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HW6_Solutions

# HW6_Solutions - 1021 The polemro plots are all shown in...

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Unformatted text preview: 1021. The polemro plots are all shown in Figure 510-21- 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[nA-5], 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 1|nj= (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’I4-lz—llil u=-s “a m a = zany-224+: = (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: ﬁnk-n (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} lam-m 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")/[1-a ‘l. 091 s yTﬂ] {£“(1—a-Nya—a-1). 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"'”u|rt 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 z-t 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) :z-l—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 Nyqu-st 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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