Exam#02_Solutions

# Exam#02_Solutions - ECSE-2410 SIGNALS AND SYSTEMS SPRING...

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Unformatted text preview: ECSE-2410 SIGNALS AND SYSTEMS SPRING 2009 Rensselaer Polytechnic Institute EXAM #2 (110 minutes) March 04, 2009 NAME: Sowﬂowg SECTION: 8:30am 10:00 am Do all work on these sheets. One page (both sides) of handwritten crib notes allowed. Calculators allowed. Label and Scale axes on all sketches and indicate all key values Show all work for full credit \ Total Grade for Exam #2: Problem Points Grade 1 9 I } 2 10 _l 3 l— 10 L 4 10 5 10 6 0 7 l 9 8 0 9-12 8 13-15 6 16-18 6 19-21 L_ 6 22-24 6 25-27 6 28-29 4 TOTAL 100 Exam#02 w Spring’09 - p.1 m NOTE. You must use properties and tables. Solving problems by directly integrating either no woo X (60): J-x(t)e‘j‘”’dt or x(t)= I X (w)e"’”da) is not acceptable. Ze—ij 10). Given the complex function H = J (0 1a) , find the value of T at a) = 2 when the angle Exam#02 — Spring’09 -— p.2 X(a)) 2(7). Find the inverse transform, x(t), of signal 275 - w W:- 27? w) : Afggkcéﬁ A,» w 2t 3:sz ELCQ3~Z)A-» 94:9: 6-96 49hcizt9 l: x 32151 3 my: Afmcézeje Exam#02 - Spring’09 —- p3 3(7). Find the equation of the Fourier transform, X , of the periodic waveform x(t). T143 Exam#02 —~ Spring’09 ~— 13.4 4(8). For the system shown, find the input signal, x(t). x(t) h(t)= [mum 331w) Kamagmw :7 law}: HIM) y(t) = 4e_2("1)u(t — 1)-— 4e'3("1)u (t — 1) {it Him): 2. \ Exam#02 — Spring’09 — p.5 x(t) 5(8). Find the Fourier transform, X , of e" . -1 1 t A, .1; M; é”? Lava—MAD 2:- 6 u£i+9-€ Law-é) “(ii—H) M {ﬁn-i +9) If 6 LLEEH? “- 8 {lbw} .. M) Exam#02 — Spring’09 — p.6 6( ). Problem omitted. Exam#02 —— Spring’09 — p.7 7(9). Find K and the values of b , so that the output of the filter is y(t) = 005(21‘). x(t) = cos(2t)cos(4z‘) wt): (253(sz 60.96%: i We ~th 324: 48* __~ t { nzrt<e +8, tg +3, » é‘CoSéH—Lééosié‘b); E @a msg-Ej Maw 3%- Wild: :mw){w[£zw~a+gml Exam#02 ~ Spring’OQ — p.8 8( ). Problem Omitted Exam#02 — Spring’09 ~— p.9 Multiple Choice (2 points each) 9(2). For the periodic signal what is the value of the exponential Fourier series coefficient, a2 ? i 1 7 1 (a) 1 (b) g (e) Z (e) ~37; 10(2). The Fourier transform of x(t) = e"u(t — 1) is (a) X(a))= 1 1+ja) 1+ja) 1+ja) e~<1+jw> [mo—1) X (w)= . (e) X(co)= . 1+ Ja) 1+ Ja) 11(2). The Fourier transform of the periodic signal, x(t) = Z 5(t ~ 4k) is knew (a) X(a)) = 2am“) (b) X(a)) = l m e’jk'i‘” (c) X(a)) 2 ~75 00 (“km k:~oo 4 k=—°c 2 k=—oc (d) X(a)):% Slam-we) «0):; i 6&4?) X (60) 1 12(2). The inverse Fourier transform of w is — 4 — 2 2 3 4 2 . 2 . . 2 . (a) x(t) = Esmc(1)cos(t) (b) x(t) = ---smc(t)sm(3t) (C) x(t) = Esmch‘ - 3) if @(r) = -2—Ev{sinc(t)ej3’} (e) x(t) = gilie{sinc(t)6 (t - 3)} 7r Exam#02 - Spring’09 — p. 10 13(2). The Fourier transforms of the signals shown are given below. Which X (60) is the Fourier transform for signal (2) above? Only four answers. ((0) = 3Slnc{l co]sinc(—3~ (0] (b) X(a)) = 13(1- sinc(w)) (c) X (60) = 1' 2812mm) 2 2 a) a) —1 (d) X (60) = j8sin(a))sinc(a)) 14(2). Find the equation for 32(1) when the input is the periodic signal shown. (a) W) : i ~§Sin0[k—2—37—r—jejk€t (I) 2—:- (C) YO) : ~2—-;[—/§~cos[~gtj (d) y(t) = «3003(375 t] (e) y(t) = —cos[-~t] 15(2). Sketch Y(a)), where w(t)= x(t)c(t) and x(t) <——> X Four choices. Exam#02 — Spring’09 — p.11 f 5 16(2). Find 2 = “36% (1+ ﬂejl} L (a) 2=£+j1 22' (I?) 2—3): (c) 2:} (r1) Z=0 17(2). 2(6) = 9"” + 8” factors into (0) 2(9): 448’?” sin(‘29) (13) 2(8): Ze'jw 605(29) (0) 2(6): 28-39 (305(29) (6?) 2(6) = 26]“? sinW) (6}) = 29% 003(6) “3(2). The value Of The exponential Fourier series; coefficient 00 is \$09532 y». N a0 = a0 =g , 1 A (‘7) 60 =": do = —— 1. 5 Exam#02 -— Spring’09 ~— p.12 "'1 19(2). The exponential Fourier Series coefﬁcient, a“ of fr J: , s1gna} x0?) = 1+ 6 * cos(‘a)0r), Where (00 IS the fundamental frequency, is ‘— j: (a) a1 = e" 3 (1)) (:1 =1 . l \ 1 (C) az=—J§ (’1 =2 20(2). C ire-16 the correci statement. X; (1’ I) H “k ’2 V W: I (a) bk = (Ike 1 (b) bk = gakeﬂm —xk7? < k=akeJ (chbk=ak+3 (e) bk =aA 21(2). The Fourier transfonn of x(I) = e‘muﬁ) is (3:10)) Give-11th:: plot of Pﬁmﬁ as shown, ﬁnd ‘0‘ . (I) (a)a=4 (Z7)a=3 =2 ((1)0:1 (e)a=% Exam#02 — Spring’09 — p.13 22(2). The F ourier transform OfxU) = J e‘f21(z')dz‘ is '1 f “x , X = £3 .— -r (a) (w) {fa-+27 ((0))(1 e , i ‘ \ (E7) Jam) = [ f,“ + “(QUJW Jm , : V 1 L w ( y _‘ = m- 5 ,t (w) ij + jtv} + m (‘0) ! ~93 (c!) X{a))=2m)‘(m)~ 1 (ye) X(ca}= 1 1+ ja) 1+jm 23(2). The Fourier transform 0fx(f) =6‘1MI is V > 26—349 (a) X m = W ( ) (i-é—jm)‘ (12) 2d ) 2‘ a) ‘ (law-1»- 4 ‘ 281m) c X (7 = () (L) 1+j(,(0——1) @ﬁ ) x" > a) = , 2 J (a =w-H4— ‘\ 1+0)“ (a) ( 1+](c0—1) 24(2). Find git) when x0) = 1 + c03(2.t) (a) W) = 005(27) (b) .1'-'(I‘)= 6030‘) @z) = 1 + 003(21) (d) 32(1) = 1 (63 )’(0= 0 3’0‘) Exam#02 — Spring’09 —— p.14 25(2). The Simulin diagram medals the equation d3}: 61)) , d}? l a ‘ —2 “ -— 2f =, I ‘ +-v+uz =0 0 di; dt M) M) (b) d? 2. i) 1 1 d2); adv . (C) Hf =HU){W“*—1) ~, +—+‘+ 2? =2; 1 ) .52 25 @512,“ 2d: 3 U U dzv ldv \. " -—-; 91 =/ I (E) (112 Zak-FM) H0 E U6} ¢ 1 1 ' " 3 5 2: m mm mm V1 «'2 Gatnt 26(2). The Simuiink diagram medals the equation Sl [E n. Wwe d) dt (‘9) HI) = Sill(i){i _ \45 d2 ‘1’ 0712 l > 2 - 4 w 9) dz“ d; . d 2 1) {)df‘ dz (a) —— Abra) = sin(z‘) + 4% 1) = Sing) 52’ 2 v (1y 4% 4 + ya) = 5mm + y({) = slum 27(2). Given the complex number 2 2 5+} *7, which Matlab command would give the magnitude of this complex number? (a) magfz) @ 3113(2) (c) magnimddz) (d) both a and Z) waulci work (9) none 0f the above Exam#02 - Spring’09 — p. 15 28(2). Given the complex number 4 :— 3 + j *5, which Matlab command waulé give the phase of this complex number? (a) phasqz} (b) angldz) (c) arg(z} ((1) both a and 1') would work (9,} none of the above 29(2). \Vliich 8f the following represents Ni _ 1 in Matlab? (a) z‘ (b) (c) sqrt(imag(100 —— 1' (d) sqrt(real(l—— 1 +1 000 * z‘ n QfIhe above End ! Exam#02 —— Spring’09 — p.16 ...
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## This note was uploaded on 05/03/2009 for the course ECSE 2410 taught by Professor Wozny during the Spring '07 term at Rensselaer Polytechnic Institute.

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Exam#02_Solutions - ECSE-2410 SIGNALS AND SYSTEMS SPRING...

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