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Unformatted text preview: SGLUTKON ANSWEK \CE7 NAME: 30 Nov. 2011
Digital Signal Processing I Exam 3 Fall 2011
Session 40 30 Nov. 2011 Cover Sheet WRITE YOUR NAME ON EACH EXAM SHEET
Test Duration: 60 minutes.
Open Book but Closed Notes.
Calculators NOT allowed.
This test contains two problems.
All work should be done in the space provided.
Do not just write answers; provide concise reasoning for each answer. Problem 1. Let be a discrete—time rectangular pulse of length L = 5 and Mn] be a
discretetime rectangular pulse of length M 2 3 as deﬁned below: = — u[n — 5] Mn] = —— u[n — 3] (a) With XN(/<:) computed as the 5pt DFT of = — um — 5] and HN(k;) computed
as the 5—pt DF T of Mn] 2 — u[n — 3]. The 5—point sequence y5 is computed as
the 5—pt inverse DFT of the product YN(k) : XN(k)H Write out the 5 numerical
values 0f .9sz in sequence form as {ysl0l7 315l1l7 95l2l: ysl3l, ysl4l}~ {94% = W3 @ “6 (/léﬁ circmlaa‘l‘ “deb/‘X: \ ( z x l ‘( E
l l l s ) :3 3
l 3 l L Cl) 3
i x I k l 0 §
L\ L t l l K
" _. 0L
0V (Ase Jr;w\eﬁ§\0moﬂh allmgxng §O'(W\CL [\D (b) With XN(k) computed as the 8pt DFT of z — u[n — 5] and HNUC) computed
as the 8pt DFT of Mn] 2 — u[n — 3]. The 8point sequence 348M is computed as
the 8—pt inverse DFT of the product YNUC) : XN(k)HN(k). Write out the 8 numerical
values of y8 in sequence form. Sehg—gk 5+$~) =7 ngeoxv ConvoM/Vg‘ion 0“" gx'nce NT 8 > j =‘7no 'ﬁvne Aomqfh ah‘msnmg (c) With XN(/<;) computed as the lO—pt DFT ofﬂn] = ~—u[n——5] and HNUC) computed
as the 10pt DFT of Mn] 2 — u[n — 3]. The 10—point sequence glow is computed
as the 10pt inverse DFT of the product YN(k) = XN(k:)HN(k). Write out the 10
numerical values of y10[n] in sequence form. szv‘l “:> /910[h—1 :{(>Z>3>3>3> 2> L>O)O>OB Problem 2.
For all parts of this problem, the reconstructed spectrum is computed according to the
equation below: Nal sin A]: w__2_7_"li .N71 27* Let be a ﬁnite—length sinewave of length L = 8 and Mn] be a discretetime rectangular
pulse of length 1V! 2 5 as deﬁned below: = 63%" {um —— Mn — 8]} Mn] 2 — u[n —— 5] (a) With XNUC) computed as the 16pt DFT of and HNUC) computed as the 16pt
DFT of Mn]7 the product YN(k) = XN(k)HN(k) is used in Eqn (1) with N = 16. Write
a closedform expression for the reconstructed spectrum Kira). @thth 8+s~1 = \Z \\'\/\eo\v Convolmlvumn f$ 0% N ‘5 >11 :2) he 13‘:V\«\e—clovv\o\{v\ ﬁx\l0\S\An% re (“a hs’l—VMC+‘G «A :> P?r’$9§l W—M—J 6
saw» W (b) With XN(I<:) computed as the 12pt DFT of = 81%” — um — 8]} and HN(/€)
computed as the 12pt DFT of Mn] 2 ——u[n—5], the product YNUC) = XN(k)HN(k)
is used in Eqn (1) with N z 12. Write a closed—form expression for the reconstructed
spectrum length o’F “meow (chvduu‘tt'oﬂi Agm'n> sz =\1 \(V 1 Y SAMQ answev 013 Z (a\ (c) The answer to this part will be useful in determining the answer to part X N(k:)
computed as the 8pt DFT of z 69%" {15M —— u[n —— 8]} and HN(k) computed as
the 8—pt DFT of Mn] 2 — u[n — 5]. Develop and delineate your answers to each of
the four steps below in the space below. Simplify each answer as much as possible. ll 63%" — u[n — {Ulnl  Uln * 5]} (i) Determine a closedform expression for the 8pt DFT, X N(k), of
)
(iii) Determine a closedform expression for the product YN(k) = XN(k)H
) ll (ii Determine a closed—form expression for the 8—pt DFT7 H N (k), of Mn] (iv Determine a simple, closedform expression for yg equal to the 8—pt inverse DFT of YN(k) = XN(k)HN(k;). Note that sin = f and sinﬁf) = —%. Z, 8 $t7€c6 a'\ (“>9 M «£5?
“0” 8 "IT q? "(T 1; 2w S‘>’&L 5»— ,9.“ (d) With XN(k) computed as the 8—pt DFT of = 63%" — um — 8]} and HN(k)
computed as the 8—pt DFT of Mn] 2 —u[n— 5], the product YNUc) = XN(k)HN(k)
is used in Eqn (1) With N = 8. Write a closedform expression for the reconstructed spectrum Yr(w). Note that sin : % and Sin I _%o g‘p'xce YNWQ 3* ‘ 8 Qﬁ“? Cy (’Rﬂl—B Tkeve‘g ohg one howzero JHarm {A the \ 1 9
Spectra) V€C0h3+rmc+soh D ’Tke ,Q' 2 + rm ...
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