Chapter 9 - cm. P182. ‘1 -"Ehh ‘wamb e5” °...

Info iconThis preview shows pages 1–12. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: cm. P182. ‘1 -"Ehh ‘wamb e5” ° oefiéN-rl H—l In . __‘ ‘T "Ll—h "" (.5) = Z__ C_{) jig-“k : M: (at a o N u} - 2. -' L": Fax NOJQ /C"[+Q’ M h) , ' . I Q :6 :‘r! FOR. MKV’Q— b I: N I h : .-- 2 ) {a d('£'={r\$£f2_ 'CJ U: (Q... (513......1 N hQfl“ 1&_ M (I m V: —1‘:._--h g.--. .. .54..“ ’\ (a) '- Q. 1“ ‘ _ 3;;- g‘r J N “=0 " .75; A! _ _‘ - hara /% "-O' 3- QFEhfiJ 5Q ‘12. ‘ I “45". " jafn(u-g) XCN— It) .. L 10‘) a " n=0 7 _ I "-1 zfl—nk * d‘ = 2: zone ~ J=XC2) "fil- . 2n... .1 b 2n. ’ -J._Kn 33(1) ‘fl (3;) =- Z 7; c N Lib m=% ":0 have." N", -: -Ekm 7.0:) = Z zen) c N ,an) Nah: 41mg 3301) '57 q égrgagky gainer-sci "coffin- pcr'md 1N; 0355‘; XCRXI 1}. ‘9171'0-611; Lawn P-cTt'oé N. 7501- {.5 NIKE) I. cons‘sg. of. I 49,9 ¢,.;,P:¢L_¢_ ‘ Fan'on o4 x01) -- “."~ ‘ . _ _ . ' N~I ‘ I §§K(N-n-I C“). $2.01) = xCN—n-I) 4%”. = Z x(m)¢, N .. - _ ‘n=¢' I- _ .' .- "1:0 ' . “'1; 3: 2 6"“: XC‘K) . w, m .jgjimk I *(‘0 pp. 3- ‘U-oll . 'Elfik ~—-—-]¢- " ...h 1* 7-1 N]; =- 2 1C3“) 7-— 0") mm = 2: xCZn) a. fi- We . n=o i ‘ . "-1 "‘ “2sz cm): J x szcmaw 2 an. I; ,. .Lx K ' z_+L‘- l ( ) ¢X( 2_ “-1 “if-1173)” "" ill—"k 11(11):: "‘ ” J (W) Eton) = Ea flouc 1M 7 his; ME) For k m [ g, Haggis: 219 :5 7 a 15 9.4 YCk) = z x(n)W§:§ = 2 x(n)Wi”5‘ + E x(n - 3)ng 11:0 11:0 n=8 7 5(3) 7 “('13) _. k = >3 mm - + [Em—8m 21a 1* n=0 "=0 so that I ‘ ‘ “0) = 2' m) = Y * (14) = “i4, ’14) = Y..* (6")? Y(10)= Yf.(12)= 2—??- m) = 2 with all other m) = 0. 0 s k s 15. ' ‘ ,1 ,2x A: 5: -;—k -J-:k+e—jzk -J-k "13* 95mg} X(k)==l—e 3 -e 2 so that X(0) =0, X(l)= 1_+ NE: X * (5), X(2) =3+ij= X *(4), X(3)=—2 Also Jr .2}: _4.1 . ,5: “J'— -J_-._k ..'_k a _ __ n H(k)=1+28J3 +J€j3 +3.2 Jflk+4€ 3 +3 _. which gives H(0) -__- 12, H(1) = —-3 — jfi: H * (5), H(2) = 11(3) = 11(4) =6 - . so that m) = —j4~fi= Y * (5). no) = Y9) = m) = NO = 0. -- Hence ‘ mu) = % 5: my?“ ' i=0 2 {0, 2, 2, O, -2. -?.} 220 t .38 _ - _-i- . - -——h 93-5 (i1) Kat) :_. ink 12: _a-jzk+£ J ._, so Lhax M» =- 1. X(1)=2+j3 = X *(3). X(2).=1 .3: .1..— - Since HUc) = l 47 e 2 , we have {2) =0 13(0):;H(13=1+j=H*<3).-‘1 . and ‘ ~g,‘-nme:zyaywi+fi=r4axnm;o‘ II Inset-hie” - I I - ' l 3 - 12:51:21. new; game - '={-i,°—3.0,2]_ 1, 9.6 (Elli) Zero-pad both sequences to length 11. Then, we have mm = 12 HQ) = H * (10): 1.2326— 3.5729 X(0)=0 X0) = x * (10).: —0. 70363-103345 Xe) =‘X * (9) = 1.2635+j1.4979 Ha) =‘ H * (9) = —2.4:356— j0.7210 XG) = X4 (8) = 1.0211 +j1.4i)§.6 ' “'HG)‘: H 3 (3) = 0.1103 — 10.2426, 15(4) = X * (7) _= 4.07133 + 10.4167 C m4) :3}! * (7): 0.2335 + 10.1822 - ” X6) '= X *' (6)‘='—' o. 0944- j2;9141 ' 33(5) =_7H * (6) = 0.2338 '-— 10.3793 ma) = H(k)X(k) gives Ho): (5, m; = Y * (10) = 7. 1438 + IT. 1838 1’0) = Y * (9) ‘= — 2. 0227 — 1'4} 5894,‘Y(3) = "1’ * (3) = 0. 4534 - 'jO. 0923, Y(4) = Y * (7) = l.062+j0. 3494, Y(5)‘= Y * t6) = — 1. 1365—- 10.9221 so-that : 1 "0 3.". - - mm) = —- 2 me’ n’“ ' . 113:0 ={1",1;o,‘-91,—3.—2,—1, 1,2, 1, 1} 221 9.6 '(iijZero-pad both sequences to length 7. Then Km) =- 1 11(0) = 2 m) = 0. 92.54 +19. 1047 = X * (6) Ha) = 0.099 - j4. 3339 = H * (5) x0) = 2. 9695+ j2. 2978 = X * (5) m2) = 1. 625 + jO. 7818 = H * (5) X0) = 1'. 9559— 10. 88990 = X * (4) 11(3) = o. 7775 7- j0._ 9749 = H * (4) m) = H(k)X(k) gives Y(0) = ~2,Y(1)_= 0.8216 + 10.61 = Y * ('6). " Ya); 3. 0245 + j6. 0521 = Y * (5), Y(3) :0. 654 -— 10. 25981 = Y(4) so that ' ' ' 2): 6 . . mu) -—- 1 2 m2?“ ' 7 i=0 ={1,-2,—-1,2,-—2,—I,1} A” . =0 — CL: 9111);; xtn): Lu:- n -=- g"??? ' .7 w _-.n_(gm+p) x024: 2 2 Z Burr-410:" [3:0 "1:0 - - 7 3 amp _- .211 (Em-r}? WEE-31W): Z Z (9%) rs“ “7 ) {9:4 mm: 1 *3 m ' __- 15'.- u, f. I P25.’[ ) (éJPJC-J s 2': ‘;L%J1(%)P;g2.-§E‘hla P54 I_J3 Let .3Cf0: '“UfsJ? f... v3 7 _-2r- “Um Magma) = Z 3933?“? = 290:) P—-° k: p’l’..- 7 222 2” U) 9.10 9.11 , th Kgo N—l —| kn * s A '2 [ 2mm, x02) N has. ":0 54‘! a .... ‘_ 2 x02»)? (L) 1’" Lu (i) IfM > N. augment x(n) by adding M -— N zeros and take the Mfipoint DF'I‘. . I I ' - - ' ' " (ii) If M'<.N. let: M: NLl‘where'L is ad integer ahdJaJ n 2 mM + p. Then ' ' 2: ML 2:: MIL—I “’ Hawk) ‘= 2 my"? = f; z me + p)e5"i‘?‘“f"*”’* n=0 _ [:0 m4) , m=0 - M—l L—l I .2: f - ' = Z 1(mM + p)]e_"fit-Pk _ L—l - = M—poimDFI‘ of Z x(mM+‘P) 7:: 15:: ' — 128 samples iriuzhe range [— --—-) correspond to 512 poian in 16' 16 the range {0, 21:). Hence do a 512-poinl DPT. The-n k = 112 cor- responds to Q = 72/16 rads and Ic = 239 cofiesponds to Q. = (239x21r)1512 = [1521116 -— (21:)I512] fads. Use the proce- dure in Problem 9.9 to determine the 512—point DFI' of x(n) for the cases in (i) and (ii). , " - Use the results ofProbIem 9.9. _ ' XCh) = Z."- xcn) Hr“ Lg»,— n: sun-r? > F... . 7 _ L v L I MK PR. x0e.) 2 Z "a. x(5m+}=). w: ]w‘ - Z“ GPCK) M; 2- - us} __I 9:» I \v' ‘ " ‘ ksth V? 7" fix c144 ’39 =.L<.o-zs- 'T.>4»s¢= T. ' .15- c‘ T - NT 3 4.41: x u ° +09; - ' fir. ' (Ci) .1- 3 2+ ’ 2 "l— - _l_ h T. 2+ (b) N T ‘ ._‘- .. FT, . :11 'r ‘3” M“ “ "'2" t 4'9- : if r ‘2“- YqJI‘afiS N 102.4. (a) ‘FD’ nu qlPASEAj, analog 533mg: mus? be. Lam-J Era-{ska} 224 347(4) 101) = ‘71:- + c+'—J4)c5%§3" _. 293.55" fécjggffinuzaj‘BI-‘g—run* (4 +4.“) Egan ] = '13 [2. + 4J3: Co; %fl'3n -45‘) .— 2Cos g-"n I. . (2a)? Cc)“: f: s 4;; 1:030 with” ¥¥3u4ncy fix} A“; no}- cause aliasinj = “563) AF .. .Linr-L —;,‘.. .LL" ‘ ;.25..-c.' ' . ' " 27F? 2" J 45F ' 74‘“) 3F ;.kcn>' T5- wi': ~ 3%) = HC-k) H‘ae) =gfi£g>+§5uyw> ‘__ ‘3 ' J ' fé'H_(Ak)+é.H(flis)_=“Hg-(K) __ - MICK) 1+: rigid-k) u " I'1”r(“i5-éutn)_- éflfm) _ ' ' .fi'_Jfiic§)Il-géc—gl - mag) a 4: Hod-ii} (fink) in I; Lu) .1; ‘ Funk! 'hQS-morj" Hark) :74“ ('4‘) HKCk) .— Ho(k) = i H(k)..i H(IV_R)- IHIO‘) I:_ HeCk) : : “(Ky-fii-HOr-k.) 225 Ct) LII-b ztn) ‘ {ice-1 *33‘3‘3’ ‘cc'flhm'q t "man W) = XKCfiffi} x;— Ga) = Xiwm xi, 0'.) +1 [2mm #:0ch Ais»! xcn) = Ffik) + H04) = r: (r. )w‘ F5201) 17- 11310:) + _} Rift.) Sim: Rn) is mi, FACL) 1s amijztx.) is «H- '5"~cc Acny s magma?” HRCL) :3 can, JHICL) us even -i~ 'FKCk) .- xficoc) I jabs) =J Km (K) Pinon) . = x“ (in) , J'Hflk) =j2<¢¢00 The I; ka = X — i ) £(k);xe(” K)+J X:(K)-—>"~ICU-KJ a J— H01) =' - ‘ 7 10m) - XngD—XRW-khjxrchhxrw-LJ 2- 2— O? G ) __. x Ck ) ' I _ z " x: (V- k -' 1 J -J XLCR) 1x2 (AA—k) 9.19 The Fourier transform of signal :90) is given by a 9.20 1 1 .Xu(f) = 2[5(f- 3) + 5(f + 3)] and consists of two 5-functions at f = 4.: U3 Hz. I Figures (a)w(d) below show the resulm obtained using the DFT on a 32—point data sequence obtained by sampling xu(t) at intervals of T =—15/16 5, corresponding to = I6HS Hz. At this value of T, the duration of the analog Signal used in computing the spectrum is 30 5. Figures (a) and (CI) show plfiots'of IX(k)i Versus k for the rectangular and Hamming windows respectively, while Figures (b) and (d) Show plots of IX ( f )1 versus f for 0 .<. f < f, for the rectangular and Harntrting Windows. a sampling frequency of f, 226 Figures (e)—(h) below Show the corresponding 13511115 obtained using the DPT when 2:30} is sampled at intervals of T =0.1 s, coutsponding to a sampling fre- quency of f, = 10 Hz. The duration of the analog signal is 3.2 s or roughly IllOth the duration in the previous case. figures (6) and (3) show plots of IX (k)i versus 1: for the recalangular and Hamming windows respectively, whjic Figures (f) and (h) Show plots of IXU')! versus f for 0 g f <. j; for the rectangular and Hamming windows. Ali the figures show two peaks, 2: fraquencics of 1.13 H; and f, -— 1.13 Hz. I E E . . i f 2 E '1 :i as; ‘ :5} If}: - i; - E m? l "'4' f‘: F i 5 Ii :2 : ‘ : 5'. M f' ; sII" " t I! 5' l E I' I! I E 9! :i I W a u u o.- u . .; llllll 0 an- -! I :51- ii mi- I s:- #9 o Oa==== Illlliaa_n: o s m :5 29 a so as 227 “(l .ll 9 o H" 3.1 :J. l . i s to Is an a :n 35°1llln5al (g) (11) Since xu(r) = Scos(l900:rt)+ 5cos(21007:r). of = 100 Hz. Thus To > 0. 1 See. The sampling rate of T = 0.4 msec corresponds to a sampling fretiuency of f, = 2500 Hz and satisfies the Nyquist criterion. \Vnh this value of T, the minimum number of Samples required to resolve the two frequencies is N = T011" 2 25. To use a radix-2 FFI‘ algorithm, we therefore need at least 32 samples. Figures (3) and (b) show the spectra obtained using a data length of 32 using a reeatangular window. Figure (a) plots lX(k)l vs I: while (b) shows a plot of iX,(f)l'.vs f in the range 0 S f < f,. Figures (c) and (d) show the corresponding results using a Ham- ming window. As can be seen from the plots. for a data length of N = 32, we are unable to resolve the two frequencies using a recamguair window. ‘ Figures (e) through (Ci) show corresponding results -for data lengths of N = 128, 256 and 512 respectively. The figures on the left correspond to the recatangular window and those on the right to the Hamming window For these data lengths, we can resolve the two'frequencies with both windows. The peaks in the Spectrum are narrower for the Hamming window. As the data length increases, the peaks become sharper. 228 229 230 ...
View Full Document

Page1 / 12

Chapter 9 - cm. P182. ‘1 -&amp;quot;Ehh ‘wamb e5” °...

This preview shows document pages 1 - 12. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online