This preview shows pages 1–10. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ...
View
Full
Document
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.
 Spring '10
 BillurBarshan

Click to edit the document details