fall07Midterm_II

fall07Midterm_II - Autumn 2007 ECE 161A Midterm || Tuesday...

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Unformatted text preview: Autumn 2007, ECE 161A, Midterm || Tuesday, November 20, 2005 ECE 161A Midterm II Tuesday, 20 November, 2007 Name: (Last) (First) _____ (MI)— Problem 1: Problem 2: Problem 3: Problem 4: Problem 5: TOTAL: [100 ____—___—____—_—u.—._ Rules: 1. By writing your name above, you certify that the solution is your own work. Cheating Chm-PLUM \J in any form (copying another student's work, using un-authorized materials, etc.) results in an immediate score of zero for the exam, and further disciplinary actions by the School of Engineering. . The exam is open book and notes. . The exam time is exactly 75 minutes. . No calculators or any other electronic devices. . Write clearly. That is necessary to ensure that you receive partial credit. . Cross out any work you do NOT want to be graded. Wrong answers cause you to loose points. . Partial credit is given ONLY to correct solution techniques, not to correct answers from wrong procedures. . The exam is 12 pages and 5 Problems. Autumn 2007, ECE 161A, Midterm ll Tuesday. November 20, 2005 Problem 1 When an input x[n] = (5[n] + 0.85[n — 1]— 0.26m — 2] is appiied to a causal linear time-invariant system, the output is y[n] = 65M] a) Find the transfer function H(z) of this system. Answer: b) Use the inverse z-transform to find the impulse response h[n] of this system. Answer: c) Find the difference equation corresponding to this system. Answer: _ —'2_ CL) Y(2):é ) X(Z): [+0.32- holE 4,, . 6,1 HQ): 6 —2_: ‘2' i+-52"_.2_2 22+.82-.a¢ is H Z” + 5% i (2) (Z “2) (2+0 Autumn 2007, ECE 161A, Midterm ll Tuesday, November 20, 2005 0 M2): iii); 5% way—Mama) W Y(2:)(M+o,<gi'--2f) A m) 133%] +‘ 9 354-13 .., 91501-23ch m]. Autumn 2007, ECE 161A, Midterm Ii Tuesday, November 20, 2005 Problem 2 Consider a causal filter with real and positive M impulse response h[n] and transfer function H(z). Let y[n] be the output of this system when the input signal is x[n]= 1:322]. Show that y[n] is increasing with n and limy[n]= H(1). Autumn 2007. ECE 161A, Midterm || Tuesday, November 20, 2005 Problem 3 Consider x[n]= aLmJfln], for a such that |a|<1, and where H is the floor functionl. Find the DTFI' of this signal. (Hint: first draw the signal.) Answer: lafl <i (217W Xz(2):X|(Z)£—l =___Z—__ midliadfia 12317194 23¢ PM? I 2 E X43(%)" 22,0L + Zia . 23a ‘33“ 1.. ___e__fl_fl.__ 1. if. X60.) :: X(%) 2:331 “' (aid-fl ._ 0L ezjfiru 11f you do not recall the definition of this function, see the last page of the exam. -5- Autumn 2007, ECE 161A, Midterm [1 Tuesday. November 20, 2005 Autumn 2007, ECE 161A, Midterm || Tuesday, November 20, 2005 Problem 4 a) Draw the impulse response of the system with the following impulse response: b) What is the DTFI'? Answer: Tm om MDHMAE‘ \OMJL ““1300 SDI"RK] | FF“ lad; Fig/i kzm we ‘W/‘i H(JL):%£: S<~Q_%E(_L) by : nLgrfl-“Q K: Autumn 2007, ECE 161A, Midterm || Tuesday, November 20, 2005 Autumn 2007, ECE 161A, Midterm H Tuesday, November 20, 2005 Problem 5 Consider a filter whose frequency response is of the form 11(9) = 1- 4005(29) + 2a sin(SQ). a) What is the length of impulse response of this filter (notice that this is a non- causal FIR)? ‘5: HM): Answer: b) Find a such that the energy of impulse response is 11. (Hint: there are more than one such a). Answer: c) Find a such that 1) the energy of impulse response is 11, and 2) the filter has real coefficients (impulse response). (Hint: there are more than one such a). Answer: d) If the input is x[n]= cos(%n), what is the output of the filter in part ((2)? Answer: Autumn 2007, ECE 161A, Midterm ll Tuesday, November 20, 2005 0L) HUL): I— 4C3) 1J1 +élq 350511 *5le “'9'“ 9331 _ 33‘“ HQ.) :29... a. a e +l w a e a +8.37%“; J 1‘ T r T 1?. M135] h [—2] h [.9] M1] 1“ L9] I l ”Em. \mgttl 049% WW ~44 i ‘ I] r Cum Arqu' 5‘“ T. Hag.» “ng ) A) ~Fbr gym? Pic-L (1:3 HM): lflq_c'3&_§L+2J'§l-,p‘3.fl_ H(.Q='_E_): I’Ll‘chzt—jt Ajfl‘nig a |-—9\+J. 3 “4+3 Autumn 2007, ECE 161A, Midterm || Tuesday. November 20, 2005 -11- Autumn 2007, ECE 161A, Midterm ll Tuesday, November 20, 2005 f[n]=a"u[n]#~F(z)= , lz|>|al z-a f[n]=—a“u[—n—1]e—£—>F(z)= Z , lzl<lal z—a Some useful Z-transforms: f[n]= a"u[n—1]‘#>F(z)= a. , lz |>| al z—a fln]=—a"u[-n]+—~Z——»F(z)= a , Izl<lal z—a Z f[n]= all #Fm = “z + , |al<lzl<llfa| l—az z—a f[n] a “[n](%F(Q) -—1 1 ”n ._. lZ I>l a — (16 we".:2 1 ',a 1 F 9 = + . Some useful DTFT-transforms: f[n]= a < ( ) l—aem l—ae'lQ fln]= 36m —kN]<—”»F(9)=— 26(Q——) Ion—co k._oo f[n*no]%e‘j""‘aF(9) Energy of a signal Ex = Elfinf Floor Function: [x] = maximum integer smaller than or equal to x -12- ...
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