信号与系统奥本海姆英文版课后答案chapter2.docx

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1、Chapter 2 answers2.1 (a) We have know that yx = xn*hn = hkxn-kThe signals xn and hn are as how in Figure S2.1 2 xn1 .3 e o 124 n-lol2 nFigure S2.1-1From this figure, we can easily see that the above convolution sum reduces to yn = h-lxn + 1 + hixn -1=2xn + 1 + 2xn -1This givesy1n = 28n + 1 + 4况+ 2bl

2、 1 + 2dn-2-2dn -4 (b)We know that 00y2n = xn + 2*hn = Z 网+ 2-k k=-gComparing with eq.(S2.1-1),we see that+ 2(c) We may rewrite eq.(S2.1-l) as 8 yxn =*hn = xkhn-kk=-x)Similarly, we may write 00 y3n = xn*hn + 2= xkhn + 2-k k=_gComparing this with eq.(S2.1),we see that y3n = yn + 22.2 Using given defin

3、ition for the signal hn, we may write (ihk= - tik + 3-uk-WThe signal hk is non zero only in the rang hn = hn + 2 . From this we know that the signal h-k is non zero only in the rage 9 A: =0(e _/例)/(0_a),(，) .y(t) =Thenteatu(t) r a = 0(b) the desire convolution iscooy(t)=X(T)h(t-T)dT0052 h(t-T)dT= h(

4、t-r)dT _This may be written as y(t)=i*5/(=)- e2(tr)dT,l=t=3J2J 2一= / = 60,6 ZTherefore(1 / 2)/口 _ 2/4+/10/=(1 /2)e2te22f+e-10-2e-49l=t 2ze-10 -e2-2tl3= t = 60.6 t(c) the desire convolution isE广2x(r)/z(Z-r)6/r = J sin(7zr)(，,This give usy(t尸0/ 1(2/)l-cosM/-l),lZ 3 (2/万)cos开a一 3) 1,3v 1 5 0,5(/-2)wher

5、e4 (0=4/3,0 =/=10, otherwiseNowY(t)=h(t)*x(t)= h(t)*x(t)- - x(Z 2)We havect 44 ii (t)* x(t)= j t -(ai + b)di = -at2-a(t-1)2 + - b(t -1)33 22Therefore4 i i1y(t)= -at2-a(t-)2 +bt-b(t -1)a(t-2) + b = at + b = x(t)(f) x(t) periodic implies y(t) periodic determine 1 period only. we havey(t)= Jj ： (C dz +

6、 J (1 - / + T)dv = +Z-124The2.23 Y(t) is sketchedTherefore1/p (1 f + j (Z-l-r)t/r =? 3Z + 7/4厂iJ22periodic of y(t) is 2.in figure s2.23 for the different values of T.3/ %1=3h2= 42+ 241+ 40n %=3h3= 43+ 242 + %ln%3=2h4= %4+ 2A13 + /z12=/z14=l h5= 4 5 + 24 4 + . 3 = 5 =0. h1n =0 for n=5. (b) in this ca

7、seYn=xn *hn=hn-hn-1 .2.25 (a)we may write xn asxn= (1)l,|!Now the desire convolution is yn= hn *xn -i00=Z (1/3厂(1/4广，左 + 3 + (1/3)(1/4)， 左 + 3 k=-gk=0By consider each summation in the above equation separately .we may show that f(124/ll)3n,/2= 3(b) now consider the convolution%同=(1/3)刎*(1/4)讥+ 3 We

8、may show that必 =必 =0, n -33(1/4)+3(256)(1/3) 2-3Also consider the convolution/卬= 1 * / 4) +3. We may show that以=f(124/ll)3/7-3Clearly , yn + y2n = yn obtained in the previous part. 2.26 (a) we have必n = xnx2n=Y X.kx2n-k k-= (0.5)“叩?+ 3口 &=oThis evaluates toy1n = xn x2n =,“30, otherwise(b) nowyn=13加*凹

9、=弘-,-1Therefore2(1 (1 / 2)-3)+ 2(1 (1 / 2)-4 ) = (1/2)3, 2 一2 yn= i 2J J S 1,77 = 30, otherwiseTherefore, yn= (1 / 2)/z+3 un+3.(c) we havey2M = x2nx3n =un+3-un+2= 8 n+3(d) from the result of part (c), we getYn= %* Xjh= xxn+3= (1 /2)*3un+32.27 the proof is as followsAy=y(t)d(t)J -ccneo30neo30x(r)h(t

10、-t) dr dt,808=x(r) h(t-T)dtdT J-X)J00=J 00812.28 (a) causal because hn=O for n0 stable because () =o 5(b)not causal because hn W 0 fbr n0 stable because5= 004(0.8) =5 0 unstable because (1/2) = oo3625(d) not causal because hn 0 for n0 stable because 乞(5) = ooM=-0C4(e) causal because hn=0 fbr n0 unst

11、able because the second term becomes infinite as n f oo .(f) not causal because hn W 0 fbr n0 stable because 力卬1=二一88(g) causal because hn=0 fbr n0. stable because 帆=1 oo二8305 0032.29. (a) causal because h(t)=Ofor t0. stable because ,依=es / 4 oo .(b) Not causal because h(t) W 0 for t0. Unstable beca

12、use /i(t) = oo.Joo(c) Not causal because h(t) W 0 fbr t0. a Stable because二 /J 00(d) Not causal because h(t) 0 for t0. stable because(e) Not causal because h(t) W 0 fbr t0. stable because(f) Causal because h(t)=O fbr t0. Stable because(g) Causal because h(t)=O fbr t0. Unstable because h(t)dt = ooJoo

13、2.30. We need to find the output of the system when the input is xn=8n.Since we are asked to assume initialrest ,we may conclude that yn=0. for n0.now, yn=xn-2yn-l?Therefore, y0=x0.2y-l=l,yl=xl-2 y0= -2,y2=x2+2y2= -4and so on. In closed form,yn=( - 2)心).this is the impulse response of the system.2.3

14、1. Initial rest implies that yn=0 for n 5.2.32. (a) If ,then we need to verifyAflYa 2Clearly this is true.(b) We now require that for n 0Therefore, B= -2.(c) From eq.(P2.32-1), we know that y0+( 1 /2)y-1 =x0= 1. Now we also have y0= A+B = A=l-B=3.2.33. (a) (i) From Example 2.14,we have know thaty(0

15、e2t (ii) We solve this along the lines of Example 2.14. First assume that yp (Z) is of the Form Ke1，For t0. Then using eq.(P2.33-l).we get fbr t02Ke2t+2Ke2t=e2t =k= 14We know that y = Le2t fbr t0. We may hypothesize the homogeneous solution to be of the form%=2Therefore,必二人一，, for t0Assuming initial

16、 rest. we can conclude that =0 fbr t A= 4Then,y2(0=-(iii) Let the input be x3 (t) = a e3lu(t) +0 e2lu(t) .Assume that the particular solution yp (?) is of the formyp(t) = Ki ae3tu(t)-K2e2tu(t)For t0. Using eq.(P2.33-1), we get3K ae3tu(t) +2K2e2tu(t)+2K1 ae3tu(t) +2K2e2tu(t)=a3t +pe2z.Equating the co

17、efficients of e and e . On both sides, we getK =1/5 and A?2=l/4,we getNow hypothesizing that yp (t) = e 2,%(%)= a3zw(Z)+ pe2fu(t) + Ae2fFor t0. Assuming initial rest,凡 ) = 0 = A + a / 5+ 0 / 4 =A= -(a / 5+0 / 4)Therefore,y3 (，) = ae3z + pe2z + y4e_2z Clearly,+0%).(iv) for the input-output pair x (/)

18、and y (?), we may use eq.(P2.33-1) and the initial restCondition to write虫包+ 2必，=。加长4(S2.33-1)dtFor the input-output pair x2 (t) and y2 (/) , we may use eq.(P2.33-1) and the initial rest condition to write虹。+ 2%。)=%2()%(。= 02(S2.33-2)dtScaling eq.(S2.33-1) by a and eq.(S2.33-2) by 0 and summing, we

19、get d孙(0 + 例2 ) + 2孙(0 + 例2 =(0 + 此，atAndy1 (t) + y2 (?) = 0 for tmin( tt2)By inspection, it is clear that the output is y3 (/) = a yl (t) +0 y2 (/) when the input is x3 (/) = a%1 (f) +px2 (t). Furthermore, y3 () = 0 for tT. Then using eq.(P2.33-1), we get for tT2Ke2(tT) +2Ke2(z-r)= e1t = K=l/4.We k

20、now that for tT.Assuming initial rest, we can conclude the y2 (/) =0 fbr t A=-e.2T 44Then,(f(f乃=_e-2u-D+e2(z-r) 44Clearly, y2(t) = yi(t-T).(iii) consider the input-output pair x2 (t) y (t) where (l) =0 for t t0 . Note that+ 2%=x1 ， =0 for t t.dtSince the derivative is a time-invariant operation, we

21、may now write-+ 2%(t T) = x (t- T), 必(0 = 0 for tt0 dtThis suggests that if the input is a signal of the form x2 (/)=玉 Q T) , then the output is a signal of the form y2 T). Also, note that new output y2 (Z) will be zero for t t() +T. Thissupports time-invariance since x2 (/) is zero for tyx (Z) and

22、x2 (/) y2 (/) .We know that yx (/) = y2 (Z) =1. nowconsider a third input to the system which is(，) = (，) + x2 (，). Let the corresponding output bey3(Z), Now , note that 歹3 (1) = 1 W yx (1) + y2 (1) . Therefore, the system is not linear. A specific example follows.2,Consider an input signal (t) = e

23、.Form Problem 2.33(a-ii), we know that the corresponding output for t0 is%+2Using the fact that yx (/) =1, we get for t0/+（1予-2（1）Now. Consider a second signal x2 (/) =0. Then, the corresponding output is力=Be-21For t0. Using the fact that y2 (1) = 1 ,we get for t0为*)Now consider a third signal x3 (Z

24、) = xx (t) + x2 (t) = xx (/) .Note that the output will still be 3 (，) = y (Z) for t0. Clearly, y3 (0 W 乂 (Z) + y2 (0 for t0. Therefore, the system is not linear.(b) Again consider an input signal x2 (t) = e2fu(t) . From part (a), we know that the corresponding output for t0 with (1) = 1 is乂=*、(1_：*

25、(一)44Now, consider an input signal of the form x2 (t) = xt(t -71) = e2(fr)u(t - T). Then for tT,1)+ Ae-2tUsing the fact that y2 (1) = 1 and also assuming that T1, we get for tT. Therefore, the system is not time invariant.(c) In order to show that the system is incrementally linear with the auxiliar

26、y condition specified as yx =1, we need to first show that the system is linear with the auxiliaryCondition specified as (1) = 0.For an input-output pair x (/) and 凹(，)，we may use eq.(P2.33-l) and the fact that yx (1) = Oto write汽0 + 2% ，*1) = 0(S2.34-1)atFor an input-output pair x9(?) and y2 (0 , w

27、e may use eq.(P2.33-1) and the initial rest condition to writeatat%=Scaling eq.(S2.34-1) by a and eq.(S2.34-2) by 0 and summing, we get+ 月% + 2 町。)+ -% ) 二 ax) + %(/)And%(1)=%+ %=0By inspection, it is clear that the output y3 (t) = ay (/) + py2 (Z) is when the input is x3(z) = ax1 (Z) + (3x2 (t) .Fu

28、rthermore;乃=0 =% + %therefore, the system is linear.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 auxiliary condition alone.(d) In the previous part, we show that the system is linear when =0 . In order to sh

29、ow that the system is not time-invariant, consider an input of the form (/) = From part(a), we know that the corresponding output will beySt) = e2 +Ae-2Using the fact that 必(1) = 0 , we get fbr t0Now consider an input of the form x2(i) = xt (r-1/2). Note that vs(1) = 0. Clearly,y2(?) W % (1 1 / 2) =

30、 (1 /4)(e e3). Therefore, y2(t) W y (% 1 / 2) for all t. This implies that the system is not time invariant.(e) A proof which is very similar to the proof fbr linear used in part(c) may be used here. We may show that the system is not time invariant by using the method outlined in part (d).2.35 (a)

31、since the system is linear, the respond yx (1) = 0 fbr all t.(b) Now let us find the output y2 (?) when the input is (Z) . The particular solution is of the form 匕,(。=匕 t-iSubstituting in eq.(P2.33-l),we get2Y=1.Now, including the homogeneous solution which is of the form yh(t) = Ae2t .we get the ov

32、erall solutiont -1t-lSince=0 , we get2) = 一y +不,For t-lSince the tow pieces of the solution fbr y2(Z) in aqs.(S2.35-l)and (S2.35-2)must match at t= -1, we can determine B from the equation-e2 = Be22 2Which yieldst-l (2 2 )Now note that since x 1 (t)= x 2 (t) for t-l,it must be true that fbr a causal

33、 system y 1 (t)= y 2 (t) fbr t-l.However the results of parts(a) and (b) show that this is not true. Therefore 9the system is not causal.2.36. Consider an input xjn such that x1n=0 for nn x .The corresponding output will beY|n =- yjn-lj+xjn, y j n=0 for nn 1.(S2.36-1)AAlso, consider another output x

34、?n such that x2n =0 fbr nn 2 The corresponding output will bey 2 n = y 2 n-l+ x2n9 y 2 n =0 for n -1since the system is linear, the response to xjn is y j n=0 for all n. Now let us find the output y 2 n when the input is x2n.Since y 2 0=0, y2l=(l/2)0+0=0,y2 =(l/2)0+0=0,y20=(l/2)y2-l+x0.y20=(l/2)y2-l

35、+x0.Therefore , y ? n=0 for n O.Now ,fbr n0,note thatTherefore , y 2 -l=-2. Proceeding similarly, we get y 2 -2=-4, y 2 -3=-8, and so on.Therefore, y2 n=- (1/2)w u-n-l.Now note that since x x n= x 2 n for n0,it must be true that for a causal system y n= y 2 nfbr n0.However the results obtained above

36、 show that this is not true. Therefore , the system is not causal.2.37. Let us consider two inputsX(0=0, for all tandx 2 (t)=e1 u(t)-u(t-l).since the system is linear, the response y (t)=0 fbr all t. Now let us find the output y ? (t)when the input is x2(t). The particular solution is of the formy(t

37、)=Y e 0tlSubstituting in e q.(p2.33-l),we get3Y=1Now, including the homogeneous solution which is of the form y h (t)=A e , we get the overall solution:y 2 (t)= A e 2z + e, 0t-1e-2z+3+1 e, 0tl,(S2.37-1)For t0,we note that x 2 (t)=0. Thus the particular solution is zero in this range and y 2。)= A e -

38、2, t09Since the two pieces of the solution fbr y 2 (t) in eqs.(S.2.37-l) and (S2.37-2) must match at t=0,we can determine B from the equation1 3e3=B3which yields/ 113、-2t72(0=(- -e )e ,t0Now note that since x (t)= x 2 (t) for tO,it must be true that fbr a causal system y 】(t)= y 2 (t) fbr t0.However

39、 the results of obtained above show that this is not true. Therefore , the system is not causal.2.38. The block diagrams are as shown in Figure S2.38.2.39. The block diagrams are as shown in Figure S2.39.2.40. (a) Note thatTherefore,7m (c)dr-ooh(t)= e u(t-2)y(t)=J-00xt(b)We haveFigure S2.38Figure S2.39h(