Slu to Ch3

Slu to Ch3 - 563120 Ho) 2. {xi-F Chapter 3 Solutions 3.1-1....

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Unformatted text preview: 563120 Ho) 2. {xi-F Chapter 3 Solutions 3.1-1. (a) E, = (3)2 +2(2)2 +2(1)2 = 19 (b) E, = (3)2 + 2(2)2 +2(1)2 = 19 ' (c) E, = 2(3)2 + 2(6)2 + 2(9)2 = 252 (d) E, = 2(2)2 + 2(4)2 = 40 3.1-2. (a) 3 _ ' 1 2 _ 1 2 2 2 _ E P, - N0 + 1 “223M111! — 6 [2(1) +2(2) +31— 6 (b) P = —1— [2(1)2 + 2(2)2 + 2(3)”) = Z ‘ 12 3 (c) No-l N N l l 0. ° - l 0. ° — l P: = —- n = —— = Nougat} ()[a—l] No(a—1) 3 2-1 (a) The energy of z[—n] is E1 given by E1 = Z FBI-4"“2 Setting n -— —m, we obtain E1 = Z livlmll2 = Z lifii'mll2 = E: mace tux—co (b) The energy E2 of z[n — m] is given by E2 = Z lzln — mu” = Z lzirll“ = E, n=—oo r=—oo (c) The energy E3 of zlm -— n] is given by E: = Z lzlm ‘nlla = Z |:I=[1‘]|2 = Z |:I=[1‘]|2 = E: n=—oo r=oo r=—°o (d) The energy E4 of K z[n] is given by & E4 = i |K:I:[‘rz.]}2 = K2 2 |:I:['n]|2 = KaE, "__m “3—” 3.2-3. Figure 83.2-3 shows all the signals. gem-6,1 Figure 83.2-3 3.3-2. (a) Trivial (b) Becaus (c) Because n(n — 1) = d Because sing— : 0 for even 11 and u[n] + (-1)"u 2 . follows. _ (e) Because cos?- = O for odd n and u follows. e sin “4;- =D~ for n = 0, the result follows. 0 for n = 0 and n: 1, the result follows. [11] = 0 for odd n, the result [n] + (-1)"+1 = 0 for even 11, the result . 3.3-3. Figure S3.3-3 shows all the signals. Figure S3.3-3 .. K 3.3-4. (a) z[n] = (n + 3) (u[n + 3] — u[n]) + (—n + 3) (u[n] — u[n e 4]) (b) z[n] =n(u[n] —u[n—4])+(—n+ 6) u —4 — — (c) 2[n] = n (u.[n + 3] — u[n — 4]) ( [n l '14" 7]) (d) 2[n] = -2n (u[n + 2] — u[n]) + 2n (u[n] — u[n — 3]) In all four cases z[n] ma _ . , I y be represented by several other (sli htl d'fi' :13212? instance, In case (a), we may also use 2[n] = (n + 3)g(u[1:,+ Mn _4] 115:;an r1] 7 uh; - Moreover because z[n] = O at n = 5:3, u[n + 3] and ace . . . apply to Other “a; also’m u[n +2] and u[n — 3], respectively. Similar observations 3.3-5. (3.) e"°'5”=(0.6065)“, (b) em" a (1.6487)”, (G) 6"“ == (e'j”)" = (-1)", (d) 6”” = (8”)“ = (-1)" Figure S3.3~5 shows locations of A and '1 in each case. By plotting these signals, we observe that when '1 > 1 (A in RHP), the signal grows exponentially. When '7 < 1 (A in LHP), the signal decays exponentially. When M = 1. (A on imaginary axis), the signal amplitude is constant. ' Figure 83.3-5 3.3-6. (a) raw“)? = (F‘s-W)" = (-%)n (b) e—(l—jw)n = (e-lejvr)n = 0%)." (c) cum” = (e eh)" = (—e)" ' (d) eu-m" = (ea-1*)" = (—e)" (e) ram)" =‘(e-1)“ e-fi" = (macs sn — jsin en] (f) ea’ji)?‘ = (el)" e‘ji" = e“[cos fin — jsin gin] 3.4—2. The net growth rate of the native population is 3.3 — 1.3 = 2% per year. Assuming the immigrants enter at a uniform rate throughout the year, their birth and death rate will be (3.3/2)% and (1.3/2)%, respectively of the immigrants at the end of the year. The population p[n] at the beginning of the kth year is p[n -— 1] plus the net increase in the native population plus i[n - 1], the immigrants entering during (71. -— 1)st year plus the net increase in the immigrant population for the year (n — 1). 3.3 — 1.3 . 3.3 - 1.3 . pin] = pin—11+ mo shuns—11+ 2x100 zln—ll .—_ 1.02p[n — 1] + 1.01-i[n — 1] or p[n] —- 1.02p[n -— 1] = 1.01i[n — 1] or p[n + 1] — I.02p[n] = _1.01i[n] 3.4-4. The input a:[n] =: u.[n], which has a constant value of unity for all n 2 0. Also . yin]. — yin 1' 1} ;. Tulnl. .Hence.the difierence between two successive output values is always constant of value T. Clearly y[n] must be a ramp with a possible constant component. Thus y[n1=(nT+ c)u[n1 To find the value of unknown constant c, we let n = 0 and obtain ylol = 6 But from the input equation yIn] - y[n 4 1] = Tu[n], we find y[0] = T [remember that y[-- 1] = 0]. Hence yIn] = (n + 1)Tu[n] : nTu[n] for T —+ 0 3.4-6. Thelnode equation at the kth node is 2'1 + £2 + £3 = 0, or v[n- 1] -u[n] v['n.+1] -v[n] _ m = R + R (112 0 Therefore - a(v[n — 11+ vln + 11 — Zvinl) — ulnl = 0 01' u[1_7.+1] — (2+3 v[n]+u['n.-1] =0 that is u[n+2]— (2+-l1;)v[n+1]+v[n]=0 3.4-8. Notice, y1[n] = -6]n] +6[n— 1] +26[n-2]. Furthermore, 2:2[11] = zl[n—1] —-2:1:1[n—2]. Since the system is LTI, ' y2[n] = y1[n — 1] —- 2y1[n — 2]. MATLAB is used to plot the result. >> y1 -'= inline(’-(n-=0)+(n==1)+2*(n-=2)’); n - [-2:8]; >> stem(n,y1(n-1)-2*y1(n-2),’k'); axis([~2 8 -4.5 4.5]); >> x1a.be1(’n’); ylabel(’y.2[n] ’); Figure 33.4-8: Plot of y2[n] = 91 [n — 1] — 2y1[n — 2]. 3.4-9. Using the sifting property, this system operation is rewritten as yIn] 0.5 (:i:[n] + :i:[—n]). ~ (3) This system extracts the even portion of the input. (b) Yes, the system is BIBO stable. If theinput is bounded, then the output is neces- sarily bounded. That is, if ]z[n]| 5 Ms < 00, then ]y[n]| = l0.5 (:1:[n] + z]-—n])| S - 9,5-(]r[n}l-+]z{‘m]fisM¢ <ool~- ~~ - - - -- - ~- - - ~-~ - - ~ '- (c) Yes, the system is linear. Let y1[n] = 0.5(zlin] + z1[-n]) and yarn] = 0.5(zg[n] +Jng[—n.]). Applying 0.21nt + bzr2[n] to the system yields y[n] = 0.5 (0.271][n] + 62:2[11] + (an [—n] + b22{-n])) = 0.5a(:z:1[n] +21[—-n]) +0.5b(:rg[n] + Sal-"ll = agilnl -!- byzlnl- ' (d) No, the system is not memoryless. For example, at time n = 1 the output y[1] = 0.5(z[1] + z[—-1]) depends on a. past value of the input, z[—1]. (e) No, the system is not causal. For example, at timen = -1 the output y[—1] = O.5(a:[-~1] + :z:[1]) depends on a future value of the input, :r[1]. (f) No, the system is not time-invariant. For example, let the input be 2:1[11] = u[n+ 10] -u_[n— 11]. Since this input isslready even, the output is just the input, y1[n] =e 2:1[n], Shifting by a. non-zero integer N, the signal 1:2[12] = 1:1[n - N] is not even, and the output is y2[n] aé y1[n — N] = 2:1[12 — N]. Thus, the system cannot be time-invariant. ' . ' 3.5-1. (3.) yin + 1] = 0-5ylnl (1) Setting a = —1 andsubstitutinggL—ll .=.10, yields y_[0] '= 0.5(10) = 5 Setting 1: =3: 0, and substituting y[0] = 5, yields y[1] = 0.5(5) = 2.5 Setting n a: 1 in (1), and substituting y[1] = 2.5, yields 3.5—4. ‘ y[2] = o.5(2.5) = 1.25 Mn 1+ 2] = -3y[n + 1) - 2y[n] + :{n + 21+ 3:1:[n + 11+ 3:1:[n] Setting n = —2, and substituting y[—-1] = 3, y[—2] = 2, :c[—1] = x[-—2] = 0, :z:[0] = 1, yields y[0] = —3(3) — 2(2) + 1 + 3(0) +3(0) = —12 Setting n = —1, and substitnt‘nrg 11.10] = —12., y[—1] = 3, z[—1] = 0, 2.10] = 1, z[1] = 3, yields y[1] = —3(—12) — 2(3) + 3 + 3(1) + 3(0) = 36 Proceeding along same lines, we obtain 3.6-5. (a) (b) y[2] = —3(36) — 2(-12) + 9 + 3(3) + 3(1) = —63 By definition, any element in the Fibonacci sequence is the sum of the previous two. Thus, 'fIn] = f [n - 1] + f [n —- 2]. Written in standard form, this yields flnl - fln- 11-fln -21=0- This is a somewhat unusual system in the fact that it has no input. In the lingo of signalls and systems, f [n] is a zero-input response that is completely driven by the amtiliary conditions. The characteristic equation is 1’ my —- 1 = O. This yields twocharacteristic roots _ 1+¢5 1-\/§ 2 2 Since one characteristic root is in the right-half planeI the system is not stable. To determine a particular Fibonacci-number, it is convenient to determine a closed form expression for f Since f [n] is a zero-input response, it has form f[n] =‘c1'yi‘ + cry; The stadium-equations yie'l'cl 'f[1] = 0. = c171 + 0272 and f[2] =4; 1 = ‘cnfcz'yg. Solving yields c; {Ti-$3133 = 3%11- 1% 0.2764 and 2 5 __ 5 1 ~ Ca _ '12 '11-‘11 ~ 0'7236' MATIiAB is used to solve for the requested values of f [n]. >> gammal - (1+sqrt(5))/2; gamma2 - (lrsqrt(5))/23 >> c1' = -gamma2/(gamma1*gamma2* (gamma2-gamma1)); >> c2 = gammal/(gamma1*gamma2*(gamma2—gamma1)) ; ' >> £510 -= c1*gamma1'50+c2*gamma2‘50 ' £50 ‘1 7.7787e+009 >> £11000 3 ci*gamma1‘1000+c2*gamma2‘1000 flood“ = 2.6864e+208 :3 1.618 and 72 = a: —0.618. 71= Thus, 1-\/5 2 f5—1 2V5 1+ 2 50 50 f[50]= «5) + ) z7.7787(109) (/5+1‘ 1000 1 3%) z 2.6864(10208) ( 3.7-1. (3.) The chatatteristic equation is '1 + 2 a1 = 2, b1 i: 1. Therefore h[n] = $6.211] + c(-2)" (1) We need one value of h[n] to determine c. This is determined by iterative solution of . (E + 2)h[n] = 6[n] or h[n + 1] + 2h[n] = 6[n] 100 Setting n = -—1, and substituting h[—1] = 6[—1] = 0, yields h[0] = 0 - Setting = 0 in Eq. (1) and using h[0] = 0 yields 1 1 0=— '=_.... 2+6 =# c 2 Therefoub 1 1 wu=gw-§emww (b) The chairmeristic root is —2, b; = 0, a1 = 2. Therefore ' ‘ m=ww -m We nee4 one value of h[n] to determine c. This is done by solving iteratiwa h[n + 11+ 2h[n] = ah + 1] ' Setting t; = -—1,'a.nd substituting h[—1] = o, 6[0] = 1, yields h[0]_ = 1 Setting in = 0 in Eq. (2) and using h[0] = 0 yields 1=c and -— 7- - ‘h[11}=(—2)”u[n}““ 3.7;3. (E2 - 6E + 25)y[n] = (2E2 — 4E)z['n.] The characterfitic roots are Seijo‘gza, b2 = 0. Therefore h[n] = c(5)" cos(0.923n + 0)u['n.] of h[n] to determine c and 0. This is done by solvixig iteratively (2) ; (1) .We need two vtiues h[n] — 6h[n — 1] + 25h[n — 21 = 26[n] — 46['n. — 1] Setting n = 0 {yields h[0] — 6(0) + 25(0) = 2(1) — 4(0) => h[0] = 2 in (2) yields V hm - 6(2) + 25(0) = 2(0) — 4 => h[1] = 8 Setting 1?. = 0) 1 in (1) and substituting h[0] = 2, h[1] = 8 yields Setting n —-‘= 1 2 = ccosfi 3} s = 5c cos(0.923 + a) = 3.017est — 3.987csin9 Solution of thiese two equations yields _ . ~ _ _ _ _ - c = 2.061 9 = —0.244tad ‘1 ccose = 2 ‘3 csinfi = ~0.493]. and ‘ 'h[n] = 2.061(5)" cos(0.923n — 0.244)u[n] 3.8-1. (-2)"u[n —11*e'“u[n+11 i z '(—2)"‘u[m - 1]e"(“"“)u[n - m - 1] i \ m=—oo 102 However, u.[m‘— 1] = 0 for m. < 1 and u[n —m+ 1] = 0 for m > 11+ 1. Hence the summation 1 We can also yin] = 6"" 2 (-2e)"‘ innits may be restricted for 1 5 m 5 n + 1. "+1 e_n [(—2e)""’2 + 2e] % m=1 -—2e-1 _ 252 fi+1_ —(n+1) ; - 2e+1[( 2) e 1‘4"] dbtain this answer by using the convolution Table and the shift property of convolution. If we advance impulse response h.[n] by one unit and delay the input by one unit, the convolution remains unchanged according to the shift property. Hence we should obtain the convolution by using h[”1=(-2)""1u[n] and 4n] =e—(n-1)u[n] Thus the desired convolution is given by yln] (-2)"+1u[n] * e'("'1)u[n] = -—2e {(-2)"u[n] III e'"u[n]} From the convolution Table, we obtain W uln] uln] _ n+1_ —(n+1) 1 _2e [( 2) ‘3 262 _. = 23 + 1 B'WH " 3 MD] "M which odnfirlhs earlier result. 3.8-3. Here let us delay a:[n] by one unit and advance hfn] by one unit to obtain y[n]. y[n] 3"“ulnl * [2"-1+ 3(-5)"+1]u[n1 1‘ - 3"+1u[n] * 2""1u[n] + 3(3)"+1u[n] * (—5)"+1u[n] fl 3mm *2"u[n1} —45 {3"uln1* (—5)"u[nl} ‘ [3"+1 — 2n+1]u[n] - 45t[w] urn] 99 n n n [#3) —3(2) ~225(-5) 114"] 1 8 3311. - (a) l M a .E. .3. *- A P C" v a 5—. .3. WI] (b) ] z[n] = 2("_3)u[n] = 2'32"u[n] = é2"ub1] From 12115 regult in part (a), it follows that yln] = gm“ — (0-5)"+11u[n1 = 11-2t2"+1 —(0-5)"+‘1u[n1 "(6) z[n] = 2"u[n - 2] = 4{2("'2)u[n — 2]} Note th 2("")u[n —- 2] is the same as the input 2"u[n] in part (a) delayed by 2 erefore from the shift property of the convolution, its response will be units. the s as in part (a) delayed by 2 units. The input here is 4{2<"-2>u{n — 21}. Therefo; yln = 4%['/-"‘+1'2 - (0.5)"H‘21uln — 2] = 2-8“ - (0-5)";‘luln - 2] 3.8-15. The equation describing this situation is [see Eq. (3.9b)] (E —— a.)y[n] = Ezr[n] a, :1 + r = 1.01 The initial con 'tion y[— 1] = 0. Hence there is only zero-state component. The input is 500u[n] — 15 6[n — 4] because at n = 4, instead of depositing the usual $500, she withdraws $10qp. To find Mn], we solve iteratively ‘ (E — a)h[n] = E6[n] or h[n + 1] — ah[n] = + 1] Setting n = ~i, and substituting h['— 1] = O, 6[0] = 1, yields ' h[0] = 1 Also, the chardoteristic root is a. and b0 = 0. Therefore I h[n] = ca"u[n] Setting n = O and substituting Mo] = 1 yields l 1 = c Therefore kin} = (a)“u[nl = (1.0mm) The (zero-statip) response is “[11] = (1.01)'3u[n]u[n1 ‘ l = (1.01)“u[n] at {5001411) - 15006 [n - 4]} = 500(1.01)"u[nl * u[‘n] — 1500(1.01)"“u[n — 4] = '%[(1.01)“+1 — 11141;} — 1500(1.01)"-4u[n — 4} = 50000[(1.01)"+1 — 1]u[n] — 1500(1.01)"-4u[n — 4] 3.846. This problem is identical to the savings account problem with negative initial deposit (loan). If M is the initial loan, then y[0] = —M. If y[n] is the loan balance, then {see Eq. (3.9b)] y[n+1]-ay[n]=x[n+1] a=1+r (E - GM"! = Exlnl The characteristic root is a, and the impulse response for this system is found in Prob.. , 3.8«15 to be h[n] = a"'u.[n] This proble can be solved in two ways. _ First math d: We may consider the loan of M dollars as an a negative input —M6[n]. The monthl payment of P starting at n = 1 also is an input. Thus the total input is a:[n] = -M n] + Pu[n - 1] with zero initial conditions. Because u[n] = 6[n] +u['n-— 1], s the input in a more convenient form as x[n] = —(M + P)6[n] + Pu[n). we can expr The loan balance (response) y[n] is Alsoa=1+r yln] = hlnl = {—Ma"+P[ a: :1:[n] n = N, the loan balance is zero. Therefore 01' M: rM -—Ma”u[n] — P [a” — 1L y[N] = _-MaN+P[ 1. yaw = ca“u[n] aN—l = a"u[n] a: + P)6[n] + Pu[n]} = -(M + P)a"u[n] + Pa"u[n] * u[n] —(M + P)a”u[n] + P u[n] an+1 _ 1 ____i—] u[n] a— : :11]u[n1 and a — 1 = r where r is the interest rate per dollar per month. At =o rM 1+a-N = 1+(1+r)-N Second method: In this approach, the initial condition is y[0] = —M, and the input is z[n] =9 Pu[n — 1] because the monthly payment of P starts at n = 1. The characteristic root is a, and The zero—input response is Setting n = 0, and substituting yo[0] = —M, yields c = —M and The zero—state response Iy[n] is yo[n] = —Ma“u[n] Here we use shift property of convolution. If we let The shift property yields The total balance is yo[n] + y[n] 111 Pa"u[n] a: u[n — 1] = :r[n — I] = P [ yoln] + W] = -Ma"u[n] + p [ a an+l _ 1 an an —1 a— —1 —1 y[n] = h[n] a: :1:[n] = h[n]‘* Pu['n.—- 1] = Pa”u[n] 2:: u[n - 1] wln] = awn] -« uIn] = [-a—‘_-1-—]u[nl 11w-” 1.1M . 3.8-26. For n > 1, u[n] = u[n - 1] = 1. Therefore a“,— 1 a-l Loan balance = -Ma“,+ P [ J n > 1 which confirms the result obtained by the first method. From here on the procedure is identical to tihat of the first method. (a) Actually, tem. 0 plus Wh before e yln] = id only a first-order difference equation is necessary to describe this sys- refill n, the amount of sugar y[n] is equal to the amount added fin] ver was still in the mug. Since Joe drinks 2/3 of his'cup of coffee refill, one-third of the sugar from the previous cup remains. Thus, — 1]/3 + z[n]. Thus, a1 = -1/3,a2 =0,bo= 1,bl=0, and bg= . In stande form, the diflerence equation is y[n] — y[n — 1]/3 = z[n]. (b) Since .10? adds two teaspoons of sugar each time he fills his cup, (0) (d) The tot a:[n] = 2u[n]. solution to a difference equation is the sum of the zero-input response and the era-state response. Since Joe starts with a clean mug, y[-—-_1] = 0 and the zer put response is necessarily zero, yo[n] = 0. Thus, the total solution is ' just the ero-state solution. Tolobtai the hero-state solution, the "impulse—response Mn] is needed. In this - case, h[ = fig +37o[n]u[n], where gob] = c7". To determine c, input a:[n] = 6[n] into the Thus, That is, teaspoo One we at the s refill. T riginal difference equation to yield h[n] — h[n — 1] / 3 = 6-[11]. Since hIn] h[0] — h[-—1]/3 = h[0] = 6{0] = 1 = :70[0}u[o] = c. Thus, 14:] (= 3)“"u[n]. 1_ ~ n+1 is causal The zertigstate solution is a:['n.] * h[n] = (Egg) 2(3)'“) u[n] = 2Wufn] = (3 — 3-" u [n] . an] =' (3 — 3-") ulnl- . __ . _ -n = 5121;0th — is; <3 3 l 3- after many cups of coffee, Joe's mug reaches a steady-state of three of sugar. y-state value of three and then add two teaspoons of sugar at each t is, :rIn] = 2u[n] + 6 The addbd 6[n] “jump starts” the sugar content of the first cup to the steady-state value. If Joe dtsires a steady value of two teaspoons, which is two-thirds the original steady-s ate value, the input is simply scaled by 2/3. That is, the steady level Mn} = 2u[n] is achieved using the input 2[n] = gum] +‘§5[n]. to make Joe’s coffee remain a constant‘for all non—negative n is begin -~ . ...
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Slu to Ch3 - 563120 Ho) 2. {xi-F Chapter 3 Solutions 3.1-1....

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