(b)Tbe dierence equations relating the input and output of the systems S, and S, are 叫=h-+圳则一-*则 thes es s, and sa the ethod specied in Emample 2is to ahow that the impulse which is the desired interal. We now evaluate the value of the integral a () nn(2r)t-1=0 2. 21.(a)Tbe desired convolution h spectively. Tbe overall impulse response of the system made up of a cascade of Srand ∑ h闲h2n一对 p>(a/Br for n 20 叫叫fora≠ 钟中2 (b)From (a) 计 )6(t+3)dt=sin(6m)=0 (e) In order to evauate the integral ui(1-T)oos(2r)dr, 间-{(91 (d) The desired convolution is consider tbe signal x()=0(2x)lv(+5-u(t-5 ∑ x因一 We know that et(t)·z(t)=u(t This is as shown in Figure S2.21 (e) The desired convolution is ye)= z(r)h(e-r)dr igure $2.21 This gives us <1 2.22. (a)The desired convolution is 1<t<3 3<:<5 (d)Let dr,t≥0 A()=h(t) Then -{5 u(t)a≠B otherwise (b) The desired convolution is (=A(·()=()r(-3x(-2) We have h(t-r)dr-/A(t-rldr This may be written as (·3“-2“(-13+M-1-3(-2)+副=a+b=2 e)/ett-rdr, Ist (e)a(t) periodic implies y(4)periodic, determine I perod only, We have ≤t≤6 +/,(1-t+)dr=1+t-2 t (1-tn)+(t-1-)=2-3+7A.1<t<3 Tberefore (1/2)e-2e-2+e2-3,t≤ The period of v(()i 2. y(t)=(/2)lea 2.23. v(e) is sketched in Figure S2. 23 for the different values of T 2.24. (a)We are given that hain]=&n)+8fn-1).Therefore 11+刮n
州l △ ∑(/3)-1/咖-k+3+∑/(/h-k+时 an2元((A“*+k+4+∑)”一k+副 y) T= ys By cousider each summation in the above equation separately, we may show that 12/1)3° n<一 ^A术∧ 4)"(1/1)+-3(1/4)+3(25601/31,n=-4 2 b) Now consider the convolution Figure 52.23 y回n=(1/3)"·l/4)"n+ Since 4=h1ln内anl,haln =h1+2hin-1]+hn-2 3(1/4)”+3(256)(1/3)”,n2-3 Therefore Also, consider the convolution A=h1间0 y=[(3)”叫-n-1·/4)”w+3 h1=l1l+2h1 A2=h2+2h1+A间→h图2=3 n<-4 3=h间+2h12+h1l (1/4)"(1/11),n2-3 =h+2h3+冈 h{=1 =h例+2h14+h围→h1图=0. Clearly, yin)+ yin- yin] obtained in the previous part. 2.26.()We have hul)=0 for n<0 and n2 (b)In this case, v=al=∑x脚2一 则=响·A=nl-hn-l 25.(a) We may writ响a 5)vn+3-料 如=x,动的={241-(2-),n-3 (g) Causal because An]=0 for n <o. Subk because ∑ 叫l-1<∞ Therefore 2.29. (a)Causal because A(t)=0 for t<0. Stable because/ph(oldt=e"/4<oo 2(1-(/+)+2(1-(/)=(,2-3 (切)=∞ 0 Therefore, vn]=(1/2 +uin+ 3 (e)Not causal because A(4) 0 for t<0.a Stable because/ IA() e/2<oo (e)We have y回=x,2=如+习一w+2=的+ (d)Not causal because A(e)/0 for t<0. Stable because/ JA(old=e"/<oo (d) From the result of part(e), we get (e)Not eausal because A()#0 for t< 0. Stable because/ Ih(e)dt= 1/3< y=lxl=a1+习=1/2+ (O) Causal because A(e)=0 for t <0. Stable because/IA(e)dt=1<oo 2.27, The proof is as follow (s)Causal because h(e)=0 for t <0. Unstable because/ J(e)dt=ac system when the input is zin m an. Since we are asked 厂厂 230. w neud to hiad he q utp m ay b s nclude that yn=0 for n <0. Now, yin]=rin]-2vin-1l 厂=4a-) x()Aadr =圳--1l=1,l1=1-2=-2.=x2+2/2=-4 rAA and so on. In closed form. s叫=(-2)"un 228.(a) Causal because An=0forn<a. Stable because∑()”-5/<∞ This is the impulse re 2.31. Initial rest implies that yin=0 for n <-2. Now (b) Not canal because h≠ofrn<a. Stable becaus∑(8y=5 =叫+2a-2]-2yln-l (c) Anti-causal became Af=0orn>a. Unstable because∑(1/2)”=∞ -2=1,-1]=0,v (d) Not causal becase A[n]≠0sn< 0. Stable because∑5”=甲<∞ 圳=5s=-11.yn=-110-2)”-3om25. 232,(a)Ⅱs问=A(1/2y, then we need to verify (e)Causal beeause An)=0 for n <o. Unstable because the second term becomes intitute ()-4()-0 (n) Not causal because A(nl≠ 0 for n<a. Stable because∑=30/3<∞
b) we bow require that for n >0 ()-()”-() hypothesizing that Da(e)= Ae- we get herefore, B=-2 ()=2ax+la2+A- (e)Fror eq(P232-1), we know that for t>0. Assuming initial rest =A+B→A=1-B=3 ()=0=A+o{5+B/4 2.33. (a)() From Example 2.14, we know that (3·9) ()=2e2 n0(+a2-(+:9- toru Aea f along the lines of Example 2.14.Firs Clearly, th(t)= ay(4)+Pv(e). (iv) For the input-output pair r(e) and y(e), we may use eq.(P2 33-1)and th et condition to writ 2Ke2+2K2=2K= +2(日)=x1(t),v()=0fort<t ow know that v,(n)= te" for t >0. We may hypothesize the homogeneous For the input-output pair zn(t )and w(e), we may use eq.(P233-1)and the y(t)=Ae-2+ +2(0-(.0=0≤ Assuming initial rest, we can conclude that va(t)= 0 for tsO. Therefore Scaling en(S2 33-1)by a and eq. (S2 33-2)by B and summing, we get 4)+(t}+2{ah(t) - or()+azn(t) n(+w(e)-0 for t min(tn, t, fii)Let the input be z(t)=oeu(e)+Beu(e) Assume that the particular solution t(t)is of the form By inspection. it is clear that the output is g(t)=ay(e)+ By(n) when th is Eo(e)=a(e)+ Ari(t). Furthermore, y(t)=0 for 4< t,, where t, deno w()-,ae+K2Be2t time untll which sa(4)a0 fort >0. Using eq (P233-1), we get (b)(i) Using the result of (a-ii), we may write 3Ka+2K2e2+2K1a+2K2/2=a+Be2 (=x2-c2(0 Equating the coefcients of e f and e on both sides, we get 2Keat-7) We now koom thu w(t)=4et-n for t >T. We may hypothesise tbe bou t>0. Urin the fact that n()=l, w get for t>0 wales Ae-a+kex-r Now consider a third signal i(t=r still be y( )a v(e) for t>0. Clearly, ya(t)f v(t)+m()for Assuming initial rat conclude that wa(t)=0 for t< T. Therefore (7)=0=Ac-打+K (b)Again consider an input signal a,(e)=e" u(e). From part(a), we know that tE 给Am--e corresponding output for 4 >0 with y(l)s I is (o=le G u(t-7) Now, consider an input signai of the form I2(e)=a(t-T)=ete-nu(t-T). Th Clearly, va(t)=n(t-TI for t> ( ai) Consider the input-output pair a(n-yi(4 where z(t)=0 for t Lo Note that v()=2c--n+A-2 +2(=+(,-0,b:< Using the fact that ta()=I and also assuming that T< I, we get for t>T Since the derivative is a time invariant operation, we may now write +2(-7=1(一T,(0=0<如 ow note that y(t)* v(r-T) for t>T. Therefore, the system is not time invariant (e)In order to show that the system is incrementally linear with the auxiliany conditio his suggests that if the input is a signal of tbe form an)=r,(t-T), then the specified as y(i)= l, we need to first show that the system is linear with the auxiliar (t-T). Also, note that tbe new output for t< to+T. Therefore, we may conclude that the system is time-i For an input-output pair =n(t) and y(t), we may use eg. (P233-1) and the fac y()=0 to write 34. (a) Consider r(o 5,y(o) and x, y(o). We know that u(l)=w()=. Ne s234-1) output be n(t). Now, note that n(l)=l y n(1)+ y(l). Therefore, the system is For an input-output pair zx(4 and va(e), we may use eg.(P233-1)and the initial rest Consider an input signal z(0)=eu(e). From Problem 2.33(a-ii), we know tha the corresponding output for t0 is 2(4),v(1)=0. (s234-2 ()=e2+Ae eq(52. 34-2)by d ing tbe fact that n(l)e l, we get for t>0 {mh()+(t)}+2{ov(t)+B()=a1(t)+B{) (=2+(1-)- v(1)=v(1)+v(1)=0 ow, consider a second signal zz(4)m0. Then, the corresponding output Is Therefore, the overall system may be treated as the cascade of a linear system with an adder which adds the response of the system to the auxilary conditions alone
(d) In the previous part, we sb is Linear when v(l)= 0. In order to Now Doue that aDoe a(t)a r(n for t<-l it must be true that for a causal system how that the system is not tim der an input of the form z1(4)=e"u(e). output will be rue. Tberefore, th 4. However the results of parta( a)and(b)show that this is not ()=c2+Ae2 2.36.(a) Consider an input zn such that zfn)=0 for n<n). The corresponding output will ing the fact that n(l)=0, we get for t>0 v-习+,y=05n<m v()=2c2-e2-2 (52.36-1 Also, consider another input zn)such that zn-0 for n n. The corresponding output will be (e) A proof which is very similar to the proof for linearity used in part bere. We may sbow that the system is not time invariant by using the Iel moy be lined vain)-ivin-1+zfn] vain=0 for n <n (s236-2) in part (d Sealing eq($2.36-1)by a and eg ($2.36-2) by B and summing, we ge 35.(a)Since the system is linear, the response n(t)=0 for all t. (b) Now let us find the output v(t) when the input is I?(o). The particular solution is of cy+间=一1+号m一+间+Ba w()=Y,1>-1. By inspection, it is clear that the output is vain a ovin)+ Ayin wben the input is Substituting in eq(P233-1), we get x+B间mbm)=0=p()+(mhm 2Y=1 (b)let us consider two inputs 1问]=0. for all ra Now, including the bomogeneous solution which is of the form yA(4) the overall solution y2(t)=A-2+2.t>-1. Since yto)= 0, we ge Since the rystem is linear, tbe response to zIn] is vinl a 0 for All n. Now let us find -32+}t>- (S235-1) the output vn wben the input is rain). Since yaoj-0, y(1=(1/2)0+0=0.v|2=(1/2)0+0=0 For t<-l, we note that r(t)=0. Thus the particular solution is zero in this range t)=Be,t<-1. (s235-2) y间=(1/2)-1+间 Since the two pieces of the solution for ya(t) in eqs (S235-1)and (S2 35-2)must match can determine B from the equntion Therefore, yaf-lla-2, Proceeding similarly, we get ya[ a-4. y2l-3]=-8, and Now note that since r in]= tn] for n <0, it must be true that for a causal system vain]= vInl for n <o. However, the results obtained above sbow 3 (4)=0, for all y(o] x[n output va(e) wben the input is zn(e). The particular solution is of the farm Substituting in eq.(P233-1), we get 3y=1 Figure S2.38 ()=Ae2+ 0<t<1 了 have v(l)=0. Using this -e3 237-1) For t <0, we Dote that z?(t)0. Thus the particular solution is zero in this range and (s237-2) Figure S2.39 inoe the two piece of the solution for h(t)in eqs. (S237-1)and(S237-2) must match at (=/h()a(t-)d which c-t-2t-+1)-u(t-7-2 Now note that since an(e)= I(t)or t 0. it must be true that for a causal A(r) and r(t-T)are As hown in the figure below. v(t)=w() for t 0. However, the results of obtained above show that this is not true Therefore, the system is not causal 38. The block diagrams are as shown in Figure S2.38 e→t2)dr=1-e(-1), t<t<4 2,39. The block diagrams are as shown in Figure S2.39 -fr-adrse(-ofI-e,t>4 2.40. (a) Note that v=/c-(r-2)d=/en(r)d 2. 41. (a)We may write Therefore a°un-a"un-1
0-广b-2m/ (a)If=*, then y(o)=0 b)Cea, our answer to part(a) is not unique. Any w=2km,k∈了andk≠0w山 Figure $2, 40 z()(o)g(t-o-r)drdo (b)Note that sin] a xin.([ n]-adm-1ll. Therefore, from part (a), we know that x·{4叫-a5n-1=n, Using this we may write mn·{6-1-o6n-2}=6n-l 问{n+1-al)=刮n+1 叫回,{+2-on+1=+2 厂厂 a()h(o)g(t-a-t)drde ow note that 的=4m+2+26+1+列+和 The equality is proved Therefore l=4·{5+2-p+1 w=叫hnl +·{-o5n-l =则h何=(n+ +(1/2)-f·{6n-1-on-2l} () We first have This may be written as 刚=hhl问= 到=圳·{4n+2 ()+2宁 adn+ +(1/2)5n 二中 +26n+1 =州n=叫=(n+i)n The same result was obtained in both parts (i)and (nl (e) Note that =4b四+2+(2-4o)n+1+(1-2o) (ha问hn)(rlhn),hn +(1/2-a)5-1-(1/2)6n-2 xjn]. hln]=a un] -(n-1=on]. Therefore 叫ht问l=·sn8n=sin8n r(t), h(4)/zr)h(t-r)dr-a(r)A(t-rydr Note that A(--)= for It> T2. Therefore, A(t-r)0 forT>1+I andt<-* Them如t>+ (b)() We have =圳∑峨一 Note that zl-k≠0fo-N3≤nS-N, Therefore, r-k+叫≠0for-N3+n≤ Figure $2.45 的+2 No. Therefore,, wn is nonzero for n≤N1+ N, and n2N+N2 (ii) We can easily show u at. the o tpr of te o e ith- cg. he response ut) a wo (e)An]=0 for n>5 we may interchange their order as own in Figure $2.45. d) Fron the figure it is clear that ben,y(o)=x(). A(e)and p(e)av(r). Since a(() and p(t) have to be the same M0,4-层x-)+ may conclude that r(e).A(e=y(t) A( in tbe earlier proofs and they would all still hold. Tay由ad Therefore vo)=/=(lds+a(-6) (ii) Consider This imples that r(e)must be known for Ists2 and for t=-6 v(t)r(t).t()(日 =x(t)·u()*()*h() 15.(a)(i) we have r()-(-l4v)-y= Taking limit as A-+0 on both sides of the above equation This shows that (4).u(o)Jw (e) is equivalent to x(t).h(t). Now the same thing x()2y(n) v(t)=r()叫t)*h( x(t)ut(tl.h(t)lu (i) Differentiating the convolution integral we get ()·(t)u(t r(t-T)A(r)dr =z(1·/M(r)dr z(t-Th( (c)Note that r(e)=& -su(o. Therefore, the output of the LTI system to r(e) be M(t)-5sin(ant). Since this has to be equal to y(o=w cos(uot), we have