A#12_Solutions-1 - Assignment#12 — Solutions ECSE~24 I(j Sigaals& Systems Spring 2007 Due Tue 1 ja 1(20 A linear time-invariant system has a

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Unformatted text preview: Assignment #12 — Solutions ECSE~24 I (j) Sigaals & Systems - Spring 2007 Due Tue 03/16/07 1+ja) 1(20); A linear, time-invariant system has a frequency response given by H = ————-———_~—- . . 1w Ja) 1+~—— [ 10 j The system respanse to the general input x(t) = M 008(6001‘ + 6) is 320} Mifflwg )1 cos(a)ot + 6 + AH(a)O )). 1+ j0.2 _ 1.0198eJ'0-‘974 _5'0986_fl..3934 j0.2 {a} Fain“ input x6 :10005(O.2t), H (0.2): 10.2[1 + ~16.) 0.2g”?1.0002e1'0-02 {amgieml s3€334 $79.8 °) [331611 jag: {z} 50.98 cos(0.21——79.8° WWMMMWWV 7%mewa ~ ‘ j1,3734 ’ {in} 1%? input 32;}(r)m10c:os(5t), H(5) = H 155 = :0 e = 0912261066“) 15(14- E) 56131.118ew‘4636 imngass-2:41:36: 10 £37.88 °) i , iii-‘2}? izzgzui m lOsin(5t)= 10008[5t —%], then since the frequency is the same, H (5) does not ms 11(5) = 0.9122e—10‘6610 ' ' err-«£3,636,310 $37.88") _\ {g} as; 9.122cos(51~37.88° —90°) = 9.122 cos(51-137«88{J - r 7 11.4711 m: <r>=wcosaor>, H(lo)=~—1i—Ji9;5-m339i3§99 ” j10[1 + 117;] we; J27? = 0.7106e'1‘0-8851 43,3255} m—50.7159°) ' 7.106cos(IOt-50.72°) ,_ (NMV Assign #12—Solutions. Spflng Big}? 9.1 Assignment #12 - Solutions ease—24 ; a Signals & Systems - Spring 2007 Due Tue 03/1 6/07 M 1. Continued. Matlab COdC. nummfi i}; den‘fiOJ 1 0]; Boddnumfien, {0. 1 , 10}) grici W2:11[O.2 :3 19]; {mag phasd‘bOddnumadenaw) Bode Diagram {:3 \_\ V \IL g a? ‘ System sys ‘ y Frequency (rad/sec): 0203 System: sys g} ‘ magmmdews): 14 ' Frequency (rad/sec): 4.92 System sys » é ‘ ‘ \ Magnitude (dB): ~0.767 rad/sec>:10 . , . . wraguuuug-l‘dBk ‘ _ , \j' // ,,,,,,,,,,,,,,,,,,,,,,,,,, ‘_'\ System: sys Frequency (rad/sec): 4.92 ; I “a. 3 ; / Phase (deg): “377 System sys % ' i / Frequency (rad/sec):10 PhaSe (deg): -50.7 ’It y ’ System: sys ‘ ‘ _ Frequency (rad/sec): 0.203 ' ' L ’ ’ 1 Phase (deg): -79.7 3 10 Frequency (rad/sec) iiéeigmi ih::r§1 the following portion of the program: " fi :0]; , pig/3%]mbode(num,den,w) Assign #12~Solutions. Spring; 2:}87 p.2 Assignment #12— Solutions —- p. 123C313~241i3 Signals & Systems - Spring 2007 fifl" ’ ’1‘“ w, / Wx M =. com-W W 31 {7 /. a»; )6], Yfl’) [2: ,.. [1) “w/ MJ ‘ ’7 # “W; / w. s z /#(»)/= W; , z , ,f/fa .. zféw"/'z’:/ ’ A/MZ’) 3 a/ dun/y HM; 2 C 0 ,1. wM” "cf/I) Assignment #12— Solutions - p. f1 ECSEéEde Signals & Systems — Spring 2007 zrgé 3 .{‘0[’ . ‘4 KM : c: (2M Assignment #12— Solutions — p. { ECSE~2410 Signais & Systems - Spn’ng 2007 %%gg§1;&é. Cygy; ,Wécfi7 w M X/M: { / / iajPC fff. 47, ‘“ fl‘ /w/‘/ flue/fl ‘6“? g 0 / /zy/?/ W ’ .26 fill/M4136, l/flI/J W : g/fuM " <3, 6 a at?!) ‘w I / « eigjL—w L933, w/ (Lu @W/yw /y“ \W r "5% w lv‘jw 7. 1w Assignment #12— Solutions —— p. é ECSE~24 18 Signals & Systems — Spring 2007 /%V The signal x(z‘) is sampled by multiplying it by an impulse train. The sampled signal is £190) = COMO t Whereggfi) is the sampled version of x(t) and C(t) is an impulse train described by it“; {:th z 5(1 ~ kT) . If x(t) = 1 +100c0321+0.01 sin4t, what should be the requirement on ism-we the peried, T, to permit full recovery of x(t) from x: (t) ? ~ fl? ,5 cc. Requirement on T: 0 L r L g Assignment #12— Solutions — p. 7 iiiCSEw2410 Signals; 85 Systems - Spring 2007 5 (20). Two signals are multiplied as follows Aft) +®——> y(t) 4‘ C(13) 27: where {:{g} is the: impulse train, C(t) = 250 — kT) , with T = 3000 k=~oa sec. {2;} {1:3} the spectrum of x(t) be X (‘0) A -lOOO 1000 Skeieh ihe spectrum of y(t) , Y (60). Can x(t) be recovered from y(t) using lowpass fiiiez’éng‘? Why or why not? U to = flL—e 1 3,060 M/peg T- fllrgago W») == 55; amazed), car 3;; 5(a)- 21$. 0 ’6 ‘fi/E :ffipp: N 32,2») krw fxfié') canch Mane/LuQ flan lat/i“) M, arm/ma Fact/La (mm. 5" L“M£MWR 0; mm I’d/MK" /"\ 2’ ‘9’ Puma; [Led/mu. m. MAT/1d Ava/mil Aye/um a?» My" Map J fie.) fit/«72911;? MO) Wflfiwuj k3 WW (444v Mgu‘uf-m/e. Assignment #12— Solutions -— p. X 3353324 16 Signals & Systems — Spring 2007 flMMa‘ufé W (0) {1(3) Repeat part (a) if the spectrum of x(t) is X (‘0 ) A, -2000 2000 ‘0 and {:(z) remains the same. Can x(t) be recovered from y(t) using lowpass fiiteying? Why or why not? /‘/\7/\\—7 , \\ //\ low Saw 3600 @5030 66007000 (530m fx(t)M_ be. Wet/EM {ng 6mm? WW mm J [2Lqu WW) £26,000 gem-WM. 0L4; as 4m. iv £414,700? Kggmgt W,» Maw «'42. A/L/gw'sf' m (haw 1;: flow Wm fiv we; mmg ) J M (slaw/7} away. Assignment #12— Solutions — p. 7 ECSEQMO Signals & Systems - Spring 2007 of 3:333 in sampled signal xp (t) is generated by sampling the original signal xéz’) @3622?) every 0.1 seconds as shown in the figure, where C(t) : 2 5(1 — kT), k=—°o with (3.1. X(t) —>®-——> xp(t) t C(t) {3){43 ESketch x p (t) . {133(3} What is the sampling frequency in radians/sec. ? - {eke} What is the highest frequency (rad/sec) in the on'ginal signal x(t)? (eke) Is it possible to recover the original signal x(t) from the sampled signal xp(t) ? Answer Yes or No and state why? a?) Kym; 605(5176, “U/ icon, 2/ -,- T:- oil ...
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This homework help was uploaded on 04/10/2008 for the course ECSE 2410 taught by Professor Wozny during the Spring '07 term at Rensselaer Polytechnic Institute.

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A#12_Solutions-1 - Assignment#12 — Solutions ECSE~24 I(j Sigaals& Systems Spring 2007 Due Tue 1 ja 1(20 A linear time-invariant system has a

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