1、158Chapter 9 Answers9.1 （a）The given integral may be written as (5)0tjedIf -5,then the integral does converge (b) The given integral may be written ast0(5)tjeIf -5 ,then the function grows towards as t decreases towards - and the given integral does not (5)te converge .but if -5 ,then the function g
2、rows towards as t decreases towards - and the given integral (5)tedoes not converge If -5,therefore, the given internal converges when 5 ,then the function grows towards as t decrease towards - and the given integral (5)tedoes not converge .but if -5.Res(b) By using eg.(9.3), we can easily show that
3、 g(t)=A u(-t- ) has the Laplace transform 5t0G(s)= 0(5)stAeThe ROC is specified as max(-5,Re ). Since we are given that the ROC is esRes-3, we know that Re =3 . there are no constraints on the imaginary part of .9.4 We know form Table 9.2 that , Res-11 1()sin(2)()LtxeutX We also know form Table 9.1
4、that x(t)= X(s)= s159The ROC of X(s) is such that if was in the ROC of , then - will be in the ROC of X(s). Putting the 0s1()Xs0two above equations together ,we have x(t)= (-t) = L X(s)= 1(=- , . From property 5 in section 9.2 we knew that x(t) can not be a right-side signal (d) Yes . Since the sign
5、al is absolutely integrable, The ROC must include , the -axis . Furthermore ,X(s) jhas a pole at s=2 .therefore, one valid ROC for the signal could be - (ii)-20 because both are right-hand signals.9.16. Taking the Laplace transform of both sides of the given differential equations ,we obtain ).()1()
6、()23 sXYtherefore, 223ssXH(a) Taking the Laplace transform of both sides of the given equation, we haveG(s) = s H(s)+ H(s). Substituting for H(s) from above,.1)1()() 2223 ssGTherefore, G(s) has 2 poles.(b) we know thatH(s) = .)(22sTherefore, H(s) has poles at If the system has to be stable, then and
7、j,31,).231(jthe real part of the poles has to be less than zero. For this to be true, we require that i.e.,02/.09.17 The overall system show in Figure 9.17 may be treated as two feedback system of the form shown in figure 9.31 connected in parallel. By carrying out an analysis similar to that descri
8、bed in Section 9.8.1, we find the system function of the upper feedback system to be 162.82)/(41)ssHSimilarly, the system function of the lower feedback system is./2The system function of the overall system is now.16023)()(21ssSince H(s)=Y(s)/X(s), we may write . XYTaking the inverse Laplace transfo
9、rm, we obtaindtxtydtty)(3)(60)(29.18. ( a) From problem 3.20, we know that differential equation relating the input and output of the RLC circuit is2)()(.dyttxTaking the Laplace transform of this (while nothing that the system is causal and stable), we obtain2()1().YsXsTherefore ,2,H1.2e(b) We note
10、that H(s) has two poles at and . It has no zeros in the finite s-plane. 3sjsjFrom Section 9.4 we know that the magnitude of the Fourier transform may be expressed as 113(Lengthofvcrmt-+j)(Lengthofvcrmt-j)22We see that the right hand side of the above expression Increases with increasing | | until |
11、| reaches . Then it starts decreasing as | | increasing even further. It finally reaches 0 for | |= . 12 Therefore is approximately lowpass.|()|Hj(c) By repeating the analysis carried out in Problem 3.20 and part (a) of this problem with R = , 31we can show that2()1,YsX0.5es(d) We have 33(Vect.Lnfro
12、m -0.5+j)(ect.Lnfro -.j)2We see that when | | is in he vicinity 0.0005, the right-hand side of the above equation takes on extremely large value. On either side of this value of | | the value of |H (j )| rolls off rapidly. Therefore, H(s) may be considered to be approximately bandpass. 9.19. (a) The
13、 unilateral Laplace transform isX(s) = 20(1)tsteud= ts= .(b) The unilateral Laplace transform is2(3)0()1)1tstXsteud(tst6s.(c) The unilateral Laplace transform is163240()()ttstXseueds12.9.20. In Problem 3.29, we know that the input of the RL circuit are related by ).()(txydtApplying the unilateral La
14、place transform to this equation, we have.0ss(a) For the zero-state response, set .Also we have ()L = . ux)(2te1Therefore,y(s)(s+1)= .sComputing the partial fraction expansion of the right-hand side of the above equation and then taking its inverse unilateral Laplace transform, we have ).()(2tuetty(
15、b) For the zero-state response, assume that x(t) = 0.Since we are given that , (0)1y.1)(0)(1ssyTaking the inverse unilateral Laplace transform, we have ().tyeuFigure S9.21(c) The total response is the sum of the zero-state and zero-input response. This is2()().ttyeu9.21. The pole zero plots for all
16、the subparts are shown in figure S9.21.(a) The Laplace transform of x(t) isX(s)= 230tstd= () (3)00/|/|s stee= 2156(b) Using an approach similar to that show in part (a), we have4(),Lteus 4.Also, 51,5tj j and. 5,LTtjeuessj From this we obtain ,555215sin2LTt tjtjtuswhere .Therefore,e-2-3 -2.5ImRa-2 RI
17、me ImRfImRg ImRh-2 24-4ImRdRb Im cRIm164.24531570sin,549LTtt seutes (c)The Laplace transform of is x023tstXed300/|/|s ste.2156The region of convergence (ROC) is .2(d)Using an approach along the lines of part (a),we obtain. (S9.21-1)2,LTteues Using an approach along the lines of part (c) ,we obtain .
18、 (S9.21-2)21,2t From these we obtain, . 224t LTtt seue 2esUsing the differentiation in the s-domain property , we obtain .2 2284tLTds (e)Using the differentiation in the s-domain property on eq.(S9.21-1),we get .2 21,LTteuess Using the differentiation in the s-domain property on eq (S9.21-2),we get
19、.2 2,LTtd Therefore,.2224,2t LTtt seuees (f)From the previous part ,we have.2221,LTtt s (g)Note that the given signal may be written as .Note that 1xtut.,0LTutes Using the time shifting property ,we get .1,LTtTherefore , 1xAll .,sLTeutNote that in this case ,since the signal is finite duration ,the
20、ROC is the entire s-plane.(h)Consider the signal .Note that the signal may be 11xtutxtexpressed as . We have from the previous part2, All .sLTeut Using the differentiation in s-domain property ,we have , All .1 21sssdextt Using the time-scaling property ,we obtain 165, All .121sLText sThen ,using th
21、e shift property ,we have ,All .12stTherefore , All s.21 211ssssLTeextt(i) The Laplace transform of is .xtu/0X(j) Note that .Therefore ,the Laplace transform is the same as the result of the 3tuprevious part.9.22 (a)From Table 9.2,we have.1sinxtt(b)From Table 9.2 we know that .2co3,09LTstue Using th
22、e time scaling property ,we obtain s,ts Therefore ,the inverse Laplace transform of isX. co3xttu(c)From Table 9.2 we know that . 21s,9LTt see Using the time scaling property ,we obtain . 2co3,1tutss Therefore ,the inverse Laplace transform of is X.txet(d)Using partial fraction expansion on ,we obtai
23、n s.2143XsFrom the given ROC ,we know that must be a two-sided signal .Thereforext.ttxeu(e)Using partial fraction expansion on ,we obtain Xs.213sFrom the given ROC ,we know that must be a two-sided signal ,Therefore,xt.3txteu(f)We may rewrite as Xs212/3s22/1Using Table 9.2 ,we obtain . / /cos/sin3/t
24、 txtetuetu(g)We may rewrite as X166.231sXFrom Table 9.2,we know that . 2,0LTtues Using the shifting property ,we obtain .1,tes Using the differentiation property ,.2,1LTtttdsueue Therefore,.3ttxt9.23.The four pole-zero plots shown may have the following possible ROCs:Plot (a): or or .2es2es2esPlot (
25、b): or .Plot (c): or .Plot (d): Entire s-plane.Also, suppose that the signal has a Laplace transform with ROC .xt XsR(1).We know from Table 9.1 that .33LTteXs The ROC of this new Laplace transform is shifted by 3 to the left .If is absolutely 1RR3txeintegrable, then must include the -axis. jwFor plo
26、t (a), this is possible only if was . 2esFor plot (b), this is possible only if was .For plot (c), this is possible only if was . For plot (d), is the entire s-plane.R(2)We know from Table 9.2 that .1,LTteues Also ,from Table 9.1 we obtain2,1tXxRes If is absolutely integrable, then must include the
27、-axis.te 2jwFor plot (a), this is possible only if was . For plot (b), this is possible only if was .esFor plot (c), this is possible only if was . RFor plot (d), is the entire s-plane.R(3)If for ,then the signal is a left-sided signal or a finite-duration signal . 0xt1tFor plot (a), this is possibl
28、e only if was . 2esFor plot (b), this is possible only if was .For plot (c), this is possible only if was . RFor plot (d), is the entire s-plane.R(4)If for ,then the signal is a right-sided signal or a finite-duration signal 0xt1t167For plot (a), this is possible only if was . R2esFor plot (b), this
29、 is possible only if was .For plot (c), this is possible only if was . For plot (d), is the entire s-plane.R9.24.(a)The pole-zero diagram with the appropriate markings is shown Figure S9.24.(b)By inspecting the pole-zero diagram of part (a), it is clear that the pole-zero diagram shown in Figure S9.
30、24 will also result in the same .This would correspond to the Laplace transformXjw, .12s1es(c) .Xjj(d) with the pole-zero diagram shown below in Figure S9.24 would have the property that 2s .Here , .jwj21/Xs(e) .21/(f)From the result of part (b),it is clear that may be obtained by reflecting the pol
31、es and zeros 1in the right-half of the s-plane to the left-half of the s-plane .Therefore,.1/sXFrom part (d),it is clear that may be obtained by reflecting the poles (zeros) in the right-half 2Xsof the s-plane to the left-half and simultaneously changing them to zeros (poles).Therefore,21/s9.25.The
32、plots are as shown in Figure S9.25.9.26.From Table 9.2 we have 211,2LTtxeuXses and.311,3tUsing the time-shifting time-scaling properties from Table 9.1,we obtain2211,sLTsextX and.3322,sst Therefore, using the convolution property we obtain .2312sLTeytxtY 9.27.From clues 1 and 2,we know that is of th
33、e formX.AXsabFurthermore , we are given that one of the poles of is .Since is real, the Xs1jxtpoles of must occur in conjugate reciprocal pairs .Therefore, anda1bjand .1HsjFrom clue 5,we know that .Therefore, we may deduce that and08X6A. 26sLet denote the ROC of .From the pole locations we know that
34、 there are two R168possible choices of . may either be or .We will now use R1essclue 4 to pick one .Note that .22LTtyexYX The ROC of is shifted by 2 to the right .Since it is given that is not absolutelyYs ytintegrable ,the ROC of should not include the .This is possible only ofjwaxisis .R19.28.(a)
35、The possible ROCs are(i) .2es(ii) .(iii) .( iv) .1s(b)(i)Unstable and anticausal.(ii) Unstable and non causal.（iii ）Stable and non causal.(iv) Unstable and causal.9.29.(a)Using Table 9.2,we obtain 1,Xsesand ,2.H(b) Since ,we may use the convolution property to obtainytxht.1YsXsThe ROC of is .e(c) Pe
36、rforming partial fraction expansion on ,we obtainYs. .12sTaking the inverse Laplace transform, we get.ttyeu(d)Explicit convolution of and gives us xhthtd20efor t0t2.u9.30.For the input , the Laplace transform isxt1.XseThe corresponding output has the Laplace transform ttyteu.22,0YssTherefore, 1,.HeX
37、sNow ,the output has the Laplace transform 3ttytu1236,0.YssTherefore , the Laplace transform of the corresponding input will be16916,0.3YsXesHTaking the inverse Laplace transform of the partial fraction expansion of we obtain1,Xs3124.txtue9.31.(a).Taking the Laplace transform of both sides of the gi
38、ven differential equation and simplifying, we obtain.21YsXThe pole-zero plot for is as shown in figure S9.31.Hb).The partial fraction expansion of is s.1/32(i).If the system is stable ,the ROC for has to be . ThereforeHs12es.tthteu(ii).If the system is causal, the ROC for has to be .Therefore .213tt
39、(iii)If the system is neither stable nor causal ,the ROC for has to be .s1esTherefore ,213tthteu9.32. If produces ,then . Also, by taking the Laplace transform of both sides of 2txe/6ty2/6Hthe given differential equation we get.4sbSince ,we may deduce that .Therefore 1/H1.24ss9.33.Since ,tttxeue.11X
40、eWe are also given that .2sHSince the poles of are at , and since is causal ,we may conclude that the ROC of jht Hsis .Now1e.21YsXsThe ROC of will be the intersection of the ROCs of and .This is . XsH1esWe may obtain the following partial fraction expansion for :Y.2/56/sWe may rewrite this as. 2/14Y
41、s0-1 2 ReImFigure S9.31170Nothing that the ROC of is and using Table9.2,we obtain Ys1es24coin55tttyeuu9.34.We know that 11,0LTxtXs Therefore, has a pole at .Now ,the Laplace transform of the output of the system with s 1yt1xtas the input is 11YHSince in clue 2, is given to be absolutely integrable ,
42、 must have a zero at which cancels Hs0sout the pole of at .1Xs0We also know that 22,0LTxtuse Therefore , has two poles at .Now ,the Laplace transform of the output of the system s 2ytwith as the input is2t2YHXsSince in clue 3, is given to be not absolutely integrable , does not have two zeros at Hs.
43、Therefore ,we conclude that has exactly one zero at .0s 0From clue 4 we know that the signal2dhtptis finite duration .Taking the Laplace transform of both sides of the above equation ,we get .2PsHsTherefore, .2Since is of finite duration, we know that will have no poles in the finite s-plane .pt PsT
44、herefore, is of the forms, 12NiiAzHwhere , represent the zeros of .Here , is some constant.iz,. sAFrom clue 5 we know that the denominator polynomial of has to have a degree which is Hsexactly one greater than the degree of the numerator polynomial .Therefore,.12AsSince we already know that has a ze
45、ro at ,we may rewrite this as 0s2AsFrom clue 1 we know that is .From this ,we may easily show that .Therefore,H0. 1.2sSince the poles of are at and since is causal and stable ,the ROC of is .1jht Hs1e9.35.(a) We may redraw the given block diagram as shown in Figure S9.35.From the figure ,it is clear
46、 that .1FsYTherefore, . Similarly, .Therefore, ./ftdyt/etdft21/etdytFrom the block diagram it is clear that .21166yeftTherefore171. 2116YssYNow ,let us determine the relationship between and .This may be done by concentrating on the lower 1ytxhalf of the above figure .We redraw this in Figure S9.35.From Example 9.30,it is clear that and must be related by the following differential equation :1t. 21d
