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Solucionario sistemas y señales 2edition oppenheim

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Solucionario

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Solucionario sistemas y señales 2edition oppenheim

  1. 1. Chapter 1 Answers 11 1.4. 19. 1 10. 1.11. 1.12. = cosn! - . e = —r : - Ci! );v'ertin§£5omP°l; lt0ClflBil¥l= _t': IIliilI; °l: .:( ) § ~*: cos(r)+: 'si-(= i)= :- ¢""--=0:/ Li)-2‘-inii)--1 at? -r-Ii= ,. r/ ‘ie’i= 2(anr(§)+jiui(§))= I+; ' . /2.= 'r= =m= i=. i-, Converting from Ciirtesiui to polar coordinate: 5 = 5¢I°. -2 - 29', -35 - se-Ii % _y= ~:e"'; . 1+j= /'21-". (l—j)’-1e"; , ()_, )-¢-Ii, ]. g.—. ¢1i_ ‘17—; t’r%= ¢-I1‘: (a) E_, : f°°r"“dt = L P. =0.becauu E. ..<eo (b) 72]! ) = .. JI'*"%', l: ;(t)] = 1. Therefore. 5.. = r]z: (t)]'fl = dl = co, P, = . . . . T T _'ii4rr; n,i, ~/: Tl: a(t)| ’dl - T1i. .Il: °*-l-Td2=_'! LI;1'—'l (6) 1'3“) - cos(t) Theselora. Ea, = j. |:; (t)|7d1 = Q‘ cos’(t)dl = 00. -. . . .. . T 1 2: . ..e/ .,( >)«- : an rm-I = (-'r)"«1-=1.I= .Iv-11’ = (i)"-Iv-1.*ri-ever-um. .. — 2 )2.)v-1)’ X ’%)" = t P, ‘ -- U, lx-muse E_, < on. “an H (c) Z1["I a“P‘i‘. min)’ :1 mrcfor= .r: ..— 2 1:21:-1)’ = w. P r ‘ N 7 li ‘ “E; 1 -i ““' ~T'ur2N + I _§N"‘["“ - N-Eo2N 4 I ___N ' ' . . . . (r) : _.)n) scorrgu). Therefore, 5.. - 2 )r, )n))’ = Z m'(’; 'u) Eu). .. ... 1 " ir I " I H: os('n) 1 ”= -== ~"L". .2w7r )3 °"": "’= ..“. "'. .z~+r_£ (—”‘L) ‘a . ..-iv -—u 2 (is) The signal z| n] is shined by 3 to the right. The nhllied signl will he mm (or n < 1 and vi > 7 (h) The signal : ]n] is shifted by 4 to tho lull. The Ihifled signal will be zero for in < -6 And vi > 0. 1.) Re(xi(l)) -. -2-a! 'cu(ot+r) V (h)Re{x1(l)) : r/ ieoo(§)uu(: i + 2:) - mm) = e"cos(lt +0) (1') Re{z3[l)) = e"sin(3l + x) = e“eu(3£+ Q) (d) 1tr(x. (r)) - <e"'o'in(lwt) = r-"i'ru(ioor + r) - z"‘cos(100t + 1;) (a) r, (l) is a periodic compllz exponentisl. nu) = .v'e"°' = =’‘“‘“‘’ The fundamental period oi z. (t) is R = (b) xgil] is 1 complex uponential multiplied by s decaying exponential Therefore. nit) is not pirnodic. (r) z; [n] is A periodic signnl. 3.)») = a’- = 4"‘ r, ]n] is a complex exponcntisl with I fundrnrntriital period 01' 77" = 2 (.1) 1.1.. ) to A periodic signal. The fundltrrientnl period is given by N = m(537'; ) —— mail). Ry ehousiiig in = 3. we obtain the fundntnennl period to he 10. (i-) x, ]n| is not periodic. :: ;]n] is n complex utponentlsl with we = 3/5- We cemml find any integer m such that m(: —:) is also so inhegu. Therefore. zsinl 35 "01 P¢Ii°<1i¢~ z(t) = 2om(l0l + 1.) — sin(4t — 1) Period of firsi term in RHS = fi = g Period of wand term in RHS = 3} = {- Tliizreiorc. the overall signal is periodic with I period which is the Inuit common multiple of the periods or the first and second terms. This is equal to ii. 2]n]. —1+e“V"" — 2'! “ Period of the first term in the RBS = 1 Period of the second term in the iu-is = m(, I3,) = 1 (when m = 2) Period ur the third term in the R118 = m(5-S3) = 5 (when m ; 1) Tlierefore, the rweinll signs] zln] is periodic with a period which is the lust rommon multiple of the periods oi the three terms in : ]n]. This is equal to 35. The signal rlri] is :5 shown in Figure 51.1.2. sin] can he obtained by flipping u[nl and then sliiltirix the flipped signs! by 3 lo the right. Therefore. z[vt] 2 u]—n + 3]. This implies that = -1 and vi], = -3. 1.5. 1.6. 1.7. 1.13. 1.14. 1.15. i‘)'n"li1|l|3|'| li| fliW0¢~'l'hnfiiwednhuvinheseioInrn<—4uu. in>2, (d)Tb¢tisMl []‘ nlppedsnd lie ' ' . ).m, ¢ » s'ud“, u:. "m‘: bm<_2:nd: ln: :d-pd-s iry2tothenghi. 'ri. ur. r.-w (e)T| -Uianlxlv-)'aipped. ur1iiie‘ - i - ~ “‘n‘l'iub"a': ‘°rn<‘s‘ndn5l>PPdo. lI5lIIlIalh: fl0dby2l4othele1t Thisnlrw ('1 :3‘ r ‘) Eiwlia; -:3‘) ; nf; hiltin5 the flipped signal by l in are right. (lI)F1rvm(I). iinomthat l- ' _, _ . ._ The ': u_”+z(; (_‘)t)’; :o: :2;‘ : fizruiuriyrztz t)lb: erol'oil; l (‘=1 ‘($0 "H °Wimd by line-rly compressing 2(1) by n tutor or 3. Therefore. am) will be [ED for I < 1. (4) ‘W31 is 05*-ined by “nail: stretching 2(1) by . um. or 3. Therefore. :(i'/3) «an br are for t < S. (I) 31(1) is not periodicbouuse it is zero for t < 0. (5) zzln] — l Sow Ill vi Therefore, it is periodic with I fundamental period or 1 (c) 2,[ri] is its shown in the Figure 31.6. Figure 81.8 Therefore, it is periodic with I lundunenul period of 4. (I) l 5"i‘I| "ll = gizilnl + xii-nl) = §(u[n] — u[n - 4) + u[—n] — u[ 71 — 4)) Tl*= '=f°-= - 5-rizilnll is zero {or In] > 3. :5)’ 5”“ *2“) *5 M 04d I-‘ml. £vIrz(t)) i: Ian on all values at L C 5"i= :l'| ll = $12.)»-1 + zil-nl) - ; [(%)"u[rI — 31- [%)"'uf-rr .1 3)) W '"*"‘°| '°- fvizilv-1) u no when In] < 3 sad when ]n| —i B). 5”l1¢(lll‘%(= ¢(1)+ 14(4)] = ,1z[r"‘u(t +2) - e5‘u(—l . 2) 'l'hi. -retort. Eu(z. (t)] is new only when ]1] —) m‘ -I-"OILI ‘rt Figure si. i2 u. . -2 wt) ‘I’; ,(r). u jrwta, ., 2) _ 3), _ 2)). u . { )_ 5 , 5 2 ' 0, z > 2 Therefore, 2 E5. = dt = 4 -2 The signs! :(t) and its daivnlive 3(1) an slimrn in Figure $1.14, -5 1 ‘((1 1(4) Figure $1.11 B 8: 9(1): : 2 5(t—2k)—'. i Z 6((-2!: -l) ks-n In-a; This imyliu that 4) as J, t] = 0, A, - —3, and la -‘ l. (I) The signal z: In]. wind: is the input in 5,, is the urine us win]. Therefore, min] = ziln - 2) + gzgln — 3] = min " 214' ; vi[" ' 31 = hrln — 2)+¢zr[n - 3) + %(2z. [n — :4) + 43.)). « 4)) ~ 2xi[n — 2] + s2.]n — 31+ Zzi[n - 4] The input-output relationship for 5 is yin] - 2z| n - 21+ 5x[n - II] + 2r[n — 4]
  2. 2. (h) The mputoutput relatiooshipdoea not change lltbe ads in tvhié S; and $2 8!! mnomtaiinuuiubnvuaed. Wccaneuilyptovethishyaasurningthats. follows 5, in [NI me, the aiytal min]. vhidt is the input to 5.. l‘ "'9 "M -5 win]- Thzreforr. vtlfll 1Itl"l * 4-til" - ll - 1v: l"1+4w["- ll 1 = z(: ,[. . _ 21+ gait. - 3]) + uz. |- ~ 31+ 512]" — 4;) _ zz, ]t-i — 21+ 52g]f| — 31+ Zulu - 4] The input-Output relationship (or S is once again , ,{. .y = 24.. — 21+ 511.. — 31+ 2.4.. — 4] : 16. (a) Tho system is not mctnot-ylas because yln] depends on past Values of z1vt| . (b) The output of the system will be ylrt] - 6(rt]6|1t - 2] = 0. (r) Ham thr rt-suit of part (b). we may conclude that the system output always zero for inputs of the form 5]n — k], I: E Z. 'l'herr. -fnrre. the system is not invertible ‘ if (it) TM system ts not causal because the output W! ) 1‘ 3°31‘ um‘ ml)’ d¢P'1nd 0“ l“‘“| ’¢ values of r(! ) For irutancc, y(-tr) = :(0). (b) Considct two arbitrary inputs x; (£) and salt)- : ](t) —-~-0 y. (l) = 1; (nin(l)l 82(1) —‘ MU) - =2(Iin(l)) In . -3(1) no : i linear combination of 31(1) and x7(l). Thu is. fail) ' fl-Ii(¢) 4* 522(4) where o and b are arbitrary scalars. If am is the input to the given nytttem. then the rorrettpondtng output y; (l) I: v: (1l - It (5530)) ’ 0-It (Siam) '* 5!: (iiflllll = Wt“) + (MU) Tlrercforo. the system is linear l 13. (n) (Tonxidcr two ttrbitrary inputs : i[rt] and zrfn]. nfll zilnl a mini = 2; mm k-a-Io t. ~. i 19 (a) (-l Consider two arbitrary inputs mm and 33(1). xt(¢) — vim = Fair -1) am -4 mm - 1'17“ ' 1] Let x; (l) be a linear combination of am) and n(f) That H. mt) v- a-nu) +bx2(rl where n and b are arbitrary acalan. l{x; (l) in the input to tho givt-it w-. trt. . ‘but the wrrusponding output yg(t) in nu) = ¢‘: .(t - I) = t'(nz, (t — t) + mu — 1)) «vi (1) + 5142(1) I. 'I'herefort: . the ryrtaern is linear. fit) Consider an arbitrary input 21(1) [A1 mm = 82.0 — I) be the oorruponding output. Consider a second input 12(1) nhtntnv-: ht ]I| ’H| 'lK z, (t) in time: 12(1) = zt(l - lol The output aomsponding to thin input its “((1) = l7x3(l - l) = 1721“ - l - lol Also mttr rhnt um — 1.) = (I A to)’z. (t - it ‘til 94 mm ': 'he. -cfora-_ the system it not time-invariant. (b) (I) Consider two arbitrary inputs z. [n] and z; [n]. 1l["] “‘° vtl'1l= xii" ' 2l :11”! —* min] = 1'31" - 1] Let 1.121;] be a linear combination of z, [n] and : g|rt] That it. 8.11"] ' utlfll + 5¢2l"l where ti and 5 are arbitrary acalarl. I! x,[rt| is the input to the given . <y. ll‘Hl. than the corrtstpondtug output yglnl H4 vxlnl = xiin - 21 - (min — 21+ min - 21)’ = 3:31.. — 21+ 9231» — 21+ 2nh¢. (n — 212,1" 4 2: 1‘ ayIl"l * hnlfil Therefore, the ryrtorn in not linoar. 2:l“l —' Ml"] = “:7 nlk] not-ii. Let xg[rt] be a linear notnblnation o{r. [rt] and :1[vt]. That is. z3(vt] = a-r. [n] + b: ;[n] where n and b are arbitrary acalars If :3]: -i] is the input to the given nystcm. ll| l‘lI the mrrtspandtng output p; (rt] is -an , ,,[. .] = 2 . ,p.1 l= I|—I| . 0*“! non. rung = 2 (ozx[k]+ltx: [lt])= n 2 zmq. .. 2 him, tan-Ia t_n—. i. A». -7.. = Ovtlnl + btalfll Therefore. the ryrtem is Ltoear. (b) Consider an arbitrary input z. [rt]. Let nvtte mlv-I = Z ztlkl t. ... .n. be the corresponding output. Consider a anmrtd input 2,[n| oblnined by xhiftlxiu Z| :Tl] in time; z, ]rt] = zxfn ~ n, ] The output mrraponding to this input is was -on. .. —n. u.. mlr-l= Z x2U= l- E zilk~n(]: ): ., ;tv. tan--. ti--t-rt. t. .—-. -. .i Also note that I'| lOV0 vt]u—m]= Z z. [It]. l'_n-ll| -rig Thrrrforr, tnlnl - vtln — nil This implia that the ryuem is timeinvariant. (C) lH: [vt]] < B. then vi’-l S (200 + 08 Therefore. C S (Inc -6- U3. (ul Consids an arbitrary input xt(n]. lat VIl"l = ill" " 2! be the mnuponding output Conaider a tooond input l’: ]7I] obtaint-ti by shifting x. [vt] in titntz 12l'| l = Ztlfi - flol The output ntirruponding to tho input ‘a mini = azin 421 = riln - 2 - «.1 Also not: that Vll" - nal = :i[. . — 2 — no] Therefore, vilnl - viln ’ no] 71155 5111956 lhll the system is tltneinvariant. (c) (i) Consider two arbitrary inputs z. [n] and z; [n] = r['| l -4 tnln] = riin+ 1] -z, [n— If hlfll —t y7]n] = z; -[vi + 1] — zg| n — I] Let x; [vt] be a linear combination of x. [rt] and . rq[n]. That is. xaln] - uxxln] + bz: [n] where n and D an: arbitrary acalara. l! x;]rt] it the input to the given ~'VSl£llt, then the corresponding output y; [rt| in vain] " 2:1" + 11- zaln -1] * ua. ')[n +1]+ iu. ]n+ 114 a2,[n 11 -hr‘; -[n if = ailtlfl + I] — ri(n -1]]-v b(x, [n-1§- x2]n-1]] = Wt! ’-l + bvtlnl Therein», the system is linear. (ii) (‘nntidcr an arbitrary input x. ]n). Let y1[n] = a; |n+ 1] — z; [n — 1] be the corresponding output. Consider a Iamnd lnpttt x; |rt] obtained by shifting : .[n] in tune: Szlfll = xtl" - no] The output oormpondinx to this input in vain] = =2l'I+ ll ’1[" - ‘I = xi]-2+ i -no] --til" ~l no]
  3. 3. Aiaoootethn y, [n~n°)= zg[n+1-"o)‘3I('3"‘"'41) Therefore. vrifl) = min ~ no] This implies Ihn the uyslem is time-iirvnriunt. (d) (i) Consider two arbitrary input: 21(1) and 11(0- : .(t) —o y, (:) = Od(zi(l)) £3“) —. mm = a.4{z, (:)) 1.: :; (i) be a linear uunbination of x. (¢) -nd z2(¢)- That is. 23(1) = ax| (t) +&:3(l) where n and 6 are orhitrnry srahn. l{z; (r) ‘H the input In the given wm-an thw- lhc corresponding output y; (l) in ma) . — 0d(z: (t)) = 0d(¢xI(1)+ 532(1)) : aOd(: ,(i)) + bOd(2;(t)) = ¢yI(l) 4' W2“) Therriore. the xyxtctn in linear. (ii) Consider an arbitrary input x; (t). Let mm = oaizmn = ’-———“" ‘, "“" he the corresponding output. Consider I second input 21(1) obtained by shining z, [n] in time. 3:0) ’ 1I('- '0) The output corresponding ID “IL! input is nil) - !2(-1) nit) = 0d(52(‘)) : 1 z; (l - In) - 2I(”‘ ‘ 50) — 2 Also note that V, “ _ go) = i'l"_"‘. ‘;l ,4. ("(1) Therefore. thz system ‘a not time-invu-iuit. It--ti Cl 113,]. u[n-1] = flu] (9) mo. (. ) cu. .. 2(3) = e‘ -4 14“) = cl" x(t) u 9"" #9 pm = 5!“ Since the system is linear, 2.1:) = gt? ‘ + W") —» mm = gm + e-W) Thaeioru. =i(¢) —- cool? !) —- xn(t) = cm(3t) (I2) We know um _ z. (¢) = cos (zu — 9) = Using the linearity prnrpaty. we may once again write me) = %(e-Ia” + eje"") ~. mu) = ;(c"e’" + end") = ms: 1, Therefore. '1“) = Wl(3(l-1/'1))'—‘* VI“) = €05(31 “ 1) 1.21. The signals are Sktlditd in Pigulv SLZL g(c-I) 1 10-1) 1. 1.(8**') " 4 o 1. 1. 4= or f _. _' _1 I (“#2) 3 tmiuwlum 1 o~. r a_; 4. 5 t I * o I * —_-‘Q 3/, _ '5 Figure 31.21 1.22. The signal: are sketched In Figure SL22. 1.23. The even and odd port: or: Iketuhed in Figurv SL23. i0 11-) 1 -I 3 -vi. ° -It "' 3 I“ left} ‘I , "I U‘ .3 a I 1 It I 3/3 114 1 -6 Figure S1.1d 1.244 Th: even And odd puts on: ahetdied in Figun SL24. 1.25. (II) Periodic. period = 211/(() — 1/2. (b) Periodic. period = 2n/ (r) : 2 (c) :0) : (1+ ooo(« — 2:/3))/2. Periodic, pa-iod = Zn/ (4) «/2. (4) 1(1) = oos(41rt)/2. Periodic. period = 2:/ (hr) : l/2. (G) x(: ) = [s| n(lin)u(t) — sin(4rl)u(—l)]/2. Not periodic. (I) No)! periodic. 1.26. (A) Periodic. period _ 7 (b) Not periodic. (c) Periodic. period - R. (d) rlnl I (I/2)| ms(3tvt/4) + cx)u(xn/4)]. Periodic. period = 5. (e) Pa-iodic. period 2 )6. L27. (1) Linear. suble. (b) Memo: -yl2sa_ linear, aiuml, stable. (c) Lima: (d) Linen. usual. ntable. (2) Time "Invariant. linen. aural. stable. (I‘) Linear. stable, (1) Time invu-iuit. Iinmr, causal.
  4. 4. 1.28. (a) Llneanstable. (h) Tame invariant. linear. null. mote. (c) Nlunorylua. Iinuntanul. (d) Linear. ltahle. (e) Linear. stable. (f) Mculoryinn. linear. canal. ntahic. (g) Linear. stahie. l 29. (a) Consider two input: to the ltystem such that , .[n} 5. p, [,.1= 1z. (z.1n]) and zrlnl 5» min) = Retain! )- Nmv amide: a third Input nln] -— xii") + -'-: l'| )- 7'‘ °°"“P°“dl“5 ‘Y"'"“ °“"““ will be y; [n] = Re(z; [n]) “ R¢l= ll“l * lllfll) - ‘R. e(zg(n]) + ‘R. e(z; [n] = iIil"l ‘l’ lhl"l Therefore, we may conclude that the system is ndditivm Let us now assume that the input-output relationship is dianged lo 1/In) = ‘)2e(¢"/ ‘3l"l)> Also, consider two inputs to the syluntn nuch that z; [n] 5. g. |n1— 1zr(rI'/ ‘zlinli and . 1‘g[1ll 39 y; [1l] = 1Ze(e""¢2l'| l)- Now consider a third input z, [n] = zllnl + 22I’| l- 759 °°“’“i>°"dl"K 5)’»"-VI" °“‘P“‘ will be y; [n) r ‘Re(c"'('2:[vt]) - a. n(im/4)'R. e(: n|n| ) - . arl(m/ I)In-izalnii + cnn(m/4)1i‘. e(zl[fl)) - Iifllml 4)1’"(1 Il'‘)) + not(rn/4)Re(: ;(n]) - Sill! "/4)'l'nl= :l"1) - p. (.. :-I4z. [n; ) + lzrirv-/ ‘z, [n]) — till") * Vii") Therefore, we may wnclude that the symun it additive. 13 1.30. (a) lnvertihle. Intuit system‘ vi! ) - sit +4). . (ll) Nnrl invutlhlc. The sipsls : (i) and zl(t) = z(I) + 2- rive the “W °"W“' (c) Non invertible. 6[n] and 26in) give the urne output. (di invertible. lnvnnc -yvteul: y(t) -= d: (t)/ dz. (e) invertible. llwene system: pp. ) = zln +1] for n 2 0 and via] - rival for H < 0 (f) Non invertible. :[n] and —: [n] give the name residi- (g) lnvt-rtibk. lnvene system: yfn] 2 - II]. (ll) invertible. lnlverne tystem: ya) - = (¢) +d= (t)/ it (l) invertible. inverse system: vlnl = zlnl - (1/2)= in - 1|- (j) Nun invertible. if 3(1) is any eons: -.nt. than v(t) = 0- (ir) Non invertible. ll[n] and 2£[n] rsult in y[n] = 0- (l) invertible. lnvene system: y(t) — ¢(t/2). (m)Non invertible. mu} = cpl] + an — I] and nlnl = Kl’-I rive vlnl = 61"} (n) invertible. Inverse system: yfn) = z[2n]. i 31 (a) Note um. z1(f) = ml) — ml -2). Therefore. u-ins line-my wad tr") 4' In”) - y. (l — 2). This is as shovrrl in Figure S151. (is) Note that 13(1) r. (t) + zlu + I)- Therefore. wins linearity we 5" N’) = '1-40+ (“(1 4 i). This ls as shown in Figure Sl.3l. ha) V-1‘) Figure 81.3! LS2. All statements are tnle (l) z(l) periodic with period '1'. mm periodic. wind TI? - (2) 311(1) periodic. period 1‘; z(t) periodic. period 27'. (3) x(i) periodic, period T. ;-; (t) periodic, period 21‘. (1) ma) paiodtc, period T; ¢(t) periodic. period T/2 1.3:. (1) 'liut~. 2[n] = 21.. + Nlivllfll = min + Nol~ is. periodic with No - N/2 If N is M“- and with period N. = N ifN isodd. 15 (h)(l) Couiderr-niapnutotheryuenuncllthu : .m —‘»r. l=l - fl and may . ‘., ,.m = fm “fl ' Nov eolnider a input z; (t) = z. (r) + an). The wrrupondinx system output will be _ 1 d= e(l) ’ ww - [T] = 1 [dizrul + z. (lll]" 21(1) + 23(1) dt 9‘ vl(l) 4- v2(I) Therefore. we may conclude that the syatmr is not additive, Nov: onnnida a {nurth input 34(4) = nn(t). The corrspondlng output will be d: .(l) ’ dt 1 [dlullxll ’ “((1) dt _ ; _ 1=_<3! ’ _ = l(‘)l 4.‘ l = will) Therefore. the intern in holnogeneoun. (ii) This system it not additive. Consider the following elrunpir-. Let 2.(vl: 21+ 26[n + l) + 26[n] and : t2(n] = 6[n + 2) + 26[n + l] + 36[n) Tile cor outputs evaluated at vi = 0 are l veil) = ll 26[n + pending vl[0] = 2 and y; [0] = 3/2 NW Oonlider I third input x; [n] : z, [n] 4 tgllll - 3£| n + 2] 4 ml: -l- l] + 56(n]. The corresponding output evaluated at I1 is 0 is “[0] la is/4 Ck-. -lrly. y; [0) 9! vl[0] + "[0]. Th": implin that the system in not additive. No ennnideran input : .[n] vhifi leads to the output y4[7I| . We know that x -2 . .l. l . { Let us now consider another input z. [rr] u a. :.| n]. The oorrupondiug output is . ,.. , . { Thertlon. the intent is homogeneous. 2.(n — I) ; E 0 otherwise ‘ : nn--i]; £0 we 5 nmlrll. M (2) Fab. "(oil pa-‘Indie dounn lnpu zlnl is paladin i. e. kt : (n] = p[n] -0- h[n) when »r»»= {.'. : 2;: Then min] = sin] in puiodic but 2(a) is nlurly not puindic. (3) 'n"I¢- 2l'| 4' NI - : [I-I): ygln + No) ' lreln] where "Q - 2N (4) Thu. mln + N] = xnfnl; xln + Na] = x[n] where N9 = N/2 L31. (a) Consider ""' """= {ll'/21'-. 23:1“ 2 z[n) -= 2(0) + ihlnj + z| —n]). n= l lf 2(a) is odd, 2|nl + 213:)‘: I). Tlierefiowe, the final summation evaluates to usrv. (5) L9‘ Ill") = 3Il"l3Il7|l- Then y(-rt] = a: ([-n]zg[-vi) 2 —z. [n]z, [n) 2 -y[n] This inipliu that y[1I] il odd. (c) Consider 2 fin] = f: (z. [rr]+z. {nn’ . .-. . . ... r H N O - Z z3Inl+ 2 =21-1+2 )3 z. lnlz. lr-l. Using the result of port (h), we knew: that z, [n]z. (n] is an odd signal. Therefore. using the ruuit of part (a) we my conclude that 2 E z. [n]: .[n] - 0 Therefore. : 2 2'1»-I= = )3 zilr-1+ )3 rlini (.1) Consider = ’l! )dt = r x3(l)d1 + / W ztmac + 2/“ z, (l); t.(t)r1l. —nn —aa -an —nn Again. since : .(l)z. (t) is odd. / “:: (t)z. (t)dt = o / _:= 'ul¢c= r': ul. u+ / _:: :u). ll. 15 Thuulore,
  5. 5. 1.55. W: Innttofindtheunllldflonchl-Intm(21/N)N. -hkwNa= kN/ nylhcnkian integer. L| ‘No hutobeuuini£g: ',thnNmtDtb¢| uu}tIpleal’m/ Isandm/ k muslbnu: inn: -get. This implici tint m/ k is A divhnr ofhoth m And N. Alnn. irwe wmt Lb! smdlest possihk No, than In/ I: should helbe GCD ofm And N. Thudore. No = N/ ;od(m. N) ; 35. (a) If z| n] is periodic ¢4*~*"*"! T = am’. whanuu = 2:/1.. This iinplie um ? l—: Nr=2xk = %-—:7=sntioInlnumbcr. (5)117/1. = y/ q then xln] = eI*'~W0. Th rmmnenul period is q/ ;cd(p. q) Ind the fundamental frequency is 2' - 3_'z = '2 __ fl 73¢-i(p. q) — , v: °d(M) Pscdimi , 5¢d(r. q) (c) p/ gcd(_p. q) petiods M’2(l) In uoodnd. 1.31. (I) From the definition oi ¢, .,, (t). we have ¢, ,,(z) = I-: ¢(l+r]y(7)df = /; y(—l+T)z(7)dr = M-t>. (h) Note from pan (ii) that ¢u(¢) = ¢u(-C). Thu implies that ¢. ,(I) is even Thmefons. the add put of ¢, ,(l) ls uvrn. (c) Herc. o, ,(t) = ¢. .(t - T) and ¢. .(t) = ¢. .(t) 1.35. (I) W» know um 2a, ,(2z) = .s, ,,, (z). Thndote. , 4 1 A| _ff1o5A(3‘) '- A'_¥P0§5A/2(‘)- This implies that I ma) - 55(4). (19) The plots are as Ihovn in Figure S118. 1.39. We have A;5.nu, (i)s(¢; = AiE°u¢(0)6(! ) = o. Alto. gap-. (c)a. <:) = gm). )4]. (1) yin] - 2:[n]. Thueismthsyflamkliunhvrhm. (b)y[1II _— (2n — l)z| n]. 1-iii. ii not time-inmflut bum. .. yln - N. { it an - ml» -Vol» (i. -) y[n} = ;-[. -.[(| . (_| )- +1+(—1)'--I) - 2.z[n]. Thereton, the Iystum is time invarilinl. 1 42. (a) (‘guide two symms S. and 57 connected in series. Axum: that u‘x. (t) and 2,11) . m- the inputs to 5.. than yi(t) and y; (l) tn: the outputs. rtspectivcly. Aiw. assume that it mm and y. .(¢) are the inputs to 52. then : .(z) uni am Are the outputs. l'epef'lI"‘! y Since 3. 5 linear. we may write . .z. (:) + new) it uyi(f) + knit). when n ma b [N constants. Since $1 is use Linux. we my W5" . ,,. (¢) + mu) & nz. (t) 4 him). We may therefore mnclude um mm + mu) “—*‘~’ aim uzim Therefore. the seriu combination of 51 And 51 is linear. Sine: S. is time invurinnt, we my write x. (-- 73) iwi(: —r. .) and mu -1-») -‘-’+ zit: — TD)- Thrrcforei no — To) am —m Therefore, the acne: combintiiinn of Si and S9 is time innriani. (5) I-‘alga, 1,. -1 y(£) _- 2(1) 4 I and 1(1) -— y(: ) — 1. These con-upond in iwu nn. I1iu‘. AI systems If thise systems Ill‘ mnnectud in series. then z(t) = :(l} whu-h is in luxtw system. (e) Let us mum the output of system i u w[n] And the output of system 2 as z[nl. Than. , ,;. .] — 42"] . m[2n] + éwfln - I] + %. ..[2u — 2} 1 1 -- x[n] + 2~: [n -114» 34:: — 2] The ovexzll system is Iinua: and timoinvariant. 1 43. (a) We have z(t) -it y(t). Figure 51.3! We have N am - N umm ~ our = fa umau - 7)d1. Thcmfore, 0. L<o .6(l—7)—0 at! ) = 1. t) 0 'u(1)li(! — 7) : au — 1) undefined kw 1 is 0 1.40. (n) it’: system is additive. then D -= 2(l) - Ill) ~» y(l) — y(t) = 0. Also. if I tysttxn is homogeneous. then 0 = 0.z(t) —~ y(z). u = 0. (b) y(l) = 27(1) is such a system. (c) No. Fox enmplmoonsider y(t) = /‘ z(r)d1 with 2(1) - u(t)—u(I— 1). ’i‘h: ~n1(I! — U forl>l, buty(l)-1fart>l. ’°° Sun: S b x(t-T) -59 y(l—T'). "Or. ii 10) is vaiodic vith period 112(2) -= :(x — 1‘). Thurefon. we uny conclude that W) —- y(t — 1'). This inpiiu um u(: ) is also periodic with period 1'. A pimilu ugument may be nude in dinuvte tinn. (b) 144- (I) Assumpf-our If 1(1) = 0 hr! < Io. uienyu) = 0 for i < 1.. To prove um Thr sysmm is uusal. Let us eonlidet an uhitnry aipal z. (t). Let us wnsidcr mother sign] 1;“) which is the sun: at 3.0) For 1 < lg. But for t > :9. 21(1) 9% 1.-. (t). Since the nystem is Imuu’. 11(9) - 82(1) —t M! ) ml! )- Sincu z. (¢) - 32(1) : 0 fox i < in. by our Isumption y. (¢) - mu) - o (D. 1 < .0 'm. ;, implies that mm = y; (l) For I < to. in attic: wands, the output is not aficctrd by input values to: 2 2 lg. Therefore. the syliem is mind. Anumptiotr The system is mm]. 1b psove tlut: ll x(l) — 0 for l < in. than y(l) = 0 for I < to. bet In anume that the signal : (t) = 0 for! < to. Then we may cxpnxi z(l) as 3“) ‘-‘ 31(3) - 22(1). '51!! ! = ;(t) = 23(1) For L < lo. Since the system is linezu-, thr- output to gm p-rill be ya) = yi(t)—y. (¢). Nov. sinua the symiu n causal. mu) - nu) for I < to unplug um mu) = y. (:) for 2 < 1.. '1'haefore_y(l) - 0 for z < 1.. (b) Consider v(¢) = z(t): (r + 1). Now, 2(1) = o for 1 < 1. iinplu: um y(l) , o for z < 2., Now that tin system in nonlinear And non-csu: AL (c) Consider y(l) as 2(1) + 1. Thin span: is nonlineu And causal. This does nm. satisfy the condition of put (.1). (d) Assumption: The lystnm is invertiblc. To prove tint: y[nI : 0 for all nonly if: r[n] = 0 is: All 1:. Considet - 0 ——b y{n] Since the mum is linur. 2z[n] I 0 —u Since the input hu nol dinngeii in the two shove cqunionn. It requirr that yfnl = 2y[n]. This implies that u[n] = 0. Sine: we have nauumed that the iynzrn IS mvmihli-, only one input cuuld ham led to this puticulnr output. Thai. input must he zln‘, = 0 Annunptian: y[n] = 0 for all n il’ 1.-[n] = 0 far A“ 11. Ta pmw that: The system is invcrtihln. Suppose than xilnl —v ml") Ind zzln) -4 y. [n]. 20
  6. 6. stat! the systan ‘n linsar. zil"l - ¢2I"l -4 VIl"l - vil"l = °~ By the origins! nuurnption. we must oontludr: tlsnt : n[n] = xqlnl. That is. any patie- ular y; [n] an be pvrorluned by only one distinct input min). Therdnrc. the system is invertible. (c) vi»! —= =’lnl» 1 45. (I) Consider 1:0) -5-» mm -¢. ..m and S 11(1) -t WU) 3 ¢As. (¢)- Now. consider : g(t) = a. :.(l) + lt: ;(t). The corresponding system output will he yg(l) 1 / an z; (r)h(t + ~r)dr = BF3|(T)h(I + 7)d'r + b 1‘g(? )A(! 9 v)dr " ¢¢Ax. (‘l +5¢Iur(‘) -W = din“) '* Wall) Therefore. 5 is linear Now. consider z. (t) - z. (t - '1‘). The corresponding system output will be will — ]°‘: .mnu + no = fwzflv -’I')h(r + 1)d-r = Fz. (r)A(r + 1 A 714. who + T) Clearly. y. (z) ; é g. (t — 1'). Therefore. the systun is not tirneinnriam The system is definitely not auul bectune the output at any time depends on future valuts oi the input sign! 2(1). (b) The system will then be linear. time invnrlnrn. and nun-causal. 1.45. The plots nn: as in Figure 51.46. 1.41. (n) The overall rmpanne oi’ the system of Pigusn Pl.4'I(s] = (the rapnnne of the system to x[u} + . t.[n]) — the nnponse ol’ the system to x; |n] = (Response ol l linear system L to zln] + z; |n| + zero input Iurponsc of S) — (Raponse oln linear system L to : st[n]+ tern input rtsponse cl .5‘) = (Rzpum: ol 1 linen system L to z[n]). 21 Still non-linear: n; .: lfx. [n| = -611:] And 1-‘1l"l = '''“l'‘l- “'51 VIM = -5In]+ 5|" ~ ll — 5 and valfll = '35l'"l+16l'I - ll -5 ! ‘ Wulnl» (iv) lncrexnetnlly linear. x(t) —~ 2(1) 1- td. z(t)/ dt-1 Ind y¢(t) = l (v) lncrenientnlly lineu : r[n] # 2oos(1rn)z| n] and yolfll - nns’(vm) (.1) Let z[nl .5. um and sln] 5. 4"]. Then. my - zlnl-I-c For time inwiuzoe. we require that when the input is 2|n - no], the output he ylv--no] r x(n—nvl * ¢ This implit: that we require xr»-«-. .1i- an -no] whinh in turn implia that L should be time intmrinnt. We nlso require that yolnl = c Aconsunt independent of n - L48. We have lg = rue-"' = raoos0n + mains. = 1.) +]yti (I) : . = zn—1'w (bl 1-2 = / =3 *v3 (C) E: = —zo-Jun = -to id) 3| - -: m+i9o (0) z-. = : o ivo The plots (‘or the points An Is shown in the Figure 51.43. 9"“ . : 1-(ti 3,3‘ C-run aw Y, l1-,9.’ "/2. yfvfl o l. S45’€? " (5) Figure SI. -to r--——-——-. .-. . _. .- 'l— — - — — — — _ 1s. NE4Ll. )§'rer4 I/ ' 3 l I Figure 51.41 (b) li'z. |rt] = 0 for In n. then y; [n] will be the zeroinpul responae yglnl 5 may then be redrnwn as shown in Figure SL41. This is the tune as Figure 1 48 (c) (i) lncxetneutnlly linear. xlnl —+ : [n] +2z[n+ 1] and yo[n] _ n (it) lnctementnlly linear. 0. vi even (II-I)/1 ’["]_'{ E: 2[k]. nodd I-—so And ’°l"] " { um. 22:: (iii) Not incrementally linear. E3. choose ynlnl = 3. Then “"1-wI~1={ : £:: ::%: : 1}‘- . . :13: :3. 22 ‘'‘°' W "”‘- ' ' -I - III. sins u «in. This input. thnr. o = ./3. (h) 5:-" (C) 5‘/ fags-/4 (d) 5,n-"(#3) = “(airy (0) 8c"' m M. »-/ « (s) 2»/ ic"""’ (H) Z-,2-/1 (E) HI/ G G) , /fenn-/ It (R) 4‘/55-, -/I2 (1) yr, -/x Plot depicting those points is as shown in Figure s1.49. Figure $1.49 1.50. (I) z -r rcos9.y = rsino (b) We hnve ’ r = /2‘ + y7 and 9 t -l V -1 z _ _ 3 -sm _, r+y go; [= ,%2+vz] _m I L] Disunddinedilr =0A. n«dAlsoirrvelevuit. flisnnt uniquestncc9and9+2m1r (me integer) give the sum nsnlts. (c) 0 and 0 + 1: hnve the tune value ot tnngent. We may know that the complex number ‘u either are" or :1 = ro’('"’ = -z. . 24
  7. 7. 151. (. )w. have : "=un0+jIin0. And e-i‘ - no - jdnfl. Summing eqa (51.51-x) ma ($151.2) we 331. mil = ,1‘(. -*' + F5’). (b) Suhlrncting eq. (51.51-2) em (sun) we get Iin0 - %(e" — e-I"). (c) We now hnve . _-5"“? 5 ever». Thudhn. cos(9+¢)+jsin(9+o) :2 tcosfiuauo-nin0s‘n¢) + j(dn0eu¢+ux0un¢) Puum; 0 = 4» in ex]. (31.51-3). we get mew - en: ’0—I' :0. Pulling 0 = —d in eq. (SLSI-3). we as I I an’ F -P sin’ 9. Addmg this "In above equation Md limplifying no-*0 = ;(1+ooez9). {s1.s1—1) (s1.s| —2) (S15!--3) (d) I-Iqunting the red put: in eq. (SL51-3) with ugulmenb (I + O) and (0 -— a) we at ua(P+ d) = cotOa: Id — linllinfi And eoa(9 - ¢) = wn0cu¢+ Ilnflnixu. Subtracting me. two about: equations. we obuin ninu-no = -12-[coo(0 — a) — un(0-O- 4-)]. (e) I-‘. qu. 'v. in; irmginu-y pins in in cq. (SL5!-3), In: get u’n(0 + 0) = sin9eoa¢ + eouflnixw. 25 (5) Sinner. >0.n>0&nd —l5cn| (0, -l¢)$ 1. am — I221)’ v? + -*3 - he If 4- V: + 27u’2°°I(ln - 9:) III + ‘II, IIIAII and (In| + Ibzl)’ I I’? +r' +? n"z Z | l1+ HI’- 2 lat (0) For a= 1. nl uhlrlyobviousthu. N-I Z a“ II N. no For a #1. we any write N—I N-I N-I (l-a): a" nu : a"— 2%" - I —a"‘ no no as! ) Therefore, no l'° (b) For | a| < I. lim a” =0. Il—~w Therefore. from the vault ol‘ the previous pan. lim o‘ + 0" - —. N40: “Q “o l - a (c) Diflerunlnaling both aid af the mun of pm (Is) wrl a. we get £(i«-") = .%(r1:. ) nil? .m_n_. I H-0)’ (d) W0 may write :7 an up‘ 212'‘ = a‘Za" I 1-: for la] < l. nut n: lI 1.55. (3) The desired mm is I ’ flan]: ’ Z‘/ _,. n/'I= T:, "—fl—u1+: . 21 1.52. L53. L56. (1) u‘ = re"r¢"' = r’ (b) 2/1' *— re"r"e" = R” (c) z+z' = z+)'y+. -.-jy-2:-2‘Re[: ) (d) z - z‘ -z+jy—: +jy=2;y= 21m(: ) (9) (1: + 12)‘ ' «II “‘ 8:) +i(vn + 192))’ = SI - iv: + in '- ill: 3 3; + *2 U) Gunsidnr (az; z¢)' ‘GI’ ¢ > 0. (dlnlzr I (llfnral-'(""”)' = un¢""V2¢"" = a: ;z; . For a < 0, n = lull". Therefore. (611317 - (|0|'I': €'"'”""’)' = |0I¢"'1'I¢""'7€"" ' 437*} (0 PM Ital at 0. : _,‘_fl_ . _"¢-11:‘, (=2) 'n""”"-In-:3 ~«: —:»+{<: -;>+<': >'4 Uting (5) on this, we get. ~-«*; '»= ;{<: —;>%+< (I) (c')' = (I. ‘e"’)‘ = z‘e‘" = 9"" —- 9". (5)115 is = 21:, ‘ and H 7 flu. Then. (5) FWD (c). I! an .3‘. L3‘. )]= %["—’i, —*. ,-"'—"]' 3|’; 4' Pf‘: ' 8; +1; = TR, a(: ;) = 2Re[z; :;) = 2: +2. = 212242.: - mama) (c) | :| = ["9"] : r = 115'”! = | ;'| (d) level = lnne""“"| = lrml = Inllrzl = lllllh (6) Sinw -1 = 2 + in. lxl = /2’ + 55. By an. triangle mqulizy. ’Ze(= } =2 5 J2‘ + 9’ = III and 1"I(¥)= v S /2! + v’ = lzl (0 I'll; * ‘H135 |7R¢(3I3iH '—’ 12"I"2°0‘(9A " 92" S 2'0’: = zlin-:2l~ (6) Th! ddllli III: '- I E 8'-/1 = ¢‘1""’ia'-/1 - -(I + 5). 18-2 Ilvd [c) The dabed sun in "' . 1 4 2;“/ ” °"“” ‘ 17%? /5 - a (d) The dfiiltd sun: is fa/2)"e"-I’ = (1/2)’e"'"'iu/2)‘-"'/ ' = —} HE] . i=2 nfi (c) The duixad sum is .2 +15. 2 o 9 2“cou(rn/2) . g2,»-/2+ £2, ml: = an +1) + §(1_, ') = u = I1 (f)1."h¢deti| -ed mm is fju/2)"eo-(-n/2) = éfu/2)'e"""+gin/2;~. —:-~/ = n= U nub an 4 2 d 3 __°5 ' fi»*"fi*: ‘o‘n-o-. -;- (: ) The desired integml is II ‘ £4"/ ‘m . 5% I = o. (5) The desired Integral is 6 = (2/m1e"" — I1= . . up e""’dt = '. —- lo‘ III? (e) The daired Infill is . -v''/ ’.: : = -—'f"" /2. 1*/2 (d) The duhtd inlqnl ia 33:’; 28 U = (2/, .)| e'“ A 2''] = -'7’. an
  8. 8. (e) The duuod inugtl is Chapter 2 Answers I an , -(u, 'y, ,_, -u—. -)- _ 1/2 + 1/2 _l 2.1. (I)W¢knoIIthat [: c'cos(¢)a-Io 2 —#—; -+7 ‘—'1_, ~-2‘ _’ y. [n]= :{n| -n[. .}= Z . ‘|[k]: [I-1-1:] (52.14) (4') The duirod integml 2; "--°° ag 1‘ 3 ¢-(1-l))l_e-(3+31')l d'_ 1/21‘ * 13 Tbetignah : [u]andI| [n]uumuhownin I-‘igu: eS2.l. _ = _: _T__ _ _. L. , /0 e sin( I)! “ In 2, 2-3: 2+3; :3 | 1 gm 1. 2- hr") 3 . ° I I 4: " -I o u 1. n -I Figure 52.1 horn this Sun. we can wily sac that the above oonvolutson mm redum 1.. - v| [7I] = h[~I| z{n + 1] 4 h| l]z[n ~11 = 2z[. .+1}+z; [.. -1] This giva min: = 261.. + x1+ 46[n] + 26[n — 11+ 261.. — 21- mn — 4] (b) We know (hut y, ,[n] - 2[n + 2] a Mrs] . -_ E ; .[1;], [,. + 2 _ )4 k'—ee Computing with aq. (52.1-I]. we see that gala] = y| [n + 7] (13) We mny rewrlu: nq. (S2.l—l) L1 min! = zlnl - hlnl = Sf am»-t»— :4 Similuly, we my wriu: L-m y; [n] = .. [..1. : .1.. + 2] : ti: x[k]h[n + 2 _ k] Comparing "ID Vlllh eq. (32.1). In in that valnl - min + 2| 29 30 La. ..‘ nu given definition so: the -59-! Mn). we may VIM I-I 1, Am 5 G) (u(: +3l-v-[k— Ion ”“"‘ -- -- ["1 The sngnnl Am '5 non am only in the range —3 5 Is 5 9. ho}: this ye know that the siplnl , 1 9 n q. I5 7‘ h[—k] 1: non new only in the range -9 S 5 S 3- "V! I0?’ I513 ‘'2 “SW” M''‘‘] '7! " W “"3 fig. " 52,4 nghl. llwn the resultant Iiyul h| n - kl will be non Iuo u. the muse (n - 9) S I: S (n + 3)- Thercfnrc. ,4 . 7. » 9, 3 _. ,, + 3 2.5. The algal y[n] is on = I. = I: — I: Let us define the siyuls n W‘! 1W ' ["1 kgwxl N" I ‘IW 1 (5) "("1 In this one, this uununnion Iedunu so and I I min! = u 1/[n] - : z[E]h[n — 1.] = EH11 — t] We nnu [Ml ‘= " '= " xi"! = ruin - 21 and '-I’-l = "II" + 3| am this an is elm um y[n] is . sumnntion of shifted replicas of nan]. 3.. ... . -». .~ nm N“, replica will begin at n = 9 ma Mn] is zero for n > N. yln) is mo for n > N . 9. Using = _ 2 A 2 this And the (act that yll-1] = 0. we may eondudn that N cu: at man be 4 I-‘unhrrmorc, W“ ‘Kg ' ‘W = “In I. ‘In ‘P 1 since y[4] = 5. we can conclude that Mn] has at least 5 non-veto polllli. The only value of , 2 hp. _ 2]I. .[n — k+ 2] N which utifie both these conditions is 4 l= —m _ . 2.6. Hum :11: given information, we have: By rcplu-ing k with m + 2 in the abovr mnunatton. we obtain W] _ zxllmwln _ ml _ “W _ MM y[n] _ x[n] . n[-:1 = ‘= z_: [A]n(y. — 1.} Usmg the rtsults nf Example 2.1 in the lat book. we my write — E: (1)"u[-In — 1]u[n — k — 1) 3 1 an k= .—- ylnl = 2 [‘ ‘ (E) 1"“ ‘ Z‘ (%)"uln - I: -1] As—a= W: know that N . . I .1.. . = .1.. . am: Exlklhln — kl = gtgvutn + k — u no ssmm :1»-I -finial are as sum -n Fiat‘ 5“ ‘'*°‘'‘ "“ ““"" "" "' ""‘ "" Rlpluins I: by » - 1. above summation um: to y[». } . .131I. [.. — 3] + x[4|"[" — 41+ zlfilhln - 51+ zffilhln - 6| + x[71hln - 7| * TIE! ’-I" ' 31 ylnj = E(§)"'u{n + y] (s2.6~ 1) This Kim I I :5‘-' 5- :25S""55'§s For n 3 0 the Above equuion reduusz Ya, y n = _ tn 2? "- 12.5.2.5.” . ... =;; .;»«- = g,_: §=; 3! 32
  9. 9. pa _~t For n < 0 ea ($2.64) nsrluuu to. 11-1 ~— '2 151"‘ -1;)-"*"):1§)' , .-. ‘ no ‘ - 1 -nl - E2 = (5) ‘1T§"(3) 2‘ 2 Theretotc. am. < o “"1 = { :1/2). ) :2 o (11) Given that z[n] - 5|’: -1]. we see that y1n1= i = [I=19ln — 2r=1- 91» - 21= «1»- — 21- nun — 61 II—$ (b) (hwn that [n 617: — 2]. we see that utv-1= f :1-s1s1n — 21:1 = 91» — 41= u1- — 41- 1-1n - s1 A-—no (c) The input to the system in part (12) is the same as the input in part (.11 1.11.1--u In 1 to the righc. 115 is time inn: -{ant than the syukm output obtained in 1:: -:1 111- 1.. s.- ti: thr he the name as the system output obtained in part la) shifted by I 1-1 1111' right <:1e;11-y, this is not the ram. Thexeforc, the synu-in is not LT1. (d) If z[n[ - uln], lbnn an 2 z1I=1y1n - 21:1 than = )5.. . _ 2;] la: The signs! gln — 2k] is piulled 60: k = 0.1.2 in Figure 52.7. H-um this figurr :1 Is clear 1: - 0. I that 1 yfnl - { 2: n > 1 II 3/1"] ~— 2u[n] —d[n] - up. . 11 0. ntherise Therdore. A= ¢-5. HI! -4. 2.10. From the given inlonnninn, we nan shtdt z(l) Ind Mt) u shown in Figure $2.10. (I) With the aid at the plots in Figure S210. we un about {Int y(! ) I 2(1) - Mt) in nu shown in Figure $2.10. 1 1- 1.01 I I I I he) -1* o 1 * o W o -t. 1 In t Figaro 52.10 Thu-cfare, 1, 0 S I 5 a __ on. a 5 t 5 1 ‘M’ 1+0-L. 1g: ~_: (1+a1 0. otherwise (1-1) F-om the plot ofy(l). i1.is clear out fig! ) has aasmunuim at 0,1», 1. and 1 + a. 11 we want 5%“ m have only that dinoontinuities, than we need to ensnrr 11m 0 = 1. 2.11. (1) I-Yom the given inlonmtion. we see that M! ) isnnn aeroonly far 0 S I '5 no. Themfolc. y(l) = 11:) . M1) = /_:1.(1): (¢ - r)dr . /0”: -"(u(n r— 1 — 31- u(t — 1 — 5))dr Wtzcaucuilythowthnt(u(t—1-3)-u(l-7-5))lsnonumonlyinthtrlngc (2 5)<1<(t—:1). Tberefore, £or¢5:. t1nabonintegra1eva1uauswzem. For 3<t55.t11cnbav: iutqg-Ali: 1-: _ -:4:-3) y(t) = j e"'d7 = I E n 3 For 1 ) 5. me inugnl is :4 _ -I —: u— 1 11(1) = I c"'d-r = (-—————] c k 3 1-: 3 2.8. 2.9. 2.11. 2.13. 1 1- .1. 11-0 “I41 I I I I I I I Iifivd 5131 " :34! R 45;; V1 FlglIleS2.'l' Uling the convolution tnngnl. =10 - 1-1:) = f” zmnn - rm _ / "° 1.1.111: - 7')d1', Given that 11(1) 4 J(t + 2) + 25(I + 1). the above integral reduces to =1!) - v(¢) - z(¢ + 2) + 2z(1+ 1) ‘Hit signals z(t + 2) and 2:“ + 1) are plotlad in Figure 52.6. , _ 1. J‘ 1&1-L) I 11%? !) -1. -1 o *_ .1 A 1 * FixureS2.l Using these plots. we can unity show that 1+3, —2<tS—l __ 2+4, 1<t50 W)" 2-2:, o<zg1 0. otherwise Using the given definition lot the signal Mt). we may write h n‘7". 7 >5 11(7): : u(--r+4)+¢-". .(r_5) = 8', 7 < 4 0, I < r z 5 Thaefintt, e". r -: -5 h(—1)= e"', 1 > -4 0» *5 ( 1 < -4 Xfwe now Ihift the signal It(—1') by t to the right, then the resulum iaignll h(z — . ~) will In- ¢'""". 7 <1-5 M1-7)= em’). 7>! —4 0. ft—s)<r<(t—4) Therdor¢. tIelatltalI. l|i: unvaInt'| nnnqybeupIusedu 0. —aa<t53 . -in-n 3<lS5 1(1) I 5-"1-. "_, .a)| ,_-u. .n1' 5 ( I S on (b) By 3(1) with respect to time we get ‘‘--: ‘7‘1 —_a(1—a) -a(: —s) Thaefan. 5(1): ‘% - 11(1) = e-“'-"nu — 3) — e-V‘-51.4; — 5) (c) 1-hm the «suit olput (a). we may compute the . -1m-ivuive of y(l) to be 0. —oo < t 5 3 %= ¢"("". 3<l$5 (e. " - 1)e"""’. 5 <1: on This 1: uutly equn In gm. Therefore, y(l) = 549. The signal y(t) may be written in ya) : ---+z"'“*u(z+c) + c‘“'°’u(t+. '1)+c"u(t) +r<"‘). .(1 — 31+ e *"‘1uu 1;) . 1n therange0sc<3,-ozmny wriu. -y(l).1s v(t) = » -+¢'"“’u(1+e)+e*“‘”u(: +3)+¢"u(: ) _ Z-1 +£41.21 ‘ ‘-12.51 + = e“(1+¢’+¢"+~) _ —x 1 P E :5 ThereIorc. A=, —_£q. (n) We require that (§)nu[n] — A (%)(n>| )u[n — 11- 5|n| Puttingn—'1u1dloIvin5foI'AglvaA- (b) Rom part (a). weknow that 1-1»: -- gm — 11 —~ s1n1 I-1n1- (av-1 - gun - 111 : 61v-1 Prom lb! definition clan hwu-sc xyuuzm, we my 35.; um ch) = a1~1— 31- ~ 11 36
  10. 10. 2114 (u) t'e&rstdetern: ineifh. (t)inLhooluulyintg-nhleulollo-I _”| h;(r)| dr = If e"d1 = 1 Therefore. bin) is the ixnyulte response 04’ A stable LT] system. (b) We dtternuno if hq(t) is absolutely intqnbk nu follow: 1n, (y)1.41. / re-'| u(2¢)| uv This intcgnl is dull, Bnluewnlund beuuac e"| axt(2l)1 is in exponentially deaiying futtdion in the nnge O 5 I 5 no. Therefore. him is the impulse raponsr of n xuhle LT! system. . ‘ 15. (n) We determine if h. [n] is Ibnoluhaly ntmmnhlc :5 [allows E Imtkll = §k1cos(§k)I k: -on no This sutn does not have n finite value humane the funetion k| mt(: k), Int-tunvs as the value of It inc: -cue Thetefiote, lulu] cannot be the impulse response --I’ A «able LT1 system (b) We determine if h, [n) is Absolutely tunumhle I: follow: fj | h2lk}l= £3 = !‘=1"/2 k-—m A--an Therefore. hfinl is the impulse response of a table 1.1‘! system 2 la. (.1) ‘Rue, This may be easily ugued by noting that mnvolulion may hr )v'htv'¢l x. nhr pm-es oi carrying out the superposition of; number ofeclm of Mn’ ‘ - Iu - ‘ah echo will occur Al the loution of the firlt Mn ten: sample 0! xiu}. In vI| ;. - -- iw fizxt echo will occur at N. . The echo or / t[n] vhid: occurs It u = N, WI” um. il. ‘- lust nun wn: sample at the time location N; + N1. Tltcreforc, for nll v. |m-s of H which urn lt': W‘r that N, » N3, the output y| n] is new (it) like Consider vi») = =['Il-hiv-I e 2 : [k]h| n - 1;} In-In From this. on yln — x] — 2 x[k]I| |n - 1 - t] has a : [n] with -1] This shows that the ‘Min statement 3 tube. 37 This gm: 1 (-1 +3j)K-MK: 1. : K= “T3 Thetefonw. e("‘*‘l", t> 0 Mt) — 3“ +1,’ in order to determine the homogeneous nolution. we hypothesize thu viii) - Ac“ Sine: the homogeneous solution has to ratify the following tliifutcntinl Pqlllhhon 4!/ AU) dz + (“(1) = 0, wt obtun Au" + Me" - Ae"(i 4 4) - n This u1lP1I¢5 lhll J ‘-4 for Ito’ A. Tlw overall mlutian to the diflt. -tcntul rqunlinn now beouma ___ -4: l (-1031): y(t) A4: + 3————-U *j)e , (> 0 Now in attic: to detettmne the constant A. wt: use the fact that the systcm satisfies lhr cottdition or initinl rest. Given thn y(0) = 0, we may conclude that 1 -I A - - = ‘stun °' ‘ 30+)? Therefore {at I : - 0. , L _ -4: («+3 ): y(t) 3(l+J)le +5 ’]. t>0 Sinrr (hr system satufia the condition of initiul mt. y(t] = 0 for l < 0 lltrr--Inm- y(l) " [—a'“ + e('‘’’’)‘] um (b) The output will now he the real part olthe Ln: -er obtained in put (I). y(! ) : 2 [: "cos3t + e"nn3t - r‘"] um. (e) 1!-tn. Calida- vm - 2(1) - Mo = /°°z<v)r-u - vldv. hom thin. 1/(-1) = Fz(v)h(—t—1)dv = lwz( —1')h(-t 4 r)dr : x(-I) 0 It(—t) This show: that the given ntumient is true. ('1) ‘D-ue. This may be ntgucd by nnnnida-Eng Vi‘) ‘ 3(1) t h(t) : /wz(r)h(l v r)d1. '5 F501" 3115- W PM 3i’) AM Ml — 7) under the assumptions that (1; 2(1) = 0 5" ‘ > T1 lid (3) M‘) = 0 for I > T1, Clnnrly. the pmdurt r(v)h(! ~ 1) is zero if 1(7) lab! -1:) 11 ‘Y -t. -‘r, _ -( rsgun 52.10 I-1‘-; >1‘i, 'l‘he: e$'on. v(t)= oforz>1‘. +T, . 2.11. (n) we thnt 1/(t) is the sum of the pnrtiuulu And homogeneous aolutionx to the given diflerential equation. We lint determine the pu-ticulu solution y, (t) by using the method specified in Entnple 2.14, Since In an given that the input i. ~ -: (t) = ¢""-""’u(t) for t > 0, we hypothesize nut to: I > 0 W“) = Kz(-I051)! Substituting for 2(1) and um in the given difiaentinl oqunbon. (-1.. . 3j)1(¢(-H11)! + ‘K£(-I031)! _, ((-1.3;): 3-1°~5=n= =rhur-antac--r. I.ypi; -otau<n. tao-. um i §ytoz+= m=o+: =» v[2] — %y{l]+z[2]-%4-0: rm : }v|21+z1a1=$+o. AI- L 16 ulml = (%)"" Therefore. vlnl = (£)"'u[n— 1] 119- (I) Consider the difiertnot nu-Hon reluina y[n] And u/ {I1} re: 5,. vi"! = av! " - 11+ Min] Hvtn this we tnny write and W585‘-ifll ‘hf Vrcvioui equation by I /2 and subtracting from the our luelnrr, we obtain I I 1 wlv-1- 5-vln -11- an-1 - gun -11- W1» - 1; . 225,1“ . 2; Substituting this in the diflexence equation reluing u. -[n] and x[u] for 5;. I n | Elli"! ‘ Elli“ ‘ ll " fill! !! — I] + ; Ey[n — 2] = zfn] Thu is. vi») = (a + %)vln ~» 1) — 5;-yln - 2] + 134"] Computing with the given equation mining yin] and x[n]. we obtain a-%. 11:]
  11. 11. (h)Tbrdiflaencreqinliourelningtheinyutandantpuldflielyiltm-15iIIId5:Ir¢ . .,i. .i= ;.. .(. .—ii+z[. .i and yin1=§vIn—xJ+wIv-J fioin Lilac. we can we the method specifed in Example 2.l5 In about that the impulse raponsc: of S, Ind 53 are I I I-ilnl —- -lnl "Il"l = (§)"«-iv-i. respectively. The overall impulse rsponseof the system made upol u-mud» 4.! $. md S, will be and Mn] = ;. ,[n]. n,[n]= iIi. [k]Ii, [i. —k] g. ..-in : §¢§>*<§i"*-«In - ‘=1 [:0 ; fi<§i'<§)~* = i: (%)""‘" I30 I10 = mg)" Q)": -Ir-I 320 (I) / W ug(t)rni(l)dl = /M sum = i . .¢ . . (la) 3 / ain(2x()6(l + :04! = sinfliw) = o 0 (c) In order to evaluate the integral /5 u. (l - 7) ixn(2rr)dr, 4 consider the uigul 2(1) - cou(2xt)[u(£ + 5) — u(¢ — 5)]. We know tha& = u. (l) 9:0) I [. u1(t — 1): (r)d1 4 —oo : [’u. (l - f)ea(21rr)dr 3 ll o I to :0 n Figure $2.21 2.22. (a) The ai-. -iin-d rnnvnlullon is :40) = /wx(1)lI(l—-r)dr . fr°'e-’<"')4i», 220 ii Thrr: ,m= { ow le"‘u(£) a = [7 (h) Thr rlr-sired convolution is y(l) = /-“z(r)h(£ "nay 3°" 5 =2 foh(l—r)dr-A h(t—r)dr This may be written as 2 5 / .= <--rm- / ¢"""d1. .51 H I 7 9 f c"""di- —‘/ ' z"""'a‘1, 1 5 I 5 3 V“) — i—i 2 5 -I e? ’-‘-Wv. 35156 (-l o. 6 <1 Therefore. (i/2)[c" — 299") + :74‘-‘>1, i 5 1 W) U/ me: , , m—si _ ggu-2)]. is , 5 3 (l/2)| e7“’°’ -:7). 35156 0. 6 < 1 43 am 7''“ = ‘/_: u.(I — r)oon(2in)d-r which is the duind inupul. We not evtluue the value ofthc integrnl ma #1:“ z xii-i(2n)i, _, = o 2.21. (I) Thedesirud wuvolution la vlnl z| n| 0 Mn] ; )3 z[k]I| [n — 1:) II-H fi": (o/3)‘ for 7| 2 0 1-0 oi_ no : (%1-I-Irnrz-its ii (b) Rom (a). vial : 0" [:21] u[n] = (vi + 1)a"u[nj . .n (C) H): n 5 6, vlv-I = 4" {f: (—§i* - ivy‘ ‘ } 1:0 [:0 For 7| > 6. an 7|-I v1nx- 4" {z¢—§i- — £<—§i‘}. I50 t-I7 Tharelonr. , (8/9)(-I/ B)‘4". so “"1 {(5/w<—1/2)". :>s (d) The desired convolution ll! vlnl - z2[k)A[ii-Is) I: --an ’ l[0]lI[n] + zfl]II[n — ll + x[2]h[n - 3] + x[3]II[n — II! I 2-[l]h[n — 4] = hInl+h[vI~l}+h. [n-2]+h[n-3]+h[n—I]. This is as shown in Fugue S2.2l. (2 (C) 'I'bedaindmuIdulion'u )/ (I) = /¢z(1)I| (¢-1)dr -on —' [°, Iill(l1)’I(l - r)d1. Th": givu us 0. K < 1 W)_ (2/I)[l —m{. (x , l)]], l< : < 3 (2/: )[m(¢(r ~ 3)) — 1]. 3 < i -: 5 0, 5 < I (U) I-=1 Ml) = h, (2) - : —l‘6(l A 2), where M 4/3. 0 5 15 l mu) ' { 0. olherwisc Now. ya) = nu) - zm = Ir-mi - mi] - gm — 2) We have I I-irx) -x(t) — / ‘_‘§(n7 +b)n'1 = g[; al7 A %11(! ~ ii’ +0: — an I! Thanforc. y(z)-5114:’-1a(: -i)’ +M- b(t— 1)] ~ 1[a(¢ _ 2) o 1:‘ — rt +b — 'll-‘ 3 2 2 a “‘ ‘ (e) x(t) periodic implim y(l) perindlc , .decermirie 1 period only. We have I-§(l-1—l)dr+fl(l—1+r)dr: l+¢-t"_ —. “_-:14‘: im — '3' ; * [(1 lO7)d1'+[§(I-l-7)d1-17-3!+7/4. ; <I<.2 I-l Tbeperiod oly(l) I52. 2.13. y(() is sketched in Fixure 52,23 [or che diflcnanl vnluts of T 2.24. (I) We nre given llul hg[n] = 5[n] + 6[n - 114 Thexefalc. h; [n] - lulu] in 6|n] 4 2A[n — 1} + 6[n - 2] 44
  12. 12. l-‘igun $2.23 Since bl») = I-ilnl - lhzlnl ' Mlnll» we gen M0] = ri, [n] + 2Ii. ]n - I] + Iiiln - 2]. Therefore. up] = ).. [0} = A. [o] = 1, h[l] = ii. [i]+2Ii. [o} as hil1)= 3. M2) . — M21.» 2h. [l] + me] = I-. |21- 3. M3] - ii. [:1+ zii. |z] -4- mm -2 M3] = 2. P114] = me] + zIi. [:] + mm = > M4! = 1 M5} = ms) + 2n. [4] + 5.43] = » ms] - 0. lulu]-_0forn<0andn25. (b) ln this case. gin] = z[n] 0 Mn] = Mn] — II[n - l]. z[n] = (am. 2.25. (1) We may write z[i-i] u (h) NGI. vlnl = zalnl - inlnl = ml’-l — ml" -11- Thndnre. 2(i - (lI2)'"") +2(i - (i/2)“) - U/2)"*’. n 2 --2 yin! ., { L n = -3 . 0. other-due Therefore. flu] = (1/2)'"'u{n + 3!- (e) WI. ‘ rim y; [n] = :>, ,|n| v : :;[n] = u[n + 3] — uln + 2] = i[n -9- 3]. (a) mm the iauie ol pm (c), w. get yln] = “[91]: :; [n] = ziln + 3] = (lI1)''‘"u[n 1- 3]. 2 21. The proof is u lallovu. A, - ]°'yt-)4: . j_: f_: ,m. (.-, ,.. ,;. = /_ 12(1) X : h(l - 1)d: dr - = (-r)A; .dr . . - A. A. . . 2.25. (A) Caiusal because up. ) - o far It < o. suiin bcunae 2(g)- = 5/4 < on. 530 D (b)Notaux<a| bei; au| eh[n]; £0l’nrn<0.SublehecAIne z(0.B)"= .'><oo II (c) Anti-cuinl beuiiac Mn] = 0 la: n > 0. Unable because 2 (1/1)" -‘ 0° 3 (El) Notcnunlhecu. L1l. 'II[n]#0lotn<0. Stohlebuaiue 2 5-=9‘? <00 an--m (e) Causal heciiuae h[n] = 0 {or n ( 0. Untnble because the leeward Ienn becorru-5 infinite )I II -0 X. . . (r) Not uuui heame Mn] is o {or II < :1. Stable beuue xvii: -11 - 305/: < an -—an 47 No-. the duiud convolution B vlnl - Mn] ‘ rlnl -—I oo - E (1/3)"(i/4)“"i4n - k + 314» £(l/ :)‘(i/4)"-‘min — k + 3] I30 his-an = (i/ i2)§j(i/3)*(i/4)'*"uin + E + 41+ i(i/3)*(i/4)“-*. .|n - A: + a] I-0 Inn By consider each suninnlon ‘in the above equation nepantdy. we may show thn 0.2‘/ ll)3'. n < -4 y[n] = (1/m4‘, n = -4 . (i/4)-(i/ ii) + -an/4)" + 3(256)(l/3)‘. n 2 -: i (b) Now consider the mrnvolutinn vilnl = [(1/3)”! -[fill - ll‘/4)"ul'I + 311- We any show um "("1 = { ‘: .3(1/4)‘ +3(256)(l/3)". :3 ‘ Also. consider the convolution lfll"l = [(3)"ul-n - ll] 0 l(1/4)"vl'I + 31]- We may show chm. 2‘ I 3'. -I "‘l"l = { :1/4;'l‘(ll/ ll). -3 ' Clearly. ml") + yg| n| = y[n] obtained in the previous pun. 1.26. (A) We hove vulv-l= =ilv-1-nlnl = 2 = il'= l=: |n~kl - §(o.5)*u[n + 3 - k]. ks! ) Tri'n imlunus to Hi — (l/2)"“), n 3 -3 0. yi[n] = 2.1"] . z,i. .i . { M be (dcoualheesinehlnl-0!otn<0.Suh| ehuzu: a i | h|'n| |i: I<rxi. 239- (a) causal bouuu I-(1) = 0 {or r < 0. Stable became 1Ii(¢)| iu = a-'/4 < co. (h) Notcunolheume II(()fl0lorl<0. Unllilbkbtuulc/ mIlI(1)]: oo. (c) Not nun] belauu II(l) )4 0 lot I < 0. A Stable because ‘| h(l)| dt - e‘°°/2 < on. —eo (4) Not uunl beam: A(r) ; £ 0 Sex I < 0. sum. hem. .. | ii(i)| .u - .4]: < m. (e)Notan-. |heuii: eh(t)g¢0£nrt<0.Stub| ebeunuseF| h(r)| d¢= l/3<m. (f)Cuiulbecu. iIeh(t)=0lott<0.Sublebecousef. -|h(t)| d1-l<ao. (g) emu because no) :0 cm i < 0. Unable been-iiae V/ ”|h(2)| dl = m. 2.30. We need to find the output of the system when the Input ‘L: z[n] = 5[n| . Since we are asked Iooslulnnlnitinlrut. wInuyennclndeth. nty[n]=0forn<0 Na-we vlnl = 2|”) - 21/! " -1]- Tbanfore. 340] = =l°l - 2yl-1l= 1. And so on. In closed form, vlil I‘ Ill] - M01 ‘ -'3. vi? ! = II? ! + 2v[2l = -4 lrlfll = (-1)'“l"]- This is the inlpulne 199090 of the nyuem. 2.81. liiititl res: implies thu y[n| - 0 for n < -1 Now y[n] = z[n] + ‘Bin - 2] — 2y[n — l]. Tbudwfi vl-3| = 1- vl-1] = 0‘ v[°l = 5~ vl4l - ~'»6.vl51- -|10- vlnl = -1l0(-21"" tom 2 5. 2.22. (. ) ii y. [n] = A(l/2)". then In need to verify 4;)“-; A(; >"'= «»i Clearly this la true. 43 1
  13. 13. (b)WenouvnquiiethatlonI30 1" 1 t "“ 1 " 5(5) 'z"(i) “(s) - Therefore. 3 : -2. (C) Hum ff}. (P232-1). we kiwi! thu yfo] = z[0]+ (I/2)y[—l] I x[0] : 1 NOW W0 also haw- y[0]: 'A+B no A= l-B= J. 2 33. (e) (x) Plum Example 1.14, we know that , ,.m = E. “ -- §e"'] u(t). (ii) We solve this along the lint: o! Example 2.14. Fin! assume thnt ; ,,, n: .3 1 the ronu Kc" for I > 0. Then ualng eq. (P233-1), we get for 1 > 0 2Ke"+2Ke""= e" = > K= ‘-: We now know that y, (t) = fa“ fort ) 0 We may hypothrsirt the liumut', u-nmus solution to be of the form ml! ) = A¢"‘- The-rrfore. y; (t) = A6“ + 413', (or t > 0 Assuming initial rust. we can conclude that y; (l) = 0 for t 5 0. Tht‘P¢": Ir. - 1 yg(0)—D—A-+3 = 4-7 Thrn. 4 (in) LN. the input be : ,(t) : o¢‘'u(t) +fle"u(t), Axsume that the particular solution y, (l) IS of the form mm = [—£e" +1e“2‘]u(t). y, (t) -= Kine‘ + K1133‘ for: :- 0 Using cq (P133-I). weget 3x. a¢’* + 2K, ,1e" + 2mm" + 2k, pJ = a" + 58' Bquatin; the ooelficients of e" And e" an both sides. we get in»! and K, 5 (9 Wznovhn>ulhnty, (l)I§J"'nIIt>T4W¢| ll1lI7P°‘h“h“5‘h°'%'-“ neousnolntinnloheofthehrni mm - Ag-7'. Therefore. mm = Ar" +154‘-”, an 1 > 1. Anuming initial test. we can conclude that y; (t) - 0 for t 5 T. Therelmr. mm = D= Ae'”+l‘—( = > A- —; e". Then. Wu) - [— fi—(e‘7("T) + §e’("n] u(t — T). Cllfl-1'1)’. ml! ) '— ml! - T)- (1.i)Considcr the Input-output pair ha) -0 mt) where em) = 0 fort < to Not» that §y‘+u—) + Zy; (I) = x. (t). y. (t) = - 0, for t -1 to. Since thv drrivative is a timu. -invariant operation. we may now unite that - T) 41 THLN yuggatu that if the input ‘at I sign] oftbc [nan 23(1) = -. inf! - 7‘), than the output 5 a signal of the form ma) = y, (t - 7‘). Alan. note that the new nutput y; (t) will be new for t < 39 +7‘. This supports timbinvuianw since 23!! ) it zero for I -: to 4 7‘. Therefore. we may conclude that the ayitcm 'L< tlme-invuumt. +2y: (t—T)-e. (:—T). y, (¢)= o. forl<la. 2 34. (7.) comm nu) 3» mm and nu) -5. mm. We know um mm : mu) . -« 1 New mnsidz-r is third input to the system which is; -;(t) = 2. (1)4-22(1)‘ Let the corresponding nntpul be y(: ,’ Now, note thnt ygu) - 1 ,1 mm + nu) Therefore, the . ~y. -u-m LI not lunw. A tpccifll example follmvs. Ccmidct an input sign! z; (t) = ¢"u(1). Rom Problem 2.33(e-ii). we ltnaw that the oonponding output for t ) 0 is y. (t) - ée" 4 A67‘. Using the fact that y; (l) -2 I. wage! for t > 0 1 . ,, .(t) 3 E93‘ + (1 - 42"“ " Noun Lurwldfl A second signs! :': (I) - 0. Then. the oometpondmg output r y, (t) . . BF“ 51 New hypothu? -?n5 thu, nu) - Ag-"_ E F; y; (l) = gm” + + AK" for t > 0. Assuming initial rut, vg(0)=0-A+a/54-fl/4 2 = e(-+-). Therefote. mm = {§o¢" + £5?‘ — 9- + 2) 6”} u(! ) 5 4 Cl-w|1.v: (t)= will) + fiIr. ~(l)- (iv) For the input-output pnir 2.(t) and y; (t), we may use eq. (P2 33-1) and th tut condition to write 5% +2mm = xx(t)- mm -otou < : . For the input-output psi: z; (t) nnd y; ((). we may uu. ‘ L-q (P2.3:H) and the rest condition to write 554:3 v 2;/7(1) s 22(1), mu) = 0 far 1 < :7 (S2 Scaling eq. ($2.33-I) by a and eq. ($2.33-2) by 13 and summing. Inc gct ; d;l0vI(1) + iim(t)) + Ila: /.(I) + sum; = m: ,(: ) + mm). and mu) +m(t) no for t < inm(t. ,:, ). By inspection. it is clu: that the output 15 y, (t) = ag; (t) + HWU, ‘ when the is nu) = a2.(t) +Bx¢(t). Hirthennore, mm = 0 for t < 1;, where L‘ denot time wall] which :30) - 0. (h) (i) Using the result of (arii), we mly writ»: (ii) We nalvt: this Along the this of Example 2.14. Pint nasume that y, ,(t) Ls 4: {arm KYe’“‘" {or t > 1'. Then uningnq. (P2.3H), we get to: t > T 2Ke""T'+2Ke7"‘n -8‘ -» K l'orl>0 Uuia¢thnEa¢they, (x).1_. .,. ¢[w¢)o n(1)= e"""’ Now consider A thin! signal an) = nu) +z3(t) = :10). Note that the output I- -till be mm - mm to: t > o. Cleuly, ,, ,(t) # mm 4- mm for c : - 0. Therefore. 2 system is not liner. (b)Ax-in consider In input signal 2.0) : e"u(: ). a-om put (.1). we know that ll corresponding output to: t > 0 with mm: 1 h _ 1 2 _ f ~2(l—l) m(! )—‘¢ +0 ‘)¢ Now. consider an input signal of the form 23(1) = z. (l — T) = em‘ '7u(t — T) The for I > 7‘, mm — £3“-"1 + 45"‘ Usingthefact t. haty; (1): lendaleoauu. mingthit7‘<1,. wegnt far I : -'1' W“) _ leztuf: + (, _ 1¢2(i—1-y)‘. -21:». .. I I Now note that y; (t) ; £ m(t - T) for 8 > 1'. Therefore. the system is not time mvnrlnnt (C) In order to show that the system I! inaetnentally lint-M with the auxilinry rondxtiu: For an input-output put mm and mm, we may use eq. (P2 33-1) uni the face Lhl! mu) = O to write % + mu) - mm. mm = o (s2.34—n For an input-output pair 21(1) And mu). In: any use eq. (P233-I) and the initial rat condition to write 13%! + 2y3(l) = - 11(1). y; (l) n D. ($2.34-2) Scaling eq. (52.34-I) by a uni eq. ($2.34-2) by 5 And summing. we get ; ;la: /ill) + fl1n(! ))+ 2(«-mm + fiv2(I)) = mm + lkzlll nnd v: (1) = ml! ) +v2(| ) -= 0- By iupedion, it is clan that the output is y; (l) = a-y. (l) 4* flmlt) when the input is 13(3) = ¢fl| (') +3Z1(t). Fhrthtrmom, 15(1) = 0 r y; (l} 4 y; (l). Thetelurr. the‘ wystetn is linear. '1‘herel'ote. the overall system may be trmuxl as the cneade ofa llllI‘Af system with an adder wind: adds the (spouse of the system to the eunihaxy onndltiom flout‘ 52
  14. 14. 2 35. 2.38. 2.39. 2.50. (d)ln thepeeriousperuwesho-edthettheuyetaniIlineuwheny(l) =0. lnorderto show that thexyeta: isnottinn-inva. rient. a:nIide‘uninput'ot‘the £or1n: .(t) = r"u(t). Hum part (8). we know that the eurrupending rrutpret will he y1(t) = §e*'+Ae“"_ Using the fact that y, (l) I 0, we get for t > 0 mm __, %¢ZI _ §‘»ru-z)_ Now mnlidcr an input of the torn: 22(1) - zi(! -1/2). Note that m(l) : 0. Clruly, will :4 vifl -1/2) 7 (I/4)(¢ -¢’l~ 'I'hervf0N- val! ) 7‘ v1(l-1/3) for -3131 Thi! implies that the intern in not tirne invariant. (r-, A proof which is very Iiniilnr to the proof for linearity used in part (c) may ho used hrrr We may show that the system u not time innnent by using the method outhnnl in part (d). (11) Smru the system 1: linear. the response y; (l) = 0 for all: (h) Now let u« find the output y, (l) when the input is xq(l). The particular solutinn is of tho form y, (t) = Y. t> -l. Siilittizuting in eq. (P2334), we get 2Y s. l. . '. m», including the homogeneous solution which it of the form yiltl ' Ar‘7‘. we get the uveull mlution. 1 mm = Ar? ‘ + -. t> -1 2 Suite y(O) = 0. we get W11) ; —ée 7' + 1> -1. (s2.Js- n For I < -1. we note thet rill] = 0. Thus the puticular iolution is zero In this nnge 331." mm : Be"'. l< —l. (S2 35-2) Slnct‘ the turn pines of the solution (or y, (t) in cqs ($2.35-I) and (52 J52) must match Al 1 e I, w can dnai-nine B from the equation é - gt’ - Be’ lA'lllCll yield: — 2-:3) Em"). t< -1. 53 21(1) - 0, for All I :10) = c‘[u(t) - u(t - 1)]. Since the eyetetn is linear. the ruponse y1(t) = 0 for all t. Now let us End the output y, (l) when the input is 19(1) The particular solution is of thr fnnn y, (1) = Ye‘. 0 < r < 1. Substituting in eq. (P1334), we get 3)’ V I. Now, including the homogeneous Iolution which in of the term y1.(r) — Ae"". we gel thi- overall solution- n(t)»-Ae‘2‘+-gt‘. D<l<l. Aswmirnr. linnl mi. we have v(l) = 0. Um; this we get A = -8/3 Tim-rm mu) = —§e"‘*’ + go‘. o < I <1 ($2.37-1) For 1 < O. we note that z7(t) 2 0. Thus the pu-ticulu Iolulion 'u zero in this range And “(:1 — Br“. t< o. (52 37-2) Since the two piece! at the solution for ma) in eqs. (52.37-1) and ($2.37-'2) must matrli at 1 —- 0. we our determine 8 from the equation 1 1, 3-5 =3 which yields mu) ~ - gel) 12"‘. t < 0. Now note that since ¢. (1) a 21(1) inc 1 < 0. it must he tnte that (or a causal «ystan mm = y; (l) (mt < 0 However. the result: of Oblatned above llllflll that this it nor mu-. Therefore. the system is not causal. The block diagrams ere as shown in Figure 52.38. The block diagrams are As shown in Figure 52.39. (a) Nule that K I—3 ’ 11(1) = ] e‘("”: (-r — 2)dr = [ (‘"7" ’r(r')dr‘ -00 -00 Therefore. 11(1) — e‘“'"u(t — 2). 55 Na-notethauinntx1(t)= x-; (t)I’crt<-l. itn1urthetruethat[orucan. i.el: y:te-nu y| (K)ly[(l)[0I'! < -l. Bowwverthruu. lt| ufparh(a)Ind(h)eht1wthetthi: rircnot tnie. Thaeinre. theI; m1m: inutt-atul. Z 36. (I) Considei an input zilnl end: that z1[n] = 0 for n < II]. The an-nsponding output will be lIIl"l"%lI1l" ‘ 1l+ f1l“l. i1x["l= 9 5°’ " < "1 (S'L30—l) Also. connder another input xaln] ouch that : r3[n] = 0 for n < 113. The mrruipundrng output will be inlv-1" gm ~11+z.1«-1. v.1«-1- 12 an n < "z— (52 36-2) Sr-Itllnp_ rq. (S2.. 'l&l) by o and en. (52.36-2) 5)’ 5 and summing. we get avxlnl + Mn: 2 $111»: — x1+ ‘; 'n1n — 11+ an1n1+ 9221:4- By inspection, it ieclur that the eutput la y, [n = a-y, [n] + flyglnf when the Plpu! ls z; [n] - n: ,[n| -o / fr-; [n] Hrrthemlon. yell) = 0 2 y1(l)+yi(l). Therefore. thr system is linear (h) Let us consider two inputs . n[n] = 0. for all vi, rind “W S 0. n < -l l. 1:2 —I Since the system is lillL‘Al', the response to z, [n] is y1[n} : O for all n Now 1»! -. < find tho output m[n| when the input is x; [n). Since m[()] r U. m[ll= (l/2)0+0=0. yz[2)= (1/2)D+0=ll. ’I'herd'nre: , “[11] -Ofnrnzo Nnw, l‘arn<0, notethat 1nl°l = U/1)v2l-1l+ =19]- Therefore. ml-ll = -2. Proceeding nitnilarly, we get y, [»2} = -4, y; [—. 'SI = ~8. and an an. Therefore. v; [n| I: -(1/2)"u[—n -1]. Nail mu thlt xiria : ,[n] = 2,[n] for n < 0. it must be true mu for . causal system y, [n] : y, [n] for n < 0 Hawenr. the results obtained above nliow that tlm. is not trut- Therelore. the Iyrtem it not cauul. ’-(Kl {(A1 in) D G B it“) o yen Plgure 52.39 (b) We have y(l) 2 mh(1): (l — -r)d1 . '/ “e'("2l[u(l -7 + 1) -u(t-1 - 2) I M1) and 2(r -I 7) Are u shown in the figure helow. Using this figure, we niqy write 0. t < l ‘I '('-T) -ll-l) v“)_ 2 e d~r= l-e , I<r<4 41 e'‘''7)dr - e ""’[l — (‘L 1:- 4 I’! 2.41. (e) We may write nlnl
  15. 15. Mr) 114-1) 5 3, g 4-1. 0 1+1 1‘ Flgun 32.40 (b) Note that g[n] - z[n] - (6[n] — aJ[n - 1|). Therefore. from part (a). we know tlm : [n] - (6|n] — a6(n - ll) - 6[n]. Using this Ire my write zlnl - 161» -11- oéln - 21) - (Sin -1). z[r1] - (5[n + 1| - a6[n]) : 5[n 1 1]. z[n] - (6[n + 21- o5[n + 1]) = 1S[n + 2) Now note that 1 21711. I| [n] — “[11 + 21+ 26111 + 1] + 6|») + §6[n ~ 11. Th¢re(o1'¢_ 1:[v1] 0 Mn) 4:(n] 0 (5|n + 21- o6[n -1- 1]} 2z[n] - {6[n + 11 — a5[n]) 2|") ' (‘I'll — 05)" - 1)) (1/2)z[n] 0 (Ski — I] - u6[n - 2]) 01+-ll This may be Irltten N5 x[n] o A[1-1] = 411] - (“In + 1] ~ 406171 +1]-1» 1611-1 +1] 2a1l[n| + J[n| — od[n -1] 4' (l/2)‘('' - ll ' (1/WW" '7} Tliexelore. 111111 a 46[n + 21-+12 - 4a)61n +11+(1- 2n)J[n] + (1/2 — 1.151» — 11- (1/2151» — 21 2.42. We hnvc ‘ y(t)= z(()ah(l)= _/0 e*~("')dr. -0.5 57 2 44 (I) We hm! Th 2(1) - M1) = ./ _m: (r)I1[t — r)d7 - If : (r)h(¢ — r)dr. Note that h(-7) = 0631171) Ta. 'l'lienIore. h(1—1)n oh: -r > H-T7u1d1 < -1'; +1. Th2refcn. tl1e| bovIint$IIeVI| u.IlaInxq'ocilheI’ifT1<-73+-loI'7i+t( -7‘) This implia that tluuonvolutiou ham-Al in unit! > | ‘1". +T, |. (b) (i) We have N1 y[n] = 1.1.. ) . 21.11 = 2 h[lt)z[n — 1:1. I-No Nocechxt z| —It} #01211-N; 5 ll 5 —N; . 'l'bucSou. z[-l: +n) ; € 011;: -N; H1 3 It s -N2+fl. Clnrly. theeonvolutlonluminot : erol! —Ng+u 5 N1 and —N2+n2N. .. Therd'om. y[n]isnomem(o1-n5N. +N; andnzN¢+N, (ii) We uneasily show that M, = Mnd-M. - I. (c) h[n] = I) (or 71 > 5 (1!) Hum the figure it is clan that y(l) - 11(1) . :(1) = /:41 — my +_z(t - 6) 1‘hem(ote. -1 y(o) = / z(1)d7 + z(-6). -2 Thu implies that 2(1) mint be known fee I 5 I 5 2 ma for 1 = ~15. 2.15. (a) (1) We haw- ¢(t) - : (t - M 1.1;: v(t) - 11(4- (I) In In ' “(flung limit &< l1 -9 0 on both sidzs ofthe above oquatiorr 2'11) ‘S’ 1/(t) (ii) Diflmminting the convolution inlays), Va get v’(t) - § :1: — rwvidv] — j-‘; %[z(t—1)]h(1)d7 / °z‘(1 — r)h(1')d-r -n z'(t) 0 MI). 1101- e-1-'11 = _§. an(u. /2). (I) Hug — 21. than 11(0) 3 0. (b)$Iy. ourwwawp. n(t)'uno1unique. A| Vuo=1k11.k€In11dIr¢0will &. 2.43. (n)We§n1have 1:11)-I-(1)1-gu) : f" [ z(r)I1(a'~1)g(t--17’)d7do' = rrz(v)h(u)g(t - a — r)d-Ida -B -D l: [_: z(l — a’)I1(r)g(a' — 1-]da'd1 I: °f_: :(a)h(-r)g(l — 7 - a)d1do / ::/ >:z(7)It(a)g(t — a — r)drda Also, z(I) n [I1(t) sg(t)1 ll The equllity is proved. (b) W! fusl lnvc " 1 ‘ 2 1 _ . w[n] - 11[11] . 1.1111: g (-5) — 5[1—(—§) "'1u‘_v1( Now, y[n] = uI[n} - Iufn] — (71 + l)u(n]. (ii) We firs: have 91»-1= 1.1-1 - 1.1:-1 = 2 (—g)‘ 1 ; E(-; ,~ = “(.1 11:0 1.1: Now. vlfll = "("10 9(0) - "I'll ' “I'll = (II + 1)! -(Ill The same result tn: obtained in both puts (i) and (ii). (e) Nate that = ['| l ' (M0) ° “1("l) = (2('| l - kiln! ) - hilr-l~ Also 21011: that : [n) - I1-‘[11] = a"u[n] — a"u[n — 1] -_- flu)‘ Tlwrsfore. = ('I] ' (I1 ('1) ' (Mn) = 6(n| - sinan = sin 811 58 1-3’) I g((’) I pa): 7 '60 Figure 52.1; (iii) M "3 WM the output of 11:. lyltcm V111. ixnpulu response -1.11) M w(l). Tm. 111(1) = 2(1) v «((1) ax : ’(t) and z(1) - 2(1) - 11(1). Sinne bath system in tin cuude IN LTI. we may intudnnge their ordn u shown in Figure S215. Then. W) = 2(1) - 11(2) and 12(2) - 1/(1). Since I“) and p(2) hm to be :11.» tune. we may conclude that z’(l) - I1(! ) = (/ (l) (b) (i) We have already 1>-we-1 that Wt) - = '(2)-11(1). Now w my inlerthangr 2(1) tad MI) in the nxliu proofi Ind they would all lllll hold. Therelore. we may A1-gun um. 1/(1) - 2(1) o l1’(¢), (ii) Consider y(! ) = 12(1) 1 11(1)] - A’(1) — 1(1) - ("(3) - 141(1)) - M! ) = 2(1) I Ml). This shows that [z(t) v u(¢)]h'(t) is equivalent to 2(1) o M1). Now the same thing my be written as: U“) 5 [1’(‘) ‘ "(Q1 ' ("(1) ' [(39) ‘ ux(f)l ‘ 5(9)) ' “(O 1 ' / _ '. "[r)h(l — 1')d1 = z'(l)-| l1(1)-u(t)] 1 = z'(1)-/ Ii(r)dr (= ) Not: thll x‘(l) = M! ) - 5e"‘u(1). Thcntote. the output or the L1‘! system to z'(£) will be h(l) - 5sin(wal). Since this has tn be equal to 1/(I) 2 wn cou(q1t). Irv have W) = woooslu-at) + Ssh-(«-hi)» 60
  16. 16. (4)11) We Inn vi! ) 8(1) ' IUIU) ' “(OI ' M! ) [#0) - vn(¢)l ° 10(1) ' M0] z’(l) - 1(2) : '(1)s(¢ - 1]dr -ea (ii) Also. gm = :(x). a(: ) . 1:(: ).u. (z)]. ..(x) A / 2'(7)u(t-r)d1 (c) In this mu: {(2) = e'u(l) . 6(1). Theretart. y(1) = 1(1) + e‘u(2) A s(l) This may be written is [e-‘‘ - 2:" + 11.41) 1 li(el _ ¢—M) W) 1 — §(¢' - 6?‘) — e‘ — l| u(t). (1') Using the Luci Ihah Mn) — 6[1I — 1]] I u[n] I: £[n] fives. yin] = [z(1I] - x[n - 1]] - .1[n] = Z[z[k] - : l[I: - l]]s[n — H I and z[n] _- [: [n] - x[n — l]] - u[n] = E — :11: — l]]u[n - k]. n--- 2.46. Note that ¥ _— —s¢-"nu — x) +260 — 1) : —3:(z) + mi 7 1). Gxvpn iluu z(t) = 2¢"‘u(4 - 1) —+ yfl) we know mu 8,91 = —3:(¢) -mu -1) must yield -3y(l) +2h(z — n u. up mupnl From the given information. we may conclude thu 2JA(l — 1) = ¢""u(l). Tlmeforv. M2) = ;r"'*"u(z + 1). 51 (3) ran. .. For annplc. nu) - I"u(t). an : (I)-(1 - = “MI) nod ‘/ :1: -e"[d1- ¢+¢"[‘, ',‘° = oe. Although the system in cubic. the Ilep lupaue is not absolutely inlzgnhlc (h)1ue. We may write u[n] - inn — k]. Tbaefore. A-=0 . ;..1 = fun — 1:}. Duo l| ’;[n]=0lorn<U. thénhln]-0lurn<Dmdtheuyttemkuusal. 2 :9. (u) 1: is nhoundcd lnpm. |z[7|" S 1 a B. so: all n (5) Consider vi‘? ! = " zl—I= lhlkl KI*fi = 2 wk): -om Therdara. the oulput )5 not bounded Thus, the sysbcm IS not stahlr and J'V'li‘)'. P sumnnbilizy is ncclmry. (e) Let mi(—: ) . -.- o 0. ‘m '{ iuipz) go Now, |z(t)| 5 l for :11 t. Therefore. 2(1) is nbuundnsd input Now. v(0) -— / ' z(—v)h(r)dv _ W) , ‘ . .., u-ml‘ w - I | nu)| au — on ~- Tlmeforc. the system i: unstable if the impulse rupome is not ubsoluwly integrable. 2.50. (a) The output will be 011(1) + b¢z(t) (h) The oulpul will he ¢. (z — v) 3 ‘7- (I) W) = 3&0} (5) W) = 10(1) - vol! - 3)- (C) v(! ) = volt - 1)- (d) No! eooud: inlu-nation. (9) VI‘) = 10(4)- (U 34(9) = lh'(¢)r The signnk for I“ pan: of {hi problem are plotted in)lhe Figure 52.41. am 1. 1 o 2. t (4) gm 1 0 5 e -1. W n. ‘.. °3. 32.41 2.48. (0) ‘hue. If He) periodic and uonurn. than no [ | h(t)| dl = co. Therefore, ha) is unstable (b) F-lu» For example. inverse of Mn] - E[n — 1:] in g[n] = flu + k] whkh is nonrnmal (6) False. For Example hln] = u[vI] implies that if II-lv-ll = no This is an unstable system. (d) ‘hue. Assuming that Mn] is bounded And nonwo in the range m 3 n 5 n7, 2 n, |h[k}| < oo l= I|, Tlus implies that the system is stable. (e) Pulse. For «sample. h(! ) = :'u(¢) in canal but not xuihle. (f) Pulse. For example. the uaude of a mural Iyuern with 'rmpulsr response Ii, [n) r 6[u — I] mind 3 non-causal system vlth impulse ruyonw hglrll ~ J[n + l] leada m n xyxlem with wunll impulse rupouc given by MA] - lulu] v hg[n} — 6[n]. G2 1.51. (I) Pan-I. hnayItnnafPkInP1.5l(n) Ihetcpoueuouunitimpulnns ml»-1= -<; >-«tr-1. For Lhnymmoll-”:5urePL51(b) theresponaeuun unit. impubeis m[1|] = 0. Chill. min] 5‘ m['| l~ (b) For ihz ayuua cl Pkuu P2.s1(| ) the rwponsu to an unit impulse ii an} - gm»: + 2. For the synem of Figure P2.5l(b) the Isponoe to an uni! impulse is utn1=(; )"-Ir-I + 4. Clearly. !lI[’‘i :4 vzlfliv 2 52. We gcl - Jlnl . = 1.17.] . u[u] = §n("* “"5 " 3 ° 0, vthcrvmc. Noting that I IIOI d d 1- a""’ (k+na*= — a/ '=_ .2“ dog da 1 - n we gel 4,, = , ,I, ,, I [u_1u), - u _au)1u" + #01-+ 1)c"] uln] 2.83. (a) Let us assume Ont N 2:3“: 7 0. DID Than. II t. N Z-. E;(Ae~‘) - 2Aa. e*"s: - 0- ha A-0 Tlurrefinre. A65 it 1 solution of cq. (P153-1), 64
  17. 17. (D) Comidu N ‘. II N ): ..F(.4na') - $n. .u*. -'+2Aa. ue"a"‘ kuo use he I N I N d I Ah‘ Zn: +A¢"£Ik; (:‘) up no N d N = Mfzqa‘ + Ae“‘-‘zap’. I30 RIB ll tr 5. is . solution. than Em, = 0. this lnapliu that : e‘-' is a solution. kill (c) (i) Hm. s'+3s+2=O. =3 -t= -2.! ='L Theniore. y. (t) = AK" -0- Be". Since y. ((! ) = 0, ; f,_(0) = 2. A + B = 0 and 2.1 + B r 2. Therefore. A = -2. 5 = 2. , ,(¢) - 2:“ - 2:"' (ii) Hen-. J"-0-3:42-O = y(t)= Ae'7'+ Be“. Since y(0) : 1. y'(0) I —l. we have y(t) = e" (iii) y(l) - 0 because of initial rn| condition. (ivlflcxc. s’+2s+I= o=(. +I)’ = as-Lo-2. and y(l) 7 A2" + Bic". Smgey(0)= I, ‘/(0): LA: L352 Tlaadou. vu) = e" +2«-'- (V) Ilerc. _. =+, =-, -;= o- (5— | )(. +1)’ = » , ,(¢)= ,4¢' +3:-'+c: »". Sine: y(O) -_- 1, : I. And = -2, we ‘:1 A = 1/2. B - 1/4. C L 3/2. Therefore. ya) - ac‘ + gr’ + 21:" 65 ml Have. 1‘ — 2: + 1 = 0. Therefore. yin]-A(l)'+Bn(l)"= A-0-Bn Since. y[0] = I. |/ [ll - D Hegel. A = l. B a -1. and y[n]-1- vs. (iii) Only dlflcrma: (mm previous pan is initinl oondilin. Since y[0] = 1. y().0] : 21. we get A = 1. B - 2, nd VIII] = l -0-21:. (ll! ) Hut. 1 . 3 ~ mil *1) Thereforrc. ylv-I = Alfill + m" + alfiu — an". Since y[0] A 0. y[—1}= l,veget A I #5. B: ;'j; .&. nd vlnl = —7‘, (§)*sin<n-rm. 2 55. (0) y[0] = :10] = - 1. hlnl satisfies the equation hlu] = ;n| n — 1], -n 3 1 The nuxiliuy condition is h[0] 4 1, Ullng the method introduced in the previous ptoblexn. we have t - 1/2. . Illa) = AU/2)“ Using the auxiliary condition. Mn] = u[n], (h) From Figure P2.$5[b). V2 know that if : [n] = 6[n], then wlnl = M»: - (gr-lax. 'l’hn implies that .41.} ; up. ) = ulnl + 2(%)""u[n — 11. (vi) Hal. I = -1:21’ and y(I) I A: "d"' + B¢"e‘”'. Sim: y(O)= 1.. /(o)= 1. A= ;(l —; ')= s'. Tharefote. 14(2) I e“[ms21 + sin 2!]. 2.54. (I) let us mum: um N 2018’: = 0 l$ The1:. il‘y[n]= A-3. N N N zanvl" ' *1 = :anM=3"'l = -4Io': In1o ‘ ‘ 0 no z-. o A-o Therefore, A13 is 1 solution 0! oq. (P156-1) (b) If ylnl = u'‘''. llnn N N ): .., ,(n — :1 = 20.0: — 1:): --‘-'. (s2.5a-n A-II I-0 Taking the righphuud side at the equlion that we nun: to prove, N N n. n.s - x""'2a. (N - k): "'H + (n — N): a;. km k-II u _ Zn“. .. _ k)lI| —k-I ($2.54-2) i=0 Conapnring eqs. (S256-1) and (92.54-2). we conclude lhnl the equation IS proved (c) (i) Here. 3 | X 1+3:-‘+3;-‘=0 -. ;= —5. x——£. Thuuforn. ylnl - A(—§)' +B(—§)'. Since y[0] - l. y(-I] - -6, wage! A = -l. B = 2. And vlnl = 24-fir — (-9-. (C) Plugs; oq. (P1553) lflo nu. (P1551) pin" §_‘_n1.. — m| z[m] - %2h[n — vn — ll: [m] = : (; )-----z| m1 — E (%)"""z[m] = (§)"'"= Iv-1 V‘ '. '[n). This implies that on. (P155-J) utisfiu eq. P(2.5sl). (d) (i) Given Ihuoo foudthntbcsyskmobeys inizinh-at, we 5:: 1 a¢y[0] = I :9 gm = The homogeneous equation is N Eqngn A A] . . o 5-0 with list initial wnditions I-[0] = I/ am hl—II - = h[-N + n = u. (i) We have H Al»! - Zn»-. ln — kl = 0. no when M[n] is u Ibove. (e) For n > M. N 2uIs[n — :4 = 0 ill with “(OI - vl0l. - - JIlM1= vi"! - (| ') (i) We :0! Mn]_ l. n¢vwA. n20 V 0. 1: odd or n < 0 (ii) We get. 1. 7: even and n 3 0 AM: 2. nuddnndn>D 0. 11 <0 (iii) We gel. 2. n-0.2 h[n] = -1. u even it 2 4 0. else
  18. 18. m) We get up. ) = ;| .,. ?+ fisin'-62). L‘ 56 (n) in riususe. s+2:0Ivhi¢in: piiaI. |nt y(l) = M! ) = A: '". Sincc y(0+) : l. A =1 and Mt) . e"‘u(: ). Nnrw consider eq. (P156-i) L_u. s_ —. 1 F h(t—1)z(r)dv+2 h(t—7)r(r)d-r -. . . . «it on - / E-*“*')a(: - r)x(~r)d1 -on = 2(1) = 3.11.5, This impln. -s thu y(l) does solve the diflerentixl equation. (b) 'l‘¢kc 14(1) = zarw(l)- 1 Then N £n. ):a. u.. .m = am A-I I integrating bet-ten t I 0‘ and 1 — 0‘ and mnching coemcienu. we get a, .— 0 rm-:91 a_, »1 = I/ an. This ilnpliks that for 0' 515 0' W) - ; ‘,; u.~<: > and and (c) The impulse response ifi “ 4*! -.(z) = b-—. no) go. 1‘, as La 0 0 gm -um ma: 8.! -1 [3 B (D (C) um i. 6 6 ‘IN '5‘: 9 5° : : "M B9 (4) -100 (4-) Figure S2.5'I (b) The figures our-ruponding tn the remaining parts of this problem an shun-'n in the Figure $2.37, 2 58. (3) Realizing that zg[n) '— y; [n]. we mny elinzinnte than from the tuna given -irlf-renco equations, This would give us 2yq[n] - yq[vi — 1] + y; [n — 3} -V n(n| - 5:-, [n — 4|. This IS the same as the overall difference equuion. (b) The figures comsponding to the re. -mainiug put: of this problem Arc slwwu ill Figure $2.58. 259. (n) inlegnting the given differential eqlutinn once and -iniplifi-‘mg, we get _ -2 ' be ‘ '2 31(1) — an/ _”y(1)dr + 0‘ _m: (r)dr Ir all“). Therefore. A : -aq/ fl|, B = in/ oi. C - bn/ or (b) Rrnlizing that z7(¢) x y; (t). we may eliminate that from the two given integral equa- tions This would give us yz(I) *— A I y-, (v)d1 + BI‘ z; (1')d7 + C2. (1) 71 (4) (in mm; W) = }: au. (c) -2 set Zia. -.. :(t) +3a. u.. .<¢J + 2a, u.1= am This inaplia that I. .. = —-2 nnd n_; :1. Thaefoce. h(0+) = 0 and h‘(D’) r l constitute the initial mnditiom. Nw, s'+3:+2=0 = x= —2..5=-I Thetcfnn, Mt) = A: "t -4- Be". (2 0 Applying initinl oonditions. we get A = —1, E -» l. Therefore. Mt) — (¢" — ¢"‘)u»: (l)‘ (ii) Th: initini mnditions are M0‘) = 0 and h'(D‘)l. Also. s : I + J The-rr-fure Me) = [e"aint]u-, (t). M (9) From put (c), ifM 2 N, then 215%} will contain singularity terms at 1 2 0 This no nu) = 2a. u.(z) +» implies that (5) (ii N01’. §_ja. u.. .tt) +2'£a. u. = sum) +a-am Therefore. r. ... , = o. 'Also ouIn(¢) 4' 0-iuo(! ) + 2aau«(! ) = 3“x(‘) + Moi‘) This give rrq - J Ind 42.. = -5. The initul condition is M0‘) = -5 and 15(1) = 3u. .(x) - se"‘u_. (t) = 35(1) — se-'"u(: ). (ii) Here. a; = i. no = -3. 0-: < 13. 0-2 = -44 Therefore M0‘) ’ 13 And no‘) = AM And M! ) = u, (!) — 3no(t) -t 18c"'u_; (l) - 5¢"'u . (l] 2.51. (n) Raaiixing that : ;[n} = min], we may eliminate these from the two given diflortmcu equations. Thin would five us m[u] = -ay; [n — i) + box. [n] + b. r.[n — II This is the some as the ovenll dlfcu. -nee equation. 70 Figure 52.58 (c) The figures corrspondiu; to the remaining pun of thin problem nre shown in Figure S259 2.60. (3) lnwgrntin; the given difietentinl equntion am: And simplifying. we get ya) - -33 'u(v)dv—: —jj_‘m[_'__u<n)dadv -an 02 r r r + I x(c)dadr + z(1)d'r+ -b3z(t). "7 -0‘: -u: "2 —n “I Thaefore, : -3|/ O1. B = —og/ cg, C I in/ al. D = 61/”, E : bu/ oz (h) Ruining that z¢(t) = y. (t), we rnny eliminate the: from the two given integral equa- lions. (c) The liguru corresponding to the remaining parts of this problem are shown in I-‘iguru S250 2.61. (I) (i) From Kirchofl"t voltage law. I: know that the input voltage must rqunl the sum of the voltages across the inductor and capacitor. Thgmfore. : (1) = Lcfdigi’ + v(l)A
  19. 19. Figure S159 Using the values of L and C we get iii) Using the rvsults of Problem 2.53, we ltww that thr homogeneous solution n! tht’ dl! lEl¢l’Il: .tl equation w d 1 7% + a. ~'; “~’ + mu) : tau: w: !l haw (trim! of the rum: K; c"‘ +K; r"‘ where so and -vi Irv 10013 04' lht equklian .17 4- ma 4- A; I 0 (It is ansiimqi here lhll Sq st .1, ) In this problem. a. v 0 and G} . -= 1 'l'lit-rt-{om thlr mot al the eqtuuon ue .19 a j uid s. - -j. . The hotnogemzous solution is m(t) = K19"-F K1¢"", Aiitl. x», - i = to, mi) if thr vnltngt and current are rtstttcted to he rul. thrn X, = K, A K. Th». -olntr mm ~ ZKcos(t) . zkuiiu + am. 73 (c) (i) F‘mmKu¢of‘tIolu¢chv. vehoIthtthinputvoltAg2niiutaqinItlwnm ol the volugu sans the rvesistar. inductor. uid iapadtoi. Therefore, 2(1) = 1.0% + RC¥‘%“2 + y(t). Using the values of R. L. And C W: gut d—%: -(: —‘)- -14%. ) + 5y(t) s 5z(t). (ii) Using the results of Problem 2 53, we know that thc homogeneous solution nf thi- diflcrontinl equation d J-# v a, —-%(“—) + ugy(t) = l>x(t). will have terms of the torn K. e"‘ +Kg: "‘ than 1.; And 3, an root: of tht- i-qu: i|um . i‘+a. .i+a, -0. (It is assumed here thiit so st 5;. ) In this problem. a. as 2 and B1 = 5. 'l’lmeluvv- the too: of the equation are .1. 2 -1 + 2; And 1. = -I - 2;‘ Th: hom0K"YW0|lF «aliitzon is yp. (t) A Ki¢"c2" t K3e"e'7". And, a — l . ::‘, if the vultaitzr and current arr rrstricted tu he rul. then K; = K; :1 K Then-lure. y, .(z) ~_ 2Ke"co: (2t) = 2Kz"iiu(2z » «/21 . ‘ 52. 1:) Th. » fnrre 2(2) must equal the sum of the {orcc ruquimd ta displatc the man Aim thi- inn 1‘ iuquixed to stretch the spring. Thmfote. zur m"—; ",§-” + Kym - zm siibstitutiug the vtlua of m and K. we gut fvttl 4:7 Using the icsults of Problem 2.53. we know that the hoino(, rnmi. is solution of the lllll'XI'll'-Id-I vquation 4* 4v(! ) = 22(1) '%§‘—’ + a. “;§" ~ mm = nu) mu haw mm of the [arm K. ¢'°' + Kn‘-‘ when so and .1. are tools of (hr vqimlion . s’+oia+n7=0 75 Figure 32.60 (b) (i) Rom Kir-: bofl"s voltage lnw, we know that the input voltiigc llluhl i—qu. i.l the sum of the vvltaga ncrots the tuistor and capacitor Then-fort. d ) my _—. nC~"d—("~ 014(1). Using the vain»: of R. L. And C Ill‘ get d% + y(l] x(t). (ii) The natural rupons: ol the system I! the lwoinogeneou: solution of thr abnw diffu- entinl equation Using the results 0! Problem 2.53. we know that the Iiuinogcnoc-us tolutton of the djfierentitl cquuion dlgi) + . .., (t) = l>z(t). will hxve terms of the form A? “ there 39 is the root of the cquatiuii a + n; = 0. In I-55 ploblfln. ai = l. Therelmt-. the toot of the equation are ‘Q = — i Thv homogeneous solution is xni(l) = K0" And. :2 ~ 1. 74 (It-sa-tun-db: nIhtuisi. )lnthi-ptobIein. o.= oinda, =4 Therelorqthe '°°‘°“N¢1l| l£'l-'l'BIlIIo= +2j| ndIi = —2). Tbebomogeneous solution ‘ts y, (t) = Km, " -+ K; r"", Asummt that W) is tall. II: have K. = K, _—. K. Theulorc. nu) = 2K mom) Clearly. mt) its periodic. (b) The Sam x(t) must eqtnl the sum olthe force required to dhplm m mA< um thr fame required to stretch the spring. Tliarefore. 2(2) = 111% 0 by(t). Substituting the ulna of m and 5. Va gct 41(3) * ill! ) : dz l00m 1000 Using the ruults of Problem 2.53, we know that the honing: -nmiu . ~inliitit. ii iif the differential equuioii 411(1) T + NV“) = 52(1)- will have terms ol the Sam A? “ Illltff :9 is thr mat of thr equntioii I + on ‘ 0. In this problem. a. - 1/10000 Themtom, the root or the cquation an An — ll)“ The homogeneous solution is . . vn(l) = K: '° '. Cleaxly. yt(t) decteues with incteuing 2 (c) (i) We know that the input for-or 2(1) - (Pam required to displace nia. '~ by gm; — (Force requtred to displace duhpot hry y(t)) + (Force Mquimi to displace spring by y(t)). Thensiorv. x(t) = mJ# + 1% + Kylt) Using the values of in, b, And K we get I’ 71:9 + 2% + 2y(l) - : (t) 76
  20. 20. 1invUun¢thelunluelPlubkn2.$l. whavIhn&thh_cD0svnemunolI1liondth difierrnhlalcqlllliofll div") -6- a. d—yi‘(t—) + l¢y(l) = h1(¢)— 71"- will have lermlolthe form K. ¢"‘+K; e"' vberuoand II In roots 0' *1’-4‘ NI“-I-'I0n I’-O-o1.I+og-0. (1: Li assumed here that so a! 1.. .) In this problem. a. = 2 ud 02 = '2 Thmfmc. :12 met ol the equation in so -_- -1 +; ' uni : . - -1- 1- Th? h«>-noeenuvu-I solution is yA(t) = K. e"e" + K¢e"¢"'. And. a = 1. (in) 1mm [owe is mmcaed to he real. then X. = K7 = K — Therelorc» y. (:) = 2K2“ cnu(l) = 2K3" s'1n(f + w/2) 2 63. (A) We have y[n] . Amt horruwud — Amt. paid+ Compoundnd Arnl (mm pres rnunlh - l00,(ll)6{n] + l.0ly[n — l] - Du[n -1]. Thcmlorc. y[n] = 1.0ly[n - 1] — D. n > 0 and y[0] 4 1oo, ooo Ind '1 = 1.0!. (5) We hm y, [i-1] = l.0ly, [n - I] - D. This irnpluss that u, [n] = IMD. Also the homogeneous solution is 0! the form VAN = 4001)" Then. -lore, yln] = y, _|u] + v, |n] 2 A(l.0l)“ + IND Using the initial condltiun y|0] = IWWJ, we have A = IUIXSO - IND. Therefore. y[n] = (IIXXJM — l(l‘JD)(l.01)“ + IWD. (c) We hnvr y[360] = o = (P — 1ooD)(1.o1)““ +1000. Therefore. D — 51028.60 Also. ¢. .Ir-1 = Zonin — kl¢~«l'= l~ 1 Themlo1e. ¢,, {n] may he viewed as oulnl -* '= l"1' ‘ll-TI] '4 4*nl"| ~ (d) d>, ,|n] And Q” are as shown: in Figure $2.55. 2.66. (n) The plot of rm) ‘:1 31 shown in Figure 52.86. (b) The plots of 29(1) and z; (l) an L! shown in Figure $7.66. 79 (6) ‘Dual pnyizn I 8110.5. (2) TM ton. -ghux question in {hit hook! ! 2164- (t) We have vii) - 2(1) - hm and 2(1) x 14(1) - g(t). Timsm. _q(t) . M1) - 5(2). Now. Mr) - 9(¢)l. =.. -r = ii-.9.-. I(: — -T). A= D Therefore we want nu: Therefore. 1 _li, _ 1 -1.’ I1, Q>= ho. 91 E50. — h°[h‘+h°. (Is) In thk rm. yo = 1.1;. = -1/2. 9, . — (-1/2)’. 9, = (-1/2)‘. and so on T111. lrupllts mu m I k g(() = 5(2) +2 (-5) 5(t — I: ‘l'). tr! (e) (1) Hm. Mt) - i¢*m—1~). K-0 (11) 110 < n < 1, am. if < 1. Thexelure. ha) is bounded And nbsolutely integrable and corresponds to A sable syslem. If n > I. then Mi) is not absolutely integrahlo innhing the uyttcm unsuble. ('ii'1)Here. 9(1) : I — a5(t — T) The inverse Iystan is as shown in the figure below it” * H “‘*’ E! ."fl I Figure 52.04 (.1) u : ,[n} : 6(0). y[n) = h[n]. 1tz, [n] = gnu] + Mn — N), ,,[. .1= 24.4. 2.65. (n) The nutnenrrelntion uquanns nn :5 shown in Figure S255. (I2) The auwcorrehtion sequences In in than in Figure 52.65. (c) We get ¢. .1n1= f hl-k1¢. .lv--*l~ II I ThnreI'ore. ¢., |n| may be viewed 3 ¢. .1v-1 « —» ¢. .1v-1. 78 Figure S2.“ (*3) 31(1) ' MU} ‘ 31(1)‘ '13“) = 11(1) ' 5:0) 3 D '07 ‘ = 4- 2.67. (a) The nutntnmzlation function: In: ¢. ... m={g'f"‘*‘*3' °5‘5’ and ¢. ... m-¢. .., (-u. ! >2 and 7(l-1). 05:51 1-1, 15152 1-3. Zstsl 3-. 35 54 . ».. ..m- , _;I “:55 ma ¢. ,.. m=¢. ...1—u. 5-¢. 559:6 (-7, 65157 0, t>7 (b) If the impulse Iesponae it h(! ) = :('I‘ — I), than y(t) = ¢, ,(t — T). (e) We have u(T) - L’-z(1)II(T - v)d-r s M‘/ ' [ 1° Tx’(t)dt] 1", 1' ‘/3 Therefore, y(t) is at most M '/7 U z’(¢)dt] 0 80
  21. 21. If-venowchooun nu)- —,54—z(r-a). / :7(l)dt B T ya‘) = Mmllo : ’(I)d¢}'/ ’. Clearly. y(T) is maximized for this theme choice ol‘h(! ). (d) (u) The response! are as shocked in Figure $2.67. lllcn Figure $2.07 (ii) In [he impulse response: ufbo and L; be h; ,,(! ) And l| ,,, (t). Then, x-1(2) ' ‘lI4.(1)l. :4 =00) ' hL. (l)| ... ll“) ' ’| l4(‘)| :=4 = I(1J ' '| I.. (l)| :-4 II III] own. » To mxke the job of the receiver easier, modify : o(l) As shown in the figur-> below 81 (b) We have x[n]u; |n) — z(0]6[n] - xmatn - xi + lzlfllfiln - 11- rl0l6l-- * II! = 2&0!--lnl - (slll - = l0n6In - 11 = :[0]6[n) — : [l}6[n — 1] + z[l]J[n] ~ zII}6lv-I s z| l]u. [n] — (z[l] - : [0])E[n] (r) W: have 021"] —- -ulnl - min) = 61»! - 26!» v 11+ ll" — 21 and U: ["l - 5lv-I - 361» -114-341" - 21- ‘In - 3]’ The plots {or these signals u: as shown in Figure 32.70. u, :-q U) Pigun 52.70 (d) we hm u_; [n] - (n +1). in 2 0 ad . ,_, [,. ] : n 3 o. The plots for these xigmls Are u shown in thc Figure 5170. 83 ZOO. “'9 ha‘? ¢. .(r) = ]»<rmm)a 5 1j»'ma+1'/ ’1fp'(« + v)-211'” s jp‘(r)-iv Therefore. ¢n(1)S¢‘n(0) —‘v ¢p(0)= mf*¢»(i)- Also. 4.42) : ¢. .(c — a. ) = ¢. .<: o> = ¢. .(n) : m_uo. .m 2 69. (I) Let. g(7) : z(l — 1). The f" y(r)v: (v)vl1 = -a‘(o> W —z'm ~59 (5) Cdflsidfl 7(1) 1' g(l]/ (I), Then, f: r(nu. u>«u r —r'(01 ; -9'(0)! (0) — 4(0)/ ‘(ru- Aha, / °°a(: ).v(o)u. u)«u — °°a(= )1'(o)uo(c)d: = -910)/ <0) A no)/ ‘(ox which is the sum as above. (c) / m 9(1')n2(f)d* =9"(0)» (d)WchAw: /_ Nah)/ (v)u2(7)d1' — £'I.4(-‘)I(-‘HI: -so = -§, lo'<-on-1) +9(—I)l'lv 1)]: -o : I(o)r(o> — mo)/ '(o) + no)/ "<0: Thcrefuns. J(l)v2(f) - l(0)u2(l) - 2/'(0)u: (1) + f"(0)vo(l) 2.10. (I) We have 2 slmlvnlml — £z4m1(-s(-v-I-6(»-- In I z[0]-zfl). 32 (G) Thesuknmnbxrnclotk-1,2,3. A-IIn: itisI. rueIc: k.1‘hen. fotl: >0 Ihgfn] - u| [n) o ug[n] 2' u. [n] — u, [n —~ I}. By induction. we may not: claim that an stalament is true for all k > 0. (T) For k - I. u_. [n] = u[n] which about: that the statement In mu» Fur k = 2. u_. [n] = (" : !”'. .[. .1 = (n + l)u[n] which again shun that I-hanuununl. is true. Assume um. i: is mm for k — 1 ‘. ~ 0, Then. u_(. -,, [n] = u-. ln] — u-. [n — 1] (S'Z.70- 1) Also. “-(~—nI"l = _ (n+k— (n+k-2)! ' TEETH! "" ' ' *1 Using the abcve equulon wlth eq. (S1701). we get A (n + I: - 1)! "-*l"1* —. mT)«""" By induction. we my now cldm chm. flu: statement u~ true for all k > 0 2.71. (1) We buve 2(2) - [u. (t)ou(2)| - 2(1) = 1, for all I. [1(l) 0 u, (l)| 0 14(1) = 0 v u(l) ’ 0 for all I, And 11(1) 0 '41)} ' HIM = w - vn(l) — undtfintd (I5) We haw. z(t) = e". M! ) -— e"u(¢). and 9(9) = u, (t) + 5(1). Therefore. 2“) ' [M0 '19)] = 1(3) = G"- [x(: ) o g(z)] - M! ) = 0, 1nd 3° 9(1)-[: (:)-h(: )]= g(1)-("j my - undefined U
  22. 22. (c) We have n I =1»-I-U-lamest-n= (Q) -at-1-; . u :21»; - mun ~ um = o - hl-1 - o. and (gm - Mun - 9!"! = ugrffu -91'-l = no- RSV - 0. the output will be y; (t) : 0 NW il (:1) Let h(t) = um) Than ilthr input is 2.(t) the Intern in not tnmtflhlm 12(1) = constant. then y3(t) = 0. 'I‘hudore. Now note that I/ _‘~xg(1)d1'| = { o The; -atom, it If an‘ 7‘ co. then only z; (t) - 0 will yield mu) . - 0 Th-rrlnrv -no the system is inv«nabi. ‘°° 3.2. 0 @ if 21(1) -- OVI ll 23(1) 3! 0 We hm , _,_ 64(1) . = . ;—. .(c) . [am — .5(: — 1-)]. Diflcmntiating both sides II! get $5‘: _— %u'(t)-[5(t)—6(l-Tl] — ism . mt) — au - 7)] film) - bu — T» 3.4. For I: = 1, u_. (:) — u(t) Therefore. the given statement is true for I: a l. Now assume that it is lfue for some I: > 1. Then. "(ll ' '--:0) / :mu-. (t) - j‘. ._. m«1 II-4:+r)(*) HI: -1)! “:31: Both z. (| -t)mdx. (t-1] uepedodie-Inhrundamuaulperhdr. - 3;‘. Sim: v(l) ll Ilinu: eornbinationol: .(l -t)and: .(t—l). it lanhn periodic with firndamarulpu-ind T, = Thcr: fou. w, : u, 3.5. Since z1[t) 0550 G3, ruling the ruultt in hble 3.1 we have z, (nl 1) £m. e”“"”" M, _ 1; . F_5, M-rm-IT. ) , ,| (_, 4, , ) gs, n_‘, -rm-rrn 3.5. Therefore, z| (t + I) + 2.0 ~ t) «'30 age’‘"'”‘) + a-. e"""/7‘) = ¢"""(ag + u-. ) (I) Compuing mm with the Fourier Ierim rynthuis eq. (3.38). we obnin the Rluriex series coefllcic-nu olz. (t) to be -= {:. .**‘- Rom ‘hhlc 3.1 wt know elm if : ,(t) is real. than a. has m be mujuguesymmetric. re. in = a'_k. Since thin in not lruz (or 21(2). tin signs! is not rul valued. Similarly. the Fourier series coefldcrns cl. -:10) IN ax(l:1v). lwskslflll n. ={0. ollm-vise l~-om hhle 3.1 we know: that if z3(t) is real. than A. luu we be oonjlrgawsymrnetrk. Le, a. = n'_, . Since this is true for 2;(t). the xigml is rel! vllund. Similarly. the Fourier series anafliciaents of z; (t) are a _ jsin(Inr/2). [00 s k 5 ll! ) * " o. new-au l-Yam Table 1| -0: know! that if :50) is real. then A. In: In In conjugtleuymmotric. Le, a, s a'_. . Sine: this is true for 23(1). tin cigml is renl nluod. (b) For a signal tn be even. its I-burier atria eocficienk must be evzn. This Car 23(1). Givcu that 36. 05145100 otbcrvhe is true only 3,9‘ 3.7. F z(t) n—§o ti. we law 4:“) 2 . 2: gm = T «Hi I). = gkfak. Thcruforflr 5‘ “* '" " "° 87 Chapter 3 Answers Using the Harris nariu rynthuis ca. (138). 3“) = ,, ,u(2-/ ru . , , _., —m-Inc + , ,,, i=(2-nv , . , _,, -ma-ms . 2¢r(’~/ I)‘ + 3,-an-/ -)1 + 4,-, :a<2~/ on , ‘j, -jam-/ us _ Z _ - “_" - (uu(‘I) sm. ( 8 1) I 31 I 2 deos(-‘-I) +8eu(Tt + -5) Ulin‘ the Fourier Iain | ynl. lin'u sq. (3.96). ,[, .] = .0 + , ,¢, mr-/ rn- . , , ,_, ,-xx: -/Nvn + , ‘er4(? -/N)- , _ , _‘, —,u2-/ mu x. 1., ,«-/ c), m-Is>v- + . -i(-m, —2:(= -/->n +zE7(l/ l)¢fl(3'IN)n + 2‘-)(I/ .|)a_‘£-;4{2I/ N)n _ -in I 8: I — l+1cos( 5 n+ ‘)4-dooe(? n+ E) = l+2ain(-‘sin 4» ¥-) + (Iin(8?'n + 5%) Tin fiven sign! is ,3) = 2 4. ; ,.v(? IIl)l , ,_ ac-:1:-/ an __ , ,-, .(s: /:; c _, ,1-(,1:-/ :3: = 2+ éeflliilfll + ée-M? -/I): _ -‘J-c, ‘s(2:/5): + 21-¢—; s(au/ I): “Om ‘N5. ‘'5 may wndudc thu lb: fundunentnl frequrzlxry 0! : (t) is 2:/ G : 1/3 Thr- Iwn-uro Fourier lair: meflciunu o[ 2(1) me: no=2. n2=a.2=%. a5:a'__. ’=—-1] sum mg = 11, 1' = 11/35) : 2, 1‘'. ..-. {°. ,._ 1 ’ _, ,,__ A. = 5 2(3); 4: NW. 1 I 1 2 do = l 541 — — I.5& = 0 n I tnd for h # 0 a. — 5/lr. s¢‘1*'*a: — 1 1.5¢'I"'. u 2 o r = — e—jl: n] %, —:b(-/ :), h,('; _*) 36 “k: t = 0. _ ‘ 2 . . . ‘ OI — T ! <‘bz(t)dl - f using pun information 'I'lIerd'orI. M _ -3». . k = o : r57'nr~ " * ° ' Slut: x(r) is real and odd (clue 1). it: Rmriisr-uric: coeficients a. are purely imlgrnnry and odd (Sue ‘IV-hie 3-1)» Thcrefiore. A. - —a. . And a. = o. Aka. mm ax is given that a. — 0 for | k| > 1. tln only unknown Fourier -cries eoelldentn are or and :2.. . Using Panrvafs relnlon, 1 , |= (:>I'a— 23 um’. <1‘; ,__w for the given: signs] we lnve : |IXil: — k--I 1 7 , 5/“ | :(¢)| d¢= Using the infiornution jvcll in clue (I) along vith thy shove zquation. Theruforu. and Thepa-iodotunpwenaipnlind. Therdom Ia-I’ +1-z--I’ =1 = mm! ’ =1 $1 = '-H»: =—‘-: or a. = -0.1: ——l, - V/5 /2) The two possible rimdn which utirfy the given intornntion Ire l - l g = .__, :l? -I7)‘ - _ -M-/2): _ _ - zr( ) fij flje 2s| n(arl) nu) = -$'; e"'”" + %je"("”l' = 2SiXI(lU 1 ‘ 5,, at = zzzlnlc" " n=0 1 = - ' 1" ‘[4 + 3: 1 1 flo=3. flx=1'1i. oz--1. ag= l+2J 88

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