ps-5 - Let x[n] x 6M] + 26[n w 11* 5&1 w 3] and...

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Unformatted text preview: Let x[n] x 6M] + 26[n w 11* 5&1 w 3] and Mn]? 25D: «i— 1] + 28[n -« 1]. Compute and plot each of the foliowing convolutions: _ (a) yiin] m x[n] * Mn] 0)) min] : x[n + 21* Mn} (C) Mn] = x[n1* hi” + 2] I {I} We know that min] = 1nt - hln] = f hlk|z[n — 1:] {52-1 —1 l=—uu The signals at“! and hln] am an Elm-W11 in Figure 52.1. L 1. I ] 111%] Figure 52.1 From this figure. we can easily m that the aha-n: catwalutmu rum reducrb tu h[—IJ:{n 1 1; -+ h[1|:r|n — 1] 31E}: + 1] + III" - 1] In [fl] ll ill This gim ylEn} = 35kt + l] + 'Ifln] + 2.5[n — l] + 2&[n — 21d Efiln — '1] [13} We klmw that m mu] =1En+211h[ni = Z h[k'|z[n+2— H k— rm Comparing with nq. {Ell-I]. we that. WW] = HLIH + 2]; {:1} W41 may remit: m1. {511-1} as “[11] = 5M - Mn] = E z[k]h[n - k] Similarly. we may write min! = r[n] I hfn -i- 2] = E J:[Jr]h[n + 2 - 5:] k2-nu Comparing this. “I'll-I1 Eq- [511]. 1“: EH? that tulnl - ynln + 2| . Compute and plot fin] m x[n] * Mn], where - 1, 3 63 n 52 8 xln] m 0, otherwme ’ We knmv that m yin] = :[n] . Mn] = z x[k]h[n _ 1:1 he: Thn signals :[n] and ylnfi art: n5 shown In Figure 52.4. Frmn above summatinn reducfi to: = rlfilhln — 3} + IHIMI'I — 41 + zislhh; - b} + :[Elhln — 6| + 517111111 — 1*] I. 535115; - 3] this figure, we see. mm. the L'l'l.‘ This givet: n —E. T5115 1]. _ 5. 1: 5 1': 5 us “M” m-n. 1115:1523 ll otherwise Th: signal y[n| is. 1.1:: y[nj = flu! - h[n] = E :[k]h[n — .i:| Jul—w In thin am, this nummninn reduces to 9 1' Mn} = Z :[Jc]h[n — in] = 2 Mn - 1:] eun i=0 me this it is film that yinfl is. a aummat'mu {if shifted mplirm: Inf hin]. Hml'r ' hn Fast replica. will begin at n = 9 and h[n.] in ma far I: :J- N. flu] '13 mm far n 1 N 1' 9. Using this and the {act that fill} = UL, 1m: may wuclude that N can a! mas-E he *1. Furlhr'rmfiffl1 since HR] = 5:. we can mnclufie that Mn] has at 1m! .5 unmanam paints. The mil} mint! (if N which satificfl! bath than! tanditinns is l. Example 2.3 "2 Consider an input x[-n] and a unit impulse response h[n] given by x{n] 2 Mafia], #2sz = “[n], with 0 < a: < 1. These signals are iilustreted in Figure 25. A150, to heip us in visuafizing and calculating the convolution of the signals, in Figure 2.6 we have depicted the sigma} ch} followed by hEWk}, Mm} w k], and h[1— k] (that is, h[n — k} for n m 0, — 1, and + 1) . and, finally, h[n M k} for an arbitrary positive value of n and an arbitrary negative value of :1. From this figure, we note that for n < 0, there is no overlap between the nonzero _-, points in ch} and lab: —— k}. Thus, for n < O, x{k}h[n M k] m 0 for all values of k, and hence, from eq. (2.6), we see that y[n] m 0, n < 0. For n a 0, xik}h{n_k}m{ak, Ofikfin. 0, otherwise {a} (b) n Figure 2.5 The signals X[n] and hfn] in Exampie 2.3. Thus, for n 2 G, yin} m Zak, I ka and using the result of Probiem 1.54 we can write this as I I? k 1 m art-+1 I _ 2 13 m m 339' fin} g0: lwa fern—.0. (. ) 3-5-5? Thus, for an n, I __ am” fin] W ( 1W a )Min} The signal fin] is sketched in Figure 2.7. r Figure 2.6 Graphical interpretafion of the ca¥cu¥ation of the convolutiohif sum for Example 2.3 meflw-fio“ a G Example 2.4 As a further example, consider the two sequences x[n]:1,01€n$4 0, otherwise m a", 0 g n E 6 Mn] N [ 0, otherwise ' oted in Figure 2.8 for a positive vatue of a > 1. In order to caioulate These signals are depi 0 consider five separate intervals for the convolution of the two signais, it is oonvenientt n. This is illustrated in Figure 2.9. Interval 1. For n < 0, there is no overlap between the nonzero portions of x[k] and h[n w k], and oonsequenfiy, y[n] m 0. Interval 2. For 0 es :1 a 4, h wk 3 . ch] [n 1 {0, otherwise I 012345 n {a} Figure 2.8 The signaés to be convolved in Example 2.4. __ Thus, in this intervai, Yin} * Ear". (2'14) kmo We can evaluate this sum using the finite sum fermula, eq. (2.13). S pecifically, changing-xii; the variable of summation in eq. (2.14) from k to r m n — k, we obiain ' H 1 __ afl‘i‘l fin} m Zar a... W rm") 1—0:- Interval 3. For n > 4 but :1 - 6 S O (i.e., 4 < n S 6), rawk Thus, in this intervai, 4 an} m Fwd} Once again, we can use the geometric sum formula in eq. (2.13) to evaluate eq. (2-15 Specifically, factoring out the constant factor of 0:” from the summation in eq. {2-15 yields 1 M (awifi anw‘i M and-i {woe—1 Ema: x1 y[n] = anZ(a—l)k m an 12:0 Interval 4. For n > 6 butn -- 6 E 4 (Le, for6 < n E 10), uw-k w x[k]h[n- £1213 . gag: k g 4’ Figure 2.9 Granhicai interpretation of the cenvuiution performed in Exampie 2‘4. so that -' We can again use eq. (2.13 20w“ [Own 1 Nfli] n~4 7 H u _ —~ 0: a w a: y[n]méa6’maég(al)’=a6 #1 =W. we we 1“ a 1"" a: Interval 5. For n w 6 > 4, or equivaientiy, n > 10, there is no overlap between then} nonzero portions of ch] and h[n - k], and hence, yEn] m 0. Summarizing, then, we obtain 0, n < 0 1 3+1 (1 , 0 g n ‘1 4 1 w a HWY”. n+l MM“ 05%me ,4<n$6 , 1 w a: rt-4 _., '1' a a , 6 < n e 10 1 w a 0, 10 < n which is pictured in Figure 2.10.- " 0 4 ---6 1o n' " Figure 2.10 Resuit of performing the ceevolution in Exampie ...
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This note was uploaded on 06/18/2010 for the course COMPUTER S EEE 391 taught by Professor Billurbarshan during the Spring '10 term at Bilkent University.

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ps-5 - Let x[n] x 6M] + 26[n w 11* 5&amp;amp;1 w 3] and...

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