 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 2.1) Continued.” No credit will be given to a solution that does not meet
this requirement. . Do not remove any pages from this exam. Loose papers will not be accepted and a grade of zero will be assigned. . 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) Let
f(t) = 6cosz(4t). o (3 points) Determine the fundamental frequency we of f(t). «a: €[é‘* “7556(2an = 3 + 3 (“C”) we = s mil/Sec. o (2 points) Determine the fundamental period T0 of f(t). To ‘' 27/000 =3 77/“! 52" o (5 points) Determine the trigonometric Fourier series coefﬁcients (0.0, an, 17,.) for °° to
Z. “n (03 (ﬂwo‘LB 4‘ "1. Ln Sin "wo%
5'" n=l ‘FC‘Q = “a + 2. (15 points) Consider the periodic signal f(t) shown in Figure 1. f(t) (+4) (4) (+4) (4) (+4) (+4) {4) (+4) (4) .0. 0.. 20 17 —10 —7 0 3 10 K To —>I
Figure 1: Periodic signal f(t). 13 20 23 e (2 points) Determine the fundamental period T0 of f(t).
To : I 0 Sec. 0 (2 points) Determine the fundamental frequency «1,, of f(t). We 3 277/79 73 775' rag/J2 o (6 points) Determine the exponential Fourier series coefﬁcients Dn for Ta 7 ‘anat +5 —d,ngl.'h
3).. = 'Ff‘ae— a“: = f; [qsmH sagsﬂe. 11:
~12 D: = 1017?”) = yum) =2/,o
[3—5 = 1063119“) =7}(I~‘D “’X/IO Problem 2: (25 points) 1. (10 points) All the nonzero exponential Fourier series coefﬁcients for a. periodic signal with fun
damental period wo : 10 rad/ sec are speciﬁed by the amplitude and phase spectra shown in Figure 2. lull ADS —90 —50 —20 20 50 90 Figure 2: Amplitude and phase spectra of a periodic signal f(t). o (2 points) What is the average value of ? 0‘:
AF “it?”L Alex4.39.. V0,)”, 7% mo = so = = 6 o (3 points) Is f(t) purely real, purely imaginary, or a complexvalued function ? Justify your
answer using the appropriate Fourier series property. (~ .
Because. NE?) 35 am Que/x {wooLion w) drug. LDn IS 4” cit/Q ‘Fvncézon % wJ {ll—EA is fvreg megvocle. o (5 points) Find an expression for f(t) in terms of cosine and sine functions (do not use complex
exponentials). Beau/Se, +953 75 {pageantcg: Q .FH.) _._ 2 bn e’d‘nwdb ,. D0 + Z, ZanlCodnwat a—XJDS) =3 2. (15 points) Once again consider the signal f (t) with exponential Fourier series speciﬁed by the spectra
in Figure 2. The signal f(t) is passed through a LTI system with the frequency response function “’ }
— —1r
H(Jw)= 3e J{50 25<Iw<75
0 otherwise to produce a zerostate response y(t). o (8 points) Specify all nonzero exponential Fourier series coefﬁcients D}; of the signal y(t). by? c ngan) D: . Gala the. 5H. hawmvnk pensan ﬁrm;qu
the 'Ftl'kev' (SW9 :1 So YWﬂ/Seo) '9 11 ' ~17
[32? = “95‘” 15$ = 3e 4 Ge 0 ?e.# o (2 points) Find an expression for y(t) in terms of cosine and sine functions (do not use complex exponentials).
 8‘04:
30:) = Dj’lgeaL + 1356 o (5 points) What is the power of the signal y(t) ? Pa = 3103‘!" 6— lbilulmz
nz—w Problem 3: (25 points) 1. (10 points) Consider the aperiodic signal f(t) = 26(t + 27r) — 26(t — Zr). 0 (2 points) Show Whether f(t) is an even or an odd function of time. If it is neither, state why.
{364:} = 2 s <— i: + 211‘)  2901 277)
= 2 80% ~217)  z 8' (i4 217) (becque. $4 ) even) = _.F.(£) =3, {ft} B an moi, 'CanEOA % '1':er o (2 points) Based on your result above, and the fact that f(t) is a realvalued function of time,
what statements can be made regarding the Fourier transform of f(t) '? 3am +19 3.: map/“ML ML an ,m ﬁnc‘ém‘n a} "fume?
Ffvu) Is Pure? Manama»; and. an oﬂﬁ. ‘Rnciluﬂ gw. o (6 points) By direct integration, ﬁnd the Fourier transform F(w) of f(t), and sketch F(w) in
Figure 3. a w 09 .. “fl:
FWD =jr(t3? if: ..—. Zj; (SCl: + 21739;). anL
.— 60 00 w * w't
— j 365217)e# a“ .6w217 .. w
1 2e. J66t+zﬂlﬁ ' 2e? 2'” w&(+:—zw)iL
sir—.1 ~°°
—_:\ “f; cl = Z i ed’wzn‘  e~dwzu3 : 2, z 29' Sm Z1709} F'CQ =' 5m Z’U’w F(rn) Figure 3: Sketch of 2. (15 points) The zerostate response of a LTI system to the input is f(t) = 3e_2t 'u.(t) y(t) = [9 am  s e4t] u(t). o (10 points) Determine the frequency response function, H ( gm), of the system, and express your
answer in standard form bm(Jw)’" + Inn—1(Jw)"“1 +    + My) + be H(Jw) —_— (Jwy;+an_1(]w)n—l+H.+a1(]w)+ao
From Tabb °t. e‘ “t wtt) (—e 9—5:». a uni So
'7 G
.. 3  .
F'Cw)  dw+z anﬂv YCw) — dW+L ﬂw‘rv’
w 'i‘ " ‘ y
“(09) : Ylw) _. 3  Ll:— ,, ‘3 9' w ‘1» 2%“ ﬁ‘”+7 o (5 points) Find the ODE representation of the system, and express your result in standard form 7" ?” m—l
:%+Gn 1:“?+...+aay=bm%?$+bm_1:tm_f+...+baf_
Y6») 2. 2w +8’
Fa") dov 1‘ ‘1
{m + 9 WW) w Pu.» + 8F1w3 Problem 4: (25 points) Narrowband phase modulation (NBPM) is a form of modulation that is similar in some ways to amplitude
modulation Consider the bandlimited signal m(t) with Fourier transform M (w) shown in Figure 4,
note that M (w) is zero for lwl > W. It can be shown that when m(t) is the input to a narrowband phase modulator, the modulated waveform has the form
W) = cos(wct) — ﬂ m(t) sin(wct), where the carrier frequency wc is much greater than 2 W. M(oo) W 0 +W Figure 4: Fourier transform of the input signal 1. (10 points) Sketch and dimension the real and imaginary parts of P(w), the Fourier transform of the
modulator output, in terms of the constants W, ﬂ, and we, using the plots provided in Figure 5. P0”) = {I i (as Wat} " Q Slim“) .Smwo'ls'k = 1r [$(w+w¢)+ swwa’j j 13,: fmwwa — mlwwa]
W W
[laZ “‘53 In it’lw)3 Re{ P((0) } IIn{ P60) } C m m) Figure 5: Sketch of P(w). A simple demodulator for p(t) is shown in Figure 6. H(co) pm a i W) w0w°) D sin(mct)
Figure 6: A simple demodulator fo a NBPM signal. 2. (10 points) Sketch and dimension Y(w), the Fourier transform of y(t), in terms of the constants W,
,6, Lac, and D, in Figure 7 on page 10. WW) : Ht«;) (1’ i [5 91453 :mwot'ﬁ ll HM») ' £2 E PCwrWJ ‘ P(W'Wo3} 2.. (M) $2? Pugrule.) Re{ Y(m) } IIII{ Y((D) } 'w  6‘ D
Figure 7: Sketch of Y(w). 3. (5 points) For what value of D will y(t) be exactly equal to m(t) ? 10 ...
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This note was uploaded on 03/17/2008 for the course EE 350 taught by Professor Schiano,jeffreyldas,arnab during the Fall '07 term at Penn State.
 Fall '07
 SCHIANO,JEFFREYLDAS,ARNAB

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