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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 hlkz[n — 1:] {521 —1
l=—uu
The signals at“! and hln] am an ElmW11 in Figure 52.1.
L 1. I ] 111%] Figure 52.1 From this ﬁgure. we can easily m that the ahan: catwalutmu rum reducrb tu h[—IJ:{n 1 1; + h[1:rn — 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. {EllI]. we that.
WW] = HLIH + 2]; {:1} W41 may remit: m1. {5111} 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:]
k2nu Comparing this. “I'llI1 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” mn. 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! masE 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. (. )
355? 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 meﬂwﬁo“ 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, changingxii;
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. (215 Speciﬁcally, factoring out the constant factor of 0:” from the summation in eq. {215
yields 1 M (awifi anw‘i M andi {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), uwk 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:
rt4 _., '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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 Spring '10
 BillurBarshan
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