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Unformatted text preview: 1021. The polemro plots are all shown in Figure 51021 The Fourier "mom does um aim beam the ROC dos not ' lud
. 1m: e If}? unit circle (a) For :rln] = o'In + 51‘ XV) ; 25 A“ z. (s) Consider 11in] = 2n":_n]
The Fourier transform exists because the ROC includes the umt circle. as
(b) For 1hr] =5[nA5], X1“) : Z zlln}z_n
X(z) = 2'”, All 1 except 0. “We
The Fourier transform axis”! because the ROC includes the unit. circle.  i n _n
(c) For :{n1=<u“u[n1. — “mum z
on
W
Xlz} = .2 ﬁnk—n z 2(2}_nz"
no?” “a
= Shun,” = 1/(1 _ Wm). M < 2
“=0 : _21_l/(1~2 _l
= 1/(1+z"), m>1 C I z ). rz<2
The Fourier transform does not exist because the ROC does not include the uan circle‘ onb'der id”: : (1/4)"“!” _ 1].
(d) For 1nj= (l/2)“”u[n+3], xiii) I Z IQIRJ‘z'"
n:_°°
w 00
x”) = 2 "EM?" = Emmy”
"=_°° n:l
w DD
= E(1/2)n+lz*” : :(l/4J’I4lz—llil
u=s “a
m a
= zany224+: = (z 1/4)f]f(1_{1/4)zrlﬂg M > W
=° Th .
= :23!“ — (l/2)z"]. 12!) 1/2 c z lramrerm D: the overall sequence 4"] = Irlnl + 12%} is.
2 *1 —1
The Fourier transiorm exists because the R00 includes the unit circle. Xlz) =  (1 Azzqu + 1 _‘(U;;‘]lz_l ’ (1/4) < ill < 2 (e) For zln] = {—1}3)"u[—n — 2}, mhp F . _
. r curler transform exists because the ROC includa the unit circle X“) = f: ﬁnkn (11) Consider :r[n]=(1/3)n—:uln _ 2]
..2 — w n
= Z (—usrw X ‘1) ~ £004ng
as g m n— —n
: E(—1/3)‘"z" ‘ EN”) 22
n=2 on
= §(_U3)_n_zzm = "gm/3W“?
"=0 _ — _
= 9211(1+ 3;). z<1/3 ‘ ‘ 1 ail/[141”): ')}. lzl>1/3
z 31/“ + (“skilh '31.: 1/3 The Fourier transform exists because the ROC includes the umt circle The Fourier transform dam not exist because the ROC does not include the unit circle.
(f) For :[n] = (1/4)"u[—n + 3], X(z) i Iﬁnlz‘" m l. £M n 1,».
n=—m 5 "d" 5440 5 Maﬁa
3
EM“ 1 h 2e,
Z (1/4)‘”z" (A, n=—J a: ) (c) on 1 4 n+3 n—3 I.“ h
E3: / ) z 4 M?”
moors/(1 — 4;), [z] '< 1/4
(1/16Jz"/(1— [1/4)z"). Izi < 1/4 ‘*
(d: (‘3) it {‘1} Figure 510.21  — . Therefore.
10 32 (a) We are given that Mn] a a”u[n} and ﬂu] = uinl u[n N] ylﬁl = 2in} ‘ hlﬂ]
on
= )3 h[n e k]z[k}
lamm
N—l
= Ean’kuin A k]
is“
Now. yln} may be cvﬂutcd to be
0 n < O
P.
2.2%", 09:qu
“'4 = r3
za“a"“ n > N —1
:0
Simplifying,
0 n < 0
‘ ' N 71
= a" a")/[1a ‘l. 091 s
yTﬂ] {£“(1—aNya—a1). n>N—l
(b) Using Table 10.2, we get
1
m2)  harp M > in!
and 1 7 f”
Xlz)=1_z_1, All:
Therefore, I
l 2””
Y z =X(Z)H(z)= — " l—nz“1 _(1—z’l)(l—az’1)
(3 z ]( The ROC is {2] > In]. Consider Pu) : (1— z"l)(i — at") with ROC la} ‘2 luL. The partial fraction expansion at Hz) is = 1/(1 me) + 1/0 a")' Pa) 1— 2‘1 1— n2"
Therefure. 1
ﬁn] = l w ﬂu[n} + l _ a_1n"u[n]. that A
Now. note Y”) = Pb)“ _ z N] Thercfnre, 1 1 via] = pin] pln — N] = mlulnl ~ urn  Nil + i e r, §a“u[n] — a"'”urt N]}. This may be written as 0 "<0 N l
_ In, 1 1__ 71% (35115 n.
ylﬂl _ {3(1 —au_)N/)(/(i. fa"), n > N —: This is the same as the result of part (a),  n the zt arm of both ties the van d. a ct: equation and simplifying,
) alt g ransf st of g} IE3! n
1 l 33 a I l we get ya) v 1 X0!) — 1—?” If??? ROG has to he
The poles at le) are at {1/4) ijh/i/tl} Since hln] IS causal. the
Izi > ill/4) +ilﬁ/4ll = (1/2)
(b) We have H(z) = l II>‘
V_._, z ,
l—%z“l xcz) 2 Titanium, 1 I
(1%z'l)(lw%z‘l+%z'2) 1‘
his imp res
' ' R065 of X[z) and H(z). T
’11 be the intersection of the I I ‘
abet :{lgckoofggijl/vllzl is 121 > 1/2. The partial fraction expansion of Y(z) JS
a l z'lf2 '——— P. 172'
t—lz" l—%2'+zz {41): meu) = Y(z) = Using Thble10.2 we get yin] = (%)“u[n] + 3—5 (%)nsin (1;)ulnl. ' ' (1 sim [ﬂyingl \w
10 36 Taking the z—transform of both sides of the given diﬂerenue equatwn an p
—l
a“ Y(z) 1 1 HM: X(z) :zl—yl6+z =1— ::"+z“2 The partial fraction expansitm of Hit) is 3/8 SIB
H“) = ‘Wr + 173;?
! Tl ( cfnr‘c.
Since H(z) corresponds to a. stable system‘ the ROC has to be (1/3) < lzi < I r 2 “ 3 n m? I
we 2 —§ utni— gm ul 1i 7.1 From the Nyqust sampling theorem, we know that only if X(jw) : 0 for {my > tun/2 will
be signal be redaverable from its sampies. Therefore. iju) = l) for lul > 50001.
7.2. From the Nyquist theorem. we know that the sampling frequency in this case must be at
least 0. = 20003. in other words, the sampling period should be at most T = 2w](w,) =
1 X 10". Clearly, only (a) and (c) satisfy this condition
7.3. (a) We can easily show that XUu) z 0 for M > 40001. Therefore. the Nyquist rate for
this signal is my = 2010(1)!) = 5000:.
(1)) From Table 4.2 we know that, Xl'ju) is a rectangular pulse [or whirh iju) : CI for
[w] > 40001. Therefore, the Nyquist rate {or this signal is my = 2(40001r) = 8000:.
(c) Mon: Tables 4.1 and 4.2, we know that X(jw] is the convoiution nf two rectangular
pulses each of which is acre for lu! > 40001ri Therefore. X(jw) : 0 for lwl > 80001
and the Nyquixt rate for this signal is MN = 2(80001) = 160001‘
7.4. If the signal 2(2) has a Nyquist rate at «In.
[H] > “0/2. (5) “cm chapter 4. then its Fourier transform X01») = 0 for W) = 2(2) + :rft — l) 3 Y3“; = XLM + Emmy). Clearly, we can only guarantee that Y(jw)
rate for y(t) is also M). (1’) From chapter 4. = (J for M > tun/2r Therefom riir Nyquist 5"" = 11—2—0 *1 You) 2 ijow; Clearly, m: can only guarantee that
rate for girl is also «:0 (C) From Chale 4, Y(ju) = U for M > (ml/2. Therefore. the Nyquxsi. 9“) = 22m +3 Yljul=l1f2rllXUwJ . Xl’jwl, Clearly. wean guarantee that:
y(:) is 2w“.
(d) Fi'orn chapter 4. YUM) = U for [co] > uo. Therefore, the Nyquist rate for y“) = WWW) 35 You) = (mime — vol) + (1/2) Clearly, we can guarantee that Yum)
rate for ﬁt) is Slug. Xiilw +wo))‘ = l] for [mi > 1.20 + Lou/2. Therefore the Nyquist ...
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 Spring '07
 Chakrabarti
 Digital Signal Processing, Harry Nyquist, Nyquist–Shannon sampling theorem, Nyquist rate

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