{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ex2sol - Name gaigghCflfs Student ID ECE 360 EXAM 2 No...

Info icon This preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
Image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Name: gaigghCflfs Student ID: ECE 360 EXAM 2 November 20, 2002 No textbooks, notes or HW solutions. One page of hand-written notes. Calculators are allowed. Exam is 50 minutes. To maximize your score on this exam, read the questions carefully and write legibly. For those problems that allow partial credit, show your work clearly. 0 Good luck. Part A- Multiple choice and short answer questions. It is not necessary to Show work, no partial credit will be given. 1. [5] A continuous-time LTl system is described by the corresponding input/output differential equation d2y(t)_ £3192 +3y(t)= 4am) — x t dt2 dt dt ( ) Which one of the following statements is true for this system? 2 a) H (s) = 12—61—132 , and the system is stable. S _. 2 — 4 b) H (s) = ill—S12 , and the system is unstable. S __ 4 —l c) H (s) = Tim, and the system is stable. S — 4s + 3 , 4 -1 . H (s) z 71...“! and the system IS unstable 3 -— 4s + 3 2. [5] What is the Fourier transform of the periodic signal x(t) = Z rect(t — 2k)? k=‘—oo a) X(f) = sin6(f) b) M) = sinctf)e'j7‘”f ® X0”) = iO.55inc(O.5k)§(f - 05k) =-oo d) X( f) = i 25inc(2k)5( f — 2k) k=~oo 3. [5] The Fourier transform of x(t) : HEM is 2+87r2f2 XW=W Based on this, what is the inverse Fourier transform of X (f) = I f {e‘lf ' ? Hint: If you use one of the properties discussed in class the answer is immediate and requires no computation. X0) : M..— 1—87z2t2 +167z4t“ b) x(t) =[t1e'l‘l c) x(t) = e""u(t) 1 1+12 d) ICU) = 4. [12] Determine whether the following statements are true or false. a) If X(f) is the Fourier transform of x(t), then [X(f)| is always an even function. ”T b) The spectrum of a periodic signal is always discrete. T c) For a continuous time signal, x(t) = lOsz‘ncOOt) the Nyquist rate of sampling is 5H2. M l‘ d) For an even symmetric periodic signal, a, s are equal to zero in the trigonometric Fourier series expansion, x(t) = a0 + Zak cos(27rlgf0t) + Zbk sin(27r/g’ot) . k=l k=1 F Part B- Show all your work to get partial credit. Correct answers without complete work will not receive full credit. 1. [36] For the following system All) C1 59L 9-» Hm jLQLTQ» Gm + Y“) r30) p(t) = $50 — 0.5k) =—oo X(f) =tri(f) H(f):{1(; «Zstwl, Isfsz , otherwise G(f)={1 ~1,<.f31 0, otherwise X“) 1 //// ’ \ ‘ ‘ 1 1 » Sketch A6), B6), C(t) and Y(f). Show amplitudes and frequencies. Extra Sheet for Question 1: AU“) (‘v- “'3 0’ (”m ”‘7 1K) rr M/ \l M K (3 MB )4 raw" ,4.“ i to TV \w/ Extra Sheet for Question 1: 2. [37] For the following periodic signal, x(t) = 2 — 4cos(8m‘) + lOsin(127zt — g) a) [4] Find the fundamental frequency of the signal. b) [15] Find the exponential Fourier series coefficients. Hint: You can first find the trigonometric Fourier series coefficients and then convert them to the exponential Fourier series. sin(a — b) 2 sin(a) cos(b) ~ sin(b) cos(a) 1.7r\/§ 7r — =~—,sn—« 2—— cos(3) 2 1(3) 2 c) [8] Find the power of the signal from the Fourier series coefficients. d) [10] When x(t) is used as an input to the following ideal highpass filter, H (f) , find the steady—state output, y(t) . fl 1,} f [2 5 I H(f) I— =,{O otherwise —]f7r 0], otherwise Q3 iitqH% fZZEH% EQtGCDiufg>22wgi i) > 244003 isn't) «t» iQYSih Kt2fit)c©§(§é> » Co§(12”fi[email protected];)] b X“? : r~ 2~l§castgfit) +§S§mii2ffi>~5gcggugfi~§:3 star} @012 KEG}: 2 Q 0 XCijzo / Xt~13r© x: ' f3} V «4- {*2 2’2. bwg X123” 2/ X 3 { ere-+93) (32:44 K133 Kean-’53) X333 ”l" J bit 2 a 2"“‘38 W other Qwiflfliefie m Extra Sheet for Question 2: oi: i?— (1' ~ \ a); éWQKj C) ‘ hm ”ZXQ Z £+§> :qfi’i, )4” (q q 211+g© (A) 7”): €91H£é>}s~muzfi%-g + (mm) ‘ z 4: 43%“) ~* 40 3m\’i?~fif~'fi/3“W/g> :‘iQSH’EH YT //‘ w Extra Sheet for Question 2: Table 9.2 Operational Properties of the Fourier Transform Property 1(1) X“) X(w) i Table 9.1 Some Useful Fourier Transform Pairs Similarity X(t) Ii-fl 2MP”) , Entry 1(1) X(f) X02) ‘ . 1 f l Lu ' Time Scaling fiat) ((1)43) MAKE) 1 6(1) 1 ‘ 1 Folding :c(—t) X(-f) Xi“wl 2 rect(t) sinc(f) 51114;) , l i . . fl —j21r]oX e‘J“°X(w) Time Shift 1” a) e (f) 3 tri(t) sinczU) sinc7<_“_}.) Frequency Shift ej7"°‘x(t) X(f — a) X(w - 27m) . 7r L 4 ' g i Convolution :c(t) * hm X(f)H(f> X<w)H(w) smc< l mm ‘ mg”) l ' ' Multiplication a:(t)h(t) X (f ) * H (f) EXM * H(w) _ 5 005(27r0‘t) 0.5m; + a) + 6(f — 0)) rim} + 2m) + 5(w - 27(0)] Modulation x(t)005(2mt) 0-5lX(f + a) + "W ’ all 05m” + 2”) + X0" “ 27ml] 6 sin(2nat) 105le + a) — 6(f — all mm + 2w) - 6(w ~ 2W} Derivative 5(1) j27er(f) ij(w) : 7 €‘°‘u(tl _1 ‘ 1 , + ‘2 ~ . Times-t ~j27rt1(t) X'(f) 27rX (w) 0 J 7rf 0+1“ ‘ 1 1 8 te'°‘u(t) ' “—1 __1_ Integration / 1(t) dt jZ—fifXU) + 0.5X(O)6(f) EXM) + 1rX(0)6(m) (o + 3-27?le (0 + jw)’ -—-co L 20 20' ‘ON w___ Conjugation I'(t) X'(‘f) X‘(—w) 9 e a? + 4w9f2 a? + w? Correlation $(i) ** y(t) Xifly‘ifl X(w)Y'(w) 10 e—M2 e’”? e‘wz/‘i” Autocorrelation I(il**1'itl Xiflx‘ifl = iXifll2 X(w)X'(w) : lXil-h’ll2 11 (t) l 2 s n ,_ _ g Jfif 1w Fourier Transform Theorems 1 1 12 um 0.56(f)+ .2 «6(w) + —— Central do) = /°° X”) d; = 3.. foo X(m)a'w X(0) : foo 1(tldt ] M W ordinates —oo 2'” ~90 —oo 13 e~ut Cosizflflilum 0 + 327d :1 +Jw (0+1?er + (21%)2 (n+2w)2+ (2M3)2 Parseval’s foe 2 on 2 _ L DC , 2 E = z (Zldt = / |X(f)i d; — Wall «is» a | 2w 2w theorem we —°“ 2” '°° 14 6 sm(2rflt)u(t) (a + j21rfl2 + (2mg)2 (0' + jw)2 + (27:3)? , w °° 1 °‘ , . “x 1 °° - °‘ l ”wheel 5 / x(t)y'(t)dt = / X(f)Y'(f)df = g / A <in (wldw 15 Z 6a — n?) ,1: Z 6(/ — 5;) g; X a( - 31") theorem ~92 '00 " ‘°° n=—tx kz—oc kz—tx T 16 m2) = Z Xiklefiwv' Z Xik]<i(f — we) 2 2nXiklé(w — 1M) k=—ocr k2-“ A-z—oc Table 11.2 Operational Properties of the Laplace Transform Note: 1(1) is to be regarded as the causal signal a:(t)u(t). Entry Property 1‘01) X(s) 4 l Superposition an (t) + 512m 0X1“) + fiXfis) Table 11.1 A Short Table of Laplace Transforms 2 Times-exp e“°'1'(t.) X(s + a) J Entry a:(t) X($) Entry 1(1) X(s) 3 Times~cos cos(at)x(t) 0.5[X(s +jci) + X(s — joj] ‘ 57 “ fig . 1 6(t) 1 10 tcos(6t)u(t) W 4 Times-sin sin(ort)z(f.) jO.5[X(5 + jar) — X(s — ja)] fl 1 25 ‘l 2 u t) — ll tsin(fil)u(t) *— 5 Time Scaling 1(0‘1), a > O lx(§) ( S ,1 (52 + 52)2 a a 3 l ~i t) I 3- 12 mam) (c) “L + 2132 6 Time Shift 1‘(t-— a)-u(t — a), o > 0 e‘”‘X(s) A )~ M 5'3 u 5(52 + 452) 2 i a 23? 7 Times—t 11(1) » (11:55) 4 (211(1) 5—3 13 Sln2(fit)u(i) m n nd'”X(s) . ,1 "I -02 8+0 8 t 1(t) (4) 7;"— a t 11(1) 5"“ 14 e cos(fit)u(t) i (s + a)? + 52 9 Derivative :r'(t) 5X(s) - 1(Du) _a. 9L_ - ~m . 5 6 e u(t) s + O 13 e sm(fit)u(t) -~——<s + a)? + 52 10 Ml) was) - sx(0—) — z'(0-) _ _ L 1 -m (s + (1)2 — a? L r { ta 0 21(1) («TL—(i7 16 t5 cos(fit)u(t) W ll xifllu) s"X(s) —- s"'lar(0~«) ‘ - xn'l(0-) 71' 213(3 + 0‘) n ~01 I w." ‘ - ~01 - m t X“) 8 i e u(t) i (.~ + a)?!“ 11 te sm([3t)u(t) [(5 + (1)2 + 52]? 12 integral / 1(1) (It ‘ t , 5 v. o_ S 9 cos(fit)'u(zfi x2 A; fl: 15 sin(fir)u(t) 2 52 l3 Convolution 1(t) * [1(1) X(5)H(s) " S + S "tcl d ' d' , " l4 “1 16 perio RC 1p(i)u(t) 1A1(f27., T: time period of 1(f.) 11(t) : first period — e Laplace Transform Theorems 1:3 Initial value :r(0+) : ‘IBEJSAXSH (if X(s) is strictly proper) Hi Final value a:(t)],~.3C = ljiléstwfl (if poles oiX(s) lie in LHP) ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern