09F_210Exam3soln - ECE 210/211 Analog Signal Processing...

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Unformatted text preview: ECE 210/211 Analog Signal Processing Fall 2009 University of Illinois - Allen, Basar, Trick Exam 3 Thursday, November 19, 2009 — 7:00-8:15 PM . Section: circle one Class: . ECE 210 ECE 211 circle one Please clearly PRINT your name IN CAPITAL LETTERS and circle your section in the boxes above. This is a closed book and closed notes exam. Calculators are not allowed. Please show all your work. Backs of pages may be used for scratch work if necessary. All answers should include units wherever appropriate. Note that the problems are not weighted equally and so, budget your time accordingly. Problem 1 (25 points) Problem 2 (25 points) Problem 3 (25 points) Problem 4 (25 points) Total (100 points) Problem 1 a) The unit step response of a LTI system is go) = e"u(t-2) i) Find G(co). ‘ 00 - (Milo); poi ,iwt -(Ifiwyc + E . = e C 0N= 6 ’ 0 * ~ Z [+du) oK WRITE -1 -61”— firélw 6". ‘6 -2'_LT)€_3L G(co)= H1» ii) Find the iiri-tpillse response of the system. ' -+ . «7% h(t): 0% z 6 39(4) we “do &? 3 6’2 Ski—>5 —— gag/((794) -2 —‘5 h(t)= 8 $94 "5 “(é-é we was “— 7C 5 iii) Find the half powerMof this system. ..2 ,_- -1 ._\ s, -2 -‘wz HQ»): 6 dawz" 6 CJUZ: ' ifiw I+Jw 7. 7* —‘f . \ ‘ ‘ jHLw>1=~——-~fiw1€ ThmaMW/Mafl (awwt 0L “WM'%W a WK/s- Jt @fii‘tifiwa= 4M 5* b) For the system description below, circle whether the system is inear or not, time invariant or not. ‘ {— i—f (t)=3f(t) notL TI @ c) Impulse response of a system is given below, circle whether the system is stable or not, causal or not. h(t) = 8(t + 1) + 28(t - 1) not stable casual (not casual ) gQL) : Hut) \L 2 $6M) Problem 2 3) Given f(t) = 3cos5t + 4cos8t , find and plot F(co). Clearly label axes. 13(0)) 477 377' = 37r[S@—§)+ SQoMfl + 47TLSQ049) Jr S’éu +9“ b) f(t) is sampled at a sampling rate of (98:12 rad/s. Plot the frequency spectrum FT ((0) of the sampled signal for -cos 5 (0 5 cos . Clearly label all values. . 0-0 f _‘ w FT(CO) : eolé‘) n 5) u=~ 977/7— 1 z 3F/T T3 -ws —9 ~7_%.§_ 4 4 g 9257 9 cos (’3 -IL _,¢ (9 [7" c) An analog signal is reconstructed from the above samples as shown below. What is the output y(t) . n =2; 00 f(nT)5(t-nT) - ya) (03) Problem 3 . . ’0’l’ Cons1der the followmg system. W f0) r(t) (t) 110) (t) 6 0 - H<<D> Colet 10 CosZt Han) F (to) : mead-:1) - l . Where f(t) = 9% and fl'j -10 10 co -1 o 1 u) a) Draw the spectrums R(co), Q(co), N(co) and Y(co). Label axes carefully. __ c‘l’.‘ w :;;c[rec£(a;f/Q)+r< ( 1 R(co) '10 : Meg mt! rel/5 ~I7-ilf‘a45/ 41ml; gimp/5 ’Zr/r/s 1r/ -S".tlr/5 Stlf/J a) ' b) What is the signal y(t)? I w “£9 : [/0+ Ham 106059? flMW 2:4 j fiat/091.125- W 41 , 40/ : /0 wxzé + $@)m/o£m1£ _ y(t>= MW +L 58-)m8t— : #9me {Q} petalflmllfl Problem 4 3) Find f(t) = u(t) * u(t) * u(t) . (u(t) is unit step function) °° .i— 319:} “mes-m? = w —_ we) "co 1 f 32%): Lowe) ace—2M7; :5: [ f<t>= p. o b) The impulse response of a LTI system is sketched below. h“) w — t 0 10 low_fl [0 ’c‘ . . . . . £10" 1) If the input f(t)1s izgen as. *4 N) 1 b *i 0 75-: t T t g@j:o WW7.WM:0 0 1 aozse>va®=lx°'”'=0" find the output values y<0)= O y<1>= 0.? y<2)= 0,! y<3)= 04 11 6t) 0"l::::::lk find the output values y<0)= o y<1)= 0.) y<2)= 0 y<3)= 0 5(0) 243M 80 Ma} W9; (xxx/x! a 0.! $8): gblmfi Wz/x&.I‘xl~l><oJxl c 0 ...
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09F_210Exam3soln - ECE 210/211 Analog Signal Processing...

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