Final Exam Solutions : Wednesday, Dec 13, Prof: J. Bilmes TA: Mingzhou Song

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1 1 of 9 EE518 Digital Signal Processing University of Washington Autumn 2000 Dept of Electrical Engineering Final Exam Solutions : Wednesday, Dec 13, 2000 Prof: J Bilmes <bilmes@eewashingtonedu> TA: Mingzhou Song <msong@uwashingtonedu> Your Name: Donald Samuel Peterson Problem # 1 (15 points) Let be a finite length sequence with non-zero extent only in the range through (ie, for and ) Let also be a finite length sequence with non-zero extent only in the range through (ie, for and ) Let, which is sent through a system which has the following input (X) and output (Y) relationship in the frequency domain "!$#&%(') *,+ -!$#/0%,12)(34!$#&%6571&/'98:):;< The overall system is described in the following diagram: x 1 + x H y x 2 (a) Is it possible for it to always be the case that = >, regardless of the non-zero values of? (b) If you answered yes to part (a), give a precise and complete definition of? which achieves this effect, where 3@!A# %(' ) is the Fourier transform of? (b) If you answered yes to part (a), is? unique or are there several possible? signals which can achieved this effect? The system performs scaled periodic convolution which in the time domain corresponds to multiplication Therefore, we need simply to define any B to be zero whenever is non-zero, and B to be C (because of the scaling) whenever is non-zero That is: B ED CGF IH JF else At all other time locations, (other than where and are non-zero), we can set B to an arbitrary non-infinite values and still have a working B since it is multiplied by zero Problem # 2 (25 points) A discrete-time periodic signal? with period K is measured in the presence of a simple echo, so that the measured signal L is: L? JM? NMPOQ The period K and the delay O are known, and also OR K (a, 125 points) Determine the most general set of conditions, if there are any, on the relationship between O so that? can be recovered exactly from L If there are no conditions under which? can be exactly recovered, and K

2 H H _ D 2 of 9 determine conditions under which as much information as possible can be obtained about? Clearly justify your answer (b, 125 points) Describe a practical algorithm (ie, one that could be easily implemented on a computer) for recovering?, or as much of it as possible under the conditions determined in (a) If we think of? as the input to a system with response giving output L, then M S / T It is possible to invert this system if!$# %(' ) does not have any zeros where? has energy (thereby canceling a component of? ) Note that /? U 3 Y #&%Z\[&]:^ _ W6X - so this means that the condition on Y is that Y `!A# %aza[&](^ ) H b dc Y cek M The zeros of are at the locations of the Of$g S roots of unity Note that there is always a zero at (corresponding to Yh ) which means that the DC value of? (if there is any) is always lost In general, the zeros of!$# %(' ) are located at ic j2k O for lcjmc OnM, which can t have any overlap with the possible components of? at ic Y kik for dc Y cqk M So the condition for invertability (to recover the remaining harmonics) becomes: j&k Y0O H for mcojpc OqM and lc Y cek M ryjo, then we can recover the remaining harmonics of? using a system ds with (b) if in fact j&k coefficients s Y t>ud Ya k Ya for Yv F&wxw&w6F,K M and where ud Ya is the of L Note that we do not include Yv since the DC component is gone ( Y when Yy ) Note that if it is not the case that j&k Ya H zy0o, we can still recover those corresponding harmonics where Problem # 3 (30 points) is a sequence of length 2{ } {02~h, ie, is zero outside the range "c c2{ { From, we define the following sequences each of length {9 a t ƒ2 D F dc c{0 0F else o{9 Fn dc c{0 0F else D x i F dc c>{9 0F else As always, an -point of a sequence corresponds to samples of its DTFT taken at Š + W Œ for dc Y c M (a, 10 points) We are given the 1024-point s Y, ƒ Y, and 2 Y of, ƒ, and ` 2 respectively We would like to obtain Ya, the 1024-point of!a# %(' ) (ie, we want to sample with 1024 points) Specify an expression for obtaining Y directly from Y, ƒ Y, and x Y Your expression may only include addition, and multiplication by +1 or -1 of sequences or sequence values No other multiplications are allowed Your procedure should not require first recovering ` a, `ƒ2, or Your answer should start with Y w&w&w and be followed by justification

3 3 of 9 Y `! Ya x Y ) ƒ Y! M ) W The reason is that sampling with 1024 points corresponds to the time aliasing operation, with a time overlap of 512 points The original can be specified as: t 9 ƒ2 0Ž 9 M {02 9Ž 9 M 2 i w but we can form a time-aliased one (which results from sampling the DTFT) as follows: f a t> 9 `ƒx 9Ž 9 M {9 w When we take the 1024-point of both sides, we get the result above (b, 10 points) Let Y be the 512-point of For the system shown in the figure below, find an expression for =, the inverse of of Y, in terms of 9, `ƒ2, and Note that Ya is of length 1536, and that the upsampling is in terms of Y Your answer should start with = ` w&wxw and be followed by justification X 2 [k] 3 Y[k] where Y D Y k F JF Y ki an integer else dc Y c2{ { =! f 0 f a NM {92 f 9 M x i ) k f ` a `ƒ2 Here, Y, which is the 512-point of can be obtained by time-aliasing such that it is of length 512, and then taking the 512-point of that resulting sequence That is, Y is the of f When we use the 3-point sample rate expander for Y, we insert two zeros between every sample of Y to get Y, leading to a 1536 point =, the inverse of Y can be obtained as follows: :,, = Y (, / W š (01) 2{ } W X - because (, W6X - Y Y œ 2{ } W X - Y œ {02 W X -! f f M {9 f M 2 i ) ( / W š has period {92 with / W6š (, (02) / W š : (03) (c, 10 points) Let Y be the 1536-point of For the system shown in the figure below, find an expression for ž, the inverse of Y, in terms of 9, ƒ2, and Note that Y is of length 512 Your answer should start with ž t w&w&w and be followed by justification (04) X 3 [k] 3 W[k] Y t Ya c Y ce{0

4 4 of 9 The reason is that: so that Y t Y t ž t ƒ `!A# %65 +Ÿ (, 8 W )6!A# %65 + Ÿ, : 8 W )6 dc Y cq2{ { c Y c{9 which exactly the of the 512-point time-aliased version of, or in other words, the ž that is given above Problem # 4 (10 points) Consider the following signal processing algorithms for transforming an N-point real input sequence to an output sequence = is non-zero only between and M Both algorithms attempt to attenuate the frequency components of higher than Ck Algorithm A A1: Compute the N-point of A2: Form the N-point Ya by setting the samples of Y to zero, for frequencies higher than Ck (ie, all samples within the range "ki >Y "k are set to zero) A3: Compute = using the inverse Algorithm B B1: Create an N-point Y by setting its samples to some function Y, except for frequencies higher than Ck which are set to zero B2: Compute B using the inverse B3: Compute = by convolving B and with linear convolution (a) Can Algorithm A be implemented as a LTI system? If so, draw a block diagram with all relevant LTI components If not, explain (b) Can you choose an Y such that algorithms A and B are equivalent? If so, explain why and provide a formula for the proper Y If not, explain why not (a) This algorithm can not be implemented by any LTI system Note that algorithm A corresponds to circular convolution If the input is shifted in the normal sense, then part of the input will move out of the range of the input systems input window Note also that if the input was not finite length, and was periodic with period N, then it could correspond to an LTI system (b) An easy way to see the answer to this is that the output of algorithm A is of length, but the output of algorithm B is of length M So, the answer is no, no choice of Y will make the algorithms equivalent Problem # 5 (15 points) Consider the two outputs = and = in the following discrete-time system x v H(z) 2 y 1 w 2 G(z) It is given that S has two zeros M k S and ik S, and has 4 poles all located at the origin, (a) Find = in terms of (b) Can there be a choice of for which it can be that = = for any choice of? If so, precisely give!$sa) that y 2

5 !!! 5 of 9 (a) Since and " has zeros at S M k and S hik and 4 poles at = S ", we have=20 and by 2-downsampling SJ/ )&! M SJ/ )(S0/ " B t!as0/ M! 9 NM JM 9Ž B ` = i (b) First we look at the FT of = and =!A# %(' )!A# %(' ) y y W6X ª - ` " «!$# %65 Z / W + 8 ) SJ/ ) 9 NM )! NM JM! M JM ( is a nonzero = constant) NM ) M )!A# W6X ª - %65 Z / W + 8 )!$# %65, Z / W + 8 )!A# % Z )!A# % Z )!A# %65 Z / + 8 )!$# %65( Z / + 8 ) )!A# %(' )!$# %(' )!A# %(' )!A# %65 Z / W + 8 ) W X,ª -!A# % Z )!A# %(' )!A# %65 Z / + 8 ) To be able to satisfy!a#6%(')!$#&%(') We must have which says!a# %(' ) which is indeed periodic over C, so That is!a# % Z ) "!A# %(' ) has to be periodic over C In this = problem,!$# %(' ) " Problem # 6 (5 points)!a# %(' ) "!A# % Z )!$Sa) "!$# /0% ' M!$#&%('t)!$# %65 Z / + 8 ) # /0%(,' )!A# %65( Z / + 8 ) "!AS / M S / )!A# %(' ) )!$# /0%(' M # /J% ' ) Suppose that we wish to compute the 8-point of a sequence Assume that we have two 4-point modules available (as indicated below) Show two ways of computing the 8-point FFT using the modules available, along with a small amount of additional computation In the first method, computation after the module outputs is not permitted In the second method, computation before the module inputs is not permitted Express your answers by completing the unfinished diagram provided on the answer sheet below Make sure to label all inputs and outputs of each system

6 6 of 9 x X[k] Problem # 7 (Bonus, points) In the process of animation (such as Bugs Bunny or Mulang), there is something called an inbetweener Describe what you think an inbetweener is and how it relates to this course? At most animation studios, the best animators only sketched a few animation drawings, leaving gaps in between Later on, a person called an inbetweener would finish the scenes, by drawing in between the areas that the animator had left This must mean, of course, that the best animators provide information at a high enough sampling rate for viewing, but the inbetweeners interpolate between the samples so that oversampling occurs when viewing, making for smoother animation