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Baixe Solução Lathi 2a ed - Sistemas Lineares - chapter 03 e outras Resumos em PDF para Engenharia Elétrica, somente na Docsity!
Chapter 3 Solutions 311. (a) EL=(32+U22+2(1)2=19 (b) Es = (32 +A22+2(1)2=19 (e) Es = 23)? + 2(6)? + 2(9)? = 252 (d) Es = 22)? +24)? = 40 3.1-2. (a) 2 2 2 19 & Bo) +2(2) += (b) Pe= E [ear 2er + 2mp]=5 +32 3 (e) No-1 No — No — po E ar A [82H] qnto No & No La-1 No(a — 1) 313 [per "Ê =|pP and 1 No-1|No-1 P=— 2 — jrNon ql = 5 Ee No-1 No- = ron * erimSon = x » Dre 5 Ds, Interchanging the order of summation yields r-m)Non p= vm DE torno] r=0 m=0 n=0 The summation within square brackets is No when 7 = m and O otherwise. Hence Mot Ne P= 5 DDi= > ID r=0 r=0 > (08"ufa] +0.8-"uf-n]) + 2 (08"uln]-08"ut-n]) zeln) zofn) oo o 1 Es= 08" =S" 0.6" =-——— =2.78 (b) To find the energy of the even component z.[n], we observe that both terms 0.8"u[n) and 0.8"u[-n) are nonzero at n = 0. Hence, the two terms are not disjoint, and the energy E, is not the sum of the energies of the two terms. For this reason, we rearrange zefn] as = ó[n]+ 5 f0.8"ufn — 1] +0.8"uln —1)) AM the above three terms are disjoint. Hence E, is the sum of energies of the three terms. Thus 1 an LG um E. = 14555064 +4 >) 0.64 na nm 1<— 1 = 1+5 Gr =145 +5),06 +5 n=1 The energy of zo[n] is Ex= E 5064 + > 0 sr] = 5 > 64" = 0.8 aa a=a na Hence Er. + Ee, =1.89+0.89=2.78= E, (e: Let us first consider a causal signal z[n), which can be expressed as = atenut-n + 5 te = alentuf-n= 1) zen) zolm) The energy of the even component ze[n) is Ex =) + 4 et +; Do letongf = 0º) 455 eta? n=1 n=—1 n=1 86 3.21. (a) The energy of af-n) is Ey given by B= 5 bet-uf Setting n = —m, we obtain B=>lmp=5> ee mêZoo (b) The energy Es of fn — m) is given by E=>, mp= 5 IrhP =: n=-00 r="00 (c) The energy Es of afm — n) is given by > kmt=2. (d) The energy Eu of Ken) is given by Ej= > |Kafn]P = K? s Ieln]ê = Kº Es 3.22. Because |— z[n]/2 = |r[n)f?, it follows that the power of — For parts b, c, d and e, using the arguments in Prob. 3.2-1, we show Power of z[-n) = Ps Power of «[n — m] = P; Power of cz[n) -n=P, “This shows the time-shift or time-inversion operation does not affect the power of a signal. Same is the case with sign change. Multiplication of a signal by a constant c causes c2-fold increase in power Power of zm 3.23. Figure $3.2-3 shows all the signals. 3.24. Pigure S3.2-4 shows all the signals. 3.31. (a) Pr= limao saga DDD =1 1 Dt” (e) Pe = limpos sx Do (1)? = 0.5 (d) Pe = limy-os 3x7 Do (12 =0.5 () No= FE =6, 1 (b) Ps =limy-o 5x 3.32. (a) Trivial 88 t-rI er XTR+61 x[R6S ala fal 2 mT3KI 12 kR> 19 xEKt Dre na Te == [4 q a 1 J-6- -3 E . q Te 41 Trtand ns Figure $3.2-4 (b) Because sin 2 = 0 for n = 0, the result follows. (c) Because n(n — 1) = 0 for n = 0 and n = 1, the result follows. (d) Because sin ZE = O for even n and uln] + (=1)"uln] = O for odd n, the result follows. (e) Because cos ZE = 0 for odd n and un] + (-1)"*! = O for even n, the result follows. 3.3-3. Figure $3.3-3 shows all the signals. 3.34. (a) af) = (143) (un +43] — ufn]) + (43) (ul) — ufn — 4)) (b) efn)=n(ufn) = ufn = 4) +(-n+6) (uln- 4 =ufn = 7) (e) ti =n(uln+3]- un 4) (d) zfn] = =2n (ufn + 2] = ufn)) + 2m (ufn] = ufn — 3]) In all o cases, a[n] may be represented by several other (slightly different) expres- sions. For instance, in case (a), we may also use z(n] = (+ 3) (uln + 3] = ufn — 1) + (cn + 3) (un — 1] = ufn — 4]). Moreover because a[n] = 0 at n = 3, ufn + 3] and ufn — 4) may be replaced with u[n + 2) and ufn — 3], respectively. Similar observations apply to other cases also. 3.3-5. (a) evºSn = (0.6065)", 89 (b) Amu) = nun) + a (mada) + nun) (c) sin (ZE) is an odd function. It has no even component. (d) cos (22) is an even function. It has no odd component. Figure $3.3-7 shows the even and odd components for the four cases. xoln] (o xeini=O xelnj=X0h4 : Figure $3.3-7 3.4-1. Because yfn] = y[n — 1) + xfn), vtrj- vn] = Realization of this equation is shown in Figure $3.4-1. 24n3 A 9 nd to . 4 tra E Figure $3.4-1 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 pfn — 1) plus the net increase in the native population plus ifn — 1), the immigrants entering during (n 9 plus the net increase in the immigrant population for the year (n — 1). dé qn) 33-13 33- = pR=i 33-18 ln Pin + 2x 100 o Pl -NV+ifn-)+ = 1.02pfn —1] + 1.0li[n — 1) or pln) — 1.02pfn — 1) = 1.01ifn 1) A pln + 1) — 1.02p[n) = 1.01:[n) 343. (a) vlnj=etn)tzino+ono]+ep-a+an=4 (b) Refer to Figure S3.4-3b. aba] D represente unir delay Figure $3.4-3b 3.4-4, The input 2] ufn), which has a constant value of unity for alln > 0. Also vin] — yfn — 1] = Tufn]. Hence the difference between two successive output values is always constant of value T. Clearly y[n] must be a ramp with a possible constant component. Thus (nT +ou vim. To find the value of unknown constant c, we let n = 0 and obtain vo] = But from the input equation y[n] - y[n — 1] = Tuln], we find y[0) = T [remember that y[=1] = 0). Hence vin) = (n+WTufn)=nTufn) for T>0 3.4-5. The differential equation is d a + as + aou(t) = (8) We use the notation y[n) to represent y(nT), «[n] to represent z(nT),--- etc. and assume that T is small enough so that the assumption T — 0 may be made. We have do) = un dy yim=unol) a É T ey delega) atelça de É >> yi = inline('-(n==0)+(n==1)+2+(n==2)?); n = [-2:8]; >> stem(n,y1(n-1)-2+y1(n-2),ºk'); axis([-2 8 -4.5 4.5]); >> xlabel(ºn'); ylabel('y 2[n]'); “tm Figure $3.4-8: Plot of yo! anln — 1) = 2n[n — 2). 3.4-9. Using the sifting property, this system operation is rewritten as y[n) = 0.5 (z[n] + z[-n]). (a) This system extracts the even portion of the input. (b) Yes, the system is BIBO stable. If the input is bounded, then the output is neces- sarily bounded. That is, if |z[n]| < Mz < 09, then [yfn]| = [0.5 (z[n] + z[-n]) | < 0.5 (lz[n]l + |z[-n])) < M, < 00. (c) Yes, the system is linear. Let ln] = 0.5(zi[n) + zi[-n)) and yofn] 0.5(zo[n] 4 xo[-n)). Applying am[n] + br>[n] to the system yields yfn] 0.5 (az,[n) + bra [n] 4 (aqi [-n) + bzo[-n])) = 0.5a(zi[n] + 2) [-n)) +0.5b(o[n] + zaf-n]) = an fn) + byo[n). (d) No, the system is not memoryless. For example, at time n = 1 the output vl1) = 0.5(x[1] + [-1]) depends on a past value of the input, z(-1]. (e) No, the system is not causal. For example, at time n = —1 the output y[-1] = 0.5(z[-1] + z[1]) depends on a future value of the input, a(1). (£) No, the system is not time-invariant. For example, let the input be z1[n] = uln +10) -ufn— 11). Since this input is already even, the output is just the input, win) = a[n). Shifting by a non-zero integer N, the signal za[n] = m[n — NJis not even, and the output is y2[n] £ nn — N] = zi[n — NJ]. Thus, the system cannot be time-invariant. 3.4-10. It is convenient to substitute nº = n + 1 and rewrite the system expression as yfn afn' = 1)/a[n') (a No, the system is not BIBO stable. Input values of zero can result in unbounded outputs. For example, at n/ = O the bounded input «[n') = ó[n!) yields an unbounded output y[1) = 1/0 = 00. (d) No, the system is not memoryless. The current output relies on a past input. For example, at nº = 0, the output y[n'] requires both the current input z[n'”] and a stored past input z[n' — 1) 94 34-11. 34-12. 3.4-13. 3.4-14. (c) Yes, the system is causal. The current output y[n'] does not depend on any future value of the input. The operation y(t) = a(21) is a one-to-one mapping, where no information is lost. Any one-to-one mapping is invertible. In this case, a(t) is recovered by taking y(t/2). Since every other sample of z[n] is removed in the operation yfn] = x[2n], one half of «[n] is lost and the process is not invertible. Thought of another way, the operation yln] = a[2n] is not a one-to-one mapping; many different signals zfn] map to the same signal y[n), which makes inversion impossible. Notice that yn[n] is obtained by multiplying z[n] by the repeating sequence (sin(gn + 1) = ([...,0.5403, —0.8415, —0.5403, 0.8415, ...). In this particular case, «[n) is recovered by simply multiplying yfn] by the inverse repeating sequence 2 - fio) — (e. 1.8508, 11884, —1.8508, 11884, Im the case second, however, yo[n] is obtained by multiplying z[n] by the repeating sequence (sin(Z(n + 1))j = (...,0,-1,0,1,...). Since the sequence includes zeros, information is lost and the original sequence cannot be recovered. Using the definition of the ramp function, the system expression is rewritten as y(n] = nafejufn. (a) No, the system is not BIBO stable. For example, if the input is a unit step a[n] = ufn), then the output is a ramp function yfn] = r[n], which grows unbounded with time. (b) Yes, the system is linear. Let qa [n] = nzi[njufn] and yo[n] = naafnjufn]. Ap- plying ax[n] + bza[n) to the system yields y[n] = n(am[n] + bra[n)) uln] = anxs [nJufn) + bnxofnju[n) = ay fr] + bya[n] (c) Yes, the system is memoryless. The current output only depends on the current input multiplied by a known (time-varying) scale factor. (d) Yes, the system is causal. All memoryless systems are causal. The output does not depend on future values of the input or output. (e) No, the system is not time-invariant. For example, applying z1[n] = ô[n) yields the output y[n) = nófnJu[n] = 0. Applying zaln] = ófn — 1) yields the output gal] = nófn — 1Jufn) = ófn — 1). Note, za[n] = mafn — 1) but gafn] £ galn — 1]. Shifting the input does not produce a corresponding shift in the output. (a) Position measurements zfn] are in meters. The difference afn — 1) is the change in position, in meters, per frame of film. Since the camera operates at 60 frames per second, dimensional analysis requires k = seconds” (b) Since vf afn — 1)) is an estimate of the velocity v(t) = Sa(t), it is sensible to use a[n] = k(vfn] — vfn — 1) as an estimate of the acceleration a(t) = £u(t). Combining estimates yields a[n] = k(A(z[n] — z[n — 1) — k(z[n — 1-«[n-2))) or al (z[n] = 22/n — 1) + z[n — 2]) = 3600 (zh 2xfn- 1] +2[n— 2). This estimate of acceleration has two primary advantages. Pirst, it is simple to calculate. Second, it is a causal, stable, LTI system and therefore enjoys the properties of such systems There are several shortcomings of the estimate as well. Of particular significance, the estimate afn) lags the actual acceleration a(t). One way to see this is that 95 3.53. 3.5-5. vln] = 0.6yfn — 1] + 0.16[n — 2] Setting n = 0, and substituting y[-1] = —25, y[-2] = 0, yields vl0] = 0.6(-25) + 0.16(0) = —15 Setting n = 1, and substituting y(—1] = 0, y[0] = —15, yields ul) = 0.6(-15) + 0.16(-25) = —13 Setting n = 2, and substituting y[1] = —13, y[0] = —15, yields v(2] = 0.6(-13) + 0.16(-15) = —10.2 This equation can be expressed as 1 1 yln+2])= =autn +] -— ETA +afn +2) Setting n = —2, and substituting y[-1] = y[-2] = 0, z[0] = 100, yields 1 1 vlo) = —5(0) — 5 (0) + 100 = 100 Setting n = —1, and substituting y[-1) = 0, y[0] = 100, z(1] = 100, yields 1 1 . vit = —5(100) — 5(0) + 100 = 75 Setting n = 0, and substituting y(0) = 100, y[1) = 75, [2] = 100, yields v2) 1 1 —=(75) - —(100 =7%5 q (75) — 46(100) 4100 = 75 yln+2) = 3y[n + 1) — 2yfn] + efn+ 2) + 3x[n 41] + 3xfn) Setting n = —2, and substituting y[-1] = 3, yl-2] = 2, «[-1] = 2[-2] = 0, a[0j = 1, yields vlo] = —3(3) - 2(2)+143(0) +3(0) = —12 Setting n = —1, and substituting y[0] = —12, [=] = 3, z[-1]=0, =[0= 1, =[1]=3, yields vit) =3(-12) -2(3) +3-+3(1) +3(0) = 36 Proceeding along same lines, we obtain vio] = -3(36) (12) +9+3(3) +3(1) = —63 vin] = -2yfn = 1) = ln — 2) + 2e[n) = 2[n. 1] 97 Setting n = 0, and substituting y(-1] = 2, y[-2] = 3, 2/0) = 1, z[-1] = 0, vields vlo)=-2(2)-3+2(1)-0=—5 Setting n = 1, and substituting y(0] = —5, y[-1] = 2, a[0] = 1, e(1] = 3, vields 1 yl)= 2-5) — (2) +45) -1.= 7.667 Setting n = 2, and substituting y(1) = 7.667, y[0] = —5, z[1] = 3, a[2] = 5, vields vl2] = —2(7.667) = (-5) + *9) - E =-5.445 (E +3E+2)y[n]=0 The characteristic equation is 42 +37+2 = (y+ 1)(y+2) =0. Therefore ytn) = est)" + col 2)" Setting n = —1 and —2 and substituting initial conditions yields 0= —cy — Le, =2 a-delo, a I=a+io ulnj= HU) -4(-2)" n>o (E? +2E + Dyln])=0 The characteristic equation is 72 +2y+1=(y+1)2 =0. ul] = (ci + co)" Setting n = —1 and —2 and substituting initial conditions yields l=-c tea q=-3 1=0—2c o=-2 vn) = (3 + 2m)(-1)* 3.6-3. (E? -2E+2)yln]=0 The characteristic equation is 42-27 +2=(y-1-j1)(y— 1431) = 0. The roots are Lt jl = v2etin/4 vin) = dv3” cos(qn +0) Setting n = —1 and —2 and substituting initial conditions yields ta ca 1 —=(— cos6 + —= sin 0) ENA 2 ) E +0)= 98 and 2v5 2 2/5 1900 1900 f(1000] = 451 ( A £) + y8+1 (: 5º) a: 2.6864(10208) v(n+2) = 2.50(n 41) + v( The auxiliary conditions are v(0) = 100, v(N) = 0. (E? -2.5E +) The characteristic equation is 42 — 2.57 +1=(y-0.5)(y— 2) =0. = 3(0.5)º + co(2)" Setting n = 0 and N, and substituting v(0) = 100, v(N) = 0, yields x 100 = ate lo a = neo 0 = (0.5) + co(2)” o = ELA 100 N A “lil= qm 055 2" (0.5) -(0.5)"(2)"] n=0,1,05,N 3.6-7. Since we are looking for the zero-input response, the term n — 8) is irrelevant and the equation becomes yofn] + yoln — 1) + 0.25yoln — 2) = 0. The characteristic equation for this second-order system is y2 +7/+0.25 = 0. This yields a repeated root at y = —0.5, and the zero-input response has form yo[n] = ci(—0.5)” + con(-0.5)". The pair of equations yo[-1] = 1 = —2c, + 2c> and yo[l) = 1 = —c1/2 — c»/2 are solved using MATLAB >> c= [-22;-1/2 -1/2]4[1;1] c = -1.2500 -0.7500 Thus, voln] = —1.25(0.5)" — 0.75n(0.5)". 371. (a) (E +2)ytr) n] The characteristic equation is 7 + 2 = 0. The characteristic root is —2. Also aj=2,b = 1. Therefore 1 = ln) + (=) () We need one value of h[n] to determine c. This is determined by iterative solution of (E +2)hfn] = 6fn] or hfn + 1) + 2h[n] = ó[n) 100 Setting n = —1, and substituting A[-1] = 6[-1] = 0, yields hjoj=0 Setting n = 0 in Eq. (1) and using A[0] = O yields 1 1 0= 3 +ce => c= -3 Therefore 1 hfn) = 5óln) = 5(-2)"ufn) (b) The characteristic root is —2, by = 0, ay = 2. Therefore hn] = (-2)" (2) We need one value of h[n) to determine c. This is done by solving iteratively hfn +41) + 2hfn) = ó[n. 41) Setting n = —1, and substituting A[-1) = 0, ó[0] = 1, yields noj=1 Setting n = 0 in Eq. (2) and using A[0] = O yields l=e and hfn) = (=2)"ufn] 3.72. Characteristic equation is 4? — 6y+9= (y-3)2=0. Also az = 9, by = 0. Therefore hfn] = (e; + con)3"ufn] 0) We need two values of A[n] to determine cy and cp. This is found from iterative solution of (E? - 6E + 9)h[n] = Eófn] or hfn + 2] = 6hfn + 1] + 9hfn] = 6fn +) (2) Also h[-1) = h[-2] = 6-1) = 0 and 6/0] = 1. Setting n = —2 in (2) yields hoj-s(0)+%0)=0 => hfoJ=0 Setting n = —1 in (2) yields nj-6(0)+9(0)=1 => Apj=1 Setting n = 0 and 1 in Eq. (1) and substituting A[0] = 0, A(1] = 1 yields 0=a - as 1=3(a +02) ; com O 101 However, um -1]=0formn+1. Hence the summation limits may be restricted for 1 Sm 
