d) ii) we have 2.47.(a)yt)=2p(日) t)==()·阿()u(lA b)yt)=yo(1)-如(:-2) =trt)(t)llu(a)ha( (c)y(t)=w(-1) (d) Not enough information signal for all parts of this problem are plotted in the Figure $247 (i)Alo. 4)(t (=()·w(l…v 厂(e (e) In this case (t)=et(t)+( Therefore v(e)=a(c)+eu(t).s(e) aTr This may be written as e s2- 47 y(e)= [e--2e+1a() +(e2-) 2.48.(a)True. If A(t)periodic and nonzero, then e2-e-)-c2-1ll( 广 A(t)dt=∞ n) Using the fact tha刚-的n-1叫叫的 Therefore, h(e) is unstable. (b)False. Rot example, inverse of An]=n-k is sin]- n+k] which is noncausal (e)False. For example An]=unl implies that anll =oo 圳=-zn-1叫= This bs an unstable system. 2.46, Note that d)Thue Assuming that An] is bounded and nonzero in the range n snsng d(Q=-6-u(t-1)+2(t-1)=-3x(t)+2(:-1 n2]l< Given that E x(t)=2eu(-1)→( This implies that the system is stable. we know that um-3x(o)+26(t-1)must yield-3v(4)+2h(t-1)at the output. From (e) False. For example, A(e)eu(e) is causal but not stable (f) False. For example, the cascade of a with impulse response aiIn system with overall impulse reasons giw hlh回=列 (s) False. For exampl, H M(4)=e"w(e), then a(e)-(1-e"ju(a)an 2.51.(a)For tbe system of Pigure P2.51(a)the response to an umit impulse s -c9m=+=∞ vn=n(2)”叫 Although the system is stable, the step response is not abolutely integrable For the system of Figure P2. 51(b)the response to an b) True. We may write=∑副n- Therefore Clearly, y≠pl (b)For the system of Figure P2.51(a)the response to an unit impulse is If sin=0 for n <0, then An]=0 for n<0 and the system is causal 啊=2y+2 19. (a)It is a bounded input. I=inll s 1- B, for all n For the system of Figure P2.51(b) the response to an unit impulse is b)Consider =G3)°n+4. yo= Clearly, yi≠vlnl ∑ k=∞ k+1)a,n20 ∑ A]→∞0 otherwise. Therefore, the output is not bounded. Thus, the system is bot stable and ahwlnte if(-)=0 ifA(-t)≠0 r(e) s 1 for all t. Therefore, r(t) is a bounded input Now, -a-0”+a如a+1」 53.(a)Let us assume that ∑嗜=0 n)a=∞ Then Therefore, the system is unstable if the impulse response is not absolutely integrable. de es=0. z.50. (a)The output will be ar(t)+ brz(e) (b) The outpul will be ai(t-T) Therefore, Ae is a solution of eq.(P253-1
(b) Consider (A")·∑m“+E和广 y(t)= Ae-e2v+Be-te-2jt Since y(0)=1,v/(0)=1, Ate anr+Ae v(e=e"loos zt +sin z4] If s, is a solution, the ∑=0 This implies that te is购um Then, if yin]= Aza (e)(i) Here, =-2 和一4=A)= 0. Therefore, vA()a inoe A(0)=0, yA(O)= 2, A+B=0 and 2A+ B-2. Tberefore, A -2, Therefore,Az is a solution of eq(P254-1) b)If yin= na"", then 2e=2e (-k)x (i)Here. 2+3+2=0=y()=Ac+Be- Since y(0)= l, y(o)m-l, we have y(t)=e-t Taking the right-hand side of the equation that we want to prove, i)v(0)=0 because of initial rest condition 一kxN一k (iv]Here. 2+2s+1=0m(a+1)2=s=-1,g=2 ylc)=Act Since v(o)= 1, y(o)=l, A=l,B-2 Ther Comparing ens. ($2.54-1)and(S2 54-2), we conclude that the equation is proved ylo)=e+2te. (e)(i) Here, + 1 (v)Here. +2--1=0=(-+p2yg=4+B+C eA2”+B-p Since y(0)=1,r()=1,andy"(0)=-2.wegA=1/2.B=3/4.C=32 Siuce yol -l, v-1=-6, we get A=-1,B-2, and (e)Plugging q(P255-3)into eq.(P255-1)gives 22x+1=0 y]=4(1)°+Bn(1)=A+B1 Since, ylo=l, vl]=0 we get A-I. B=-l, and 圳=1-n (ii)Only diference from previous part is initial conditions. Since v(o]=l, y(lo]=21 This implies that eg.( P255-3)satisfies eq. P(2. 55-1). (d)(i Given that ao o and that the system obeys initial rest, we get vm=1+2 a=1→=a (iv)Here, The bomogeneous equation is An-=0 A0++ with the initial conditions Since yfo=0, vl-1]=1. we get a-d, B= =a. and 小=1/A1m…=A-N+l=0 (i)We have h叫=∑bh1n-=0 :55.(a)yol=ro= I. An satisfies the equation where hin is as above A=2h-1,n21 T=计1间断厘 agh(]=0 hQ=s3…,lM=vM (n(i) We get b)From Figure P2. 55(b), we know that if ain]= n], then eve,n≥0 odd arn<o ai) We get even and n≥0 n=02
iw)we ge (d)(a) Taking = ileo 256. (a)In this case, s+2=0 which implies that ∑a-+)+30+)+2aul=6 y(e)=A(e)=Ae-a his implies that fmw =-2 and a-2=I. Therefore, A(0+)=0 and h Since y(o+)=l A-land M()=ea() 1) s3+3s+2=0ss=-2,;=-1 Therefore, h(4)=Ae t+Be",t20. Applying initial conditions, we get A=-1, B-1. Therefore, -)d(t-r)r(r)d h(t=(e-e")u-a(t) x(t=RHS Ga)The initial conditions are h(o+)-0 and w(O)l, Also, a=-It3. Therefore This implies that v() does solve the differential equation (b)Take v(t)=∑awt小 (e)From part (e). if M 2 N, then>A aao will contain singularity terms at t-0 This mples that Integrating between t= 0" and :=0* and matching coefficients, we get aiwa (n() Now. aN=1/aN. This implies that for0≤ts0° ∑au()+2∑u=31(0+-( (t)=二u-N() Tberefore, far =0. Also a()+a-o(t)+2ata(t)=3a;()+o() y(0+)=y(o)=…=y2(0+)=0 This gives ao=3 and a-1=-N. The initial condition is A(0)=-5 and t)=3u()-5e-2y-1l()=36(0)- 5e"t(t) 2 (ii)Here, a 1, ao =-3, a1 =13, a-z--44. Therefore h(0*)=13 and h(0·)=-44mnd (e) The impulse response is h(t)=()-3(1)+18u-1(4)-5e-a-t() e间 ve may eliminate these from the 问=-gl=习+bx/n]+bzn-1 This is the same as the overall difference equation. Yin] x,(- D Figure $2.58 (b)The figures corresponding to the remaining parts of this problem are shown in the Figure 52.37 (e)The figures corresponding to the remaining parts of thin problem are shown in Figure 2.58. (a)Realizing that rfn)- y(n], we may eliminate these from the two given afference equations. This would give ur 2.60. (a)Integrating the given dierential equation once and simplifying, we get 2y四-mn-1+v-习=an问-5r/n-4 v--=2/-//厂 yo)dad This is the same as the overall diference equation. (b)The figures corresponding to the remaining parts of this problem are shown n Figure 厂厂o+句广m0 2.59.(a) Integrating the given differential equation once and simplifying, we get Therefore, A=-a1/an, B=-oo/a2. C= b/an, D=b/en, E-b/az (b)Realizing that zz(t)=v(t), we may eliminate these from the two given integral equa- 厂)+广x20 (e) The fgures corresponding to the remaining parts of this problem are shown in Figure Therefore, A=-aolai, B=b/a1, C-bo/ai (b)Realizing that s (0)- v(n), we may eliminate these from the two given integral qun. 2.61.(a)(i) From Kirchott's voltage law, we know that the input voltage must equal the sun tions. This would give us w(e)=A/w(rldr+B/ a(r)dr +Ca(n
键甏 + (d) Figure S2. 59 Figure $2.60 Using the values of L and C we get (b)(i) From Kirchoff's voltage law, we know that the input voltage must equal the sum the resistor and pacitor, Therefore. (an) Using the results of Problem 2.53, we know that the homogeneous solution of the Using the values of R, L, and C we get 2+a+ay()=如(1 +pa=( will bawe terms of the form Ke l+Kxe t where ao and a are roots of the equation (Hi) The natural response of the system is the bomogeneous solution of the above differ. ential equation. Using the results of Problem 2.53, we know that the homogeneous x2+a1+02=0 solution of the differential equation (It is assumed here that &e*an)In this problem, at=0 and az m I. Therefore. dy(t) +ay(e)= br(e). the root of the equation are Mo si and ai m-3.. The bomogeneous solution will have terms of the form Ae where sg is the root of the equation ()=K11+K2e And,=1=的 (inn)If the voltage and current are restricted to be real, then Kt- K,-K. Therefore In this problem, at= I. Therefore. the root of the equation are so=-1. The yA(s)= 2K cos(e)= 2K sin(t +*/2) (c)(a) Phoe Kircos volue law, we know that the input voltage must equal the sum (It is assumed bert that ao f an- )Ia this problem, 1 =0 and az=4. Therefore, th of the voltage across tb root of tbe equation are so=+23 and s # -2,. The bomogeneous solution is r() +RC +(t) w()=K1e2+k2e2 ing the values of R, L, and C we get Assuming that y(t)is real, we have Ki=K2=K. Therefore, va(t)= 2K cos(2t). +22+5y()=5r(t Clearly, v(f)is periodic. (ai)Uaing the results of Problem 2 53, we know that the bomogeneous solution of the (b)Tbe force z(t) must equal the sum of the force required to displace the mass and th () wil have terms of the form Ke of +kge where so and s: are roots of the egaLon Substituting the values of m and b, we get 十G;3+2= (lt is assumed bere that so*a-)In this problem, at-2 and a2-5. Therefore the root of the equation are sg-1+2j and n1=-1-2j. The homogeneous Using the results of Problem 2.53, we know that the homogeneous solution of the solution 9(t)=Kiee+ K2ee-7t dy +any()=bz(e). a=1 the voltage and cun estricted to be real, then K= k=K. Therefore, will bave terms of the form AeMf where so is the root of the equntion yA(t)=2Ke-cos(2t)-2Kesin(24+w/ C62. (a)The force z(s) must equal the sum of tbe force required to displace the mass and the In this problem, a 1/10000. Therefore, the root of the equation are so=-10. The homogeneous solution is force required to stretch the spring. Therefore, ()=K z(4)= 2+ Clearly, y,(t) deceases with increasing t (e)()We know that the input force z(t)-( Faree required to displace mms by y(e)) Substituting the values of m and K, we get (Force required to displace dashpot by y(e))+(Force required to displace spring +4y(t)=2(t) 0=n29+ +Ky() Using the results of Problem 253, we differential equation Using the values of m, b, and k we get )= br(e) will have terms of the form K eof +Keit so and si are roots of the equat on 2+15+a2=0
Iu Usag the results cf Problem 2.53, we kmow that the bomogeneous solntion of the (d)Total payment 8370, 296. 2+a+agy()-=bx() 2.64.(a)Wehawey()x()h()andx(4)=y(t)·g(t)Theresore,s(t)·h(1)=b(t).Now ill have terms of the form Keo+Kze t wbere so and at are roots of the equation A()·()l4-=∑A9-k6(t-n 32+a13+a2=0 Therefore we want (I: is assumed here that ap t) In this problem, al= 2 and az=2. Therefore he root of the equation are so =-1+i and t s-1-3. The homogeneou %={ 1,23 Therelore, yA(o)=Kiee+Kac"e> And, a =l. 角“,“2每,[+ ()If the force is restricted to be roal, then Ki K2=K. Therefore b)In this ease, g-1. 9=-1/2,92*(-1/2).9(-1/2], and so on. This implies va(e)=2Ke-tcos(()=2Ke"sin(t+*/2) 0=0+(-3)- yin= Amt. borrowed-Ant paid +Compounded Amt from prev month (e)()Here,A()=∑-7) =100004+101yn-1-Dvn-l (u) Ifo <a<l, then a 1. Therefore, A(t) is bounded and absolutely integrable Therefore od corresponds to a stable system. If a>I, then A()is not absolutel =101hn-1-D,n>0 100,000and=101 (i)Here, g(e)=I-ad(t-T). The inverse system is as shown in the figure below y问=101yn-1-D This implies that wa ln]-100D. Also the homogeneous solution is of the form B=A01) Figure $2.84 叫]=熟+叫=4(01)”+100D0 d)Ⅱxl=刷则=Ⅱm2-{叫+{n-Mm=小 Using the initial coadition yo= 100000, we bave 2.65. (a)The autocorrelation s265 vnl=(1000100D)(1.01)”+100D n同=∑川中一母 (e) We have s30}=0=(P-100D10 +10D Therefore, yn] may be viewed as Therefore, D S102860 4n→[→ il 中n () ilr Figure $2.86 一出l (e)x1().h(4)=2(t)·ht)=1().h()=0for4=4 2.67. (a)The autocorrelation functions (5 +3,0≤15 中nn 7(1-4}.0≤t≤1 ≤ 4(t)= 3≤ 4≤t≤5 d中ax(t)=4x(- 小=∑中-k (b)I the impulse response is a(e)=I(T-t), then yo=ou(t-T Therefore.] may be viewed as n→,A-n]→n w(T)-/*()A-r)dr (d)erin and oy are as shown in Figure 2.66. (a)The plot of r(t)is as shown in Figure S2.66 (b)The plots of ra(e)and zg(0)are as shown in Figure S2.66. Therefore, y4)is at most M/